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. Usesorted()when you want to sort something that is an iterable, not a list yet.For lists,
list.sort()is faster thansorted()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.
Code: https://www.30secondsofcode.org/articles/s/python-sortedlist-vs-list-sort
Sorting lists in python: sorted() vs sort()
sorted() vs sort()
min() vs 'sort() and then list[0]'
As others have mentioned, min will be quicker. min just has to walk through the list once, whereas sort must do more compares, making it more complex.
However, one case you might want to use sort is when you need more than just the minimum. Eg. you want the 5 smallest items, not just the smallest. Here "sorted(mylist)[:5]" will do the job, but another option to keep in mind is the heapq module, as "heapq.nsmallest(5, mylist)" will often perform better than sort, at least for a small number of items.
Automatically turn messy Python imports into clean sorted sections using isort.
I feel like it should have been named "impsort," because "isort" sounds like some kind of iterator sort.
Also, imps are cool.
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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. Usesorted()when you want to sort something that is an iterable, not a list yet.For lists,
list.sort()is faster thansorted()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)vslist.sort()?
list.sortmutates the list in-place & returnsNonesortedtakes 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 is one preferred over the other?
- Use
list.sortwhen 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
sortedwhen 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 be reverted to the unsorted state after list.sort() has been performed?
No - unless you made a copy yourself, that information is lost because the sort is done in-place.
Which is more efficient? By how much?
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
random.seed(0)
lst = list(range(100000))
random.shuffle(lst)
"""
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
random.seed(12122353453462456)
lst = list(range({length}))
random.shuffle(lst)
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(t)
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()
610.5015971539542
l = next(it); sorted(l)
646.7786222379655
Conclusion:
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.
