So I can start from len(collection) and end in collection[0].

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

Use the built-in reversed() function:

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

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.

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”).

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

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:

help(range)

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.

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
>>>

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

OR

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

I like the one-liner generator approach:

((i, sequence[i]) for i in reversed(xrange(len(sequence))))

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

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

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

An approach with no imports:

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

or

for i in arr[::-1]:
    print(i)

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

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]

An expressive way to achieve reverse(enumerate(collection)) in python 3:

zip(reversed(range(len(collection))), reversed(collection))

in python 2:

izip(reversed(xrange(len(collection))), reversed(collection))

I’m not sure why we don’t have a shorthand for this, eg.:

def reversed_enumerate(collection):
    return zip(reversed(range(len(collection))), reversed(collection))

or why we don’t have reversed_range()

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)

the reverse function comes in handy here:

myArray = [1,2,3,4]
myArray.reverse()
for x in myArray:
    print x

You can also use a while loop:

i = len(collection)-1
while i>=0:
    value = collection[i]
    index = i
    i-=1

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
eggs
spam
ham

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

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

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

Example:

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)))
]

Result:

[6, 4, 2]

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 ) ]

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

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]))

input_list = ['foo','bar','baz']
for i in range(-1,-len(input_list)-1,-1)
    print(input_list[i])

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

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:

>>> 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))
4.6937971115112305
>>> 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))
4.809093952178955
>>> min(timeit.repeat('for i, x in enumerate(reversed(xs), 1):\n    if 128 == x: break', setup='xs, n = range(256), 0', repeat=8))
4.931743860244751
>>> min(timeit.repeat('for i, x in enumerate(xs[::-1]):\n    if 128 == x: break', setup='xs, n = range(256), 0', repeat=8))
5.548468112945557
>>> 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))
6.286104917526245
>>> min(timeit.repeat('i = len(xs)\nwhile 0 < i:\n    i -= 1\n    if 128 == xs[i]: break', setup='xs, n = range(256), 0', repeat=8))
8.384078979492188

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

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))

finally:

for i in gen:
    print(li[i])

hope this help you.