Python Traverse list reverse order

Traverse a list in reverse order in Python

up vote 252 down vote favorite 50	So I can start from len(collection) and end in collection[0]. EDIT: Sorry, I forgot to mention I also want to be able to access the loop index. python  loops share improve this question edited Nov 9 '11 at 9:53 bluish 8,391 12 66 109 asked  Feb 9 '09 at 19:04 Joan Venge 54.3k 114 319 552
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up vote 412 down vote accepted

Use the reversed() built-in function: >>> a = ["foo", "bar", "baz"] >>> for i in reversed(a): ... print i ... baz bar foo To also access the original index: >>> for i, e in reversed(list(enumerate(a))): ... print i, e ... 2 baz 1 bar 0 foo share improve this answer edited Feb 9 '09 at 19:12 answered  Feb 9 '09 at 19:05 Greg Hewgill 472k 104 825 1007

32   No copy is created, the elements are reversed on the fly while traversing! This is an important feature of all these iteration functions (which all end on “ed”). –  Konrad Rudolph  Feb 9 '09 at 19:10 5   @Greg Hewgill No, it's an iterator over the original, no copy is created! –  André  Feb 9 '09 at 19:14 38   To avoid the confusion: reversed() doesn't modify the list. reversed() doesn't make a copy of the list (otherwise it would require O(N) additional memory). If you need to modify the list use alist.reverse(); if you need a copy of the list in reversed order use alist[::-1]. –  J.F. Sebastian Feb 9 '09 at 19:27 31   in this answer though, list(enumerate(a)) DOES create a copy. –  Triptych  Feb 9 '09 at 19:29 9   @ JF, reversed() doesn't make a copy, but list(enumerate()) DOES make a copy. –  Triptych  Feb 9 '09 at 19:55 show 13 more comments

up vote 61 down vote

You can do: for item in my_list[::-1]: print item (Or whatever you want to do in the for loop.) The [::-1] slice reverses the list in the for loop (but won't actually modify your list "permanently"). share improve this answer answered  Feb 9 '09 at 19:07 mipadi 192k 43 361 381

7   [::-1] creates a shallow copy, therefore it doesn't change the array neither "permanently" nor "temporary".–  J.F. Sebastian  Feb 9 '09 at 19:15 2   This is slightly slower than using reversed, at least under Python 2.7 (tested). –  kgriffs  Jan 2 '14 at 16:49 This works for query objects, reversed() does not. Thanks!!! –  Roman  Jun 4 '14 at 11:30 How this answer works: It creates the temporary copy from a list slice which is an iteration of the list with the parameters: start point: unspecified (defaults to end of list - len(my_list)), end point: unspecified (defaults to start of list - 0) and step: -1 (iterate backwards through the list, 1 item at a time) –  Edward  Nov 29 '15 at 16:48  add a comment

up vote 32 down vote

If you need the loop index, and don't want to traverse the entire list twice, or use extra memory, I'd write a generator.  def reverse_enum(L): for index in reversed(xrange(len(L))): yield index, L[index] L = ['foo', 'bar', 'bas'] for index, item in reverse_enum(L): print index, item share improve this answer edited Feb 10 '09 at 17:23 answered  Feb 9 '09 at 19:14 Triptych 99.7k 20 113 149

I would call the function enumerate_reversed, but that might be only my taste. I believe your answer is the cleanest for the specific question. –  tzot  Feb 9 '09 at 20:58 1   reversed(xrange(len(L))) produces the same indices as xrange(len(L)-1, -1, -1). –  J.F. Sebastian  Feb 10 '09 at 16:52 1   I prefer fewer moving parts to understand: for index, item in enumerate(reversed(L)): print len(L)-1-index, item –  Don Kirkby  Nov 5 '14 at 21:56 1   @Triptych I just had to cope with fact that enumerate from reversed() won't yield reversed indexes, and your code helped a lot. This method should be in the standard library. –  oski86  Jul 21 '15 at 18:57 1   reversed(xrange()) works because an xrange object has the __reversed__ method as well as the __len__ and __getitem__ methods, and reversed can detect that and use them. But an enumerate object doesn't have __reversed__, __len__ or __getitem__. But why doesn't enumerate have them? I don't know that. –  FutureNerd  Sep 24 '15 at 1:01  show 3 more comments

up vote 15 down vote

It can be done like this: for i in range(len(collection)-1, -1, -1): print collection[i] So your guess was pretty close :) A little awkward but it's basically saying: start with 1 less than len(collection), keep going until you get to just before -1, by steps of -1.  Fyi, the help function is very useful as it lets you view the docs for something from the Python console, eg: help(range) share improve this answer answered  Feb 9 '09 at 23:18 Alan Rowarth 671 1 4 6

For versions of Python prior to 3.0, I believe xrange is preferable to range for large len(collection). –  Brian M. Hunt  Feb 9 '09 at 23:26 I believe you are correct :) iirc, range() generates the whole range as an array but xrange() returns an iterator that only generates the values as they are needed. –  Alan Rowarth  Feb 9 '09 at 23:57 5   This just looks too weird with so many -1's. I would just say reversed(xrange(len(collection)))–  musiphil  Sep 7 '13 at 1:18 add a comment

up vote 10 down vote

The reversed builtin function is handy: for item in reversed(sequence): The documentation for reversed explains its limitations. For the cases where I have to walk a sequence in reverse along with the index (e.g. for in-place modifications changing the sequence length), I have this function defined an my codeutil module: import itertools def reversed_enumerate(sequence): return itertools.izip( reversed(xrange(len(sequence))), reversed(sequence), ) This one avoids creating a copy of the sequence. Obviously, the reversed limitations still apply. share improve this answer answered  Oct 11 '11 at 6:28 tzot 43k 14 78 141

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up vote 2 down vote	How about without recreating a new list, you can do by indexing: >>> foo = ['1a','2b','3c','4d'] >>> for i in range(len(foo)): ... print foo[-(i+1)] ... 4d 3c 2b 1a >>> OR >>> length = len(foo) >>> for i in range(length): ... print foo[length-i-1] ... 4d 3c 2b 1a >>> share improve this answer answered  Feb 15 '14 at 5:04 sapam 4,708 2 16 32
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up vote 1 down vote	Use list.reverse() and then iterate as you normally would. http://docs.python.org/tutorial/datastructures.html share improve this answer edited Nov 9 '11 at 9:51 bluish 8,391 12 66 109 answered  Feb 9 '09 at 19:07 Bill Konrad 289 2 4
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up vote 1 down vote	The other answers are good, but if you want to do as List comprehension style collection = ['a','b','c'] [item for item in reversed( collection ) ] share improve this answer answered  Jul 27 '13 at 15:17 fedmich 3,358 2 18 40
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up vote 1 down vote	def reverse(spam): k = [] for i in spam: k.insert(0,i) return "".join(k) share improve this answer edited Apr 25 '14 at 1:27 dreamcrash 10.1k 12 34 61 answered  Apr 25 '14 at 1:02 Jase 11 2
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up vote 1 down vote

>>> l = ["a","b","c","d"] >>> l.reverse() >>> l ['d', 'c', 'b', 'a'] OR >>> print l[::-1] ['d', 'c', 'b', 'a'] share improve this answer