What the f*ck Python!
A collection of tricky Python examples
Python being an awesome higher level language, provides us many functionalities for programmer's comfort. But sometimes, the outcomes 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 see what exactly is happening under the hood! Anyways, I find it a nice way to learn internals of a language and I think you'll like them as well!
- If you're an beginner to intermdediate level Python programmer, I'd personally recommend you to go through all of the examples below, as being aware about such pitfalls may be able to save a lot of debugging time in your future.
- 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
- 👀 Examples
- Example heading
datetime.time
object is considered to be false if it represented midnight in UTCis
is not what it is!- The function inside loop magic
- Loop variables leaking out of local scope!
- A tic-tac-toe where X wins in first attempt!
- Beware of default mutable arguments!
- You can't change the values contained in tuples because they're immutable.. Oh really?
- Using a varibale not defined in scope
- The disappearing variable from outer scope
- Return in both
try
andfinally
clauses - When True is actually False
- The GIL messes it up (Multithreading vs Mutliprogramming example)
- Be careful with chained comparisons
- a += b doesn't behave the same way as a = a + b
- Backslashes at the end of string
- Editing a dictionary while iterating over it
- Minor ones
- "Needle in a Haystack" bugs
- Contributing
- Acknowledgements
- 🎓 License
👀 Examples
Environment: All the examples mentioned below are run on Python 3.5.2 interactive interpreter unless explicitly specified.
Example heading
One line of what's happening:
setting up
>>> triggering_statement
weird output
Explanation:
- Better to give outside links
- or just explain again in brief
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 returnsTrue
then the equality is definitelyTrue
, 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 already257
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 alist()
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
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
)
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])
Explanation
Quoting from https://docs.python.org/2/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 changed; however, the collection of objects directly referenced by an immutable object cannot change.)
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.
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 theexcept
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 toException()
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 of1
,8
and15
, only the count of8
is greater than0
, the generator only yields8
.
The GIL messes it up (Multithreading vs Mutliprogramming example)
Be careful with chained comparisons
>>> 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.
False is False is False
is equivalent to(False is False) and (False is False)
True is False == False
is equivalent toTrue is False and False == False
and since the first part of the statement (True is False
) evaluates toFalse
, the overall expression evaluates toFalse
.1 > 0 < 1
is equivalent to1 > 0 and 0 < 1
which evaluates toTrue
.- The expression
(1 > 0) < 1
is equivalent toTrue < 1
and
So,>>> int(True) 1 >>> True + 1 #not relevant for this example, but just for fun 2
1 < 1
evaluates toFalse
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 setsa
's reference to that new object, leavingb
unchanged. -
The expression
a + =[5,6,7,8]
is actually mapped to an "extend" function that operates on the object such thata
andb
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 usingis
andnot
separated.is not
evaluates toFalse
if the variables on either side of the operator point to the same object andTrue
otherwise.
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
The functions like append
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, raisesValueError
if value is not found. -
del
removes a specific index (That's why firstlist_1
was unaffected), raisesIndexError
if invalid index is specified. -
pop
removes element at specific index and returns it, raisesIndexError
if invalid index is specified. -
Why the output is
[2, 4]
? The list iteration is done index by index, and when we remove1
fromlist_2
orlist_4
, the contents of the lists are now[2, 3, 4]
. The remaining elements are shifted down i.e.2
is at index 0 and3
is at index 1. Since the next iteration is going to look at index 1 (which is the3
), the2
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.
Minor ones
-
join()
is a string operation instead of list operation. (sort of counter-intuitive at first usage) Explanation: Ifjoin()
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.
-
[] = ()
is a semantically correct statement (unpacking an emptytuple
into an emptylist
) -
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))
- No multicore support yet
"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',)
ort = 'one',
(missing comma) otherwise the interpreter considerst
to be astr
and iterates over it character by character. ()
is a special token and denotes emptytuple
.
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!
- https://www.youtube.com/watch?v=sH4XF6pKKmk
- https://www.reddit.com/r/Python/comments/3cu6ej/what_are_some_wtf_things_about_python
- https://sopython.com/wiki/Common_Gotchas_In_Python
- https://stackoverflow.com/questions/530530/python-2-x-gotchas-and-landmines