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What is the difference between `sorted(list)` vs `list.sort()` ? python

Posted by: admin November 29, 2017 Leave a comment


list.sort() sorts the list and save the sorted list, while sorted(list) returns a sorted list without changing the original list.

  • But when to use which?
  • And which is faster? And how much faster?
  • Can a list’s original positions be retrieved after list.sort()?

sorted() returns a new sorted list, leaving the original list unaffected. list.sort() sorts the list in-place, mutating the list indices, and returns None (like all in-place operations).

sorted() works on any iterable, not just lists. Strings, tuples, dictionaries (you’ll get the keys), generators, etc., returning a list containing all elements, sorted.

  • Use list.sort() when you want to mutate the list, sorted() when you want a new sorted object back. Use sorted() when you want to sort something that is an iterable, not a list yet.

  • For lists, list.sort() is faster than sorted() because it doesn’t have to create a copy. For any other iterable, you have no choice.

  • No, you cannot retrieve the original positions. Once you called list.sort() the original order is gone.


What is the difference between sorted(list) vs list.sort()?

  • list.sort mutates the list in-place & returns None
  • sorted creates a new list from the old & returns the new one, sorted.

sorted is equivalent to this Python implementation, but the CPython builtin function should run measurably faster as it is written in C:

def sorted(original_list):
    list_copy = list(original_list) # make a copy
    list_copy.sort()                # sort it
    return list_copy                # return the copied and sorted list

(To copy, we could have used a slice or in Python 3, list.copy, but list is clear enough for our purposes.)

when to use which?

  • Use list.sort when you do not wish to retain the original sort order
    (Thus you will be able to reuse the list in-place in memory.) and when
    you are the sole owner of the list (if the list is shared by other code
    and you mutate it, you could introduce bugs where that list is used.)
  • Use sorted when you want to retain the original sort order or when you
    wish to create a new list that only your local code owns.

Can a list’s original positions be retrieved after list.sort()?

No – unless you made a copy yourself, that information is lost because the sort is done in-place.

“And which is faster? And how much faster?”

To illustrate the penalty of creating a new list, use the timeit module, here’s our setup:

import timeit
setup = """
import random
lists = [list(range(10000)) for _ in range(1000)]  # list of lists
for l in lists:
    random.shuffle(l) # shuffle each list
shuffled_iter = iter(lists) # wrap as iterator so next() yields one at a time

And here’s our results for a list of randomly arranged 10000 integers, as we can see here, we’ve disproven an older list creation expense myth:

Python 2.7

>>> timeit.repeat("next(shuffled_iter).sort()", setup=setup, number = 1000)
[3.75168503401801, 3.7473005310166627, 3.753129180986434]
>>> timeit.repeat("sorted(next(shuffled_iter))", setup=setup, number = 1000)
[3.702025591977872, 3.709248117986135, 3.71071034099441]

Python 3

>>> timeit.repeat("next(shuffled_iter).sort()", setup=setup, number = 1000)
[2.797430992126465, 2.796825885772705, 2.7744789123535156]
>>> timeit.repeat("sorted(next(shuffled_iter))", setup=setup, number = 1000)
[2.675589084625244, 2.8019039630889893, 2.849375009536743]

After some feedback, I decided another test would be desirable with different characteristics. Here I provide the same randomly ordered list of 100,000 in length for each iteration 1,000 times.

import timeit
setup = """
import random
lst = list(range(100000))

I interpret this larger sort’s difference coming from the copying mentioned by Martijn, but it does not dominate to the point stated in the older more popular answer here, here the increase in time is only about 10%

>>> timeit.repeat("lst[:].sort()", setup=setup, number = 10000)
[572.919036605, 573.1384446719999, 568.5923951]
>>> timeit.repeat("sorted(lst[:])", setup=setup, number = 10000)
[647.0584738299999, 653.4040515829997, 657.9457361929999]

I also ran the above on a much smaller sort, and saw that the new sorted copy version still takes about 2% longer running time on a sort of 1000 length.

Poke ran his own code as well, here’s the code:

setup = '''
import random
lst = list(range({length}))
lists = [lst[:] for _ in range({repeats})]
it = iter(lists)
t1 = 'l = next(it); l.sort()'
t2 = 'l = next(it); sorted(l)'
length = 10 ** 7
repeats = 10 ** 2
print(length, repeats)
for t in t1, t2:
    print(timeit(t, setup=setup.format(length=length, repeats=repeats), number=repeats))

He found for 1000000 length sort, (ran 100 times) a similar result, but only about a 5% increase in time, here’s the output:

10000000 100
l = next(it); l.sort()
l = next(it); sorted(l)


A large sized list being sorted with sorted making a copy will likely dominate differences, but the sorting itself dominates the operation, and organizing your code around these differences would be premature optimization. I would use sorted when I need a new sorted list of the data, and I would use list.sort when I need to sort a list in-place, and let that determine my usage.


The main difference is that sorted(some_list) returns a new list:

a = [3, 2, 1]
print sorted(a) # new list
print a         # is not modified

and some_list.sort(), sorts the list in place:

a = [3, 2, 1]
print a.sort() # in place
print a         # it's modified

Note that since a.sort() doesn’t return anything, print a.sort() will print None.

Can a list original positions be retrieved after list.sort()?

No, because it modifies the original list.