Traverse a list in reverse order in Python


How do I traverse a list in reverse order in Python? So I can start from collection[len(collection)-1] and end in collection[0].

I also want to be able to access the loop index.

Asked By: Joan Venge



Use the built-in reversed() function:

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

To also access the original index, use enumerate() on your list before passing it to reversed():

>>> for i, e in reversed(list(enumerate(a))):
...     print(i, e)
2 baz
1 bar
0 foo

Since enumerate() returns a generator and generators can’t be reversed, you need to convert it to a list first.

Answered By: Greg Hewgill

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

Answered By: Bill Konrad

the reverse function comes in handy here:

myArray = [1,2,3,4]
for x in myArray:
    print x
Answered By: bchhun

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”).

Answered By: mipadi

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
Answered By: Kenan Banks

It can be done like this:

for i in range(len(collection)-1, -1, -1):
    print collection[i]

    # print(collection[i]) for python 3. +

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:


Answered By: Alan Rowarth

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:

from six.moves import zip as izip, range as xrange

def reversed_enumerate(sequence):
    return izip(

This one avoids creating a copy of the sequence. Obviously, the reversed limitations still apply.

Answered By: tzot

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 ) ]
Answered By: fedmich

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)]


>>> length = len(foo)
>>> for i in range(length):
...     print foo[length-i-1]
Answered By: James Sapam
def reverse(spam):
    k = []
    for i in spam:
    return "".join(k)
Answered By: Jase

I like the one-liner generator approach:

((i, sequence[i]) for i in reversed(xrange(len(sequence))))
Answered By: lkraider
>>> l = ["a","b","c","d"]
>>> l.reverse()
>>> l
['d', 'c', 'b', 'a']


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

To use negative indices: start at -1 and step back by -1 at each iteration.

>>> a = ["foo", "bar", "baz"]
>>> for i in range(-1, -1*(len(a)+1), -1):
...     print i, a[i]
-1 baz
-2 bar
-3 foo
Answered By: stroz

You can also use a while loop:

i = len(collection)-1
while i>=0:
    value = collection[i]
    index = i
Answered By: Yuval A.

An approach with no imports:

for i in range(1,len(arr)+1):

Time complexity O(n) and space complexity O(1).

An approach that creates a new list in memory, be careful with large lists:

for i in arr[::-1]:

Time complexity O(n) and space complexity O(n).

Answered By: Kenan

for what ever it’s worth you can do it like this too. very simple.

a = [1, 2, 3, 4, 5, 6, 7]
for x in xrange(len(a)):
    x += 1
    print a[-x]
Answered By: emorphus

You can use a negative index in an ordinary for loop:

>>> collection = ["ham", "spam", "eggs", "baked beans"]
>>> for i in range(1, len(collection) + 1):
...     print(collection[-i])
baked beans

To access the index as though you were iterating forward over a reversed copy of the collection, use i - 1:

>>> for i in range(1, len(collection) + 1):
...     print(i-1, collection[-i])
0 baked beans
1 eggs
2 spam
3 ham

To access the original, un-reversed index, use len(collection) - i:

>>> for i in range(1, len(collection) + 1):
...     print(len(collection)-i, collection[-i])
3 baked beans
2 eggs
1 spam
0 ham
Answered By: Malcolm

A simple way :

n = int(input())
arr = list(map(int, input().split()))

for i in reversed(range(0, n)):
    print("%d %d" %(i, arr[i]))
Answered By: rashedcs

Also, you could use either “range” or “count” functions.
As follows:

a = ["foo", "bar", "baz"]
for i in range(len(a)-1, -1, -1):
    print(i, a[i])

3 baz
2 bar
1 foo

You could also use “count” from itertools as following:

a = ["foo", "bar", "baz"]
from itertools import count, takewhile

def larger_than_0(x):
    return x > 0

for x in takewhile(larger_than_0, count(3, -1)):
    print(x, a[x-1])

3 baz
2 bar
1 foo
Answered By: disooqi

In python 3, list creates a copy, so reversed(list(enumerate(collection)) could be inefficient, generating yet an other list is not optimized away.

If collection is a list for sure, then it may be best to hide the complexity behind an iterator

def reversed_enumerate(collection: list):
    for i in range(len(collection)-1, -1, -1):
        yield i, collection[i]

so, the cleanest is:

for i, elem in reversed_enumerate(['foo', 'bar', 'baz']):
    print(i, elem)
Answered By: Barney Szabolcs
input_list = ['foo','bar','baz']
for i in range(-1,-len(input_list)-1,-1)

i think this one is also simple way to do it… read from end and keep decrementing till the length of list, since we never execute the “end” index hence added -1 also

Answered By: Varun Maurya

If you need the index and your list is small, the most readable way is to do reversed(list(enumerate(your_list))) like the accepted answer says. But this creates a copy of your list, so if your list is taking up a large portion of your memory you’ll have to subtract the index returned by enumerate(reversed()) from len()-1.

If you just need to do it once:

a = ['b', 'd', 'c', 'a']

for index, value in enumerate(reversed(a)):
    index = len(a)-1 - index

    do_something(index, value)

or if you need to do this multiple times you should use a generator:

def enumerate_reversed(lyst):
    for index, value in enumerate(reversed(lyst)):
        index = len(lyst)-1 - index
        yield index, value

for index, value in enumerate_reversed(a):
    do_something(index, value)
Answered By: Boris Verkhovskiy

Assuming task is to find last element that satisfies some condition in a list (i.e. first when looking backwards), I’m getting following numbers.

Python 2:

>>> min(timeit.repeat('for i in xrange(len(xs)-1,-1,-1):n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))
>>> min(timeit.repeat('for i in reversed(xrange(0, len(xs))):n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))
>>> min(timeit.repeat('for i, x in enumerate(reversed(xs), 1):n    if 128 == x: break', setup='xs, n = range(256), 0', repeat=8))
>>> min(timeit.repeat('for i, x in enumerate(xs[::-1]):n    if 128 == x: break', setup='xs, n = range(256), 0', repeat=8))
>>> min(timeit.repeat('for i in xrange(len(xs), 0, -1):n    if 128 == xs[i - 1]: break', setup='xs, n = range(256), 0', repeat=8))
>>> min(timeit.repeat('i = len(xs)nwhile 0 < i:n    i -= 1n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))

So, the ugliest option xrange(len(xs)-1,-1,-1) is the fastest.

Python 3 (different machine):

>>> timeit.timeit('for i in range(len(xs)-1,-1,-1):n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', number=400000)
>>> timeit.timeit('for i in reversed(range(0, len(xs))):n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', number=400000)
>>> timeit.timeit('for i, x in enumerate(reversed(xs), 1):n    if 128 == x: break', setup='xs, n = range(256), 0', number=400000)
>>> timeit.timeit('for i, x in enumerate(xs[::-1]):n    if 128 == x: break', setup='xs, n = range(256), 0', number=400000)
>>> timeit.timeit('for i in range(len(xs), 0, -1):n    if 128 == xs[i - 1]: break', setup='xs, n = range(256), 0', number=400000)
>>> timeit.timeit('i = len(xs)nwhile 0 < i:n    i -= 1n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', number=400000)

Here, enumerate(reversed(xs), 1) is the fastest.

Answered By: wonder.mice

If you don’t mind the index being negative, you can do:

>>> a = ["foo", "bar", "baz"]
>>> for i in range(len(a)):
...     print(~i, a[~i]))
-1 baz
-2 bar
-3 foo
Answered By: jss367

I think the most elegant way is to transform enumerate and reversed using the following generator

(-(ri+1), val) for ri, val in enumerate(reversed(foo))

which generates a the reverse of the enumerate iterator


foo = [1,2,3]
bar = [3,6,9]
    bar[i] - val
    for i, val in ((-(ri+1), val) for ri, val in enumerate(reversed(foo)))


[6, 4, 2]
Answered By: CervEd

you can use a generator:

li = [1,2,3,4,5,6]
len_li = len(li)
gen = (len_li-1-i for i in range(len_li))


for i in gen:

hope this help you.

Answered By: AXin

I’m confused why the obvious choice did not pop up so far:

If reversed() is not working because you have a generator (as the case with enumerate()), just use sorted():

>>> l = list( 'abcdef' )
>>> sorted( enumerate(l), reverse=True )
[(5, 'f'), (4, 'e'), (3, 'd'), (2, 'c'), (1, 'b'), (0, 'a')]
Answered By: Suuuehgi

As a beginner in python, I found this way more easy to understand and reverses a list.

say numlst = [1, 2, 3, 4]

for i in range(len(numlst)-1,-1,-1):

ie., for i in range(3,-1,-1), where 3 is length of list minus 1,
second -1 means list starts from last element and 
third -1 signifies it will traverse in reverse order.

print( numlst[ i ] )

o/p = 4, 3, 2, 1

Answered By: Sajid Khan
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