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Function with dictionary works unexpectedly [duplicate]

This question already has answers here:
How do I initialize a dictionary of empty lists in Python?
(7 answers)
Closed 2 years ago.
I came across this behavior that surprised me in Python 2.6 and 3.2:
>>> xs = dict.fromkeys(range(2), [])
>>> xs
{0: [], 1: []}
>>> xs[0].append(1)
>>> xs
{0: [1], 1: [1]}
However, dict comprehensions in 3.2 show a more polite demeanor:
>>> xs = {i:[] for i in range(2)}
>>> xs
{0: [], 1: []}
>>> xs[0].append(1)
>>> xs
{0: [1], 1: []}
>>>
Why does fromkeys behave like that?

Your Python 2.6 example is equivalent to the following, which may help to clarify:
>>> a = []
>>> xs = dict.fromkeys(range(2), a)
Each entry in the resulting dictionary will have a reference to the same object. The effects of mutating that object will be visible through every dict entry, as you've seen, because it's one object.
>>> xs[0] is a and xs[1] is a
True
Use a dict comprehension, or if you're stuck on Python 2.6 or older and you don't have dictionary comprehensions, you can get the dict comprehension behavior by using dict() with a generator expression:
xs = dict((i, []) for i in range(2))

In the first version, you use the same empty list object as the value for both keys, so if you change one, you change the other, too.
Look at this:
>>> empty = []
>>> d = dict.fromkeys(range(2), empty)
>>> d
{0: [], 1: []}
>>> empty.append(1) # same as d[0].append(1) because d[0] references empty!
>>> d
{0: [1], 1: [1]}
In the second version, a new empty list object is created in every iteration of the dict comprehension, so both are independent from each other.
As to "why" fromkeys() works like that - well, it would be surprising if it didn't work like that. fromkeys(iterable, value) constructs a new dict with keys from iterable that all have the value value. If that value is a mutable object, and you change that object, what else could you reasonably expect to happen?

To answer the actual question being asked: fromkeys behaves like that because there is no other reasonable choice. It is not reasonable (or even possible) to have fromkeys decide whether or not your argument is mutable and make new copies every time. In some cases it doesn't make sense, and in others it's just impossible.
The second argument you pass in is therefore just a reference, and is copied as such. An assignment of [] in Python means "a single reference to a new list", not "make a new list every time I access this variable". The alternative would be to pass in a function that generates new instances, which is the functionality that dict comprehensions supply for you.
Here are some options for creating multiple actual copies of a mutable container:
As you mention in the question, dict comprehensions allow you to execute an arbitrary statement for each element:
d = {k: [] for k in range(2)}
The important thing here is that this is equivalent to putting the assignment k = [] in a for loop. Each iteration creates a new list and assigns it to a value.
Use the form of the dict constructor suggested by #Andrew Clark:
d = dict((k, []) for k in range(2))
This creates a generator which again makes the assignment of a new list to each key-value pair when it is executed.
Use a collections.defaultdict instead of a regular dict:
d = collections.defaultdict(list)
This option is a little different from the others. Instead of creating the new list references up front, defaultdict will call list every time you access a key that's not already there. You can there fore add the keys as lazily as you want, which can be very convenient sometimes:
for k in range(2):
d[k].append(42)
Since you've set up the factory for new elements, this will actually behave exactly as you expected fromkeys to behave in the original question.
Use dict.setdefault when you access potentially new keys. This does something similar to what defaultdict does, but it has the advantage of being more controlled, in the sense that only the access you want to create new keys actually creates them:
d = {}
for k in range(2):
d.setdefault(k, []).append(42)
The disadvantage is that a new empty list object gets created every time you call the function, even if it never gets assigned to a value. This is not a huge problem, but it could add up if you call it frequently and/or your container is not as simple as list.

Why can't a list be constructed and modified in the same line?

For example, why is a not equal to b?
a = [1]
a.append(2)
print(a) # [1, 2]
b = [1].append(2)
print(b) # None
The syntax for b doesn't look wrong to me, but it is. I want to write one-liners to define a list (e.g. using a generator expression) and then append elements, but all I get is None.

It's because:
append, extend, sort and more list function are all "in-place".
What does "in-place" mean? it means it modifies the original variable directly, some things you would need:
l = sorted(l)
To modify the list, but append already does that, so:
l.append(3)
Will modify l already, don't need:
l = l.append(3)
If you do:
l = [1].append(2)
Yes it will modify the list of [1], but it would be lost in memory somewhere inaccessible, whereas l will become None as we discovered above.
To make it not "in-place", without using append either do:
l = l + [2]
Or:
l = [*l, 2]

The one-liner for b does these steps:
Defines a list [1]
Appends 2 to the list in-place
Append has no return, so b = None
The same is true for all list methods that alter the list in-place without a return. These are all None:
c = [1].extend([2])
d = [2, 1].sort()
e = [1].insert(1, 2)
...
If you wanted a one-liner that is similar to your define and extend, you could do
c2 = [1, *[2]]
which you could use to combine two generator expressions.

All built-in methods under class 'List' in Python are just modifying the list 'in situ'. They only change the original list and return nothing.
The advantage is, you don't need to pass the object to the original variable every time you modify it. Meanwhile, you can't accumulatively call its methods in one line of code such as what is used in Javascript. Because Javascript always turns its objects into DOM, but Python not.

Appending object to a dictionary by key is replicating over the whole dictionary [duplicate]

This question already has answers here:
How do I initialize a dictionary of empty lists in Python?
(7 answers)
Closed 2 years ago.
I came across this behavior that surprised me in Python 2.6 and 3.2:
>>> xs = dict.fromkeys(range(2), [])
>>> xs
{0: [], 1: []}
>>> xs[0].append(1)
>>> xs
{0: [1], 1: [1]}
However, dict comprehensions in 3.2 show a more polite demeanor:
>>> xs = {i:[] for i in range(2)}
>>> xs
{0: [], 1: []}
>>> xs[0].append(1)
>>> xs
{0: [1], 1: []}
>>>
Why does fromkeys behave like that?

Your Python 2.6 example is equivalent to the following, which may help to clarify:
>>> a = []
>>> xs = dict.fromkeys(range(2), a)
Each entry in the resulting dictionary will have a reference to the same object. The effects of mutating that object will be visible through every dict entry, as you've seen, because it's one object.
>>> xs[0] is a and xs[1] is a
True
Use a dict comprehension, or if you're stuck on Python 2.6 or older and you don't have dictionary comprehensions, you can get the dict comprehension behavior by using dict() with a generator expression:
xs = dict((i, []) for i in range(2))

In the first version, you use the same empty list object as the value for both keys, so if you change one, you change the other, too.
Look at this:
>>> empty = []
>>> d = dict.fromkeys(range(2), empty)
>>> d
{0: [], 1: []}
>>> empty.append(1) # same as d[0].append(1) because d[0] references empty!
>>> d
{0: [1], 1: [1]}
In the second version, a new empty list object is created in every iteration of the dict comprehension, so both are independent from each other.
As to "why" fromkeys() works like that - well, it would be surprising if it didn't work like that. fromkeys(iterable, value) constructs a new dict with keys from iterable that all have the value value. If that value is a mutable object, and you change that object, what else could you reasonably expect to happen?

To answer the actual question being asked: fromkeys behaves like that because there is no other reasonable choice. It is not reasonable (or even possible) to have fromkeys decide whether or not your argument is mutable and make new copies every time. In some cases it doesn't make sense, and in others it's just impossible.
The second argument you pass in is therefore just a reference, and is copied as such. An assignment of [] in Python means "a single reference to a new list", not "make a new list every time I access this variable". The alternative would be to pass in a function that generates new instances, which is the functionality that dict comprehensions supply for you.
Here are some options for creating multiple actual copies of a mutable container:
As you mention in the question, dict comprehensions allow you to execute an arbitrary statement for each element:
d = {k: [] for k in range(2)}
The important thing here is that this is equivalent to putting the assignment k = [] in a for loop. Each iteration creates a new list and assigns it to a value.
Use the form of the dict constructor suggested by #Andrew Clark:
d = dict((k, []) for k in range(2))
This creates a generator which again makes the assignment of a new list to each key-value pair when it is executed.
Use a collections.defaultdict instead of a regular dict:
d = collections.defaultdict(list)
This option is a little different from the others. Instead of creating the new list references up front, defaultdict will call list every time you access a key that's not already there. You can there fore add the keys as lazily as you want, which can be very convenient sometimes:
for k in range(2):
d[k].append(42)
Since you've set up the factory for new elements, this will actually behave exactly as you expected fromkeys to behave in the original question.
Use dict.setdefault when you access potentially new keys. This does something similar to what defaultdict does, but it has the advantage of being more controlled, in the sense that only the access you want to create new keys actually creates them:
d = {}
for k in range(2):
d.setdefault(k, []).append(42)
The disadvantage is that a new empty list object gets created every time you call the function, even if it never gets assigned to a value. This is not a huge problem, but it could add up if you call it frequently and/or your container is not as simple as list.

my_set.copy().add( something ) why return None [duplicate]

What am I doing wrong here?
a = set().add(1)
print a # Prints `None`
I'm trying to add the number 1 to the empty set.

It is a convention in Python that methods that mutate sequences return None.
Consider:
>>> a_list = [3, 2, 1]
>>> print a_list.sort()
None
>>> a_list
[1, 2, 3]
>>> a_dict = {}
>>> print a_dict.__setitem__('a', 1)
None
>>> a_dict
{'a': 1}
>>> a_set = set()
>>> print a_set.add(1)
None
>>> a_set
set([1])
Some may consider this convention "a horrible misdesign in Python", but the Design and History FAQ gives the reasoning behind this design decision (with respect to lists):
Why doesn’t list.sort() return the sorted list?
In situations where performance matters, making a copy of the list
just to sort it would be wasteful. Therefore, list.sort() sorts the
list in place. In order to remind you of that fact, it does not return
the sorted list. This way, you won’t be fooled into accidentally
overwriting a list when you need a sorted copy but also need to keep
the unsorted version around.
In Python 2.4 a new built-in function – sorted() – has been added.
This function creates a new list from a provided iterable, sorts it
and returns it.
Your particular problems with this feature come from a misunderstanding of good ways to create a set rather than a language misdesign. As Lattyware points out, in Python versions 2.7 and later you can use a set literal a = {1} or do a = set([1]) as per Sven Marnach's answer.
Parenthetically, I like Ruby's convention of placing an exclamation point after methods that mutate objects, but I find Python's approach acceptable.

The add() method adds an element to the set, but it does not return the set again -- it returns None.
a = set()
a.add(1)
or better
a = set([1])
would work.

Because add() is modifing your set in place returning None:
>>> empty = set()
>>> print(empty.add(1))
None
>>> empty
set([1])

Another way to do it that is relatively simple would be:
a = set()
a = set() | {1}
this creates a union between your set a and a set with 1 as the element
print(a) yields {1} then because a would now have all elements of both a and {1}

You should do this:
a = set()
a.add(1)
print a
Notice that you're assigning to a the result of adding 1, and the add operation, as defined in Python, returns None - and that's what is getting assigned to a in your code.
Alternatively, you can do this for initializing a set:
a = set([1, 2, 3])

The add method updates the set, but returns None.
a = set()
a.add(1)
print a

You are assigning the value returned by set().add(1) to a. This value is None, as add() does not return any value, it instead acts in-place on the list.
What you wanted to do was this:
a = set()
a.add(1)
print(a)
Of course, this example is trivial, but Python does support set literals, so if you really wanted to do this, it's better to do:
a = {1}
print(a)
The curly brackets denote a set (although be warned, {} denotes an empty dict, not an empty set due to the fact that curly brackets are used for both dicts and sets (dicts are separated by the use of the colon to separate keys and values.)

Alternatively to a = set() | {1} consider "in-place" operator:
a = set()
a |= {1}

Python List concepts [duplicate]

This question already has answers here:
Why do these list operations (methods: clear / extend / reverse / append / sort / remove) return None, rather than the resulting list?
(6 answers)
Closed 7 years ago.
I am new to python programming
why this assignment value to k is giving 'None' value
> >>> l=[2,3,4]
> >>> k=l.append(14)
> >>> print k
None
> >>> print l
[2, 3, 4, 14]
in above example List,I append the 14 value to list and then assigned to k but k is printing None please tell me the reason why its printing None instead of appended list?
Thanks
mukthyar

append changes the current list and doesn't return anything. Use:
k = l + [14]
or
k = l[:] # Copy list
k.append(14)

Most methods in Python which mutate their instance (.append(), .sort(), et al.) do not return a copy of the object. Thus l.append(X) does actually return "None" just as l.sort() would.
You'd probably want to use something more like:
l.append(14)
k = l[-1]

In python default return type is None so if your function is not return any value then you will get the None from that function. There are different functions for the object. Some functions are perform on same object and others are return the modified copy. For example list.sort will change the original list while sorted will return the new sorted list without changing the original one. So append is working on object so it will change the object value instead of returning new modified copy.
Hope this will make your view clear about append method.

What you are seeing is expected - the list.append method does not return the modified list. If you take a look at the relevant section of the Python tutorial, you'll notice that methods which return useful values are documented as such.
Working in the interactive interpreter, you can tell that the expression yields None by the lack of output:
>>> l = [1, 2, 3]
>>> l.append(14)
>>>
If (theoretically) list.append returned the list, you would see the following instead:
>>> l = [1, 2, 3]
>>> l.append(14)
[1, 2, 3, 14]
>>>
Using the interactive interpreter this way can save you from printing every value.

from python document
list.append(x)
Add an item to the end of the list; equivalent to a[len(a):] = [x].
in simple word append function returns nothing but None
suggested reading