How do I reverse a list or loop over it backwards?
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Chapters
00:00 How Do I Reverse A List Or Loop Over It Backwards?
00:22 Answer 1 Score 1542
00:35 Accepted Answer Score 1740
00:53 Answer 3 Score 478
01:13 Answer 4 Score 22
02:12 Thank you
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Full question
https://stackoverflow.com/questions/3940...
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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
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Tags
#python #list
#avk47
ACCEPTED ANSWER
Score 1740
To get a new reversed list, apply the reversed function and collect the items into a list:
>>> xs = [0, 10, 20, 40]
>>> list(reversed(xs))
[40, 20, 10, 0]
To iterate backwards through a list:
>>> xs = [0, 10, 20, 40]
>>> for x in reversed(xs):
... print(x)
40
20
10
0
ANSWER 2
Score 1542
>>> xs = [0, 10, 20, 40]
>>> xs[::-1]
[40, 20, 10, 0]
Extended slice syntax is explained here. See also, documentation.
ANSWER 3
Score 478
Use list.reverse to reverse a list in-place:
>>> xs = [0, 10, 20, 40]
>>> xs.reverse()
>>> xs
[40, 20, 10, 0]
Use slices to create a new list with the items in reverse order:
>>> xs[::-1]
[40, 20, 10, 0]
ANSWER 4
Score 22
I find (contrary to some other suggestions) that l.reverse() is by far the fastest way to reverse a long list in Python 3 and 2. I'd be interested to know if others can replicate these timings.
l[::-1] is probably slower because it copies the list prior to reversing it. Adding the list() call around the iterator made by reversed(l) must add some overhead. Of course if you want a copy of the list or an iterator then use those respective methods, but if you want to just reverse the list then l.reverse() seems to be the fastest way.
Functions
def rev_list1(l):
return l[::-1]
def rev_list2(l):
return list(reversed(l))
def rev_list3(l):
l.reverse()
return l
List
l = list(range(1000000))
Python 3.5 timings
timeit(lambda: rev_list1(l), number=1000)
# 6.48
timeit(lambda: rev_list2(l), number=1000)
# 7.13
timeit(lambda: rev_list3(l), number=1000)
# 0.44
Python 2.7 timings
timeit(lambda: rev_list1(l), number=1000)
# 6.76
timeit(lambda: rev_list2(l), number=1000)
# 9.18
timeit(lambda: rev_list3(l), number=1000)
# 0.46