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What the f*ck Python! 😱

کاوش و درک پایتون از طریق تکه‌های کد شگفت‌انگیز.

حالت‌های دیگر: وبسایت تعاملی | دفترچه تعاملی

پایتون، یه زبان زیبا طراحی شده، سطح بالا و مبتنی بر مفسره که قابلیت‌های بسیاری برای راحتی ما برنامه‌نویس‌ها فراهم می‌کنه. ولی گاهی اوقات قطعه‌کدهایی رو می‌بینیم که تو نگاه اول خروجی‌هاشون واضح نیست.

این یه پروژه باحاله که سعی داریم توش توضیح بدیم که پشت پرده یه سری قطعه‌کدهای غیرشهودی و فابلیت‌های کمتر شناخته شده پایتون چه خبره.

درحالی که بعضی از مثال‌هایی که قراره تو این سند ببینید واقعا عجیب و غریب نیستند ولی بخش‌های جالبی از پایتون رو ظاهر می‌کنند که ممکنه شما از وجودشون بی‌خبر باشید. به نظرم این شیوه جالبیه برای یادگیری جزئیات داخلی یه زبان برنامه نویسی و باور دارم که برای شما هم جالب خواهد بود.

اگه شما یه پایتون کار سابقه‌دار هستید، می‌تونید از این فرصت به عنوان یه چالش برای خودتون استفاده کنید تا بیشتر مثال‌ها رو تو تلاش اول حدس بزنید. ممکنه شما بعضی از این مثال‌ها رو قبلا تجربه کرده باشید و من خاطراتشون رو در این سند براتون زنده کرده باشم! 😅

پ.ن: اگه شما قبلا این سند رو خوندید، می‌تونید تغییرات جدید رو در بخش انتشار (فعلا در اینجا) مطالعه کنید (مثال‌هایی که کنارشون علامت ستاره دارند، در آخرین ویرایش اضافه شده‌اند).

پس، بزن بریم...

فهرست مطالب

فهرست مطالب
ساختار مثال‌ها
استفاده
👀 مثال‌ها
‫ مشارکت
‫ تقدیر و تشکر - ‫ چند لینک جالب!
‫ 🎓 مجوز
- ‫ دوستانتان را هم شگفت‌زده کنید!
- ‫ آیا به یک نسخه pdf نیاز دارید؟

ساختار مثال‌ها

همه مثال‌ها به صورت زیر ساخته می‌شوند:

▶ یه اسم خوشگل
# راه اندازی کد
# آماده سازی برای جادو...
خروجی (نسخه(های) پایتون):
>>> triggering_statement
یه خروجی غیرمنتظره
(دلخواه): توضیح یک‌خطی خروجی غیرمنتظره

💡 توضیح:

توضیح کوتاه درمورد این‌که چی داره اتفاق میافته و چرا.
# راه اندازی کد
# مثال‌های بیشتر برای شفاف سازی (در صورت نیاز)
خروجی (نسخه(های) پایتون):
>>> trigger # یک مثال که رونمایی از جادو رو راحت‌تر می‌کنه
# یک خروجی توجیه شده و واضح

توجه: همه مثال‌ها در برنامه مفسر تعاملی پایتون نسخه ۳.۵.۲ آزمایش شده‌اند و باید در همه نسخه‌های پایتون کار کنند مگراینکه به صورت جداگانه و به طور واضح نسخه مخصوص پایتون قبل از خروجی ذکر شده باشد.

استفاده

یه راه خوب برای بیشتر بهره بردن، به نظرم، اینه که مثال‌ها رو به ترتیب متوالی بخونید و برای هر مثال:

کد ابتدایی برای راه اندازی مثال رو با دقت بخونید. اگه شما یه پایتون کار سابقه‌دار باشید، با موفقیت بیشتر اوقات اتفاق بعدی رو پیش‌بینی می‌کنید.
قطعه خروجی رو بخونید و
- بررسی کنید که آیا خروجی‌ها همونطور که انتظار دارید هستند.
- مطمئین بشید که دقیقا دلیل اینکه خروجی اون طوری هست رو می‌دونید.
  - اگه نمی‌دونید (که کاملا عادیه و اصلا بد نیست)، یک نفس عمیق بکشید و توضیحات رو بخونید (و اگه نفهمیدید، داد بزنید! و اینجا درموردش حرف بزنید).
  - اگه می‌دونید، به افتخار خودتون یه دست محکم بزنید و برید سراغ مثال بعدی.

👀 مثال‌ها

بخش: ذهن خود را به چالش بکشید!

▶ اول از همه! *

به دلایلی، عملگر "Walrus" (:=) که در نسخه ۳.۸ پایتون معرفی شد، خیلی محبوب شده. بیاید بررسیش کنیم.

# Python version 3.8+

>>> a = "wtf_walrus"
>>> a
'wtf_walrus'

>>> a := "wtf_walrus"
File "<stdin>", line 1
    a := "wtf_walrus"
      ^
SyntaxError: invalid syntax

>>> (a := "wtf_walrus") # ولی این کار می‌کنه
'wtf_walrus'
>>> a
'wtf_walrus'

2 .

# Python version 3.8+

>>> a = 6, 9
>>> a
(6, 9)

>>> (a := 6, 9)
(6, 9)
>>> a
6

>>> a, b = 6, 9 # باز کردن معمولی
>>> a, b
(6, 9)
>>> (a, b = 16, 19) # آخ آخ
  File "<stdin>", line 1
    (a, b = 16, 19)
          ^
SyntaxError: invalid syntax

>>> (a, b := 16, 19) # این یه تاپل ۳تایی چاپ می‌کنه رو صفحه
(6, 16, 19)

>>> a # هنوز تغییر نکرده؟
6

>>> b
16

💡 توضیحات

مرور سریع بر عملگر Walrus

عملگر Walrus همونطور که اشاره شد، در نسخه ۳.۸ پایتون معرفی شد. این عملگر می‌تونه تو مقعیت‌هایی کاربردی باشه که شما می‌خواید داخل یه عبارت، مقادیری رو به متغیرها اختصاص بدید

def some_func():
        # فرض کنید اینجا یک سری محاسبه سنگین انجام میشه
        # time.sleep(1000)
        return 5

# پس به جای اینکه این کارو بکنید:
if some_func():
        print(some_func()) # که خیلی راه نادرستیه چون محاسبه دوبار انجام میشه

# یا حتی این کارو کنید (که کار بدی هم نیست)
a = some_func()
if a:
    print(a)

# می‌تونید از این به بعد به طور مختصر بنویسید:
if a := some_func():
        print(a)

خروجی (+۳.۸):

5
5
5

این باعث میشه که یک خط کمتر کد بزنیم و از دوبار فراخوندن some_func جلوگیری کرد.

"عبارت اختصاص‌دادن مقدار" بدون پرانتز (نحوه استفاده عملگر Walrus)، در سطح بالا محدود است، SyntaxError در عبارت a := "wtf_walrus" در قطعه‌کد اول به همین دلیل است. قرار دادن آن داخل پرانتز، همانطور که می‌خواستیم کار کرد و مقدار را به a اختصاص داد.
به طور معمول، قرار دادن عبارتی که دارای = است داخل پرانتز مجاز نیست. به همین دلیل ‍عبارت (a, b = 6, 9) به ما خطای سینتکس داد.
قائده استفاده از عملگر Walrus به صورت NAME:= expr است، به طوری که NAME یک شناسه صحیح و expr یک عبارت صحیح است. به همین دلیل باز و بسته کردن با تکرار (iterable) پشتیبانی نمی‌شوند. پس،
- عبارت (a := 6, 9) معادل عبارت ((a := 6), 9) و در نهایت (a, 9) است. (که مقدار a عدد 6 است)
```
>>> (a := 6, 9) == ((a := 6), 9)
True
>>> x = (a := 696, 9)
>>> x
(696, 9)
>>> x[0] is a # هر دو به یک مکان در حافظه دستگاه اشاره می‌کنند
True
```
- به طور مشابه، عبارت (a, b := 16, 19) معادل عبارت (a, (b := 16), 19) است که چیزی جز یک تاپل ۳تایی نیست.

▶ بعضی وقت‌ها رشته‌ها می‌توانند دردسرساز شوند

>>> a = "some_string"
>>> id(a)
140420665652016
>>> id("some" + "_" + "string") # دقت کنید که هردو ID یکی هستند.
140420665652016

>>> a = "wtf"
>>> b = "wtf"
>>> a is b
True

>>> a = "wtf!"
>>> b = "wtf!"
>>> a is b
False

>>> a, b = "wtf!", "wtf!"
>>> a is b # همه‌ی نسخه‌ها به جز 3.7.x
True

>>> a = "wtf!"; b = "wtf!"
>>> a is b # ممکن است True یا False باشد بسته به جایی که آن را اجرا می‌کنید (python shell / ipython / به‌صورت اسکریپت)
False

# این بار در فایل some_file.py
a = "wtf!"
b = "wtf!"
print(a is b)

# موقع اجرای ماژول، True را چاپ می‌کند!

خروجی (< Python3.7 )

>>> 'a' * 20 is 'aaaaaaaaaaaaaaaaaaaa'
True
>>> 'a' * 21 is 'aaaaaaaaaaaaaaaaaaaaa'
False

Makes sense, right?

💡 Explanation:

The behavior in first and second snippets is due to a CPython optimization (called string interning) that tries to use existing immutable objects in some cases rather than creating a new object every time.
After being "interned," many variables may reference the same string object in memory (saving memory thereby).
In the snippets above, strings are implicitly interned. The decision of when to implicitly intern a string is implementation-dependent. There are some rules that can be used to guess if a string will be interned or not:
- All length 0 and length 1 strings are interned.
- Strings are interned at compile time ('wtf' will be interned but ''.join(['w', 't', 'f']) will not be interned)
- Strings that are not composed of ASCII letters, digits or underscores, are not interned. This explains why 'wtf!' was not interned due to !. CPython implementation of this rule can be found here

Shows a string interning process.

When a and b are set to "wtf!" in the same line, the Python interpreter creates a new object, then references the second variable at the same time. If you do it on separate lines, it doesn't "know" that there's already "wtf!" as an object (because "wtf!" is not implicitly interned as per the facts mentioned above). It's a compile-time optimization. This optimization doesn't apply to 3.7.x versions of CPython (check this issue for more discussion).
A compile unit in an interactive environment like IPython consists of a single statement, whereas it consists of the entire module in case of modules. a, b = "wtf!", "wtf!" is single statement, whereas a = "wtf!"; b = "wtf!" are two statements in a single line. This explains why the identities are different in a = "wtf!"; b = "wtf!", and also explain why they are same when invoked in some_file.py
The abrupt change in the output of the fourth snippet is due to a peephole optimization technique known as Constant folding. This means the expression 'a'*20 is replaced by 'aaaaaaaaaaaaaaaaaaaa' during compilation to save a few clock cycles during runtime. Constant folding only occurs for strings having a length of less than 21. (Why? Imagine the size of .pyc file generated as a result of the expression 'a'*10**10). Here's the implementation source for the same.
Note: In Python 3.7, Constant folding was moved out from peephole optimizer to the new AST optimizer with some change in logic as well, so the fourth snippet doesn't work for Python 3.7. You can read more about the change here.

▶ Be careful with chained operations

>>> (False == False) in [False] # makes sense
False
>>> False == (False in [False]) # makes sense
False
>>> False == False in [False] # now what?
True

>>> True is False == False
False
>>> False is False is False
True

>>> 1 > 0 < 1
True
>>> (1 > 0) < 1
False
>>> 1 > (0 < 1)
False

💡 Explanation:

As per https://docs.python.org/3/reference/expressions.html#comparisons

Formally, if a, b, c, ..., y, z are expressions and op1, op2, ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z, except that each expression is evaluated at most once.

While such behavior might seem silly to you in the above examples, it's fantastic with stuff like a == b == c and 0 <= x <= 100.

False is False is False is equivalent to (False is False) and (False is False)
True is False == False is equivalent to (True is False) and (False == False) and since the first part of the statement (True is False) evaluates to False, the overall expression evaluates to False.
1 > 0 < 1 is equivalent to (1 > 0) and (0 < 1) which evaluates to True.

The expression (1 > 0) < 1 is equivalent to True < 1 and

>>> int(True)
1
>>> True + 1 #not relevant for this example, but just for fun
2

So, 1 < 1 evaluates to False

▶ How not to use `is` operator

The following is a very famous example present all over the internet.

>>> a = 256
>>> b = 256
>>> a is b
True

>>> a = 257
>>> b = 257
>>> a is b
False

>>> a = []
>>> b = []
>>> a is b
False

>>> a = tuple()
>>> b = tuple()
>>> a is b
True

3. Output

>>> a, b = 257, 257
>>> a is b
True

Output (Python 3.7.x specifically)

>>> a, b = 257, 257
>>> a is b
False

💡 Explanation:

The difference between is and ==

is operator checks if both the operands refer to the same object (i.e., it checks if the identity of the operands matches or not).
== operator compares the values of both the operands and checks if they are the same.

So is is for reference equality and == is for value equality. An example to clear things up,

>>> class A: pass
>>> A() is A() # These are two empty objects at two different memory locations.
False

256 is an existing object but 257 isn't

When you start up python the numbers from -5 to 256 will be allocated. These numbers are used a lot, so it makes sense just to have them ready.

Quoting from https://docs.python.org/3/c-api/long.html

The current implementation keeps an array of integer objects for all integers between -5 and 256, when you create an int in that range you just get back a reference to the existing object. So it should be possible to change the value of 1. I suspect the behavior of Python, in this case, is undefined. :-)

>>> id(256)
10922528
>>> a = 256
>>> b = 256
>>> id(a)
10922528
>>> id(b)
10922528
>>> id(257)
140084850247312
>>> x = 257
>>> y = 257
>>> id(x)
140084850247440
>>> id(y)
140084850247344

Here the interpreter isn't smart enough while executing y = 257 to recognize that we've already created an integer of the value 257, and so it goes on to create another object in the memory.

Similar optimization applies to other immutable objects like empty tuples as well. Since lists are mutable, that's why [] is [] will return False and () is () will return True. This explains our second snippet. Let's move on to the third one,

Both a and b refer to the same object when initialized with same value in the same line.

Output

>>> a, b = 257, 257
>>> id(a)
140640774013296
>>> id(b)
140640774013296
>>> a = 257
>>> b = 257
>>> id(a)
140640774013392
>>> id(b)
140640774013488

When a and b are set to 257 in the same line, the Python interpreter creates a new object, then references the second variable at the same time. If you do it on separate lines, it doesn't "know" that there's already 257 as an object.
It's a compiler optimization and specifically applies to the interactive environment. When you enter two lines in a live interpreter, they're compiled separately, therefore optimized separately. If you were to try this example in a .py file, you would not see the same behavior, because the file is compiled all at once. This optimization is not limited to integers, it works for other immutable data types like strings (check the "Strings are tricky example") and floats as well,
```
>>> a, b = 257.0, 257.0
>>> a is b
True
```
Why didn't this work for Python 3.7? The abstract reason is because such compiler optimizations are implementation specific (i.e. may change with version, OS, etc). I'm still figuring out what exact implementation change cause the issue, you can check out this issue for updates.

▶ Hash brownies

some_dict = {}
some_dict[5.5] = "JavaScript"
some_dict[5.0] = "Ruby"
some_dict[5] = "Python"

Output:

>>> some_dict[5.5]
"JavaScript"
>>> some_dict[5.0] # "Python" destroyed the existence of "Ruby"?
"Python"
>>> some_dict[5] 
"Python"

>>> complex_five = 5 + 0j
>>> type(complex_five)
complex
>>> some_dict[complex_five]
"Python"

So, why is Python all over the place?

💡 Explanation

Uniqueness of keys in a Python dictionary is by equivalence, not identity. So even though 5, 5.0, and 5 + 0j are distinct objects of different types, since they're equal, they can't both be in the same dict (or set). As soon as you insert any one of them, attempting to look up any distinct but equivalent key will succeed with the original mapped value (rather than failing with a KeyError):
```
>>> 5 == 5.0 == 5 + 0j
True
>>> 5 is not 5.0 is not 5 + 0j
True
>>> some_dict = {}
>>> some_dict[5.0] = "Ruby"
>>> 5.0 in some_dict
True
>>> (5 in some_dict) and (5 + 0j in some_dict)
True
```
This applies when setting an item as well. So when you do some_dict[5] = "Python", Python finds the existing item with equivalent key 5.0 -> "Ruby", overwrites its value in place, and leaves the original key alone.
```
>>> some_dict
{5.0: 'Ruby'}
>>> some_dict[5] = "Python"
>>> some_dict
{5.0: 'Python'}
```
So how can we update the key to 5 (instead of 5.0)? We can't actually do this update in place, but what we can do is first delete the key (del some_dict[5.0]), and then set it (some_dict[5]) to get the integer 5 as the key instead of floating 5.0, though this should be needed in rare cases.
How did Python find 5 in a dictionary containing 5.0? Python does this in constant time without having to scan through every item by using hash functions. When Python looks up a key foo in a dict, it first computes hash(foo) (which runs in constant-time). Since in Python it is required that objects that compare equal also have the same hash value (docs here), 5, 5.0, and 5 + 0j have the same hash value.
```
>>> 5 == 5.0 == 5 + 0j
True
>>> hash(5) == hash(5.0) == hash(5 + 0j)
True
```
Note: The inverse is not necessarily true: Objects with equal hash values may themselves be unequal. (This causes what's known as a hash collision, and degrades the constant-time performance that hashing usually provides.)

▶ Deep down, we're all the same.

class WTF:
  pass

Output:

>>> WTF() == WTF() # two different instances can't be equal
False
>>> WTF() is WTF() # identities are also different
False
>>> hash(WTF()) == hash(WTF()) # hashes _should_ be different as well
True
>>> id(WTF()) == id(WTF())
True

💡 Explanation:

When id was called, Python created a WTF class object and passed it to the id function. The id function takes its id (its memory location), and throws away the object. The object is destroyed.
When we do this twice in succession, Python allocates the same memory location to this second object as well. Since (in CPython) id uses the memory location as the object id, the id of the two objects is the same.
So, the object's id is unique only for the lifetime of the object. After the object is destroyed, or before it is created, something else can have the same id.

But why did the is operator evaluate to False? Let's see with this snippet.

class WTF(object):
  def __init__(self): print("I")
  def __del__(self): print("D")

Output:

>>> WTF() is WTF()
I
I
D
D
False
>>> id(WTF()) == id(WTF())
I
D
I
D
True

As you may observe, the order in which the objects are destroyed is what made all the difference here.

▶ Disorder within order *

from collections import OrderedDict

dictionary = dict()
dictionary[1] = 'a'; dictionary[2] = 'b';

ordered_dict = OrderedDict()
ordered_dict[1] = 'a'; ordered_dict[2] = 'b';

another_ordered_dict = OrderedDict()
another_ordered_dict[2] = 'b'; another_ordered_dict[1] = 'a';

class DictWithHash(dict):
    """
    A dict that also implements __hash__ magic.
    """
    __hash__ = lambda self: 0

class OrderedDictWithHash(OrderedDict):
    """
    An OrderedDict that also implements __hash__ magic.
    """
    __hash__ = lambda self: 0

Output

>>> dictionary == ordered_dict # If a == b
True
>>> dictionary == another_ordered_dict # and b == c
True
>>> ordered_dict == another_ordered_dict # then why isn't c == a ??
False

# We all know that a set consists of only unique elements,
# let's try making a set of these dictionaries and see what happens...

>>> len({dictionary, ordered_dict, another_ordered_dict})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'dict'

# Makes sense since dict don't have __hash__ implemented, let's use
# our wrapper classes.
>>> dictionary = DictWithHash()
>>> dictionary[1] = 'a'; dictionary[2] = 'b';
>>> ordered_dict = OrderedDictWithHash()
>>> ordered_dict[1] = 'a'; ordered_dict[2] = 'b';
>>> another_ordered_dict = OrderedDictWithHash()
>>> another_ordered_dict[2] = 'b'; another_ordered_dict[1] = 'a';
>>> len({dictionary, ordered_dict, another_ordered_dict})
1
>>> len({ordered_dict, another_ordered_dict, dictionary}) # changing the order
2

What is going on here?

💡 Explanation:

The reason why intransitive equality didn't hold among dictionary, ordered_dict and another_ordered_dict is because of the way __eq__ method is implemented in OrderedDict class. From the docs

Equality tests between OrderedDict objects are order-sensitive and are implemented as list(od1.items())==list(od2.items()). Equality tests between OrderedDict objects and other Mapping objects are order-insensitive like regular dictionaries.
The reason for this equality in behavior is that it allows OrderedDict objects to be directly substituted anywhere a regular dictionary is used.

Okay, so why did changing the order affect the length of the generated set object? The answer is the lack of intransitive equality only. Since sets are "unordered" collections of unique elements, the order in which elements are inserted shouldn't matter. But in this case, it does matter. Let's break it down a bit,

>>> some_set = set()
>>> some_set.add(dictionary) # these are the mapping objects from the snippets above
>>> ordered_dict in some_set
True
>>> some_set.add(ordered_dict)
>>> len(some_set)
1
>>> another_ordered_dict in some_set
True
>>> some_set.add(another_ordered_dict)
>>> len(some_set)
1

>>> another_set = set()
>>> another_set.add(ordered_dict)
>>> another_ordered_dict in another_set
False
>>> another_set.add(another_ordered_dict)
>>> len(another_set)
2
>>> dictionary in another_set
True
>>> another_set.add(another_ordered_dict)
>>> len(another_set)
2

So the inconsistency is due to another_ordered_dict in another_set being False because ordered_dict was already present in another_set and as observed before, ordered_dict == another_ordered_dict is False.

▶ Keep trying... *

def some_func():
    try:
        return 'from_try'
    finally:
        return 'from_finally'

def another_func(): 
    for _ in range(3):
        try:
            continue
        finally:
            print("Finally!")

def one_more_func(): # A gotcha!
    try:
        for i in range(3):
            try:
                1 / i
            except ZeroDivisionError:
                # Let's throw it here and handle it outside for loop
                raise ZeroDivisionError("A trivial divide by zero error")
            finally:
                print("Iteration", i)
                break
    except ZeroDivisionError as e:
        print("Zero division error occurred", e)

Output:

>>> some_func()
'from_finally'

>>> another_func()
Finally!
Finally!
Finally!

>>> 1 / 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ZeroDivisionError: division by zero

>>> one_more_func()
Iteration 0

💡 Explanation:

When a return, break or continue statement is executed in the try suite of a "try…finally" statement, the finally clause is also executed on the way out.
The return value of a function is determined by the last return statement executed. Since the finally clause always executes, a return statement executed in the finally clause will always be the last one executed.
The caveat here is, if the finally clause executes a return or break statement, the temporarily saved exception is discarded.

▶ For what?

some_string = "wtf"
some_dict = {}
for i, some_dict[i] in enumerate(some_string):
    i = 10

Output:

>>> some_dict # An indexed dict appears.
{0: 'w', 1: 't', 2: 'f'}

💡 Explanation:

A for statement is defined in the Python grammar as:
```
for_stmt: 'for' exprlist 'in' testlist ':' suite ['else' ':' suite]
```
Where exprlist is the assignment target. This means that the equivalent of {exprlist} = {next_value} is executed for each item in the iterable. An interesting example that illustrates this:
```
for i in range(4):
    print(i)
    i = 10
```
Output:
```
0
1
2
3
```
Did you expect the loop to run just once?

💡 Explanation:
- The assignment statement i = 10 never affects the iterations of the loop because of the way for loops work in Python. Before the beginning of every iteration, the next item provided by the iterator (range(4) in this case) is unpacked and assigned the target list variables (i in this case).
The enumerate(some_string) function yields a new value i (a counter going up) and a character from the some_string in each iteration. It then sets the (just assigned) i key of the dictionary some_dict to that character. The unrolling of the loop can be simplified as:
```
>>> i, some_dict[i] = (0, 'w')
>>> i, some_dict[i] = (1, 't')
>>> i, some_dict[i] = (2, 'f')
>>> some_dict
```

▶ Evaluation time discrepancy

array = [1, 8, 15]
# A typical generator expression
gen = (x for x in array if array.count(x) > 0)
array = [2, 8, 22]

Output:

>>> print(list(gen)) # Where did the other values go?
[8]

array_1 = [1,2,3,4]
gen_1 = (x for x in array_1)
array_1 = [1,2,3,4,5]

array_2 = [1,2,3,4]
gen_2 = (x for x in array_2)
array_2[:] = [1,2,3,4,5]

Output:

>>> print(list(gen_1))
[1, 2, 3, 4]

>>> print(list(gen_2))
[1, 2, 3, 4, 5]

array_3 = [1, 2, 3]
array_4 = [10, 20, 30]
gen = (i + j for i in array_3 for j in array_4)

array_3 = [4, 5, 6]
array_4 = [400, 500, 600]

Output:

>>> print(list(gen))
[401, 501, 601, 402, 502, 602, 403, 503, 603]

💡 Explanation

In a generator expression, the in clause is evaluated at declaration time, but the conditional clause is evaluated at runtime.
So before runtime, array is re-assigned to the list [2, 8, 22], and since out of 1, 8 and 15, only the count of 8 is greater than 0, the generator only yields 8.
The differences in the output of g1 and g2 in the second part is due the way variables array_1 and array_2 are re-assigned values.
In the first case, array_1 is bound to the new object [1,2,3,4,5] and since the in clause is evaluated at the declaration time it still refers to the old object [1,2,3,4] (which is not destroyed).
In the second case, the slice assignment to array_2 updates the same old object [1,2,3,4] to [1,2,3,4,5]. Hence both the g2 and array_2 still have reference to the same object (which has now been updated to [1,2,3,4,5]).
Okay, going by the logic discussed so far, shouldn't be the value of list(gen) in the third snippet be [11, 21, 31, 12, 22, 32, 13, 23, 33]? (because array_3 and array_4 are going to behave just like array_1). The reason why (only) array_4 values got updated is explained in PEP-289

Only the outermost for-expression is evaluated immediately, the other expressions are deferred until the generator is run.

▶ `is not ...` is not `is (not ...)`

>>> 'something' is not None
True
>>> 'something' is (not None)
False

💡 Explanation

is not is a single binary operator, and has behavior different than using is and not separated.
is not evaluates to False if the variables on either side of the operator point to the same object and True otherwise.
In the example, (not None) evaluates to True since the value None is False in a boolean context, so the expression becomes 'something' is True.

▶ A tic-tac-toe where X wins in the first attempt!

# Let's initialize a row
row = [""] * 3 #row i['', '', '']
# Let's make a board
board = [row] * 3

Output:

>>> board
[['', '', ''], ['', '', ''], ['', '', '']]
>>> board[0]
['', '', '']
>>> board[0][0]
''
>>> board[0][0] = "X"
>>> board
[['X', '', ''], ['X', '', ''], ['X', '', '']]

We didn't assign three "X"s, did we?

💡 Explanation:

When we initialize row variable, this visualization explains what happens in the memory

Shows a memory segment after row is initialized.

And when the board is initialized by multiplying the row, this is what happens inside the memory (each of the elements board[0], board[1] and board[2] is a reference to the same list referred by row)

Shows a memory segment after board is initialized.

We can avoid this scenario here by not using row variable to generate board. (Asked in this issue).

>>> board = [['']*3 for _ in range(3)]
>>> board[0][0] = "X"
>>> board
[['X', '', ''], ['', '', ''], ['', '', '']]

▶ Schrödinger's variable *

funcs = []
results = []
for x in range(7):
    def some_func():
        return x
    funcs.append(some_func)
    results.append(some_func())  # note the function call here

funcs_results = [func() for func in funcs]

Output (Python version):

>>> results
[0, 1, 2, 3, 4, 5, 6]
>>> funcs_results
[6, 6, 6, 6, 6, 6, 6]

The values of x were different in every iteration prior to appending some_func to funcs, but all the functions return 6 when they're evaluated after the loop completes.

>>> powers_of_x = [lambda x: x**i for i in range(10)]
>>> [f(2) for f in powers_of_x]
[512, 512, 512, 512, 512, 512, 512, 512, 512, 512]

💡 Explanation:

When defining a function inside a loop that uses the loop variable in its body, the loop function's closure is bound to the variable, not its value. The function looks up x in the surrounding context, rather than using the value of x at the time the function is created. So all of the functions use the latest value assigned to the variable for computation. We can see that it's using the x from the surrounding context (i.e. not a local variable) with:

>>> import inspect
>>> inspect.getclosurevars(funcs[0])
ClosureVars(nonlocals={}, globals={'x': 6}, builtins={}, unbound=set())

Since x is a global value, we can change the value that the funcs will lookup and return by updating x:

>>> x = 42
>>> [func() for func in funcs]
[42, 42, 42, 42, 42, 42, 42]

To get the desired behavior you can pass in the loop variable as a named variable to the function. Why does this work? Because this will define the variable inside the function's scope. It will no longer go to the surrounding (global) scope to look up the variables value but will create a local variable that stores the value of x at that point in time.

funcs = []
for x in range(7):
    def some_func(x=x):
        return x
    funcs.append(some_func)

Output:

>>> funcs_results = [func() for func in funcs]
>>> funcs_results
[0, 1, 2, 3, 4, 5, 6]

It is not longer using the x in the global scope:

>>> inspect.getclosurevars(funcs[0])
ClosureVars(nonlocals={}, globals={}, builtins={}, unbound=set())

▶ The chicken-egg problem *

>>> isinstance(3, int)
True
>>> isinstance(type, object)
True
>>> isinstance(object, type)
True

So which is the "ultimate" base class? There's more to the confusion by the way,

>>> class A: pass
>>> isinstance(A, A)
False
>>> isinstance(type, type)
True
>>> isinstance(object, object)
True

>>> issubclass(int, object)
True
>>> issubclass(type, object)
True
>>> issubclass(object, type)
False

💡 Explanation

type is a metaclass in Python.
Everything is an object in Python, which includes classes as well as their objects (instances).
class type is the metaclass of class object, and every class (including type) has inherited directly or indirectly from object.
There is no real base class among object and type. The confusion in the above snippets is arising because we're thinking about these relationships (issubclass and isinstance) in terms of Python classes. The relationship between object and type can't be reproduced in pure python. To be more precise the following relationships can't be reproduced in pure Python,
- class A is an instance of class B, and class B is an instance of class A.
- class A is an instance of itself.
These relationships between object and type (both being instances of each other as well as themselves) exist in Python because of "cheating" at the implementation level.

▶ Subclass relationships

Output:

>>> from collections.abc import Hashable
>>> issubclass(list, object)
True
>>> issubclass(object, Hashable)
True
>>> issubclass(list, Hashable)
False

The Subclass relationships were expected to be transitive, right? (i.e., if A is a subclass of B, and B is a subclass of C, the A should a subclass of C)

💡 Explanation:

Subclass relationships are not necessarily transitive in Python. Anyone is allowed to define their own, arbitrary __subclasscheck__ in a metaclass.
When issubclass(cls, Hashable) is called, it simply looks for non-Falsey "__hash__" method in cls or anything it inherits from.
Since object is hashable, but list is non-hashable, it breaks the transitivity relation.
More detailed explanation can be found here.

▶ Methods equality and identity

class SomeClass:
    def method(self):
        pass

    @classmethod
    def classm(cls):
        pass

    @staticmethod
    def staticm():
        pass

Output:

>>> print(SomeClass.method is SomeClass.method)
True
>>> print(SomeClass.classm is SomeClass.classm)
False
>>> print(SomeClass.classm == SomeClass.classm)
True
>>> print(SomeClass.staticm is SomeClass.staticm)
True

Accessing classm twice, we get an equal object, but not the same one? Let's see what happens with instances of SomeClass:

o1 = SomeClass()
o2 = SomeClass()

Output:

>>> print(o1.method == o2.method)
False
>>> print(o1.method == o1.method)
True
>>> print(o1.method is o1.method)
False
>>> print(o1.classm is o1.classm)
False
>>> print(o1.classm == o1.classm == o2.classm == SomeClass.classm)
True
>>> print(o1.staticm is o1.staticm is o2.staticm is SomeClass.staticm)
True

Accessing classm or method twice, creates equal but not same objects for the same instance of SomeClass.

💡 Explanation

Functions are descriptors. Whenever a function is accessed as an attribute, the descriptor is invoked, creating a method object which "binds" the function with the object owning the attribute. If called, the method calls the function, implicitly passing the bound object as the first argument (this is how we get self as the first argument, despite not passing it explicitly).

>>> o1.method
<bound method SomeClass.method of <__main__.SomeClass object at ...>>

Accessing the attribute multiple times creates a method object every time! Therefore o1.method is o1.method is never truthy. Accessing functions as class attributes (as opposed to instance) does not create methods, however; so SomeClass.method is SomeClass.method is truthy.

>>> SomeClass.method
<function SomeClass.method at ...>

classmethod transforms functions into class methods. Class methods are descriptors that, when accessed, create a method object which binds the class (type) of the object, instead of the object itself.

>>> o1.classm
<bound method SomeClass.classm of <class '__main__.SomeClass'>>

Unlike functions, classmethods will create a method also when accessed as class attributes (in which case they bind the class, not to the type of it). So SomeClass.classm is SomeClass.classm is falsy.

>>> SomeClass.classm
<bound method SomeClass.classm of <class '__main__.SomeClass'>>

A method object compares equal when both the functions are equal, and the bound objects are the same. So o1.method == o1.method is truthy, although not the same object in memory.
staticmethod transforms functions into a "no-op" descriptor, which returns the function as-is. No method objects are ever created, so comparison with is is truthy.

>>> o1.staticm
<function SomeClass.staticm at ...>
>>> SomeClass.staticm
<function SomeClass.staticm at ...>

Having to create new "method" objects every time Python calls instance methods and having to modify the arguments every time in order to insert self affected performance badly. CPython 3.7 solved it by introducing new opcodes that deal with calling methods without creating the temporary method objects. This is used only when the accessed function is actually called, so the snippets here are not affected, and still generate methods :)

▶ All-true-ation *

>>> all([True, True, True])
True
>>> all([True, True, False])
False

>>> all([])
True
>>> all([[]])
False
>>> all([[[]]])
True

Why's this True-False alteration?

💡 Explanation:

The implementation of all function is equivalent to

def all(iterable):
    for element in iterable:
        if not element:
            return False
    return True

all([]) returns True since the iterable is empty.
all([[]]) returns False because the passed array has one element, [], and in python, an empty list is falsy.
all([[[]]]) and higher recursive variants are always True. This is because the passed array's single element ([[...]]) is no longer empty, and lists with values are truthy.

▶ The surprising comma

Output (< 3.6):

>>> def f(x, y,):
...     print(x, y)
...
>>> def g(x=4, y=5,):
...     print(x, y)
...
>>> def h(x, **kwargs,):
  File "<stdin>", line 1
    def h(x, **kwargs,):
                     ^
SyntaxError: invalid syntax

>>> def h(*args,):
  File "<stdin>", line 1
    def h(*args,):
                ^
SyntaxError: invalid syntax

💡 Explanation:

Trailing comma is not always legal in formal parameters list of a Python function.
In Python, the argument list is defined partially with leading commas and partially with trailing commas. This conflict causes situations where a comma is trapped in the middle, and no rule accepts it.
Note: The trailing comma problem is fixed in Python 3.6. The remarks in this post discuss in brief different usages of trailing commas in Python.

▶ Strings and the backslashes

Output:

>>> print("\"")
"

>>> print(r"\"")
\"

>>> print(r"\")
File "<stdin>", line 1
    print(r"\")
              ^
SyntaxError: EOL while scanning string literal

>>> r'\'' == "\\'"
True

💡 Explanation

In a usual python string, the backslash is used to escape characters that may have a special meaning (like single-quote, double-quote, and the backslash itself).
```
>>> "wt\"f"
'wt"f'
```
In a raw string literal (as indicated by the prefix r), the backslashes pass themselves as is along with the behavior of escaping the following character.
```
>>> r'wt\"f' == 'wt\\"f'
True
>>> print(repr(r'wt\"f'))
'wt\\"f'

>>> print("\n")

>>> print(r"\\n")
'\\n'
```
This means when a parser encounters a backslash in a raw string, it expects another character following it. And in our case (print(r"\")), the backslash escaped the trailing quote, leaving the parser without a terminating quote (hence the SyntaxError). That's why backslashes don't work at the end of a raw string.

▶ not knot!

x = True
y = False

Output:

>>> not x == y
True
>>> x == not y
  File "<input>", line 1
    x == not y
           ^
SyntaxError: invalid syntax

💡 Explanation:

Operator precedence affects how an expression is evaluated, and == operator has higher precedence than not operator in Python.
So not x == y is equivalent to not (x == y) which is equivalent to not (True == False) finally evaluating to True.
But x == not y raises a SyntaxError because it can be thought of being equivalent to (x == not) y and not x == (not y) which you might have expected at first sight.
The parser expected the not token to be a part of the not in operator (because both == and not in operators have the same precedence), but after not being able to find an in token following the not token, it raises a SyntaxError.

▶ Half triple-quoted strings

Output:

>>> print('wtfpython''')
wtfpython
>>> print("wtfpython""")
wtfpython
>>> # The following statements raise `SyntaxError`
>>> # print('''wtfpython')
>>> # print("""wtfpython")
  File "<input>", line 3
    print("""wtfpython")
                        ^
SyntaxError: EOF while scanning triple-quoted string literal

💡 Explanation:

Python supports implicit string literal concatenation, Example,

>>> print("wtf" "python")
wtfpython
>>> print("wtf" "") # or "wtf"""
wtf

''' and """ are also string delimiters in Python which causes a SyntaxError because the Python interpreter was expecting a terminating triple quote as delimiter while scanning the currently encountered triple quoted string literal.

▶ What's wrong with booleans?

# A simple example to count the number of booleans and
# integers in an iterable of mixed data types.
mixed_list = [False, 1.0, "some_string", 3, True, [], False]
integers_found_so_far = 0
booleans_found_so_far = 0

for item in mixed_list:
    if isinstance(item, int):
        integers_found_so_far += 1
    elif isinstance(item, bool):
        booleans_found_so_far += 1

Output:

>>> integers_found_so_far
4
>>> booleans_found_so_far
0

>>> some_bool = True
>>> "wtf" * some_bool
'wtf'
>>> some_bool = False
>>> "wtf" * some_bool
''

def tell_truth():
    True = False
    if True == False:
        print("I have lost faith in truth!")

Output (< 3.x):

>>> tell_truth()
I have lost faith in truth!

💡 Explanation:

bool is a subclass of int in Python

>>> issubclass(bool, int)
True
>>> issubclass(int, bool)
False

And thus, True and False are instances of int

>>> isinstance(True, int)
True
>>> isinstance(False, int)
True

The integer value of True is 1 and that of False is 0.
```
>>> int(True)
1
>>> int(False)
0
```
See this StackOverflow answer for the rationale behind it.
Initially, Python used to have no bool type (people used 0 for false and non-zero value like 1 for true). True, False, and a bool type was added in 2.x versions, but, for backward compatibility, True and False couldn't be made constants. They just were built-in variables, and it was possible to reassign them
Python 3 was backward-incompatible, the issue was finally fixed, and thus the last snippet won't work with Python 3.x!

▶ Class attributes and instance attributes

class A:
    x = 1

class B(A):
    pass

class C(A):
    pass

Output:

>>> A.x, B.x, C.x
(1, 1, 1)
>>> B.x = 2
>>> A.x, B.x, C.x
(1, 2, 1)
>>> A.x = 3
>>> A.x, B.x, C.x # C.x changed, but B.x didn't
(3, 2, 3)
>>> a = A()
>>> a.x, A.x
(3, 3)
>>> a.x += 1
>>> a.x, A.x
(4, 3)

class SomeClass:
    some_var = 15
    some_list = [5]
    another_list = [5]
    def __init__(self, x):
        self.some_var = x + 1
        self.some_list = self.some_list + [x]
        self.another_list += [x]

Output:

>>> some_obj = SomeClass(420)
>>> some_obj.some_list
[5, 420]
>>> some_obj.another_list
[5, 420]
>>> another_obj = SomeClass(111)
>>> another_obj.some_list
[5, 111]
>>> another_obj.another_list
[5, 420, 111]
>>> another_obj.another_list is SomeClass.another_list
True
>>> another_obj.another_list is some_obj.another_list
True

💡 Explanation:

Class variables and variables in class instances are internally handled as dictionaries of a class object. If a variable name is not found in the dictionary of the current class, the parent classes are searched for it.
The += operator modifies the mutable object in-place without creating a new object. So changing the attribute of one instance affects the other instances and the class attribute as well.

▶ yielding None

some_iterable = ('a', 'b')

def some_func(val):
    return "something"

Output (<= 3.7.x):

>>> [x for x in some_iterable]
['a', 'b']
>>> [(yield x) for x in some_iterable]
<generator object <listcomp> at 0x7f70b0a4ad58>
>>> list([(yield x) for x in some_iterable])
['a', 'b']
>>> list((yield x) for x in some_iterable)
['a', None, 'b', None]
>>> list(some_func((yield x)) for x in some_iterable)
['a', 'something', 'b', 'something']

💡 Explanation:

This is a bug in CPython's handling of yield in generators and comprehensions.
Source and explanation can be found here: https://stackoverflow.com/questions/32139885/yield-in-list-comprehensions-and-generator-expressions
Related bug report: https://bugs.python.org/issue10544
Python 3.8+ no longer allows yield inside list comprehension and will throw a SyntaxError.

▶ Yielding from... return! *

def some_func(x):
    if x == 3:
        return ["wtf"]
    else:
        yield from range(x)

Output (> 3.3):

>>> list(some_func(3))
[]

Where did the "wtf" go? Is it due to some special effect of yield from? Let's validate that,

def some_func(x):
    if x == 3:
        return ["wtf"]
    else:
        for i in range(x):
          yield i

Output:

>>> list(some_func(3))
[]

The same result, this didn't work either.

💡 Explanation:

From Python 3.3 onwards, it became possible to use return statement with values inside generators (See PEP380). The official docs say that,

"... return expr in a generator causes StopIteration(expr) to be raised upon exit from the generator."

In the case of some_func(3), StopIteration is raised at the beginning because of return statement. The StopIteration exception is automatically caught inside the list(...) wrapper and the for loop. Therefore, the above two snippets result in an empty list.

To get ["wtf"] from the generator some_func we need to catch the StopIteration exception,

try:
    next(some_func(3))
except StopIteration as e:
    some_string = e.value

>>> some_string
["wtf"]

▶ Nan-reflexivity *

a = float('inf')
b = float('nan')
c = float('-iNf')  # These strings are case-insensitive
d = float('nan')

Output:

>>> a
inf
>>> b
nan
>>> c
-inf
>>> float('some_other_string')
ValueError: could not convert string to float: some_other_string
>>> a == -c # inf==inf
True
>>> None == None # None == None
True
>>> b == d # but nan!=nan
False
>>> 50 / a
0.0
>>> a / a
nan
>>> 23 + b
nan

>>> x = float('nan')
>>> y = x / x
>>> y is y # identity holds
True
>>> y == y # equality fails of y
False
>>> [y] == [y] # but the equality succeeds for the list containing y
True

💡 Explanation:

'inf' and 'nan' are special strings (case-insensitive), which, when explicitly typecast-ed to float type, are used to represent mathematical "infinity" and "not a number" respectively.
Since according to IEEE standards NaN != NaN, obeying this rule breaks the reflexivity assumption of a collection element in Python i.e. if x is a part of a collection like list, the implementations like comparison are based on the assumption that x == x. Because of this assumption, the identity is compared first (since it's faster) while comparing two elements, and the values are compared only when the identities mismatch. The following snippet will make things clearer,
```
>>> x = float('nan')
>>> x == x, [x] == [x]
(False, True)
>>> y = float('nan')
>>> y == y, [y] == [y]
(False, True)
>>> x == y, [x] == [y]
(False, False)
```
Since the identities of x and y are different, the values are considered, which are also different; hence the comparison returns False this time.
Interesting read: Reflexivity, and other pillars of civilization

▶ Mutating the immutable!

This might seem trivial if you know how references work in Python.

some_tuple = ("A", "tuple", "with", "values")
another_tuple = ([1, 2], [3, 4], [5, 6])

Output:

>>> some_tuple[2] = "change this"
TypeError: 'tuple' object does not support item assignment
>>> another_tuple[2].append(1000) #This throws no error
>>> another_tuple
([1, 2], [3, 4], [5, 6, 1000])
>>> another_tuple[2] += [99, 999]
TypeError: 'tuple' object does not support item assignment
>>> another_tuple
([1, 2], [3, 4], [5, 6, 1000, 99, 999])

But I thought tuples were immutable...

💡 Explanation:

Quoting from https://docs.python.org/3/reference/datamodel.html

Immutable sequences An object of an immutable sequence type cannot change once it is created. (If the object contains references to other objects, these other objects may be mutable and may be modified; however, the collection of objects directly referenced by an immutable object cannot change.)
+= operator changes the list in-place. The item assignment doesn't work, but when the exception occurs, the item has already been changed in place.
There's also an explanation in official Python FAQ.

▶ The disappearing variable from outer scope

e = 7
try:
    raise Exception()
except Exception as e:
    pass

Output (Python 2.x):

>>> print(e)
# prints nothing

Output (Python 3.x):

>>> print(e)
NameError: name 'e' is not defined

💡 Explanation:

Source: https://docs.python.org/3/reference/compound_stmts.html#except

When an exception has been assigned using as target, it is cleared at the end of the except clause. This is as if
```
except E as N:
    foo
```
was translated into
```
except E as N:
    try:
        foo
    finally:
        del N
```
This means the exception must be assigned to a different name to be able to refer to it after the except clause. Exceptions are cleared because, with the traceback attached to them, they form a reference cycle with the stack frame, keeping all locals in that frame alive until the next garbage collection occurs.
The clauses are not scoped in Python. Everything in the example is present in the same scope, and the variable e got removed due to the execution of the except clause. The same is not the case with functions that have their separate inner-scopes. The example below illustrates this:
```
def f(x):
    del(x)
    print(x)

x = 5
y = [5, 4, 3]
```
Output:
```
>>> f(x)
UnboundLocalError: local variable 'x' referenced before assignment
>>> f(y)
UnboundLocalError: local variable 'x' referenced before assignment
>>> x
5
>>> y
[5, 4, 3]
```
In Python 2.x, the variable name e gets assigned to Exception() instance, so when you try to print, it prints nothing.

Output (Python 2.x):
```
>>> e
Exception()
>>> print e
# Nothing is printed!
```

▶ The mysterious key type conversion

class SomeClass(str):
    pass

some_dict = {'s': 42}

Output:

>>> type(list(some_dict.keys())[0])
str
>>> s = SomeClass('s')
>>> some_dict[s] = 40
>>> some_dict # expected: Two different keys-value pairs
{'s': 40}
>>> type(list(some_dict.keys())[0])
str

💡 Explanation:

Both the object s and the string "s" hash to the same value because SomeClass inherits the __hash__ method of str class.
SomeClass("s") == "s" evaluates to True because SomeClass also inherits __eq__ method from str class.
Since both the objects hash to the same value and are equal, they are represented by the same key in the dictionary.

For the desired behavior, we can redefine the __eq__ method in SomeClass

class SomeClass(str):
  def __eq__(self, other):
      return (
          type(self) is SomeClass
          and type(other) is SomeClass
          and super().__eq__(other)
      )

  # When we define a custom __eq__, Python stops automatically inheriting the
  # __hash__ method, so we need to define it as well
  __hash__ = str.__hash__

some_dict = {'s':42}

Output:

>>> s = SomeClass('s')
>>> some_dict[s] = 40
>>> some_dict
{'s': 40, 's': 42}
>>> keys = list(some_dict.keys())
>>> type(keys[0]), type(keys[1])
(__main__.SomeClass, str)

▶ Let's see if you can guess this?

a, b = a[b] = {}, 5

Output:

>>> a
{5: ({...}, 5)}

💡 Explanation:

According to Python language reference, assignment statements have the form
```
(target_list "=")+ (expression_list | yield_expression)
```
and

An assignment statement evaluates the expression list (remember that this can be a single expression or a comma-separated list, the latter yielding a tuple) and assigns the single resulting object to each of the target lists, from left to right.

The + in (target_list "=")+ means there can be one or more target lists. In this case, target lists are a, b and a[b] (note the expression list is exactly one, which in our case is {}, 5).
After the expression list is evaluated, its value is unpacked to the target lists from left to right. So, in our case, first the {}, 5 tuple is unpacked to a, b and we now have a = {} and b = 5.
a is now assigned to {}, which is a mutable object.
The second target list is a[b] (you may expect this to throw an error because both a and b have not been defined in the statements before. But remember, we just assigned a to {} and b to 5).
Now, we are setting the key 5 in the dictionary to the tuple ({}, 5) creating a circular reference (the {...} in the output refers to the same object that a is already referencing). Another simpler example of circular reference could be
```
>>> some_list = some_list[0] = [0]
>>> some_list
[[...]]
>>> some_list[0]
[[...]]
>>> some_list is some_list[0]
True
>>> some_list[0][0][0][0][0][0] == some_list
True
```
Similar is the case in our example (a[b][0] is the same object as a)
So to sum it up, you can break the example down to
```
a, b = {}, 5
a[b] = a, b
```
And the circular reference can be justified by the fact that a[b][0] is the same object as a
```
>>> a[b][0] is a
True
```

▶ Exceeds the limit for integer string conversion

>>> # Python 3.10.6
>>> int("2" * 5432)

>>> # Python 3.10.8
>>> int("2" * 5432)

Output:

>>> # Python 3.10.6
222222222222222222222222222222222222222222222222222222222222222...

>>> # Python 3.10.8
Traceback (most recent call last):
   ...
ValueError: Exceeds the limit (4300) for integer string conversion:
   value has 5432 digits; use sys.set_int_max_str_digits()
   to increase the limit.

💡 Explanation:

This call to int() works fine in Python 3.10.6 and raises a ValueError in Python 3.10.8. Note that Python can still work with large integers. The error is only raised when converting between integers and strings.

Fortunately, you can increase the limit for the allowed number of digits when you expect an operation to exceed it. To do this, you can use one of the following:

The -X int_max_str_digits command-line flag
The set_int_max_str_digits() function from the sys module
The PYTHONINTMAXSTRDIGITS environment variable

Check the documentation for more details on changing the default limit if you expect your code to exceed this value.

Section: Slippery Slopes

▶ Modifying a dictionary while iterating over it

x = {0: None}

for i in x:
    del x[i]
    x[i+1] = None
    print(i)

Output (Python 2.7- Python 3.5):

Yes, it runs for exactly eight times and stops.

💡 Explanation:

Iteration over a dictionary that you edit at the same time is not supported.
It runs eight times because that's the point at which the dictionary resizes to hold more keys (we have eight deletion entries, so a resize is needed). This is actually an implementation detail.
How deleted keys are handled and when the resize occurs might be different for different Python implementations.
So for Python versions other than Python 2.7 - Python 3.5, the count might be different from 8 (but whatever the count is, it's going to be the same every time you run it). You can find some discussion around this here or in this StackOverflow thread.
Python 3.7.6 onwards, you'll see RuntimeError: dictionary keys changed during iteration exception if you try to do this.

▶ Stubborn `del` operation

class SomeClass:
    def __del__(self):
        print("Deleted!")

Output: 1.

>>> x = SomeClass()
>>> y = x
>>> del x # this should print "Deleted!"
>>> del y
Deleted!

Phew, deleted at last. You might have guessed what saved __del__ from being called in our first attempt to delete x. Let's add more twists to the example.

>>> x = SomeClass()
>>> y = x
>>> del x
>>> y # check if y exists
<__main__.SomeClass instance at 0x7f98a1a67fc8>
>>> del y # Like previously, this should print "Deleted!"
>>> globals() # oh, it didn't. Let's check all our global variables and confirm
Deleted!
{'__builtins__': <module '__builtin__' (built-in)>, 'SomeClass': <class __main__.SomeClass at 0x7f98a1a5f668>, '__package__': None, '__name__': '__main__', '__doc__': None}

Okay, now it's deleted 😕

💡 Explanation:

del x doesn’t directly call x.__del__().
When del x is encountered, Python deletes the name x from current scope and decrements by 1 the reference count of the object x referenced. __del__() is called only when the object's reference count reaches zero.
In the second output snippet, __del__() was not called because the previous statement (>>> y) in the interactive interpreter created another reference to the same object (specifically, the _ magic variable which references the result value of the last non None expression on the REPL), thus preventing the reference count from reaching zero when del y was encountered.
Calling globals (or really, executing anything that will have a non None result) caused _ to reference the new result, dropping the existing reference. Now the reference count reached 0 and we can see "Deleted!" being printed (finally!).

▶ The out of scope variable

a = 1
def some_func():
    return a

def another_func():
    a += 1
    return a

def some_closure_func():
    a = 1
    def some_inner_func():
        return a
    return some_inner_func()

def another_closure_func():
    a = 1
    def another_inner_func():
        a += 1
        return a
    return another_inner_func()

Output:

>>> some_func()
1
>>> another_func()
UnboundLocalError: local variable 'a' referenced before assignment

>>> some_closure_func()
1
>>> another_closure_func()
UnboundLocalError: local variable 'a' referenced before assignment

💡 Explanation:

When you make an assignment to a variable in scope, it becomes local to that scope. So a becomes local to the scope of another_func, but it has not been initialized previously in the same scope, which throws an error.

To modify the outer scope variable a in another_func, we have to use the global keyword.

def another_func()
    global a
    a += 1
    return a

Output:

>>> another_func()
2

In another_closure_func, a becomes local to the scope of another_inner_func, but it has not been initialized previously in the same scope, which is why it throws an error.

To modify the outer scope variable a in another_inner_func, use the nonlocal keyword. The nonlocal statement is used to refer to variables defined in the nearest outer (excluding the global) scope.

def another_func():
    a = 1
    def another_inner_func():
        nonlocal a
        a += 1
        return a
    return another_inner_func()

Output:

>>> another_func()
2

The keywords global and nonlocal tell the python interpreter to not declare new variables and look them up in the corresponding outer scopes.
Read this short but an awesome guide to learn more about how namespaces and scope resolution works in Python.

▶ Deleting a list item while iterating

list_1 = [1, 2, 3, 4]
list_2 = [1, 2, 3, 4]
list_3 = [1, 2, 3, 4]
list_4 = [1, 2, 3, 4]

for idx, item in enumerate(list_1):
    del item

for idx, item in enumerate(list_2):
    list_2.remove(item)

for idx, item in enumerate(list_3[:]):
    list_3.remove(item)

for idx, item in enumerate(list_4):
    list_4.pop(idx)

Output:

>>> list_1
[1, 2, 3, 4]
>>> list_2
[2, 4]
>>> list_3
[]
>>> list_4
[2, 4]

Can you guess why the output is [2, 4]?

💡 Explanation:

It's never a good idea to change the object you're iterating over. The correct way to do so is to iterate over a copy of the object instead, and list_3[:] does just that.

>>> some_list = [1, 2, 3, 4]
>>> id(some_list)
139798789457608
>>> id(some_list[:]) # Notice that python creates new object for sliced list.
139798779601192

Difference between del, remove, and pop:

del var_name just removes the binding of the var_name from the local or global namespace (That's why the list_1 is unaffected).
remove removes the first matching value, not a specific index, raises ValueError if the value is not found.
pop removes the element at a specific index and returns it, raises IndexError if an invalid index is specified.

Why the output is [2, 4]?

The list iteration is done index by index, and when we remove 1 from list_2 or list_4, the contents of the lists are now [2, 3, 4]. The remaining elements are shifted down, i.e., 2 is at index 0, and 3 is at index 1. Since the next iteration is going to look at index 1 (which is the 3), the 2 gets skipped entirely. A similar thing will happen with every alternate element in the list sequence.

Refer to this StackOverflow thread explaining the example
See also this nice StackOverflow thread for a similar example related to dictionaries in Python.

▶ Lossy zip of iterators *

>>> numbers = list(range(7))
>>> numbers
[0, 1, 2, 3, 4, 5, 6]
>>> first_three, remaining = numbers[:3], numbers[3:]
>>> first_three, remaining
([0, 1, 2], [3, 4, 5, 6])
>>> numbers_iter = iter(numbers)
>>> list(zip(numbers_iter, first_three)) 
[(0, 0), (1, 1), (2, 2)]
# so far so good, let's zip the remaining
>>> list(zip(numbers_iter, remaining))
[(4, 3), (5, 4), (6, 5)]

Where did element 3 go from the numbers list?

💡 Explanation:

From Python docs, here's an approximate implementation of zip function,

def zip(*iterables):
    sentinel = object()
    iterators = [iter(it) for it in iterables]
    while iterators:
        result = []
        for it in iterators:
            elem = next(it, sentinel)
            if elem is sentinel: return
            result.append(elem)
        yield tuple(result)

So the function takes in arbitrary number of iterable objects, adds each of their items to the result list by calling the next function on them, and stops whenever any of the iterable is exhausted.
The caveat here is when any iterable is exhausted, the existing elements in the result list are discarded. That's what happened with 3 in the numbers_iter.

The correct way to do the above using zip would be,

>>> numbers = list(range(7))
>>> numbers_iter = iter(numbers)
>>> list(zip(first_three, numbers_iter))
[(0, 0), (1, 1), (2, 2)]
>>> list(zip(remaining, numbers_iter))
[(3, 3), (4, 4), (5, 5), (6, 6)]

The first argument of zip should be the one with fewest elements.

▶ Loop variables leaking out!

for x in range(7):
    if x == 6:
        print(x, ': for x inside loop')
print(x, ': x in global')

Output:

6 : for x inside loop
6 : x in global

But x was never defined outside the scope of for loop...

# This time let's initialize x first
x = -1
for x in range(7):
    if x == 6:
        print(x, ': for x inside loop')
print(x, ': x in global')

Output:

6 : for x inside loop
6 : x in global

Output (Python 2.x):

>>> x = 1
>>> print([x for x in range(5)])
[0, 1, 2, 3, 4]
>>> print(x)
4

Output (Python 3.x):

>>> x = 1
>>> print([x for x in range(5)])
[0, 1, 2, 3, 4]
>>> print(x)
1

💡 Explanation:

In Python, for-loops use the scope they exist in and leave their defined loop-variable behind. This also applies if we explicitly defined the for-loop variable in the global namespace before. In this case, it will rebind the existing variable.
The differences in the output of Python 2.x and Python 3.x interpreters for list comprehension example can be explained by following change documented in What’s New In Python 3.0 changelog:

"List comprehensions no longer support the syntactic form [... for var in item1, item2, ...]. Use [... for var in (item1, item2, ...)] instead. Also, note that list comprehensions have different semantics: they are closer to syntactic sugar for a generator expression inside a list() constructor, and in particular, the loop control variables are no longer leaked into the surrounding scope."

▶ Beware of default mutable arguments!

def some_func(default_arg=[]):
    default_arg.append("some_string")
    return default_arg

Output:

>>> some_func()
['some_string']
>>> some_func()
['some_string', 'some_string']
>>> some_func([])
['some_string']
>>> some_func()
['some_string', 'some_string', 'some_string']

💡 Explanation:

The default mutable arguments of functions in Python aren't really initialized every time you call the function. Instead, the recently assigned value to them is used as the default value. When we explicitly passed [] to some_func as the argument, the default value of the default_arg variable was not used, so the function returned as expected.

def some_func(default_arg=[]):
    default_arg.append("some_string")
    return default_arg

Output:

>>> some_func.__defaults__ #This will show the default argument values for the function
([],)
>>> some_func()
>>> some_func.__defaults__
(['some_string'],)
>>> some_func()
>>> some_func.__defaults__
(['some_string', 'some_string'],)
>>> some_func([])
>>> some_func.__defaults__
(['some_string', 'some_string'],)

A common practice to avoid bugs due to mutable arguments is to assign None as the default value and later check if any value is passed to the function corresponding to that argument. Example:
```
def some_func(default_arg=None):
    if default_arg is None:
        default_arg = []
    default_arg.append("some_string")
    return default_arg
```

▶ Catching the Exceptions

some_list = [1, 2, 3]
try:
    # This should raise an ``IndexError``
    print(some_list[4])
except IndexError, ValueError:
    print("Caught!")

try:
    # This should raise a ``ValueError``
    some_list.remove(4)
except IndexError, ValueError:
    print("Caught again!")

Output (Python 2.x):

Caught!

ValueError: list.remove(x): x not in list

Output (Python 3.x):

  File "<input>", line 3
    except IndexError, ValueError:
                     ^
SyntaxError: invalid syntax

💡 Explanation

To add multiple Exceptions to the except clause, you need to pass them as parenthesized tuple as the first argument. The second argument is an optional name, which when supplied will bind the Exception instance that has been raised. Example,

some_list = [1, 2, 3]
try:
   # This should raise a ``ValueError``
   some_list.remove(4)
except (IndexError, ValueError), e:
   print("Caught again!")
   print(e)

Output (Python 2.x):

Caught again!
list.remove(x): x not in list

Output (Python 3.x):

  File "<input>", line 4
    except (IndexError, ValueError), e:
                                     ^
IndentationError: unindent does not match any outer indentation level

Separating the exception from the variable with a comma is deprecated and does not work in Python 3; the correct way is to use as. Example,

some_list = [1, 2, 3]
try:
    some_list.remove(4)

except (IndexError, ValueError) as e:
    print("Caught again!")
    print(e)

Output:

Caught again!
list.remove(x): x not in list

▶ Same operands, different story!

a = [1, 2, 3, 4]
b = a
a = a + [5, 6, 7, 8]

Output:

>>> a
[1, 2, 3, 4, 5, 6, 7, 8]
>>> b
[1, 2, 3, 4]

a = [1, 2, 3, 4]
b = a
a += [5, 6, 7, 8]

Output:

>>> a
[1, 2, 3, 4, 5, 6, 7, 8]
>>> b
[1, 2, 3, 4, 5, 6, 7, 8]

💡 Explanation:

a += b doesn't always behave the same way as a = a + b. Classes may implement the op= operators differently, and lists do this.
The expression a = a + [5,6,7,8] generates a new list and sets a's reference to that new list, leaving b unchanged.
The expression a += [5,6,7,8] is actually mapped to an "extend" function that operates on the list such that a and b still point to the same list that has been modified in-place.

▶ Name resolution ignoring class scope

x = 5
class SomeClass:
    x = 17
    y = (x for i in range(10))

Output:

>>> list(SomeClass.y)[0]
5

x = 5
class SomeClass:
    x = 17
    y = [x for i in range(10)]

Output (Python 2.x):

>>> SomeClass.y[0]
17

Output (Python 3.x):

>>> SomeClass.y[0]
5

💡 Explanation

Scopes nested inside class definition ignore names bound at the class level.
A generator expression has its own scope.
Starting from Python 3.X, list comprehensions also have their own scope.

▶ Rounding like a banker *

Let's implement a naive function to get the middle element of a list:

def get_middle(some_list):
    mid_index = round(len(some_list) / 2)
    return some_list[mid_index - 1]

Python 3.x:

>>> get_middle([1])  # looks good
1
>>> get_middle([1,2,3])  # looks good
2
>>> get_middle([1,2,3,4,5])  # huh?
2
>>> len([1,2,3,4,5]) / 2  # good
2.5
>>> round(len([1,2,3,4,5]) / 2)  # why?
2

It seems as though Python rounded 2.5 to 2.

💡 Explanation:

This is not a float precision error, in fact, this behavior is intentional. Since Python 3.0, round() uses banker's rounding where .5 fractions are rounded to the nearest even number:

>>> round(0.5)
0
>>> round(1.5)
2
>>> round(2.5)
2
>>> import numpy  # numpy does the same
>>> numpy.round(0.5)
0.0
>>> numpy.round(1.5)
2.0
>>> numpy.round(2.5)
2.0

This is the recommended way to round .5 fractions as described in IEEE 754. However, the other way (round away from zero) is taught in school most of the time, so banker's rounding is likely not that well known. Furthermore, some of the most popular programming languages (for example: JavaScript, Java, C/C++, Ruby, Rust) do not use banker's rounding either. Therefore, this is still quite special to Python and may result in confusion when rounding fractions.
See the round() docs or this stackoverflow thread for more information.
Note that get_middle([1]) only returned 1 because the index was round(0.5) - 1 = 0 - 1 = -1, returning the last element in the list.

▶ Needles in a Haystack *

I haven't met even a single experience Pythonist till date who has not come across one or more of the following scenarios,

x, y = (0, 1) if True else None, None

Output:

>>> x, y  # expected (0, 1)
((0, 1), None)

t = ('one', 'two')
for i in t:
    print(i)

t = ('one')
for i in t:
    print(i)

t = ()
print(t)

Output:

one
two
o
n
e
tuple()

ten_words_list = [
    "some",
    "very",
    "big",
    "list",
    "that"
    "consists",
    "of",
    "exactly",
    "ten",
    "words"
]

Output

>>> len(ten_words_list)
9

4. Not asserting strongly enough

a = "python"
b = "javascript"

Output:

# An assert statement with an assertion failure message.
>>> assert(a == b, "Both languages are different")
# No AssertionError is raised

some_list = [1, 2, 3]
some_dict = {
  "key_1": 1,
  "key_2": 2,
  "key_3": 3
}

some_list = some_list.append(4) 
some_dict = some_dict.update({"key_4": 4})

Output:

>>> print(some_list)
None
>>> print(some_dict)
None

def some_recursive_func(a):
    if a[0] == 0:
        return
    a[0] -= 1
    some_recursive_func(a)
    return a

def similar_recursive_func(a):
    if a == 0:
        return a
    a -= 1
    similar_recursive_func(a)
    return a

Output:

>>> some_recursive_func([5, 0])
[0, 0]
>>> similar_recursive_func(5)
4

💡 Explanation:

For 1, the correct statement for expected behavior is x, y = (0, 1) if True else (None, None).
For 2, the correct statement for expected behavior is t = ('one',) or t = 'one', (missing comma) otherwise the interpreter considers t to be a str and iterates over it character by character.
() is a special token and denotes empty tuple.
In 3, as you might have already figured out, there's a missing comma after 5th element ("that") in the list. So by implicit string literal concatenation,
```
>>> ten_words_list
['some', 'very', 'big', 'list', 'thatconsists', 'of', 'exactly', 'ten', 'words']
```

No AssertionError was raised in 4th snippet because instead of asserting the individual expression a == b, we're asserting entire tuple. The following snippet will clear things up,

>>> a = "python"
>>> b = "javascript"
>>> assert a == b
Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
AssertionError

>>> assert (a == b, "Values are not equal")
<stdin>:1: SyntaxWarning: assertion is always true, perhaps remove parentheses?

>>> assert a == b, "Values are not equal"
Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
AssertionError: Values are not equal

As for the fifth snippet, most methods that modify the items of sequence/mapping objects like list.append, dict.update, list.sort, etc. modify the objects in-place and return None. The rationale behind this is to improve performance by avoiding making a copy of the object if the operation can be done in-place (Referred from here).
Last one should be fairly obvious, mutable object (like list) can be altered in the function, and the reassignment of an immutable (a -= 1) is not an alteration of the value.
Being aware of these nitpicks can save you hours of debugging effort in the long run.

▶ Splitsies *

>>> 'a'.split()
['a']

# is same as
>>> 'a'.split(' ')
['a']

# but
>>> len(''.split())
0

# isn't the same as
>>> len(''.split(' '))
1

💡 Explanation:

It might appear at first that the default separator for split is a single space ' ', but as per the docs

If sep is not specified or is None, a different splitting algorithm is applied: runs of consecutive whitespace are regarded as a single separator, and the result will contain no empty strings at the start or end if the string has leading or trailing whitespace. Consequently, splitting an empty string or a string consisting of just whitespace with a None separator returns []. If sep is given, consecutive delimiters are not grouped together and are deemed to delimit empty strings (for example, '1,,2'.split(',') returns ['1', '', '2']). Splitting an empty string with a specified separator returns [''].
Noticing how the leading and trailing whitespaces are handled in the following snippet will make things clear,
```
>>> ' a '.split(' ')
['', 'a', '']
>>> ' a '.split()
['a']
>>> ''.split(' ')
['']
```

▶ Wild imports *

# File: module.py

def some_weird_name_func_():
    print("works!")

def _another_weird_name_func():
    print("works!")

Output

>>> from module import *
>>> some_weird_name_func_()
"works!"
>>> _another_weird_name_func()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name '_another_weird_name_func' is not defined

💡 Explanation:

It is often advisable to not use wildcard imports. The first obvious reason for this is, in wildcard imports, the names with a leading underscore don't get imported. This may lead to errors during runtime.

Had we used from ... import a, b, c syntax, the above NameError wouldn't have occurred.

>>> from module import some_weird_name_func_, _another_weird_name_func
>>> _another_weird_name_func()
works!

If you really want to use wildcard imports, then you'd have to define the list __all__ in your module that will contain a list of public objects that'll be available when we do wildcard imports.

__all__ = ['_another_weird_name_func']

def some_weird_name_func_():
    print("works!")

def _another_weird_name_func():
    print("works!")

Output

>>> _another_weird_name_func()
"works!"
>>> some_weird_name_func_()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'some_weird_name_func_' is not defined

▶ All sorted? *

>>> x = 7, 8, 9
>>> sorted(x) == x
False
>>> sorted(x) == sorted(x)
True

>>> y = reversed(x)
>>> sorted(y) == sorted(y)
False

💡 Explanation:

The sorted method always returns a list, and comparing lists and tuples always returns False in Python.

>>> [] == tuple()
False
>>> x = 7, 8, 9
>>> type(x), type(sorted(x))
(tuple, list)

Unlike sorted, the reversed method returns an iterator. Why? Because sorting requires the iterator to be either modified in-place or use an extra container (a list), whereas reversing can simply work by iterating from the last index to the first.
So during comparison sorted(y) == sorted(y), the first call to sorted() will consume the iterator y, and the next call will just return an empty list.
```
>>> x = 7, 8, 9
>>> y = reversed(x)
>>> sorted(y), sorted(y)
([7, 8, 9], [])
```

▶ Midnight time doesn't exist?

from datetime import datetime

midnight = datetime(2018, 1, 1, 0, 0)
midnight_time = midnight.time()

noon = datetime(2018, 1, 1, 12, 0)
noon_time = noon.time()

if midnight_time:
    print("Time at midnight is", midnight_time)

if noon_time:
    print("Time at noon is", noon_time)

Output (< 3.5):

('Time at noon is', datetime.time(12, 0))

The midnight time is not printed.

💡 Explanation:

Before Python 3.5, the boolean value for datetime.time object was considered to be False if it represented midnight in UTC. It is error-prone when using the if obj: syntax to check if the obj is null or some equivalent of "empty."

‫ بخش: گنجینه‌های پنهان!

‫ این بخش شامل چند مورد جالب و کمتر شناخته‌شده درباره‌ی پایتون است که بیشتر مبتدی‌هایی مثل من از آن بی‌خبرند (البته دیگر اینطور نیست).

▶ ‫ خب پایتون، می‌توانی کاری کنی پرواز کنم؟

‫ خب، بفرمایید

import antigravity

‫ خروجی: Sshh... It's a super-secret.

‫ 💡 توضیح:

‫ ماژول antigravity یکی از معدود ایستر اِگ‌هایی است که توسط توسعه‌دهندگان پایتون ارائه شده است.
‫ دستور import antigravity باعث می‌شود مرورگر وب به سمت کمیک کلاسیک XKCD در مورد پایتون باز شود.
‫ البته موضوع عمیق‌تر است؛ در واقع یک ایستر اگ دیگر داخل این ایستر اگ وجود دارد. اگر به کد منبع نگاه کنید، یک تابع تعریف شده که ادعا می‌کند الگوریتم جئوهشینگ XKCD را پیاده‌سازی کرده است.

▶ ‫ `goto`، ولی چرا؟

from goto import goto, label
for i in range(9):
    for j in range(9):
        for k in range(9):
            print("I am trapped, please rescue!")
            if k == 2:
                goto .breakout # خروج از یک حلقه‌ی تودرتوی عمیق
label .breakout
print("Freedom!")

‫ خروجی (پایتون ۲.۳):

I am trapped, please rescue!
I am trapped, please rescue!
Freedom!

‫ 💡 توضیح:

‫ نسخه‌ی قابل استفاده‌ای از goto در پایتون به عنوان یک شوخی در اول آوریل ۲۰۰۴ معرفی شد.
‫ نسخه‌های فعلی پایتون فاقد این ماژول هستند.
‫ اگرچه این ماژول واقعاً کار می‌کند، ولی لطفاً از آن استفاده نکنید. در این صفحه می‌توانید دلیل عدم حضور دستور goto در پایتون را مطالعه کنید.

▶ ‫ خودتان را آماده کنید!

‫ اگر جزو افرادی هستید که دوست ندارند در پایتون برای مشخص کردن محدوده‌ها از فضای خالی (whitespace) استفاده کنند، می‌توانید با ایمپورت کردن ماژول زیر از آکولاد {} به سبک زبان C استفاده کنید:

from __future__ import braces

‫ خروجی:

  File "some_file.py", line 1
    from __future__ import braces
SyntaxError: not a chance

‫ آکولاد؟ هرگز! اگر از این بابت ناامید شدید، بهتر است از جاوا استفاده کنید. خب، یک چیز شگفت‌آور دیگر؛ آیا می‌توانید تشخیص دهید که ارور SyntaxError در کجای کد ماژول __future__ اینجا ایجاد می‌شود؟

‫ 💡 توضیح:

‫ ماژول __future__ معمولاً برای ارائه قابلیت‌هایی از نسخه‌های آینده پایتون به کار می‌رود. اما کلمه «future» (آینده) در این زمینه خاص، حالت طنز و کنایه دارد.
‫ این مورد یک «ایستر اگ» (easter egg) است که به احساسات جامعه برنامه‌نویسان پایتون در این خصوص اشاره دارد.
‫ کد مربوط به این موضوع در واقع اینجا در فایل future.c قرار دارد.
‫ زمانی که کامپایلر CPython با یک عبارت future مواجه می‌شود، ابتدا کد مرتبط در future.c را اجرا کرده و سپس آن را همانند یک دستور ایمپورت عادی در نظر می‌گیرد.

▶ ‫ بیایید با «عمو زبان مهربان برای همیشه» آشنا شویم

Output (Python 3.x)

>>> from __future__ import barry_as_FLUFL
>>> "Ruby" != "Python" # شکی در این نیست.
  File "some_file.py", line 1
    "Ruby" != "Python"
              ^
SyntaxError: invalid syntax

>>> "Ruby" <> "Python"
True

‫ حالا می‌رسیم به اصل ماجرا.

‫ 💡 توضیح:

‫ این مورد مربوط به PEP-401 است که در تاریخ ۱ آوریل ۲۰۰۹ منتشر شد (اکنون می‌دانید این یعنی چه!).
‫ نقل قولی از PEP-401:

‫ با توجه به اینکه عملگر نابرابری != در پایتون ۳.۰ یک اشتباه وحشتناک و انگشت‌سوز (!) بوده است، عمو زبان مهربان برای همیشه (FLUFL) عملگر الماسی‌شکل <> را مجدداً به‌عنوان تنها روش درست برای این منظور بازگردانده است.
‫ البته «عمو بَری» چیزهای بیشتری برای گفتن در این PEP داشت؛ می‌توانید آن‌ها را اینجا مطالعه کنید.
‫ این قابلیت در محیط تعاملی به خوبی عمل می‌کند، اما در زمان اجرای کد از طریق فایل پایتون، با خطای SyntaxError روبرو خواهید شد (برای اطلاعات بیشتر به این issue مراجعه کنید). با این حال، می‌توانید کد خود را درون یک eval یا compile قرار دهید تا این قابلیت فعال شود.
```
from __future__ import barry_as_FLUFL
print(eval('"Ruby" <> "Python"'))
```

▶ ‫ حتی پایتون هم می‌داند که عشق پیچیده است

import this

Wait, what's this? this is love ❤️

‫ خروجی:

The Zen of Python, by Tim Peters

Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!

‫ این ذنِ پایتون است!

>>> love = this
>>> this is love
True
>>> love is True
False
>>> love is False
False
>>> love is not True or False
True
>>> love is not True or False; love is love  # عشق پیجیده است
True

‫ 💡 توضیح:

‫ ماژول this در پایتون، یک ایستر اگ برای «ذنِ پایتون» (PEP 20) است.
‫ اگر این موضوع به‌اندازه کافی جالب است، حتماً پیاده‌سازی this.py را ببینید. نکته جالب این است که کد مربوط به ذنِ پایتون، خودش اصول ذن را نقض کرده است (و احتمالاً این تنها جایی است که چنین اتفاقی می‌افتد).
‫ درباره جمله love is not True or False; love is love، اگرچه طعنه‌آمیز است، اما خود گویاست. (اگر واضح نیست، لطفاً مثال‌های مربوط به عملگرهای is و is not را مشاهده کنید.)

▶ ‫ بله، این واقعاً وجود دارد!

‫ عبارت else برای حلقه‌ها. یک مثال معمول آن می‌تواند چنین باشد:

  def does_exists_num(l, to_find):
      for num in l:
          if num == to_find:
              print("Exists!")
              break
      else:
          print("Does not exist")

خروجی:

>>> some_list = [1, 2, 3, 4, 5]
>>> does_exists_num(some_list, 4)
Exists!
>>> does_exists_num(some_list, -1)
Does not exist

عبارت else در مدیریت استثناها. مثالی از آن:

try:
    pass
except:
    print("Exception occurred!!!")
else:
    print("Try block executed successfully...")

خروجی:

Try block executed successfully...

‫ 💡 توضیح:

عبارت else بعد از حلقه‌ها تنها زمانی اجرا می‌شود که در هیچ‌کدام از تکرارها (iterations) از دستور break استفاده نشده باشد. می‌توانید آن را به عنوان یک شرط «بدون شکست» (nobreak) در نظر بگیرید.
عبارت else پس از بلاک try به عنوان «عبارت تکمیل» (completion clause) نیز شناخته می‌شود؛ چراکه رسیدن به عبارت else در ساختار try به این معنی است که بلاک try بدون رخ دادن استثنا با موفقیت تکمیل شده است.

▶ Ellipsis *

def some_func():
    Ellipsis

خروجی

>>> some_func()
# بدون خروجی و بدون خطا

>>> SomeRandomString
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'SomeRandomString' is not defined

>>> Ellipsis
Ellipsis

‫ 💡توضیح

‫ در پایتون، Ellipsis یک شیء درونی (built-in) است که به صورت سراسری (global) در دسترس است و معادل ... است.
```
>>> ...
Ellipsis
```
‫ Ellipsis می‌تواند برای چندین منظور استفاده شود:
- ‫ به عنوان یک نگه‌دارنده برای کدی که هنوز نوشته نشده است (مانند دستور pass)
- ‫ در سینتکس برش (slicing) برای نمایش برش کامل در ابعاد باقی‌مانده
```
>>> import numpy as np
>>> three_dimensional_array = np.arange(8).reshape(2, 2, 2)
array([
    [
        [0, 1],
        [2, 3]
    ],

    [
        [4, 5],
        [6, 7]
    ]
])
```
‫ بنابراین، آرایه‌ی three_dimensional_array ما، آرایه‌ای از آرایه‌ها از آرایه‌ها است. فرض کنیم می‌خواهیم عنصر دوم (اندیس 1) از تمامی آرایه‌های درونی را چاپ کنیم؛ در این حالت می‌توانیم از Ellipsis برای عبور از تمامی ابعاد قبلی استفاده کنیم:
```
>>> three_dimensional_array[:,:,1]
array([[1, 3],
   [5, 7]])
>>> three_dimensional_array[..., 1] # با استفاده از Ellipsis.
array([[1, 3],
   [5, 7]])
```
‫ نکته: این روش برای آرایه‌هایی با هر تعداد بُعد کار می‌کند. حتی می‌توانید از برش (slice) در بُعد اول و آخر استفاده کرده و ابعاد میانی را نادیده بگیرید (به صورت n_dimensional_array[first_dim_slice, ..., last_dim_slice]).
- ‫ در نوع‌دهی (type hinting) برای اشاره به بخشی از نوع (مانند Callable[..., int] یا Tuple[str, ...]) استفاده می‌شود.
- ‫ همچنین می‌توانید از Ellipsis به عنوان آرگومان پیش‌فرض تابع استفاده کنید (برای مواردی که می‌خواهید میان «آرگومانی ارسال نشده است» و «مقدار None ارسال شده است» تمایز قائل شوید).

▶ ‫ بی‌نهایت (`Inpinity`)

‫ این املای کلمه تعمداً به همین شکل نوشته شده است. لطفاً برای اصلاح آن درخواست (patch) ارسال نکنید.

‫ خروجی (پایتون 3.x):

>>> infinity = float('infinity')
>>> hash(infinity)
314159
>>> hash(float('-inf'))
-314159

‫ 💡 توضیح:

‫ هش (hash) مقدار بی‌نهایت برابر با 10⁵ × π است.
‫ نکته جالب اینکه در پایتون ۳ هشِ مقدار float('-inf') برابر با «-10⁵ × π» است، در حالی که در پایتون ۲ برابر با «-10⁵ × e» است.

▶ ‫ بیایید خرابکاری کنیم

class Yo(object):
    def __init__(self):
        self.__honey = True
        self.bro = True

‫ خروجی:

>>> Yo().bro
True
>>> Yo().__honey
AttributeError: 'Yo' object has no attribute '__honey'
>>> Yo()._Yo__honey
True

class Yo(object):
    def __init__(self):
        # این بار بیایید چیزی متقارن را امتحان کنیم
        self.__honey__ = True
        self.bro = True

‫ خروجی:

>>> Yo().bro
True

>>> Yo()._Yo__honey__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'Yo' object has no attribute '_Yo__honey__'

چرا کد Yo()._Yo__honey کار کرد؟

_A__variable = "Some value"

class A(object):
    def some_func(self):
        return __variable # هنوز در هیچ جا مقداردهی اولیه نشده است

‫ خروجی:

>>> A().__variable
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'A' object has no attribute '__variable'

>>> A().some_func()
'Some value'

‫ 💡 توضیح:

‫ تغییر نام برای جلوگیری از برخورد نام‌ها بین فضاهای نام مختلف استفاده می‌شود.
‫ در پایتون، مفسر نام‌های اعضای کلاس که با __ (دو آندرلاین که به عنوان "دندر" شناخته می‌شود) شروع می‌شوند و بیش از یک آندرلاین انتهایی ندارند را با اضافه کردن _NameOfTheClass در ابتدای آنها تغییر می‌دهد.
‫ بنابراین، برای دسترسی به ویژگی __honey در اولین قطعه کد، مجبور بودیم _Yo را به ابتدای آن اضافه کنیم، که از بروز تعارض با ویژگی با همان نام تعریف‌شده در هر کلاس دیگری جلوگیری می‌کند.
‫ اما چرا در دومین قطعه کد کار نکرد؟ زیرا تغییر نام، نام‌هایی که با دو آندرلاین خاتمه می‌یابند را شامل نمی‌شود.
‫ قطعه سوم نیز نتیجه تغییر نام بود. نام __variable در عبارت return __variable به _A__variable تغییر یافت، که همچنین همان نام متغیری است که در محدوده بیرونی تعریف کرده بودیم.
‫ همچنین، اگر نام تغییر یافته بیش از ۲۵۵ کاراکتر باشد، برش داده می‌شود.

‫ بخش: ظاهرها فریبنده‌اند!

▶ ‫ خطوط را رد می‌کند؟

Output:

>>> value = 11
>>> valuе = 32
>>> value
11

‫ چی?

‫ نکته: ساده‌ترین روش برای بازتولید این رفتار، کپی کردن دستورات از کد بالا و جایگذاری (paste) آن‌ها در فایل یا محیط تعاملی (shell) خودتان است.

‫ 💡 توضیح

‫ برخی از حروف غیرغربی کاملاً مشابه حروف الفبای انگلیسی به نظر می‌رسند، اما مفسر پایتون آن‌ها را متفاوت در نظر می‌گیرد.

>>> ord('е') # حرف سیریلیک «е» (Ye)
1077
>>> ord('e') # حرف لاتین «e»، که در انگلیسی استفاده می‌شود و با صفحه‌کلید استاندارد تایپ می‌گردد
101
>>> 'е' == 'e'
False

>>> value = 42 # حرف لاتین e
>>> valuе = 23 # حرف سیریلیک «е»؛ مفسر پایتون نسخه ۲ در اینجا خطای `SyntaxError` ایجاد می‌کند
>>> value
42

‫ تابع داخلی ord()، کدپوینت یونیکد مربوط به یک نویسه را برمی‌گرداند. موقعیت‌های کدی متفاوت برای حرف سیریلیک «е» و حرف لاتین «e»، علت رفتار مثال بالا را توجیه می‌کنند.

▶ ‫ تله‌پورت کردن

# `pip install numpy` first.
import numpy as np

def energy_send(x):
    # مقداردهی اولیه یک آرایه numpy
    np.array([float(x)])

def energy_receive():
    # بازگرداندن یک آرایه‌ی خالی numpy
    return np.empty((), dtype=np.float).tolist()

‫ خروجی:

>>> energy_send(123.456)
>>> energy_receive()
123.456

‫ جایزه نوبل کجاست؟

‫ 💡 توضیح:

‫ توجه کنید که آرایه‌ی numpy ایجادشده در تابع energy_send برگردانده نشده است، بنابراین فضای حافظه‌ی آن آزاد شده و مجدداً قابل استفاده است.
‫ تابع numpy.empty() نزدیک‌ترین فضای حافظه‌ی آزاد را بدون مقداردهی مجدد برمی‌گرداند. این فضای حافظه معمولاً همان فضایی است که به‌تازگی آزاد شده است (البته معمولاً این اتفاق می‌افتد و نه همیشه).

▶ ‫ خب، یک جای کار مشکوک است...

def square(x):
    """
    یک تابع ساده برای محاسبه‌ی مربع یک عدد با استفاده از جمع.
    """
    sum_so_far = 0
    for counter in range(x):
        sum_so_far = sum_so_far + x
  return sum_so_far

‫ خروجی (پایتون 2.X):

>>> square(10)
10

‫ آیا این نباید ۱۰۰ باشد؟

‫ نکته: اگر نمی‌توانید این مشکل را بازتولید کنید، سعی کنید فایل mixed_tabs_and_spaces.py را از طریق شِل اجرا کنید.

‫ 💡 توضیح

‫ تب‌ها و فاصله‌ها (space) را با هم ترکیب نکنید! کاراکتری که دقیقاً قبل از دستور return آمده یک «تب» است، در حالی که در بقیۀ مثال، کد با مضربی از «۴ فاصله» تورفتگی دارد.
‫ نحوۀ برخورد پایتون با تب‌ها به این صورت است:

‫ ابتدا تب‌ها (از چپ به راست) با یک تا هشت فاصله جایگزین می‌شوند به‌طوری که تعداد کل کاراکترها تا انتهای آن جایگزینی، مضربی از هشت باشد <...>
‫ بنابراین «تب» در آخرین خط تابع square با هشت فاصله جایگزین شده و به همین دلیل داخل حلقه قرار می‌گیرد.
‫ پایتون ۳ آنقدر هوشمند هست که چنین مواردی را به‌صورت خودکار با خطا اعلام کند.

‫ خروجی (Python 3.x):
```
TabError: inconsistent use of tabs and spaces in indentation
```

بخش: متفرقه

‫ ▶ `+=` سریع‌تر است

# استفاده از "+"، سه رشته:
>>> timeit.timeit("s1 = s1 + s2 + s3", setup="s1 = ' ' * 100000; s2 = ' ' * 100000; s3 = ' ' * 100000", number=100)
0.25748300552368164
# استفاده از "+="، سه رشته:
>>> timeit.timeit("s1 += s2 + s3", setup="s1 = ' ' * 100000; s2 = ' ' * 100000; s3 = ' ' * 100000", number=100)
0.012188911437988281

‫ 💡 توضیح:

‫ استفاده از += برای اتصال بیش از دو رشته سریع‌تر از + است، زیرا هنگام محاسبه رشته‌ی نهایی، رشته‌ی اول (به‌عنوان مثال s1 در عبارت s1 += s2 + s3) از بین نمی‌رود.

‫ ▶ بیایید یک رشته‌ی بزرگ بسازیم!

def add_string_with_plus(iters):
    s = ""
    for i in range(iters):
        s += "xyz"
    assert len(s) == 3*iters

def add_bytes_with_plus(iters):
    s = b""
    for i in range(iters):
        s += b"xyz"
    assert len(s) == 3*iters

def add_string_with_format(iters):
    fs = "{}"*iters
    s = fs.format(*(["xyz"]*iters))
    assert len(s) == 3*iters

def add_string_with_join(iters):
    l = []
    for i in range(iters):
        l.append("xyz")
    s = "".join(l)
    assert len(s) == 3*iters

def convert_list_to_string(l, iters):
    s = "".join(l)
    assert len(s) == 3*iters

Output:

‫ اجرا شده در پوسته‌ی ipython با استفاده از %timeit برای خوانایی بهتر نتایج. ‫ همچنین می‌توانید از ماژول timeit در پوسته یا اسکریپت عادی پایتون استفاده کنید؛ نمونه‌ی استفاده در زیر آمده است: timeit.timeit('add_string_with_plus(10000)', number=1000, globals=globals())


>>> NUM_ITERS = 1000
>>> %timeit -n1000 add_string_with_plus(NUM_ITERS)
124 µs ± 4.73 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
>>> %timeit -n1000 add_bytes_with_plus(NUM_ITERS)
211 µs ± 10.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit -n1000 add_string_with_format(NUM_ITERS)
61 µs ± 2.18 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit -n1000 add_string_with_join(NUM_ITERS)
117 µs ± 3.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> l = ["xyz"]*NUM_ITERS
>>> %timeit -n1000 convert_list_to_string(l, NUM_ITERS)
10.1 µs ± 1.06 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

‫ بیایید تعداد تکرارها را ۱۰ برابر افزایش دهیم.

>>> NUM_ITERS = 10000
>>> %timeit -n1000 add_string_with_plus(NUM_ITERS) # افزایش خطی در زمان اجرا
1.26 ms ± 76.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit -n1000 add_bytes_with_plus(NUM_ITERS) # افزایش درجه دو (افزایش مربعی)
6.82 ms ± 134 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit -n1000 add_string_with_format(NUM_ITERS) # افزایش خطی
645 µs ± 24.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit -n1000 add_string_with_join(NUM_ITERS) # افزایش خطی
1.17 ms ± 7.25 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> l = ["xyz"]*NUM_ITERS
>>> %timeit -n1000 convert_list_to_string(l, NUM_ITERS) # افزایش خطی
86.3 µs ± 2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

💡 توضیحات

توضیحات

‫ برای اطلاعات بیشتر درباره‌ی timeit یا %timeit، می‌توانید به این لینک‌ها مراجعه کنید. این توابع برای اندازه‌گیری زمان اجرای قطعه‌کدها استفاده می‌شوند.
‫ برای تولید رشته‌های طولانی از + استفاده نکنید — در پایتون، نوع داده‌ی str تغییرناپذیر (immutable) است؛ بنابراین برای هر الحاق (concatenation)، رشته‌ی چپ و راست باید در رشته‌ی جدید کپی شوند. اگر چهار رشته‌ی ۱۰ حرفی را متصل کنید، به‌جای کپی ۴۰ کاراکتر، باید (10+10) + ((10+10)+10) + (((10+10)+10)+10) = 90 کاراکتر کپی کنید. این وضعیت با افزایش تعداد و طول رشته‌ها به‌صورت درجه دو (مربعی) بدتر می‌شود (که توسط زمان اجرای تابع add_bytes_with_plus تأیید شده است).
‫ بنابراین توصیه می‌شود از .format یا سینتکس % استفاده کنید (البته این روش‌ها برای رشته‌های بسیار کوتاه کمی کندتر از + هستند).
‫ اما بهتر از آن، اگر محتوای شما از قبل به‌شکل یک شیء قابل تکرار (iterable) موجود است، از دستور ''.join(iterable_object) استفاده کنید که بسیار سریع‌تر است.

‫ برخلاف تابع add_bytes_with_plus و به‌دلیل بهینه‌سازی‌های انجام‌شده برای عملگر += (که در مثال قبلی توضیح داده شد)، تابع add_string_with_plus افزایشی درجه دو در زمان اجرا نشان نداد. اگر دستور به‌صورت s = s + "x" + "y" + "z" بود (به‌جای s += "xyz")، افزایش زمان اجرا درجه دو می‌شد.

def add_string_with_plus(iters):
    s = ""
    for i in range(iters):
        s = s + "x" + "y" + "z"
    assert len(s) == 3*iters

>>> %timeit -n100 add_string_with_plus(1000)
388 µs ± 22.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit -n100 add_string_with_plus(10000) # افزایش درجه دو در زمان اجرا
9 ms ± 298 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

‫ وجود راه‌های متعدد برای قالب‌بندی و ایجاد رشته‌های بزرگ تا حدودی در تضاد با ذِن پایتون است که می‌گوید:

‫ «باید یک راه — و ترجیحاً فقط یک راه — واضح برای انجام آن وجود داشته باشد.»

▶ ‫ کُند کردن جستجوها در `dict` *

some_dict = {str(i): 1 for i in range(1_000_000)}
another_dict = {str(i): 1 for i in range(1_000_000)}

‫ خروجی:

>>> %timeit some_dict['5']
28.6 ns ± 0.115 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
>>> some_dict[1] = 1
>>> %timeit some_dict['5']
37.2 ns ± 0.265 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

>>> %timeit another_dict['5']
28.5 ns ± 0.142 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
>>> another_dict[1]  # تلاش برای دسترسی به کلیدی که وجود ندارد
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 1
>>> %timeit another_dict['5']
38.5 ns ± 0.0913 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

چرا جستجوهای یکسان کندتر می‌شوند؟

‫ 💡 توضیح:

‫ در CPython یک تابع عمومی برای جستجوی کلید در دیکشنری‌ها وجود دارد که از تمام انواع کلیدها (str، int و هر شیء دیگر) پشتیبانی می‌کند؛ اما برای حالت متداولی که تمام کلیدها از نوع str هستند، یک تابع بهینه‌شده‌ی اختصاصی نیز وجود دارد.
‫ تابع اختصاصی (که در کد منبع CPython با نام lookdict_unicode شناخته می‌شود) فرض می‌کند که تمام کلیدهای موجود در دیکشنری (از جمله کلیدی که در حال جستجوی آن هستید) رشته (str) هستند و برای مقایسه‌ی کلیدها، به‌جای فراخوانی متد __eq__، از مقایسه‌ی سریع‌تر و ساده‌تر رشته‌ای استفاده می‌کند.
‫ اولین باری که یک دیکشنری (dict) با کلیدی غیر از str فراخوانی شود، این حالت تغییر می‌کند و جستجوهای بعدی از تابع عمومی استفاده خواهند کرد.
‫ این فرایند برای آن نمونه‌ی خاص از دیکشنری غیرقابل بازگشت است و حتی لازم نیست کلید موردنظر در دیکشنری موجود باشد. به همین دلیل است که حتی تلاش ناموفق برای دسترسی به کلیدی ناموجود نیز باعث ایجاد همین تأثیر (کند شدن جستجو) می‌شود.

‫ ▶ حجیم کردن دیکشنری نمونه‌ها (`instance dicts`) *

import sys

class SomeClass:
    def __init__(self):
        self.some_attr1 = 1
        self.some_attr2 = 2
        self.some_attr3 = 3
        self.some_attr4 = 4


def dict_size(o):
    return sys.getsizeof(o.__dict__)

‫ خروجی: (پایتون ۳.۸؛ سایر نسخه‌های پایتون ۳ ممکن است کمی متفاوت باشند)

>>> o1 = SomeClass()
>>> o2 = SomeClass()
>>> dict_size(o1)
104
>>> dict_size(o2)
104
>>> del o1.some_attr1
>>> o3 = SomeClass()
>>> dict_size(o3)
232
>>> dict_size(o1)
232

‫ بیایید دوباره امتحان کنیم... در یک مفسر (interpreter) جدید:

>>> o1 = SomeClass()
>>> o2 = SomeClass()
>>> dict_size(o1)
104  # همان‌طور که انتظار می‌رفت
>>> o1.some_attr5 = 5
>>> o1.some_attr6 = 6
>>> dict_size(o1)
360
>>> dict_size(o2)
272
>>> o3 = SomeClass()
>>> dict_size(o3)
232

‫ چه چیزی باعث حجیم‌شدن این دیکشنری‌ها می‌شود؟ و چرا اشیاء تازه ساخته‌شده نیز حجیم هستند؟

💡 توضیح:

‫ در CPython، امکان استفاده‌ی مجدد از یک شیء «کلیدها» (keys) در چندین دیکشنری وجود دارد. این ویژگی در PEP 412 معرفی شد تا مصرف حافظه کاهش یابد، به‌ویژه برای دیکشنری‌هایی که به نمونه‌ها (instances) تعلق دارند و معمولاً کلیدها (نام صفات نمونه‌ها) بین آن‌ها مشترک است.
‫ این بهینه‌سازی برای دیکشنری‌های نمونه‌ها کاملاً شفاف و خودکار است؛ اما اگر بعضی فرضیات نقض شوند، غیرفعال می‌شود.
‫ دیکشنری‌هایی که کلیدهایشان به اشتراک گذاشته شده باشد، از حذف کلید پشتیبانی نمی‌کنند؛ بنابراین اگر صفتی از یک نمونه حذف شود، دیکشنریِ آن نمونه «غیر مشترک» (unshared) شده و این قابلیت اشتراک‌گذاری کلیدها برای تمام نمونه‌هایی که در آینده از آن کلاس ساخته می‌شوند، غیرفعال می‌گردد.
‫ همچنین اگر اندازه‌ی دیکشنری به‌علت اضافه‌شدن کلیدهای جدید تغییر کند (resize شود)، اشتراک‌گذاری کلیدها تنها زمانی ادامه می‌یابد که فقط یک دیکشنری در حال استفاده از آن‌ها باشد (این اجازه می‌دهد در متد __init__ برای اولین نمونه‌ی ساخته‌شده، صفات متعددی تعریف کنید بدون آن‌که اشتراک‌گذاری کلیدها از بین برود). اما اگر چند نمونه همزمان وجود داشته باشند و تغییر اندازه‌ی دیکشنری رخ دهد، قابلیت اشتراک‌گذاری کلیدها برای نمونه‌های بعدی همان کلاس غیرفعال خواهد شد. زیرا CPython دیگر نمی‌تواند مطمئن باشد که آیا نمونه‌های بعدی دقیقاً از مجموعه‌ی یکسانی از صفات استفاده خواهند کرد یا خیر.
‫ نکته‌ای کوچک برای کاهش مصرف حافظه‌ی برنامه: هرگز صفات نمونه‌ها را حذف نکنید و حتماً تمام صفات را در متد __init__ تعریف و مقداردهی اولیه کنید!

‫ ▶ موارد جزئی *

‫ متد join() عملیاتی مربوط به رشته (str) است، نه لیست (list). (در نگاه اول کمی برخلاف انتظار است.)

** ‫💡 توضیح:** اگر join() به‌عنوان متدی روی رشته پیاده‌سازی شود، می‌تواند روی هر شیء قابل پیمایش (iterable) از جمله لیست، تاپل و هر نوع تکرارشونده‌ی دیگر کار کند. اگر به‌جای آن روی لیست تعریف می‌شد، باید به‌طور جداگانه برای هر نوع دیگری نیز پیاده‌سازی می‌شد. همچنین منطقی نیست که یک متد مختص رشته روی یک شیء عمومی مانند list پیاده شود.
‫ تعدادی عبارت با ظاهری عجیب اما از نظر معنا صحیح:
- ‫ عبارت [] = () از نظر معنایی صحیح است (باز کردن یا unpack کردن یک تاپل خالی درون یک لیست خالی).
- ‫ عبارت 'a'[0][0][0][0][0] نیز از نظر معنایی صحیح است، زیرا پایتون برخلاف زبان‌هایی که از C منشعب شده‌اند، نوع داده‌ای جداگانه‌ای برای کاراکتر ندارد. بنابراین انتخاب یک کاراکتر از یک رشته، منجر به بازگشت یک رشته‌ی تک‌کاراکتری می‌شود.
- ‫ عبارات 3 --0-- 5 == 8 و --5 == 5 هر دو از لحاظ معنایی درست بوده و مقدارشان برابر True است.
‫ با فرض اینکه a یک عدد باشد، عبارات ++a و --a هر دو در پایتون معتبر هستند؛ اما رفتاری مشابه با عبارات مشابه در زبان‌هایی مانند C، ++C یا جاوا ندارند.
```
>>> a = 5
>>> a
5
>>> ++a
5
>>> --a
5
```
** ‫ 💡 توضیح:**
- ‫ در گرامر پایتون عملگری به‌نام ++ وجود ندارد. در واقع ++ دو عملگر + جداگانه است.
- ‫ عبارت ++a به‌شکل +(+a) تفسیر می‌شود که معادل a است. به‌همین ترتیب، خروجی عبارت --a نیز قابل توجیه است.
- ‫ این تاپیک در StackOverflow دلایل نبودن عملگرهای افزایش (++) و کاهش (--) در پایتون را بررسی می‌کند.
‫ احتمالاً با عملگر Walrus (گراز دریایی) در پایتون آشنا هستید؛ اما تا به حال در مورد عملگر Space-invader (مهاجم فضایی) شنیده‌اید؟
```
>>> a = 42
>>> a -=- 1
>>> a
43
```

‫ از آن به‌عنوان جایگزینی برای عملگر افزایش (increment)، در ترکیب با یک عملگر دیگر استفاده می‌شود.

>>> a +=+ 1
>>> a
>>> 44

‫ 💡 توضیح: این شوخی از توییت Raymond Hettinger برگرفته شده است. عملگر «مهاجم فضایی» در واقع همان عبارت بدفرمت‌شده‌ی a -= (-1) است که معادل با a = a - (- 1) می‌باشد. حالت مشابهی برای عبارت a += (+ 1) نیز وجود دارد.

‫ پایتون یک عملگر مستندنشده برای استلزام معکوس (converse implication) دارد.
```
>>> False ** False == True
True
>>> False ** True == False
True
>>> True ** False == True
True
>>> True ** True == True
True
```
‫ 💡 توضیح: اگر مقادیر False و True را به‌ترتیب با اعداد ۰ و ۱ جایگزین کرده و محاسبات را انجام دهید، جدول درستی حاصل، معادل یک عملگر استلزام معکوس خواهد بود. (منبع)
‫ حالا که صحبت از عملگرها شد، عملگر @ نیز برای ضرب ماتریسی در پایتون وجود دارد (نگران نباشید، این بار واقعی است).
```
>>> import numpy as np
>>> np.array([2, 2, 2]) @ np.array([7, 8, 8])
46
```
‫ 💡 توضیح: عملگر @ در پایتون ۳٫۵ با در نظر گرفتن نیازهای جامعه علمی اضافه شد. هر شی‌ای می‌تواند متد جادویی __matmul__ را بازنویسی کند تا رفتار این عملگر را مشخص نماید.
‫ از پایتون ۳٫۸ به بعد می‌توانید از نحو متداول f-string مانند f'{some_var=}' برای اشکال‌زدایی سریع استفاده کنید. مثال,
```
>>> some_string = "wtfpython"
>>> f'{some_string=}'
"some_string='wtfpython'"
```
‫ پایتون برای ذخیره‌سازی متغیرهای محلی در توابع از ۲ بایت استفاده می‌کند. از نظر تئوری، این به معنای امکان تعریف حداکثر ۶۵۵۳۶ متغیر در یک تابع است. با این حال، پایتون راهکار مفیدی ارائه می‌کند که می‌توان با استفاده از آن بیش از ۲^۱۶ نام متغیر را ذخیره کرد. کد زیر نشان می‌دهد وقتی بیش از ۶۵۵۳۶ متغیر محلی تعریف شود، در پشته (stack) چه اتفاقی رخ می‌دهد (هشدار: این کد تقریباً ۲^۱۸ خط متن چاپ می‌کند، بنابراین آماده باشید!):
```
import dis
exec("""
def f():
   """ + """
   """.join(["X" + str(x) + "=" + str(x) for x in range(65539)]))

f()

print(dis.dis(f))
```
‫ چندین رشته (Thread) در پایتون، کدِ پایتونی شما را به‌صورت همزمان اجرا نمی‌کنند (بله، درست شنیدید!). شاید به نظر برسد که ایجاد چندین رشته و اجرای همزمان آن‌ها منطقی است، اما به دلیل وجود قفل مفسر سراسری (GIL) در پایتون، تمام کاری که انجام می‌دهید این است که رشته‌هایتان به‌نوبت روی یک هسته اجرا می‌شوند. رشته‌ها در پایتون برای وظایفی مناسب هستند که عملیات I/O دارند، اما برای رسیدن به موازی‌سازی واقعی در وظایف پردازشی سنگین (CPU-bound)، بهتر است از ماژول multiprocessing در پایتون استفاده کنید.
‫ گاهی اوقات، متد print ممکن است مقادیر را فوراً چاپ نکند. برای مثال،
```
# File some_file.py
import time

print("wtfpython", end="_")
time.sleep(3)
```
‫ این کد عبارت wtfpython را به دلیل آرگومان end پس از ۳ ثانیه چاپ می‌کند؛ چرا که بافر خروجی تنها پس از رسیدن به کاراکتر \n یا در زمان اتمام اجرای برنامه تخلیه می‌شود. برای تخلیه‌ی اجباری بافر می‌توانید از آرگومان flush=True استفاده کنید.
‫ برش لیست‌ها (List slicing) با اندیس‌های خارج از محدوده، خطایی ایجاد نمی‌کند.
```
>>> some_list = [1, 2, 3, 4, 5]
>>> some_list[111:]
[]
```

‫ برش زدن (slicing) یک شئ قابل پیمایش (iterable) همیشه یک شئ جدید ایجاد نمی‌کند. به‌عنوان مثال،

>>> some_str = "wtfpython"
>>> some_list = ['w', 't', 'f', 'p', 'y', 't', 'h', 'o', 'n']
>>> some_list is some_list[:] # False expected because a new object is created.
False
>>> some_str is some_str[:] # True because strings are immutable, so making a new object is of not much use.
True

‫ در پایتون ۳، فراخوانی int('١٢٣٤٥٦٧٨٩') مقدار 123456789 را برمی‌گرداند. در پایتون، نویسه‌های ده‌دهی (Decimal characters) شامل تمام ارقامی هستند که می‌توانند برای تشکیل اعداد در مبنای ده استفاده شوند؛ به‌عنوان مثال نویسه‌ی U+0660 که همان رقم صفر عربی-هندی است. اینجا داستان جالبی درباره این رفتار پایتون آمده است.
‫ از پایتون ۳ به بعد، می‌توانید برای افزایش خوانایی، اعداد را با استفاده از زیرخط (_) جدا کنید.
```
>>> six_million = 6_000_000
>>> six_million
6000000
>>> hex_address = 0xF00D_CAFE
>>> hex_address
4027435774
```
‫ عبارت 'abc'.count('') == 4 مقدار True برمی‌گرداند. در اینجا یک پیاده‌سازی تقریبی از متد count آورده شده که این موضوع را شفاف‌تر می‌کند:
```
def count(s, sub):
    result = 0
    for i in range(len(s) + 1 - len(sub)):
        result += (s[i:i + len(sub)] == sub)
    return result
```

‫ این رفتار به این دلیل است که زیررشته‌ی خالی ('') با برش‌هایی (slices) به طول صفر در رشته‌ی اصلی مطابقت پیدا می‌کند.

‫ مشارکت

‫چند روشی که می‌توانید در wtfpython مشارکت داشته باشید:

‫ پیشنهاد مثال‌های جدید
‫ کمک به ترجمه (به مشکلات برچسب ترجمه مراجعه کنید)
‫ اصلاحات جزئی مثل اشاره به تکه‌کدهای قدیمی، اشتباهات تایپی، خطاهای قالب‌بندی و غیره.
‫ شناسایی نواقص (مانند توضیحات ناکافی، مثال‌های تکراری و ...)
‫ هر پیشنهاد خلاقانه‌ای برای مفیدتر و جذاب‌تر شدن این پروژه

‫ برای اطلاعات بیشتر CONTRIBUTING.md را مشاهده کنید. برای بحث درباره موارد مختلف می‌توانید یک مشکل جدید ایجاد کنید.

‫ نکته: لطفاً برای درخواست بک‌لینک (backlink) تماس نگیرید. هیچ لینکی اضافه نمی‌شود مگر اینکه ارتباط بسیار زیادی با پروژه داشته باشد.

‫ تقدیر و تشکر

‫ ایده و طراحی این مجموعه ابتدا از پروژه عالی wtfjs توسط Denys Dovhan الهام گرفته شد. حمایت فوق‌العاده‌ جامعه پایتون باعث شد پروژه به شکل امروزی خود درآید.

‫ چند لینک جالب!

‫ 🎓 مجوز

‫ دوستانتان را هم شگفت‌زده کنید!

‫ اگر از wtfpython خوشتان آمد، می‌توانید با این لینک‌های سریع آن را با دوستانتان به اشتراک بگذارید:

‫ توییتر | لینکدین | فیسبوک

‫ آیا به یک نسخه pdf نیاز دارید؟

‫ من چند درخواست برای نسخه PDF (و epub) کتاب wtfpython دریافت کرده‌ام. برای دریافت این نسخه‌ها به محض آماده شدن، می‌توانید اطلاعات خود را اینجا وارد کنید.

‫ همین بود دوستان! برای دریافت مطالب آینده مشابه این، می‌توانید ایمیل خود را اینجا اضافه کنید.

145 KiB Vendored Raw Blame History Unescape Escape

What the f*ck Python! 😱

فهرست مطالب

ساختار مثال‌ها

▶ یه اسم خوشگل

💡 توضیح:

استفاده

👀 مثال‌ها

بخش: ذهن خود را به چالش بکشید!

▶ اول از همه! *

💡 توضیحات

▶ بعضی وقت‌ها رشته‌ها می‌توانند دردسرساز شوند

💡 Explanation:

▶ Be careful with chained operations

💡 Explanation:

▶ How not to use is operator

💡 Explanation:

▶ Hash brownies

💡 Explanation

▶ Deep down, we're all the same.

💡 Explanation:

▶ Disorder within order *

💡 Explanation:

▶ Keep trying... *

💡 Explanation:

▶ For what?

💡 Explanation:

▶ Evaluation time discrepancy

💡 Explanation

▶ is not ... is not is (not ...)

💡 Explanation

▶ A tic-tac-toe where X wins in the first attempt!

💡 Explanation:

▶ Schrödinger's variable *

💡 Explanation:

▶ The chicken-egg problem *

💡 Explanation

▶ Subclass relationships

💡 Explanation:

▶ Methods equality and identity

💡 Explanation

▶ All-true-ation *

💡 Explanation:

▶ The surprising comma

💡 Explanation:

▶ Strings and the backslashes

💡 Explanation

▶ not knot!

💡 Explanation:

▶ Half triple-quoted strings

💡 Explanation:

▶ What's wrong with booleans?

💡 Explanation:

▶ Class attributes and instance attributes

💡 Explanation:

▶ yielding None

💡 Explanation:

▶ Yielding from... return! *

💡 Explanation:

▶ Nan-reflexivity *

💡 Explanation:

▶ Mutating the immutable!

💡 Explanation:

▶ The disappearing variable from outer scope

💡 Explanation:

▶ The mysterious key type conversion

💡 Explanation:

▶ Let's see if you can guess this?

💡 Explanation:

▶ Exceeds the limit for integer string conversion

💡 Explanation:

Section: Slippery Slopes

▶ Modifying a dictionary while iterating over it

💡 Explanation:

▶ Stubborn del operation

💡 Explanation:

▶ The out of scope variable

💡 Explanation:

▶ Deleting a list item while iterating

💡 Explanation:

145 KiB

Vendored

Raw Blame History

▶ How not to use `is` operator

▶ `is not ...` is not `is (not ...)`

▶ Stubborn `del` operation

▶ ‫ `goto`، ولی چرا؟

▶ ‫ بی‌نهایت (`Inpinity`)

‫ ▶ `+=` سریع‌تر است

▶ ‫ کُند کردن جستجوها در `dict` *

‫ ▶ حجیم کردن دیکشنری نمونه‌ها (`instance dicts`) *