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README.md

What the f*ck Python! 🐍

WTFPL 2.0

A collection of tricky Python examples

Python, being a high-level and interpreter based programming language which is awesome by design, provides us with many functionalities for programmer's comfort. But sometimes, the outcomes of your code may not seem obvious to a normal Python user at the first sight.

Here's an attempt to collect such classic and tricky examples of unexpected behaviors in Python and discuss what exactly is happening under the hood! While some of the examples you see below may not be WTFs in truest sense, but they'll definitely reveal some of the interesting parts of Pyhton that you might be unaware of.

Anyways, I find it a nice way to learn internals of a programming language and I think you'll like them as well!

If you're an experienced Python programmer, you might be familiar with most of these examples, and I might be able to revive some nice old memories of yours being bitten by these gotchas 😅

So, here ya go...

Table of Contents

Table of Contents generated with DocToc

Structure of the Examples

All the examples are structured like below:

Some fancy Title

# Setting up the code.
# Preparation for the magic...

Output (Python version):

>>> triggering_statement
Probably unexpected output

(Optional): One line describing the unexpected output.

Note: All the examples mentioned below are run on Python 3.5.2 interactive interpreter unless explicitly specified.

💡 Explanation:

  • Brief 💡 Explanation: of what's happening and why is it happening. 
    Setting up examples for clarification (if necessary)
    Outupt: 
    >>> trigger # some example that makes it easy to unveil the magic
# some justified output

Usage

A good way to get the most out of these examples, in my opinion, will be to just read the examples chronologically, and for every example:

  • Carefully read the initial code for setting up the example. If you're an experienced Python programmer, most of the times you will successfully anticipate what's gonna happen next.
  • Read the output snippets and
    • Check if the outputs are the same as you'd expect.
    • Make sure know the exact reason behind the output being the way it is.
      • If no, read the explaination (and if you still don't understand, shout out! and create an issue here).
      • If yes, give a gentle pat on your back and you may skip to the next example.

👀 Examples

datetime.time object is considered to be false if it represented midnight in UTC

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:

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

💡 Explanation:

Before Python 3.5, a 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".

is is not what it is!

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

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

>>> a = 257; b = 257
>>> a is b
True

💡 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 if the is operator returns True then the equality is definitely True, but the opposite may or may not be True.

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 to just 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 actually just get back a reference to the existing object. So it should be possible to change the value of 1. I suspect the behaviour 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 integer isn't smart enough while executing y = 257 to recongnize that we've already created an integer of the value 257 and so it goes on to create another object in the memory.

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

  • When a and b are set to 257 in the same line, the Python interpretor creates new object, then references the second variable at the same time. If you do it in separate lines, it doesn't "know" that there's already 257 as an object.
  • It's a compiler optimization and specifically applies to interactive environment. When you do 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.
>>> a, b = 257, 257
>>> id(a)
140640774013296
>>> id(b)
140640774013296
>>> a = 257
>>> b = 257
>>> id(a)
140640774013392
>>> id(b)
140640774013488

The function inside loop magic

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

funcs_results = [func() for func in funcs]

Output:

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

//OR

>>> 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]

Explaination

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. So all of the functions use the latest value assigned to the variable for computation.

To get the desired behavior you can pass in the loop variable as a named varibable to the function which will define the variable again within the function's scope.

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]

Loop variables leaking out of local scope!

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 ourtside 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

x = 1
print([x for x in range(5)])
print(x, ': x in global')

Output (on Python 2.x):

[0, 1, 2, 3, 4]
(4, ': x in global')

Output (on Python 3.x):

[0, 1, 2, 3, 4]
1 : x in global

💡 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 documentation:

"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."

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

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

Output:

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

💡 Explanation:

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

image

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)

image

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

>>> 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. Examlple:

def some_func(default_arg=None):
    if not default_arg:
        default_arg = []
    default_arg.append("some_string")
    return default_arg

You can't change the values contained in tuples because they're immutable.. Oh really?

This might be obvious for most of you guys, but it took me a lot of time to realize it.

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])

💡 Explanation:

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 changed; 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.

Using a varibale not defined in scope

a = 1
def some_func():
    return a

def another_func():
    a += 1
    return a

Output:

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

💡 Explanation:

  • When you make an assignment to a variable in a 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.
  • Read this short but awesome guide to learn more about how namespaces and scope resolution works in Python.
  • To actually modify the outer scope variable a in another_func, use global keyword. 
    def another_func()
    global a
    a += 1
    return a
    Output: 
    >>> another_func()
2

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 to

    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 same scope and the variable e got removed due to the execution of the except clause. The same is not the case with functions which 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!

Return in both try and finally clauses

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

Output:

>>> some_func()
'from_finally'

💡 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.

When True is actually False

True == False
if True == False:
    print("I've lost faith in truth!")

Output:

I've lost faith in truth!

💡 Explanation:

Initially, Python used to have no bool type (people used 0 for false and non-zero value like 1 for true). Then they added True, False, and a bool type, but, for backwards compatibility, they couldn't make True and False constants- they just were built-in variables. Python 3 was backwards-incompatible, so it was now finally possible to fix that, and so this example wont't work with Python 3.x.

Evaluation time disperancy

array = [1, 8, 15]
g = (x for x in array if array.count(x) > 0)
array = [2, 8, 22]

Output:

>>> print(list(g))
[8]

Explainiation

  • In a generator expression, the in clause is evaluated at declaration time, but the conditional clause is evaluated at run time.
  • So before run time, 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.

Be careful with chained operations

>>> 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/2/reference/expressions.html#not-in

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

a += b doesn't behave the same way as a = a + b

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:

  • The expression a = a + [5,6,7,8] generates a new object and sets a's reference to that new object, leaving b unchanged.

  • The expression a + =[5,6,7,8] is actually mapped to an "extend" function that operates on the object such that a and b still point to the same object that has been modified in-place.

Backslashes at the end of string

>>> print("\\ some string \\")
>>> print(r"\ some string")
>>> print(r"\ some string \")

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

Explaination

A raw string literal, where the backslash doesn't have the special meaning, as indicated by the prefix r. What it actually does, though, is simply change the behavior of backslashes so they pass themselves and the following character through. That's why backslashes don't work at the end of a raw string.

Editing a dictionary while iterating over it

x = {0: None}

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

Output:

0
1
2
3
4
5
6
7

Yes, it runs for exactly 8 times and stops.

Explaination:

  • Iteration over a dictionary that you edit at the same time is not supported.
  • It runs 8 times because that's the point at which the dictionary resizes to hold more keys (we have 8 deletion entries so a resize is needed). This is actually an implementation detail.
  • Refer to this StackOverflow thread explaining a similar example.

is not ... is different from is (not ...)

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

Explaination

  • 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.

Time for some hash brownies!

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

Output:

>>> some_dict[5.5]
"Ruby"
>>> some_dict[5.0]
"Python"
>>> some_dict[5]
"Python"

Explaination

  • 5 (an int type) is implicitly converted to 5.0 (a float type) before calculating the hash in Python. 
    >>> hash(5) == hash(5.0)
True
  • This StackOverflow answer explains beautifully the rationale behind it.

Identical looking names

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

Wut?

Explaination

Some Unicode characters look identical to ASCII ones, but are considered distinct by the interpreter.

>>> value = 42 #ascii e
>>> valuе = 23 #cyrillic e, Python 2.x interpreter would raise a `SyntaxError` here
>>> print(value)

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

Explaination

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

In-place update functions of mutable object types

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

Explaination

Most methods that modiy 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)

Deleting a list item while iterating over it

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]

💡 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:

  • remove removes the first matching value, not a specific index, raises ValueError if value is not found.

  • del removes a specific index (That's why first list_1 was unaffected), raises IndexError if invalid index is specified.

  • pop removes element at specific index and returns it, raises IndexError if 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. Similar thing will happen with every alternate element in the list sequence.

  • See this StackOverflow thread for a similar example related to dictionaries in Python.

Explicit typecast of strings

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
>>> b == d #but nan!=nan
False
>>> 50/a
0
>>> a/a
nan
>>> 23 + b
nan

💡 Explanation:

'inf' and 'nan' are special strings (case-insensitieve), which when explicitly type casted to float type, are used to represent mathematical "infinity" and "not a number" respectively.

Well, something is fishy...

def square(x):
    sum_so_far = 0
    for counter in range(x):
        sum_so_far = sum_so_far + x
  return sum_so_far

print(square(10))

Output (Python 2.x):

10

Output (Python 3.x):

TabError: inconsistent use of tabs and spaces in indentation

Note: If you're not able to reproduce this, try running the file mixed_tabs_and_spaces.py via the shell.

Explaination

  • Don't mix tabs and spaces! The character just preceding return is a "tab" and the code is indented by multiple of "4 spaces" elsewhere in the example.
  • This is how Python handles tabs:

    First, tabs are replaced (from left to right) by one to eight spaces such that the total number of characters up to and including the replacement is a multiple of eight <...>

  • So the "tab" at the last line of square function is replace with 8 spaces and it gets into the loop.

Class attributes and instance attributes

class A:
    x = 1

class B(A):
    pass

class C(A):
    pass

Ouptut:

>>> 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
(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

Explaination:

  • 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 current class, the parent classes are searched for it.
  • The += operator modifies the mutable object in-place without creating a new object.

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

Explaination

  • 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

String concatenation

This is not a WTF, but just a nice thing to be aware of :)

def add_string_with_plus(iters):
    s = ""
    for i in range(iters):
        s += "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:

>>> timeit(add_string_with_plus(10000))
100 loops, best of 3: 9.73 ms per loop
>>> timeit(add_string_with_format(10000))
100 loops, best of 3: 5.47 ms per loop
>>> timeit(add_string_with_join(10000))
100 loops, best of 3: 10.1 ms per loop
>>> l = ["xyz"]*10000
>>> timeit(convert_list_to_string(l, 10000))
10000 loops, best of 3: 75.3 µs per loop

Explanination

  • You can read more about timeit from here. It is generally used to measure the execution time of snippets.
  • Don't use + for generating long strings — In Python, str is immutable, so the left and right strings have to be copied into the new string for every pair of concatenations. If you concatenate four strings of length 10, you'll be copying (10+10) + ((10+10)+10) + (((10+10)+10)+10) = 90 characters instead of just 40 characters. Things get quadratically worse as the number and size of the string increases.
  • Therefore, it's advised to use .format or % syntax (however, they are slightly slower than + for short strings).
  • Or better, if already you've contents available in the form of an iterable object, then use ''.join(iterable_object) which is much faster.

Minor Ones

  • join() is a string operation instead of list operation. (sort of counter-intuitive at first usage) 💡 Explanation: If join() is a method on a string then it can operate on any iterable (list, tuple, iterators). If it were a method on a list it'd have to be implemented separately by every type. Also, it doesn't make much sense to put a string-specific method on a generic list.

    Also, it's string specific, and it sounds wrong to put a string-specific method on a generic list.

  • Few weird looking but semantically correct statements:

    • [] = () is a semantically correct statement (unpacking an empty tuple into an empty list)
    • 'a'[0][0][0][0][0] is also a semantically correct statement as strings are iterable in Python.
    • 3 --0-- 5 == 8 and --5 == 5 are both semantically correct statments and evalute to True.

  • Booleans are a subclass of int

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

  • Python uses 2 bytes for local variable storage in functions. In theory this means that only 65536 variables can be defined in a function. However, python has a handy solution built in that can be used to store more than 2^16 variable names. The following code demonstrates what happens in the stack when more than 65536 local variables are defined (Warning: This code prints around 2^18 lines of text, so be prepared!):

import dis
exec("""
def f():*     """ + """
    """.join(["X"+str(x)+"=" + str(x) for x in range(65539)]))

f()

print(dis.dis(f))

  • Multiple Python threads don't run concurrently (yes you heard it right!). It may seem intuitive to spawn several threads and let them execute concurrently, but, because of the Global Interpreter Lock in Python, all you're doing is making your threads execute on the same core turn by turn. To achieve actual parallelization in Python, you might want to use the Python multiprocessing module.

  • List slicing with out of the bound indices throws no errors

    >>> some_list = [1, 2, 3, 4, 5]
>>> some_list[111:]
[]

  • The else clause for loops (yeah, it exists!). One typical example might be

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


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

💡 Explanation: The else clause is executed only when there's no explicit break after all the iterations of the loop.

  • The else clause in exception handling. An example,

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

    Output:

    Try block executed successfully...

    Such an else clause is also called "completion clause" as reaching the else clause in a try statement means that the try block actually completed successfully.

  • String concatenation interpreter optimizations.

    >>> a = "some_string"
140420665652016
>>> id(a)
>>> id("some" + "_" + "string") # Notice that both the ids are same.
140420665652016
# using "+", three strings:
>>> timeit.timeit("s1 = s1 + s2 + s3", setup="s1 = ' ' * 100000; s2 = ' ' * 100000; s3 = ' ' * 100000", number=100)
0.25748300552368164
# using "+=", three strings:
>>> timeit.timeit("s1 += s2 + s3", setup="s1 = ' ' * 100000; s2 = ' ' * 100000; s3 = ' ' * 100000", number=100)
0.012188911437988281

    Explaination:

    • += is faster than + for concatenating more than two strings because the first string (example, s1 for s1 += s2 + s3) is not destroyed while calculating the complete string.
    • Both the strings refer to the same object because of CPython optimization hat tries to use existing immutable objects in some cases (implementation specific) rather than creating a new object every time. You can read more about this here

"Needle in a Haystack" bugs

This contains some of the potential bugs in you code that are very common but hard to detect.

Initializing a tuple containing single element

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()

💡 Explanation:

  • 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.

TODO: Collect and add more such examples

TODO: Hell of an example!

Trying to comeup with an example that combines multiple examples discussed above, making it difficult for the reader to guess the output correctly 😅.

Contributing

All patches are Welcome! Filing an issue first before submitting a patch will be appreciated :)

Acknowledgements

The idea and design for this list is inspired from Denys Dovhan's awesome project wtfjs.

Some nice Links!

🎓 License

CC 4.0

© Satwik Kansal