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Assigning types to variables as values
Only the first clause of generators is evaluated immediately

The C3 class resolution algorithm for multiple class inheritance

If we are dealing with multiple inheritance, according to the newer C3 class resolution algorithm, the following applies:
Assuming that child class C inherits from two parent classes A and B, “class A should be checked before class B”.

If you want to learn more, please read the original blog post by Guido van Rossum.

(Original source: http://gistroll.com/rolls/21/horizontal_assessments/new)

class A(object):
    def foo(self):
        print("class A")

class B(object):
    def foo(self):
        print("class B")

class C(A, B):
    pass

C().foo()

class A

So what actually happened above was that class C looked in the scope of the parent class A for the method .foo() first (and found it)!

I received an email containing a suggestion which uses a more nested example to illustrate Guido van Rossum’s point a little bit better:

class A(object):
   def foo(self):
      print("class A")

class B(A):
   pass

class C(A):
   def foo(self):
      print("class C")

class D(B,C):
   pass

D().foo()

class C

Here, class D searches in B first, which in turn inherits from A (note that class C also inherits from A, but has its own .foo() method) so that we come up with the search order: D, B, C, A.

Assignment operators and lists - simple-add vs. add-AND operators

Python lists are mutable objects as we all know. So, if we are using the += operator on lists, we extend the list by directly modifying the object.

However, if we use the assignment via my_list = my_list + ..., we create a new list object, which can be demonstrated by the following code:

a_list = []
print('ID:', id(a_list))

a_list += [1]
print('ID (+=):', id(a_list))

a_list = a_list + [2]
print('ID (list = list + ...):', id(a_list))

ID: 4366496544
ID (+=): 4366496544
ID (list = list + ...): 4366495472

Just for reference, the .append() and .extends() methods are modifying the list object in place, just as expected.

a_list = []
print(a_list, '\nID (initial):',id(a_list), '\n')

a_list.append(1)
print(a_list, '\nID (append):',id(a_list), '\n')

a_list.extend([2])
print(a_list, '\nID (extend):',id(a_list))

[]
ID (initial): 140704077653128

[1]
ID (append): 140704077653128

[1, 2]
ID (extend): 140704077653128

`True` and `False` in the datetime module

“It often comes as a big surprise for programmers to find (sometimes by way of a hard-to-reproduce bug) that, unlike any other time value, midnight (i.e. datetime.time(0,0,0)) is False. A long discussion on the python-ideas mailing list shows that, while surprising, that behavior is desirable—at least in some quarters.”

(Original source: http://lwn.net/SubscriberLink/590299/bf73fe823974acea/)

import datetime

print('"datetime.time(0,0,0)" (Midnight) ->', bool(datetime.time(0,0,0)))

print('"datetime.time(1,0,0)" (1 am) ->', bool(datetime.time(1,0,0)))

"datetime.time(0,0,0)" (Midnight) -> False
"datetime.time(1,0,0)" (1 am) -> True

Python reuses objects for small integers - use “==” for equality, “is” for identity

This oddity occurs, because Python keeps an array of small integer objects (i.e., integers between -5 and 256, see the doc).

a = 1
b = 1
print('a is b', bool(a is b))
True

c = 999
d = 999
print('c is d', bool(c is d))

a is b True
c is d False

(I received a comment that this is in fact a CPython artefact and must not necessarily be true in all implementations of Python!)

So the take home message is: always use “==” for equality, “is” for identity!

Here is a nice article explaining it by comparing “boxes” (C language) with “name tags” (Python).

This example demonstrates that this applies indeed for integers in the range in -5 to 256:

print('256 is 257-1', 256 is 257-1)
print('257 is 258-1', 257 is 258 - 1)
print('-5 is -6+1', -5 is -6+1)
print('-7 is -6-1', -7 is -6-1)

256 is 257-1 True
257 is 258-1 False
-5 is -6+1 True
-7 is -6-1 False

And to illustrate the test for equality (`==`) vs. identity (`is`):

a = 'hello world!'
b = 'hello world!'
print('a is b,', a is b)
print('a == b,', a == b)

a is b, False
a == b, True

We would think that identity would always imply equality, but this is not always true, as we can see in the next example:

a = float('nan')
print('a is a,', a is a)
print('a == a,', a == a)

a is a, True
a == a, False

Shallow vs. deep copies if list contains other structures and objects

Shallow copy:
If we use the assignment operator to assign one list to another list, we just create a new name reference to the original list. If we want to create a new list object, we have to make a copy of the original list. This can be done via a_list[:] or a_list.copy().

list1 = [1,2]
list2 = list1        # reference
list3 = list1[:]     # shallow copy
list4 = list1.copy() # shallow copy

print('IDs:\nlist1: {}\nlist2: {}\nlist3: {}\nlist4: {}\n'
      .format(id(list1), id(list2), id(list3), id(list4)))

list2[0] = 3
print('list1:', list1)

list3[0] = 4
list4[1] = 4
print('list1:', list1)

IDs:
list1: 4346366472
list2: 4346366472
list3: 4346366408
list4: 4346366536

list1: [3, 2]
list1: [3, 2]

Deep copy
As we have seen above, a shallow copy works fine if we want to create a new list with contents of the original list which we want to modify independently.

However, if we are dealing with compound objects (e.g., lists that contain other lists, read here for more information) it becomes a little trickier.

In the case of compound objects, a shallow copy would create a new compound object, but it would just insert the references to the contained objects into the new compound object. In contrast, a deep copy would go “deeper” and create also new objects
for the objects found in the original compound object. If you follow the code, the concept should become more clear:

from copy import deepcopy

list1 = [[1],[2]]
list2 = list1.copy()    # shallow copy
list3 = deepcopy(list1) # deep copy

print('IDs:\nlist1: {}\nlist2: {}\nlist3: {}\n'
      .format(id(list1), id(list2), id(list3)))

list2[0][0] = 3
print('list1:', list1)

list3[0][0] = 5
print('list1:', list1)

IDs:
list1: 4377956296
list2: 4377961752
list3: 4377954928

list1: [[3], [2]]
list1: [[3], [2]]

Picking `True` values from logical `and`s and `or`s

Logical or:

a or b == a if a else b

If both values in or expressions are True, Python will select the first value (e.g., select "a" in "a" or "b"), and the second one in and expressions.
This is also called short-circuiting - we already know that the logical or must be True if the first value is True and therefore can omit the evaluation of the second value.

Logical and:

a and b == b if a else a

If both values in and expressions are True, Python will select the second value, since for a logical and, both values must be true.

result = (2 or 3) * (5 and 7)
print('2 * 7 =', result)

2 * 7 = 14

Don’t use mutable objects as default arguments for functions!

Don’t use mutable objects (e.g., dictionaries, lists, sets, etc.) as default arguments for functions! You might expect that a new list is created every time when we call the function without providing an argument for the default parameter, but this is not the case: Python will create the mutable object (default parameter) the first time the function is defined - not when it is called, see the following code:

(Original source: http://docs.python-guide.org/en/latest/writing/gotchas/

def append_to_list(value, def_list=[]):
    def_list.append(value)
    return def_list

my_list = append_to_list(1)
print(my_list)

my_other_list = append_to_list(2)
print(my_other_list)

[1]
[1, 2]

Another good example showing that demonstrates that default arguments are created when the function is created (and not when it is called!):

import time
def report_arg(my_default=time.time()):
    print(my_default)

report_arg()

time.sleep(5)

report_arg()

1397764090.456688
1397764090.456688

Be aware of the consuming generator

Be aware of what is happening when combining “in” checks with generators, since they won’t evaluate from the beginning once a position is “consumed”.

gen = (i for i in range(5))
print('2 in gen,', 2 in gen)
print('3 in gen,', 3 in gen)
print('1 in gen,', 1 in gen)

in gen, True
in gen, True
in gen, False

Although this defeats the purpose of a generator (in most cases), we can convert a generator into a list to circumvent the problem.

gen = (i for i in range(5))
a_list = list(gen)
print('2 in l,', 2 in a_list)
print('3 in l,', 3 in a_list)
print('1 in l,', 1 in a_list)

in l, True
in l, True
in l, True

`bool` is a subclass of `int`

Chicken or egg? In the history of Python (Python 2.2 to be specific) truth values were implemented via 1 and 0 (similar to the old C). In order to avoid syntax errors in old (but perfectly working) Python code, bool was added as a subclass of int in Python 2.3.

Original source: http://www.peterbe.com/plog/bool-is-int

print('isinstance(True, int):', isinstance(True, int))
print('True + True:', True + True)
print('3*True + True:', 3*True + True)
print('3*True - False:', 3*True - False)

isinstance(True, int): True
True + True: 2
3*True + True: 4
3*True - False: 3

About lambda-in-closures-and-a-loop pitfall

Remember the section about the “consuming generators”? This example is somewhat related, but the result might still come unexpected.

(Original source: http://openhome.cc/eGossip/Blog/UnderstandingLambdaClosure3.html)

In the first example below, we call a lambda function in a list comprehension, and the value i will be dereferenced every time we call lambda. Since the list comprehension has already been constructed and evaluated when we for-loop through the list, the closure-variable will be set to the last value 4.

my_list = [lambda: i for i in range(5)]
for l in my_list:
    print(l())

However, by using a generator expression, we can make use of its stepwise evaluation (note that the returned variable still stems from the same closure, but the value changes as we iterate over the generator).

my_gen = (lambda: n for n in range(5))
for l in my_gen:
    print(l())

And if you are really keen on using lists, there is a nifty trick that circumvents this problem as a reader nicely pointed out in the comments: We can simply pass the loop variable i as a default argument to the lambdas.

my_list = [lambda x=i: x for i in range(5)]
for l in my_list:
    print(l())

Python’s LEGB scope resolution and the keywords `global` and `nonlocal`

There is nothing particularly surprising about Python’s LEGB scope resolution (Local -> Enclosed -> Global -> Built-in), but it is still useful to take a look at some examples!

`global` vs. `local`

According to the LEGB rule, Python will first look for a variable in the local scope. So if we set the variable x = 1 locally in the function’s scope, it won’t have an effect on the global x.

x = 0
def in_func():
    x = 1
    print('in_func:', x)

in_func()
print('global:', x)

in_func: 1
global: 0

If we want to modify the global x via a function, we can simply use the global keyword to import the variable into the function’s scope:

x = 0
def in_func():
    global x
    x = 1
    print('in_func:', x)

in_func()
print('global:', x)

in_func: 1
global: 1

`local` vs. `enclosed`

Now, let us take a look at local vs. enclosed. Here, we set the variable x = 1 in the outer function and set x = 1 in the enclosed function inner. Since inner looks in the local scope first, it won’t modify outer’s x.

def outer():
       x = 1
       print('outer before:', x)
       def inner():
           x = 2
           print("inner:", x)
       inner()
       print("outer after:", x)
outer()

outer before: 1
inner: 2
outer after: 1

Here is where the nonlocal keyword comes in handy - it allows us to modify the x variable in the enclosed scope:

def outer():
       x = 1
       print('outer before:', x)
       def inner():
           nonlocal x
           x = 2
           print("inner:", x)
       inner()
       print("outer after:", x)
outer()

outer before: 1
inner: 2
outer after: 2

When mutable contents of immutable tuples aren’t so mutable

As we all know, tuples are immutable objects in Python, right!? But what happens if they contain mutable objects?

First, let us have a look at the expected behavior: a TypeError is raised if we try to modify immutable types in a tuple:

tup = (1,)
tup[0] += 1

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)

<ipython-input-41-c3bec6c3fe6f> in <module>()
      1 tup = (1,)
----> 2 tup[0] += 1


TypeError: 'tuple' object does not support item assignment

But what if we put a mutable object into the immutable tuple? Well, modification works, but we also get a `TypeError` at the same time.

tup = ([],)
print('tup before: ', tup)
tup[0] += [1]

tup before:  ([],)

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)

<ipython-input-42-aebe9a31dbeb> in <module>()
      1 tup = ([],)
      2 print('tup before: ', tup)
----> 3 tup[0] += [1]

TypeError: 'tuple' object does not support item assignment

print('tup after: ', tup)

tup after:  ([1],)

However, there are ways to modify the mutable contents of the tuple without raising the TypeError, the solution is the .extend() method, or alternatively .append() (for lists):

tup = ([],)
print('tup before: ', tup)
tup[0].extend([1])
print('tup after: ', tup)

tup before:  ([],)
tup after:  ([1],)

tup = ([],)
print('tup before: ', tup)
tup[0].append(1)
print('tup after: ', tup)

tup before:  ([],)
tup after:  ([1],)

Explanation

A. Jesse Jiryu Davis has a nice explanation for this phenomenon (Original source: http://emptysqua.re/blog/python-increment-is-weird-part-ii/)

If we try to extend the list via += “then the statement executes STORE_SUBSCR, which calls the C function PyObject_SetItem, which checks if the object supports item assignment. In our case the object is a tuple, so PyObject_SetItem throws the TypeError. Mystery solved.”

One more note about the `immutable` status of tuples. Tuples are famous for being immutable. However, how comes that this code works?

my_tup = (1,)
my_tup += (4,)
my_tup = my_tup + (5,)
print(my_tup)

(1, 4, 5)

What happens “behind” the curtains is that the tuple is not modified, but a new object is generated every time, which will inherit the old “name tag”:

my_tup = (1,)
print(id(my_tup))
my_tup += (4,)
print(id(my_tup))
my_tup = my_tup + (5,)
print(id(my_tup))

4337381840
4357415496
4357289952

List comprehensions are fast, but generators are faster!?

“List comprehensions are fast, but generators are faster!?” - No, not really (or significantly, see the benchmarks below). So what’s the reason to prefer one over the other?

use lists if you want to use the plethora of list methods
use generators when you are dealing with huge collections to avoid memory issues

import timeit

def plainlist(n=100000):
    my_list = []
    for i in range(n):
        if i % 5 == 0:
            my_list.append(i)
    return my_list

def listcompr(n=100000):
    my_list = [i for i in range(n) if i % 5 == 0]
    return my_list

def generator(n=100000):
    my_gen = (i for i in range(n) if i % 5 == 0)
    return my_gen

def generator_yield(n=100000):
    for i in range(n):
        if i % 5 == 0:
            yield i

To be fair to the list, let us exhaust the generators:

def test_plainlist(plain_list):
    for i in plain_list():
        pass

def test_listcompr(listcompr):
    for i in listcompr():
        pass

def test_generator(generator):
    for i in generator():
        pass

def test_generator_yield(generator_yield):
    for i in generator_yield():
        pass

print('plain_list:     ', end = '')
%timeit test_plainlist(plainlist)
print('\nlistcompr:     ', end = '')
%timeit test_listcompr(listcompr)
print('\ngenerator:     ', end = '')
%timeit test_generator(generator)
print('\ngenerator_yield:     ', end = '')
%timeit test_generator_yield(generator_yield)

plain_list:     10 loops, best of 3: 22.4 ms per loop

listcompr:     10 loops, best of 3: 20.8 ms per loop

generator:     10 loops, best of 3: 22 ms per loop

generator_yield:     10 loops, best of 3: 21.9 ms per loop

Public vs. private class methods and name mangling

Who has not stumbled across this quote “we are all consenting adults here” in the Python community, yet? Unlike in other languages like C++ (sorry, there are many more, but that’s one I am most familiar with), we can’t really protect class methods from being used outside the class (i.e., by the API user).
All we can do is indicate methods as private to make clear that they are not to be used outside the class, but it is really up to the class user, since “we are all consenting adults here”!
So, when we want to mark a class method as private, we can put a single underscore in front of it.
If we additionally want to avoid name clashes with other classes that might use the same method names, we can prefix the name with a double-underscore to invoke the name mangling.

This doesn’t prevent the class users to access this class member though, but they have to know the trick and also know that it is at their own risk…

Let the following example illustrate what I mean:

class my_class():
    def public_method(self):
        print('Hello public world!')
    def __private_method(self):
        print('Hello private world!')
    def call_private_method_in_class(self):
        self.__private_method()

my_instance = my_class()

my_instance.public_method()
my_instance._my_class__private_method()
my_instance.call_private_method_in_class()

Hello public world!
Hello private world!
Hello private world!

The consequences of modifying a list when looping through it

It can be really dangerous to modify a list when iterating through it - this is a very common pitfall that can cause unintended behavior!
Look at the following examples, and for a fun exercise: try to figure out what is going on before you skip to the solution!

a = [1, 2, 3, 4, 5]
for i in a:
    if not i % 2:
        a.remove(i)
print(a)

[1, 3, 5]

b = [2, 4, 5, 6]
for i in b:
     if not i % 2:
         b.remove(i)
print(b)

[4, 5]

The solution is that we are iterating through the list index by index, and if we remove one of the items in-between, we inevitably mess around with the indexing. Look at the following example and it will become clear:

b = [2, 4, 5, 6]
for index, item in enumerate(b):
    print(index, item)
    if not item % 2:
        b.remove(item)
print(b)

2
5
6
[4, 5]

Dynamic binding and typos in variable names

Be careful, dynamic binding is convenient, but can also quickly become dangerous!

print('first list:')
for i in range(3):
    print(i)

print('\nsecond list:')
for j in range(3):
    print(i) # I (intentionally) made typo here!

first list:
0
1
2

second list:
2
2
2

List slicing using indexes that are “out of range”

As we have all encountered it 1 (x10000) time(s) in our life, the infamous IndexError:

my_list = [1, 2, 3, 4, 5]
print(my_list[5])

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)

<ipython-input-15-eb273dc36fdc> in <module>()
      1 my_list = [1, 2, 3, 4, 5]
----> 2 print(my_list[5])


IndexError: list index out of range

But suprisingly, it is not raised when we are doing list slicing, which can be a real pain when debugging:

my_list = [1, 2, 3, 4, 5]
print(my_list[5:])

[]

Reusing global variable names and `UnboundLocalErrors`

Usually, it is no problem to access global variables in the local scope of a function:

def my_func():
    print(var)

var = 'global'
my_func()

global

And is also no problem to use the same variable name in the local scope without affecting the local counterpart:

def my_func():
    var = 'locally changed'

var = 'global'
my_func()
print(var)

global

But we have to be careful if we use a variable name that occurs in the global scope, and we want to access it in the local function scope if we want to reuse this name:

def my_func():
    print(var) # want to access global variable
    var = 'locally changed' # but Python thinks we forgot to define the local variable!

var = 'global'
my_func()

---------------------------------------------------------------------------
UnboundLocalError                         Traceback (most recent call last)

<ipython-input-40-3afd870b7c35> in <module>()
      4
      5 var = 'global'
----> 6 my_func()


<ipython-input-40-3afd870b7c35> in my_func()
      1 def my_func():
----> 2     print(var) # want to access global variable
      3     var = 'locally changed'
      4
      5 var = 'global'


UnboundLocalError: local variable 'var' referenced before assignment

In this case, we have to use the global keyword!

def my_func():
    global var
    print(var) # want to access global variable
    var = 'locally changed' # changes the gobal variable

var = 'global'

my_func()
print(var)

global
locally changed

Creating copies of mutable objects

Let’s assume a scenario where we want to duplicate sublists of values stored in another list. If we want to create an independent sublist object, using the arithmetic multiplication operator could lead to rather unexpected (or undesired) results:

my_list1 = [[1, 2, 3]] * 2

print('initially ---> ', my_list1)

# modify the 1st element of the 2nd sublist
my_list1[1][0] = 'a'
print("after my_list1[1][0] = 'a' ---> ", my_list1)

initially --->  [[1, 2, 3], [1, 2, 3]]
after my_list1[1][0] = 'a' --->  [['a', 2, 3], ['a', 2, 3]]

In this case, we should better create “new” objects:

my_list2 = [[1, 2, 3] for i in range(2)]

print('initially:  ---> ', my_list2)

# modify the 1st element of the 2nd sublist
my_list2[1][0] = 'a'
print("after my_list2[1][0] = 'a':  ---> ", my_list2)

initially:  --->  [[1, 2, 3], [1, 2, 3]]
after my_list2[1][0] = 'a':  --->  [[1, 2, 3], ['a', 2, 3]]

And here is the proof:

for a,b in zip(my_list1, my_list2):
    print('id my_list1: {}, id my_list2: {}'.format(id(a), id(b)))

id my_list1: 4350764680, id my_list2: 4350766472
id my_list1: 4350764680, id my_list2: 4350766664

Key differences between Python 2 and 3

There are some good articles already that are summarizing the differences between Python 2 and 3, e.g.,

But it might be still worthwhile, especially for Python newcomers, to take a look at some of those! (Note: the the code was executed in Python 3.4.0 and Python 2.7.5 and copied from interactive shell sessions.)

Overview - Key differences between Python 2 and 3

Unicode
The print statement
Integer division
xrange()
Raising exceptions
Handling exceptions
next() function and .next() method
Loop variables and leaking into the global scope
Comparing unorderable types

Unicode…

[back to Python 2.x vs 3.x overview]

- Python 2:

We have ASCII str() types, separate unicode(), but no byte type

- Python 3:

Now, we finally have Unicode (utf-8) strings, and 2 byte classes: byte and bytearrays

#############
# Python 2
#############

>>> type(unicode('is like a python3 str()'))
<type 'unicode'>

>>> type(b'byte type does not exist')
<type 'str'>

>>> 'they are really' + b' the same'
'they are really the same'

>>> type(bytearray(b'bytearray oddly does exist though'))
<type 'bytearray'>

#############
# Python 3
#############

>>> print('strings are now utf-8 \u03BCnico\u0394é!')
strings are now utf-8 μnicoΔé!


>>> type(b' and we have byte types for storing data')
<class 'bytes'>

>>> type(bytearray(b'but also bytearrays for those who prefer them over strings'))
<class 'bytearray'>

>>> 'string' + b'bytes for data'
Traceback (most recent call last):s
  File "<stdin>", line 1, in <module>
TypeError: Can't convert 'bytes' object to str implicitly

The print statement

Very trivial, but this change makes sense, Python 3 now only accepts prints with proper parentheses - just like the other function calls …

# Python 2
>>> print 'Hello, World!'
Hello, World!
>>> print('Hello, World!')
Hello, World!

# Python 3
>>> print('Hello, World!')
Hello, World!
>>> print 'Hello, World!'
  File "<stdin>", line 1
    print 'Hello, World!'
                        ^
SyntaxError: invalid syntax

And if we want to print the output of 2 consecutive print functions on the same line, you would use a comma in Python 2, and a end="" in Python 3:

# Python 2
>>> print "line 1", ; print 'same line'
line 1 same line

# Python 3
>>> print("line 1", end="") ; print (" same line")
line 1 same line

Integer division

This is a pretty dangerous thing if you are porting code, or executing Python 3 code in Python 2 since the change in integer-division behavior can often go unnoticed.
So, I still tend to use a float(3)/2 or 3/2.0 instead of a 3/2 in my Python 3 scripts to save the Python 2 guys some trouble … (PS: and vice versa, you can from __future__ import division in your Python 2 scripts).

# Python 2
>>> 3 / 2
1
>>> 3 // 2
1
>>> 3 / 2.0
1.5
>>> 3 // 2.0
1.0

# Python 3
>>> 3 / 2
1.5
>>> 3 // 2
1
>>> 3 / 2.0
1.5
>>> 3 // 2.0
1.0

###xrange()

xrange() was pretty popular in Python 2.x if you wanted to create an iterable object. The behavior was quite similar to a generator (‘lazy evaluation’), but you could iterate over it infinitely. The advantage was that it was generally faster than range() (e.g., in a for-loop) - not if you had to iterate over the list multiple times, since the generation happens every time from scratch!
In Python 3, the range() was implemented like the xrange() function so that a dedicated xrange() function does not exist anymore.

# Python 2
> python -m timeit 'for i in range(1000000):' ' pass'
10 loops, best of 3: 66 msec per loop

    > python -m timeit 'for i in xrange(1000000):' ' pass'
10 loops, best of 3: 27.8 msec per loop

# Python 3
> python3 -m timeit 'for i in range(1000000):' ' pass'
10 loops, best of 3: 51.1 msec per loop

> python3 -m timeit 'for i in xrange(1000000):' ' pass'
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/timeit.py", line 292, in main
    x = t.timeit(number)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/timeit.py", line 178, in timeit
    timing = self.inner(it, self.timer)
  File "<timeit-src>", line 6, in inner
    for i in xrange(1000000):
NameError: name 'xrange' is not defined

Raising exceptions

Where Python 2 accepts both notations, the ‘old’ and the ‘new’ way, Python 3 chokes (and raises a SyntaxError in turn) if we don’t enclose the exception argument in parentheses:

# Python 2
>>> raise IOError, "file error"
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IOError: file error
>>> raise IOError("file error")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IOError: file error


# Python 3    
>>> raise IOError, "file error"
  File "<stdin>", line 1
    raise IOError, "file error"
                 ^
SyntaxError: invalid syntax
>>> raise IOError("file error")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OSError: file error

Handling exceptions

Also the handling of exceptions has slightly changed in Python 3. Now, we have to use the as keyword!

# Python 2
>>> try:
...     blabla
... except NameError, err:
...     print err, '--> our error msg'
...
name 'blabla' is not defined --> our error msg

# Python 3
>>> try:
...     blabla
... except NameError as err:
...     print(err, '--> our error msg')
...
name 'blabla' is not defined --> our error msg

The `next()` function and `.next()` method

Where you can use both function and method in Python 2.7.5, the next() function is all that remains in Python 3!

# Python 2
>>> my_generator = (letter for letter in 'abcdefg')
>>> my_generator.next()
'a'
>>> next(my_generator)
'b'

# Python 3
>>> my_generator = (letter for letter in 'abcdefg')
>>> next(my_generator)
'a'
>>> my_generator.next()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'generator' object has no attribute 'next'

In Python 3.x for-loop variables don’t leak into the global namespace anymore

This goes back to a change that was made in Python 3.x and is described in What’s New In Python 3.0 as follows:

“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.”

from platform import python_version
print('This code cell was executed in Python', python_version())

i = 1
print([i for i in range(5)])
print(i, '-> i in global')

This code cell was executed in Python 3.3.5
[0, 1, 2, 3, 4]
1 -> i in global

from platform import python_version
print 'This code cell was executed in Python', python_version()

i = 1
print [i for i in range(5)]
print i, '-> i in global'

This code cell was executed in Python 2.7.6
[0, 1, 2, 3, 4]
4 -> i in global

Python 3.x prevents us from comparing unorderable types

from platform import python_version
print 'This code cell was executed in Python', python_version()

print [1, 2] > 'foo'
print (1, 2) > 'foo'
print [1, 2] > (1, 2)

This code cell was executed in Python 2.7.6
False
True
False

from platform import python_version
print('This code cell was executed in Python', python_version())

print([1, 2] > 'foo')
print((1, 2) > 'foo')
print([1, 2] > (1, 2))

This code cell was executed in Python 3.3.5



---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)

<ipython-input-3-1d774c677f73> in <module>()
      2 print('This code cell was executed in Python', python_version())
      3
----> 4 [1, 2] > 'foo'
      5 (1, 2) > 'foo'
      6 [1, 2] > (1, 2)


TypeError: unorderable types: list() > str()

Function annotations - What are those `->`’s in my Python code?

Have you ever seen any Python code that used colons inside the parantheses of a function definition?

def foo1(x: 'insert x here', y: 'insert x^2 here'):
    print('Hello, World')
    return

And what about the fancy arrow here?

def foo2(x, y) -> 'Hi!':
    print('Hello, World')
    return

Q: Is this valid Python syntax?
A: Yes!

Q: So, what happens if I just call the function?
A: Nothing!

Here is the proof!

foo1(1,2)

Hello, World

foo2(1,2)    

Hello, World

**So, those are function annotations … **

the colon for the function parameters
the arrow for the return value

You probably will never make use of them (or at least very rarely). Usually, we write good function documentations below the function as a docstring - or at least this is how I would do it (okay this case is a little bit extreme, I have to admit):

def is_palindrome(a):
    """
    Case-and punctuation insensitive check if a string is a palindrom.

    Keyword arguments:
        a (str): The string to be checked if it is a palindrome.

    Returns `True` if input string is a palindrome, else False.

    """
    stripped_str = [l for l in my_str.lower() if l.isalpha()]
    return stripped_str == stripped_str[::-1]

However, function annotations can be useful to indicate that work is still in progress in some cases. But they are optional and I see them very very rarely.

As it is stated in PEP3107:

Function annotations, both for parameters and return values, are completely optional.
Function annotations are nothing more than a way of associating arbitrary Python expressions with various parts of a function at compile-time.

The nice thing about function annotations is their __annotations__ attribute, which is a dictionary of all the parameters and/or the return value you annotated.

foo1.__annotations__

{'y': 'insert x^2 here', 'x': 'insert x here'}

foo2.__annotations__

{'return': 'Hi!'}

When are they useful?

Function annotations can be useful for a couple of things

Documentation in general
pre-condition testing
type checking

…

Abortive statements in `finally` blocks

Python’s try-except-finally blocks are very handy for catching and handling errors. The finally block is always executed whether an exception has been raised or not as illustrated in the following example.

def try_finally1():
    try:
        print('in try:')
        print('do some stuff')
        float('abc')
    except ValueError:
        print('an error occurred')
    else:
        print('no error occurred')
    finally:
        print('always execute finally')

try_finally1()

in try:
do some stuff
an error occurred
always execute finally

But can you also guess what will be printed in the next code cell?

def try_finally2():
    try:
        print("do some stuff in try block")
        return "return from try block"
    finally:
        print("do some stuff in finally block")
        return "always execute finally"

print(try_finally2())

do some stuff in try block
do some stuff in finally block
always execute finally

Here, the abortive return statement in the finally block simply overrules the return in the try block, since finally is guaranteed to always be executed. So, be careful using abortive statements in finally blocks!

Assigning types to variables as values

I am not yet sure in which context this can be useful, but it is a nice fun fact to know that we can assign types as values to variables.

a_var = str
a_var(123)

'123'

from random import choice

a, b, c = float, int, str
for i in range(5):
    j = choice([a,b,c])(i)
    print(j, type(j))

<class 'int'>
<class 'int'>
0 <class 'float'>
<class 'str'>
<class 'int'>

Only the first clause of generators is evaluated immediately

The main reason why we love to use generators in certain cases (i.e., when we are dealing with large numbers of computations) is that it only computes the next value when it is needed, which is also known as “lazy” evaluation. However, the first clause of an generator is already checked upon it’s creation, as the following example demonstrates:

gen_fails = (i for i in 1/0)

---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)

<ipython-input-18-29312e1ece8d> in <module>()
----> 1 gen_fails = (i for i in 1/0)


ZeroDivisionError: division by zero

Certainly, this is a nice feature, since it notifies us about syntax erros immediately. However, this is (unfortunately) not the case if we have multiple cases in our generator.

gen_succeeds = (i for i in range(5) for j in 1/0)

print('But obviously fails when we iterate ...')
for i in gen_succeeds:
    print(i)

---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)

<ipython-input-20-8a83a1022971> in <module>()
      1 print('But obviously fails when we iterate ...')
----> 2 for i in gen_succeeds:
      3     print(i)

<ipython-input-19-c54c53f2218a> in <genexpr>(.0)
----> 1 gen_succeeds = (i for i in range(5) for j in 1/0)

ZeroDivisionError: division by zero

But obviously fails when we iterate ...

##Keyword argument unpacking syntax - *args and **kwargs

Python has a very convenient “keyword argument unpacking syntax” (often referred to as “splat”-operators). This is particularly useful, if we want to define a function that can take a arbitrary number of input arguments.

Single-asterisk (*args)

def a_func(*args):
    print('type of args:', type(args))
    print('args contents:', args)
    print('1st argument:', args[0])

a_func(0, 1, 'a', 'b', 'c')

type of args: <class 'tuple'>
args contents: (0, 1, 'a', 'b', 'c')
1st argument: 0

Double-asterisk (**kwargs)

def b_func(**kwargs):
    print('type of kwargs:', type(kwargs))
    print('kwargs contents: ', kwargs)
    print('value of argument a:', kwargs['a'])

b_func(a=1, b=2, c=3, d=4)

type of kwargs: <class 'dict'>
kwargs contents:  {'d': 4, 'a': 1, 'c': 3, 'b': 2}
value of argument a: 1

(Partially) unpacking of iterables

Another useful application of the “unpacking”-operator is the unpacking of lists and other other iterables.

val1, *vals = [1, 2, 3, 4, 5]
print('val1:', val1)
print('vals:', vals)

val1: 1
vals: [2, 3, 4, 5]

Metaclasses - What creates a new instance of a class?

Usually, it is the __init__ method when we think of instanciating a new object from a class. However, it is the static method __new__ (it is not a class method!) that creates and returns a new instance before __init__() is called. More specifically, this is what is returned:
return super(<currentclass>, cls).__new__(subcls, *args, **kwargs)

For more information about the __new__ method, please see the documentation.

As a little experiment, let us screw with __new__ so that it returns None and see if __init__ will be executed:

class a_class(object):
    def __new__(clss, *args, **kwargs):
        print('excecuted __new__')
        return None
    def __init__(self, an_arg):
        print('excecuted __init__')
        self.an_arg = an_arg

a_object = a_class(1)
print('Type of a_object:', type(a_object))

excecuted __new__
Type of a_object: <class 'NoneType'>

As we can see in the code above, __init__ requires the returned instance from __new__ in order to called. So, here we just created a NoneType object.
Let us override the __new__, now and let us confirm that __init__ is called now to instantiate the new object”:

class a_class(object):
    def __new__(cls, *args, **kwargs):
        print('excecuted __new__')
        inst = super(a_class, cls).__new__(cls)
        return inst
    def __init__(self, an_arg):
        print('excecuted __init__')
        self.an_arg = an_arg

a_object = a_class(1)
print('Type of a_object:', type(a_object))
print('a_object.an_arg: ', a_object.an_arg)

excecuted __new__
excecuted __init__
Type of a_object: <class '__main__.a_class'>
a_object.an_arg:  1

for i in range(5):
    if i == 1:
        print('in for')
else:
    print('in else')
print('after for-loop')

in for
in else
after for-loop

for i in range(5):
    if i == 1:
        break
else:
    print('in else')
print('after for-loop')

after for-loop

Else-clauses: “conditional else” and “completion else”

I would claim that the conditional “else” is every programmer’s daily bread and butter. However, there is a second flavor of “else”-clauses in Python, which I will call “completion else” (for reason that will become clear later).
But first, let us take a look at our “traditional” conditional else that we all are familiar with.

Conditional else:

# conditional else

a_list = [1,2]
if a_list[0] == 1:
    print('Hello, World!')
else:
    print('Bye, World!')

Hello, World!

# conditional else

a_list = [1,2]
if a_list[0] == 2:
    print('Hello, World!')
else:
    print('Bye, World!')

Bye, World!

Why am I showing those simple examples? I think they are good to highlight some of the key points: It is either the code under the if clause that is executed, or the code under the else block, but not both.
If the condition of the if clause evaluates to True, the if-block is exectured, and if it evaluated to False, it is the else block.

Completion else

In contrast to the either…or* situation that we know from the conditional else, the completion else is executed if a code block finished.
To show you an example, let us use else for error-handling:

Completion else (try-except)

try:
    print('first element:', a_list[0])
except IndexError:
    print('raised IndexError')
else:
    print('no error in try-block')

first element: 1
no error in try-block

try:
    print('third element:', a_list[2])
except IndexError:
    print('raised IndexError')
else:
    print('no error in try-block')

raised IndexError

In the code above, we can see that the code under the else-clause is only executed if the try-block was executed without encountering an error, i.e., if the try-block is “complete”.
The same rule applies to the “completion” else in while- and for-loops, which you can confirm in the following samples below.

Completion else (while-loop)

i = 0
while i < 2:
    print(i)
    i += 1
else:
    print('in else')

0
1
in else

i = 0
while i < 2:
    print(i)
    i += 1
    break
else:
    print('completed while-loop')

Completion else (for-loop)

for i in range(2):
    print(i)
else:
    print('completed for-loop')

0
1
completed for-loop

for i in range(2):
    print(i)
    break
else:
    print('completed for-loop')

Interning of compile-time constants vs. run-time expressions

This might not be particularly useful, but it is nonetheless interesting: Python’s interpreter is interning compile-time constants but not run-time expressions (note that this is implementation-specific).

(Original source: Stackoverflow)

Let us have a look at the simple example below. Here we are creating 3 variables and assign the value “Hello” to them in different ways before we test them for identity.

hello1 = 'Hello'

hello2 = 'Hell' + 'o'

hello3 = 'Hell'
hello3 = hello3 + 'o'

print('hello1 is hello2:', hello1 is hello2)
print('hello1 is hello3:', hello1 is hello3)

hello1 is hello2: True
hello1 is hello3: False

Now, how does it come that the first expression evaluates to true, but the second does not? To answer this question, we need to take a closer look at the underlying byte codes:

import dis
def hello1_func():
    s = 'Hello'
    return s
dis.dis(hello1_func)

  3           0 LOAD_CONST               1 ('Hello')
              3 STORE_FAST               0 (s)

  4           6 LOAD_FAST                0 (s)
              9 RETURN_VALUE

def hello2_func():
    s = 'Hell' + 'o'
    return s
dis.dis(hello2_func)

  2           0 LOAD_CONST               3 ('Hello')
              3 STORE_FAST               0 (s)

  3           6 LOAD_FAST                0 (s)
              9 RETURN_VALUE

def hello3_func():
    s = 'Hell'
    s = s + 'o'
    return s
dis.dis(hello3_func)

  2           0 LOAD_CONST               1 ('Hell')
              3 STORE_FAST               0 (s)

  3           6 LOAD_FAST                0 (s)
              9 LOAD_CONST               2 ('o')
             12 BINARY_ADD
             13 STORE_FAST               0 (s)

  4          16 LOAD_FAST                0 (s)
             19 RETURN_VALUE

It looks like that 'Hello' and 'Hell' + 'o' are both evaluated and stored as 'Hello' at compile-time, whereas the third version
s = 'Hell'
s = s + 'o' seems to not be interned. Let us quickly confirm the behavior with the following code:

print(hello1_func() is hello2_func())
print(hello1_func() is hello3_func())

True
False

Finally, to show that this hypothesis is the answer to this rather unexpected observation, let us intern the value manually:

import sys

print(hello1_func() is sys.intern(hello3_func()))

True