Why does ‘.sort()’ cause the list to be ‘None’ in Python? [duplicate]

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Question :

Why does ‘.sort()’ cause the list to be ‘None’ in Python? [duplicate]

I am attempting to sort a Python list of ints and then use the .pop() function to return the highest one. I have tried a writing the method in different ways:

def LongestPath(T):    
    paths = [Ancestors(T,x) for x in OrdLeaves(T)]
    #^ Creating a lists of lists of ints, this part works
    result =[len(y) for y in paths ]
    #^ Creating a list of ints where each int is a length of the a list in paths
    result = result.sort()
    #^meant to sort the result
    return result.pop()
    #^meant to return the largest int in the list (the last one)

I have also tried

def LongestPath(T):
    return[len(y) for y in [Ancestors(T,x) for x in OrdLeaves(T)] ].sort().pop()

In both cases .sort() causes the list to be None (which has no .pop() function and returns an error). When I remove the .sort() it works fine but does not return the largest int since the list is not sorted.

Asked By: Btuman


Answer #1:

Simply remove the assignment from

result = result.sort()

leaving just


The sort method works in-place (it modifies the existing list), so no assignment is necessary, and it returns None. When you assign its result to the name of the list, you’re assigning None.

It can easily (and more efficiently) be written as a one-liner:

max(len(Ancestors(T,x)) for x in OrdLeaves(T))

max operates in linear time, O(n), while sorting is O(nlogn). You also don’t need nested list comprehensions, a single generator expression will do.

Answered By: agf

Answer #2:


result = result.sort()

should be this


It is a convention in Python that methods that mutate sequences return None.


>>> a_list = [3, 2, 1]
>>> print a_list.sort()
>>> a_list
[1, 2, 3]

>>> a_dict = {}
>>> print a_dict.__setitem__('a', 1)
>>> a_dict
{'a': 1}

>>> a_set = set()
>>> print a_set.add(1)
>>> a_set

Python’s 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.

Answered By: Steven Rumbalski

Answer #3:

.sort() returns None and sorts the list in place.

Answered By: Omri Barel

Answer #4:

This has already been correctly answered: list.sort() returns None. The reason why is “Command-Query Separation”:


Python returns None because every function must return something, and the convention is that a function that doesn’t produce any useful value should return None.

I have never before seen your convention of putting a comment after the line it references, but starting the comment with a carat to point at the line. Please put comments before the lines they reference.

While you can use the .pop() method, you can also just index the list. The last value in the list can always be indexed with -1, because in Python negative indices “wrap around” and index backward from the end.

But we can simplify even further. The only reason you are sorting the list is so you can find its max value. There is a built-in function in Python for this: max()

Using list.sort() requires building a whole list. You will then pull one value from the list and discard it. max() will consume an iterator without needing to allocate a potentially-large amount of memory to store the list.

Also, in Python, the community prefers the use of a coding standard called PEP 8. In PEP 8, you should use lower-case for function names, and an underscore to separate words, rather than CamelCase.


With the above comments in mind, here is my rewrite of your function:

def longest_path(T):
    paths = [Ancestors(T,x) for x in OrdLeaves(T)]
    return max(len(path) for path in paths)

Inside the call to max() we have a “generator expression” that computes a length for each value in the list paths. max() will pull values out of this, keeping the biggest, until all values are exhausted.

But now it’s clear that we don’t even really need the paths list. Here’s the final version:

def longest_path(T):
    return max(len(Ancestors(T, x)) for x in OrdLeaves(T))

I actually think the version with the explicit paths variable is a bit more readable, but this isn’t horrible, and if there might be a large number of paths, you might notice a performance improvement due to not building and destroying the paths list.

Answered By: steveha

Answer #5:

list.sort() does not return a list – it destructively modifies the list you are sorting:

In [177]: range(10)
Out[177]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [178]: range(10).sort()

In [179]:

That said, max finds the largest element in a list, and will be more efficient than your method.

Answered By: Marcin

Answer #6:

In Python sort() is an inplace operation. So result.sort() returns None, but changes result to be sorted. So to avoid your issue, don’t overwrite result when you call sort().

Answered By: ely

Answer #7:

Is there any reason not to use the sorted function? sort() is only defined on lists, but sorted() works with any iterable, and functions the way you are expecting. See this article for sorting details.

Also, because internally it uses timsort, it is very efficient if you need to sort on key 1, then sort on key 2.

Answered By: Spencer Rathbun

Answer #8:

You don’t need a custom function for what you want to achieve, you first need to understand the methods you are using!

sort()ing a list in python does it in place, that is, the return from sort() is None. The list itself is modified, a new list is not returned.

>>>results = ['list','of','items']



<type 'list'>



>>>results = results.sort()

<type 'NoneType'>

As you can see, when you try to assign the sort() , you no longer have the list type.

Answered By: PenguinCoder

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