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

python – What is the difference between `sorted(list)` vs `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 (youll 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 doesnt 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 takes any iterable & returns a new list, 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(iterable, key=None):
    new_list = list(iterable)    # make a new list
    new_list.sort(key=key)       # sort it
    return new_list              # return it

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 lists 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, heres 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 heres our results for a list of randomly arranged 10000 integers, as we can see here, weve 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 sorts 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, heres 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, heres 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.

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

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         # its modified

Note that since a.sort() doesnt 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.

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