Python Traverse list reverse order

Traverse a list in reverse order in Python

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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

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edited Nov 9 '11 at 9:53

bluish

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asked Feb 9 '09 at 19:04

Joan Venge

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13 Answers

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Use the reversed() built-in function:

>>> a = ["foo", "bar", "baz"]>>> for i in reversed(a):...     print i... bazbarfoo

To also access the original index:

>>> for i, e in reversed(list(enumerate(a))):...     print i, e... 2 baz1 bar0 foo

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edited Feb 9 '09 at 19:12

answered Feb 9 '09 at 19:05

Greg Hewgill

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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. SebastianFeb 9 '09 at 19:27

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in this answer though, list(enumerate(a)) DOES create a copy. – Triptych Feb 9 '09 at 19:29

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@ JF, reversed() doesn't make a copy, but list(enumerate()) DOES make a copy. – Triptych Feb 9 '09 at 19:55

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up vote61down 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").

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answered Feb 9 '09 at 19:07

mipadi

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[::-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

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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 

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up vote32down 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

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edited Feb 10 '09 at 17:23

answered Feb 9 '09 at 19:14

Triptych

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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

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reversed(xrange(len(L))) produces the same indices as xrange(len(L)-1, -1, -1). – J.F. Sebastian Feb 10 '09 at 16:52

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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

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@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 

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up vote15down 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)

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answered Feb 9 '09 at 23:18

Alan Rowarth

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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

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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

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up vote10down 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 itertoolsdef 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.

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answered Oct 11 '11 at 6:28

tzot

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up vote2down 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)]...4d3c2b1a>>>

OR

>>> length = len(foo)>>> for i in range(length):...     print foo[length-i-1]...4d3c2b1a>>>

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answered Feb 15 '14 at 5:04

sapam

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up vote1down vote

Use list.reverse() and then iterate as you normally would.

http://docs.python.org/tutorial/datastructures.html

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edited Nov 9 '11 at 9:51

bluish

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answered Feb 9 '09 at 19:07

Bill Konrad

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up vote1down 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 ) ]

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answered Jul 27 '13 at 15:17

fedmich

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up vote1down vote

def reverse(spam):    k = []    for i in spam:        k.insert(0,i)    return "".join(k)

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edited Apr 25 '14 at 1:27

dreamcrash

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answered Apr 25 '14 at 1:02

Jase

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up vote1down vote

>>> l = ["a","b","c","d"]>>> l.reverse()>>> l['d', 'c', 'b', 'a']

OR

>>> print l[::-1]['d', 'c', 'b', 'a']

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