Python, being awesome by design high-level and interpreter-based programming language, provides us with many features for the programmer's comfort. But sometimes, the outcomes of a Python snippet may not seem obvious to a regular user at first sight.
Here I'm sharing with you guys a fun collection of classic and tricky examples of unexpected behaviors in Python with an attempt to discuss what exactly is happening under the hood!
While some of the examples you see below may not be WTFs in the truest sense, but they'll reveal some of the interesting parts of Python that you might be unaware of. I find it a nice way to learn the internals of a programming language, and I think you'll find them interesting 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 sweet old memories of yours being bitten by these gotchas :sweat_smile:
So, here ya go...
is is not what it is!
is not ... is not is (not ...)
Note: All the examples mentioned below are tested on Python 3.5.2 interactive interpreter, and they should work for all the Python versions unless explicitly specified in the example description.
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.
💡 Explanation:
- Brief explanation of what's happening and why is it happening.
Output:Setting up examples for clarification (if necessary)>>> trigger # some example that makes it easy to unveil the magic # some justified output
A good way to get the most out of these examples, in my opinion, will be just to read the examples chronologically, and for every example:
Output:
>>> value = 11
>>> valuе = 32
>>> value
11
Wut?
Note: The easiest way to reproduce this is to simply copy the statements from the above snippet and paste them into your file/shell.
Some non-Western characters look identical to letters in the English alphabet, but are considered distinct by the interpreter.
>>> ord('е') # cyrillic 'e' (Ye)
1077
>>> ord('e') # latin 'e', as used in English and typed using standard keyboard
101
>>> 'е' == 'e'
False
>>> value = 42 # latin e
>>> valuе = 23 # cyrillic 'e', Python 2.x interpreter would raise a `SyntaxError` here
>>> value
42
The built-in ord() function returns a character's Unicode code point, and different code positions of cyrillic 'e' and latin 'e' justify the behavior of the above example.
def square(x):
"""
A simple function to calculate square of a number by addition.
"""
sum_so_far = 0
for counter in range(x):
sum_so_far = sum_so_far + x
return sum_so_far
Output (Python 2.x):
>>> square(10)
10
Shouldn't that be 100?
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 replaced with eight spaces, and it gets into the loop.
Python 3 is nice enough to automatically throw an error for such cases.
Output (Python 3.x):
TabError: inconsistent use of tabs and spaces in indentation
1.
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"
"Python" destroyed the existence of "JavaScript"?
>>> 5 == 5.0
True
>>> hash(5) == hash(5.0)
True
some_dict[5] = "Python" is executed, the existing value "JavaScript" is overwritten with "Python" because Python recongnizes 5 and 5.0 as the same keys of the dictionary some_dict.array = [1, 8, 15]
g = (x for x in array if array.count(x) > 0)
array = [2, 8, 22]
Output:
>>> print(list(g))
[8]
in clause is evaluated at declaration time, but the conditional clause is evaluated at 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.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 eight times and stops.
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]?
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 element at a specific index and returns it, raises IndexError if an invalid index is specified.Why the output is [2, 4]?
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.Output:
>>> 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
r, the backslash doesn't have the special meaning.>>> a = "some_string"
>>> id(a)
140420665652016
>>> 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
+= 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.This is not a WTF at all, just some nice things 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_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:
>>> timeit(add_string_with_plus(10000))
1000 loops, best of 3: 972 µs per loop
>>> timeit(add_bytes_with_plus(10000))
1000 loops, best of 3: 815 µs per loop
>>> timeit(add_string_with_format(10000))
1000 loops, best of 3: 508 µs per loop
>>> timeit(add_string_with_join(10000))
1000 loops, best of 3: 878 µs per loop
>>> l = ["xyz"]*10000
>>> timeit(convert_list_to_string(l, 10000))
10000 loops, best of 3: 80 µs per loop
Let's increase the number of iterations by a factor of 10.
>>> timeit(add_string_with_plus(100000)) # Linear increase in execution time
100 loops, best of 3: 9.75 ms per loop
>>> timeit(add_bytes_with_plus(100000)) # Quadratic increase
1000 loops, best of 3: 974 ms per loop
>>> timeit(add_string_with_format(100000)) # Linear increase
100 loops, best of 3: 5.25 ms per loop
>>> timeit(add_string_with_join(100000)) # Linear increase
100 loops, best of 3: 9.85 ms per loop
>>> l = ["xyz"]*100000
>>> timeit(convert_list_to_string(l, 100000)) # Linear increase
1000 loops, best of 3: 723 µs per loop
+ 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 (justified with the execution times of add_bytes_with_plus function).format. or % syntax (however, they are slightly slower than + for short strings).''.join(iterable_object) which is much faster.add_string_with_plus didn't show a quadratic increase in execution time unlike add_bytes_with_plus becuase of the += optimizations discussed in the previous example. Had the statement been s = s + "x" + "y" + "z" instead of s += "xyz", the increase would have been quadratic.
def add_string_with_plus(iters):
s = ""
for i in range(iters):
s = s + "x" + "y" + "z"
assert len(s) == 3*iters
>>> timeit(add_string_with_plus(10000))
100 loops, best of 3: 9.87 ms per loop
>>> timeit(add_string_with_plus(100000)) # Quadratic increase in execution time
1 loops, best of 3: 1.09 s per loop
The else clause for loops. 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:
>>> some_list = [1, 2, 3, 4, 5]
>>> does_exists_num(some_list, 4)
Exists!
>>> does_exists_num(some_list, -1)
Does not exist
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...
else clause after a loop is executed only when there's no explicit break after all the iterations.else clause after try block is also called "completion clause" as reaching the else clause in a try statement means that the try block actually completed successfully.is is not what it is!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 = 257; b = 257
>>> a is b
True
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.is is for reference equality and == is for value equality. An example to clear things up,
>>> [] == []
True
>>> [] is [] # These are two empty lists 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.
Both a and b refer to the same object, when initialized with same value in the same line.
>>> a, b = 257, 257
>>> id(a)
140640774013296
>>> id(b)
140640774013296
>>> a = 257
>>> b = 257
>>> id(a)
140640774013392
>>> id(b)
140640774013488
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..py file, you would not see the same behavior, because the file is compiled all at once.is not ... is not is (not ...)>>> 'something' is not None
True
>>> 'something' is (not None)
False
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.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]
Even when the values of x were different in every iteration prior to appending some_func to funcs, all the functions return 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]
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 variable to the function. Why this works? Because this 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]
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
But x was never defined outside the scope of for loop...
2.
# 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
3.
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
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."
# 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 assigned 3 "X"s or did we?
When we initialize row variable, 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)
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']
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 not default_arg:
default_arg = []
default_arg.append("some_string")
return default_arg
1.
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]
2.
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]
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.
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...
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 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.
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
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 an awesome guide to learn more about how namespaces and scope resolution works in Python.
To 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
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
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 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 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!
def some_func():
try:
return 'from_try'
finally:
return 'from_finally'
Output:
>>> some_func()
'from_finally'
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.return statement executed. Since the finally clause always executes, a return statement executed in the finally clause will always be the last one executed.True = False
if True == False:
print("I've lost faith in truth!")
Output:
I've lost faith in truth!
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 backward compatibility, they couldn't make True and False constants- they just were built-in variables.>>> True is False == False
False
>>> False is False is False
True
>>> 1 > 0 < 1
True
>>> (1 > 0) < 1
False
>>> 1 > (0 < 1)
False
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.(1 > 0) < 1 is equivalent to True < 1 and
>>> int(True)
1
>>> True + 1 #not relevant for this example, but just for fun
2
1 < 1 evaluates to False1.
x = 5
class SomeClass:
x = 17
y = (x for i in range(10))
Output:
>>> list(SomeClass.y)[0]
5
2.
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
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
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)
This is not a WTF at all, but it took me so long to realize such things existed in Python. So sharing it here for the beginners.
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
'inf' and 'nan' are special strings (case-insensitive), which when explicitly type casted to float type, are used to represent mathematical "infinity" and "not a number" respectively.
1.
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)
2.
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
+= 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.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 "", line 3
except IndexError, ValueError:
^
SyntaxError: invalid syntax
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 "", 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
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))
The midnight time is not printed.
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."
1.
# A simple example to count the number of boolean 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:
>>> booleans_found_so_far
0
>>> integers_found_so_far
4
2.
another_dict = {}
another_dict[True] = "JavaScript"
another_dict[1] = "Ruby"
another_dict[1.0] = "Python"
Output:
>>> another_dict[True]
"Python"
Booleans are a subclass of int
>>> isinstance(True, int)
True
>>> isinstance(False, int)
True
The integer value of True is 1 and that of False is 0.
>>> True == 1 == 1.0 and False == 0 == 0.0
True
See this StackOverflow answer for rationale behind it.
Almost every Python programmer would have faced this situation.
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()
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.some_iterable = ('a', 'b')
def some_func(val):
return "something"
Output:
>>> [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']
Output:
>>> def f(x, y,):
... print(x, y)
...
>>> def g(x=4, y=5,):
... print(x, y)
...
>>> def h(x, **kwargs,):
File "", line 1
def h(x, **kwargs,):
^
SyntaxError: invalid syntax
>>> def h(*args,):
File "", line 1
def h(*args,):
^
SyntaxError: invalid syntax
some_string = "wtf"
some_dict = {}
for i, some_dict[i] in enumerate(some_string):
pass
Output:
>>> some_dict # An indexed dict is created.
{0: 'w', 1: 't', 2: 'f'}
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:
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) 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
x = True
y = False
Output:
>>> not x == y
True
>>> x == not y
File "", line 1
x == not y
^
SyntaxError: invalid syntax
== operator has higher precedence than not operator in Python.not x == y is equivalent to not (x == y) which is equivalent to not (True == False) finally evaluating to True.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.not token to be a part of the not in operator (because both == and not in operators have same precedence), but after not being able to find a in token following the not token, it raises a SyntaxError.a, b = a[b] = {}, 5
Output:
>>> a
{5: ({...}, 5)}
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, it's 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
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
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 object API.
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 sequences(iterables supporting element access using integer indices) in Python.3 --0-- 5 == 8 and --5 == 5 are both semantically correct statements and evaluate to True.Given that a is a number, ++a and --a are both valid Python statements, but don't behave the same way as compared with similar statements in languages like C, C++ or Java.
>>> a = 5
>>> a
5
>>> ++a
5
>>> --a
5
💡 Explanation:
++ operator in Python grammar. It is actually two + operators.++a parses as +(+a) which translates to a. Similarly, the output of the statement --a can be justified.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))
Half triple-quoted strings
>>> print('wtfpython''')
wtfpython
>>> print("wtfpython""")
wtfpython
>>> # The following statements raise `SyntaxError`
>>> # print('''wtfpython')
>>> # print("""wtfpython")
💡 Explanation:
>>> 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.Multiple Python threads won't run your Python code concurrently (yes you heard it right!). It may seem intuitive to spawn several threads and let them execute your Python code 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. Python threads are good for IO-bound tasks, but to achieve actual parallelization in Python for CPU-bound tasks, you might want to use the Python multiprocessing module.
List slicing with out of the bounds indices throws no errors
>>> some_list = [1, 2, 3, 4, 5]
>>> some_list[111:]
[]
Well, that's all I had to share for now. I'm working on more such snippets and will be sharing them very soon. The source of the above discussed snippets is maintained at https://github.com/satwikkansal/wtfpython, feel free to point out discrepencies (if any), or suggest your own discovered tricky snippets (gems), or even ask if you didn't understood any part of this post.