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Answer 1

+30 A:

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

Greg Hewgill 2009-02-09 19:05:54

Thanks. Out of curiosity, does this reversed method, creates a separate copy of the collection or just alters the looping?

Joan Venge 2009-02-09 19:09:32

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 2009-02-09 19:10:59

The reversed() function makes a copy of the collection. To modify an existing list, use the .reverse() method.

Greg Hewgill 2009-02-09 19:11:02

@Greg Hewgill No, it's an iterator over the original, no copy is created!

André 2009-02-09 19:14:19

Thanks Greg, it's awesome method.

Joan Venge 2009-02-09 19:15:09

@André: you're right of course, but I was using the same words as the question asked by Joan, so that the answer would be more easily understood. Conceptually, it creates a copy. In implementation, it uses an iterator.

Greg Hewgill 2009-02-09 19:16:15

@Greg: Your new chunk of code to access the original index creates a second copy of the original collection. It would be better just to do something like i = len(a) - 1 - ridx

Theran 2009-02-09 19:23:05

Kinda sucks that this has to traverse the entire list before iteration... Still upvoted.

Triptych 2009-02-09 19:23:20

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 2009-02-09 19:27:12

in this answer though, list(enumerate(a)) DOES create a copy.

Triptych 2009-02-09 19:29:14

It doesn't quite make a copy, it seems to do the same logic as manually walking backwards. If you call reversed, then alter the list, then iterate over the reversed list, the changes will be reflected by reversed().

Richard Levasseur 2009-02-09 19:30:52

It absolutely makes a copy of the list. More accurately a zip of the list with it's positional indexes.

Triptych 2009-02-09 19:33:02

@Triptych: `reversed([a,b,c])` does *not* create a copy of `[a, b, c]`.

J.F. Sebastian 2009-02-09 19:39:45

I think there is some confusion here about exactly which example we're discussing. The first example only uses reversed(); the second example uses reversed(list(enumerate())). Clearly the second example makes more copies than the first. I think you're all correct on some level. :)

Greg Hewgill 2009-02-09 19:47:14

@ JF, reversed() doesn't make a copy, but list(enumerate()) DOES make a copy.

Triptych 2009-02-09 19:55:36

Thanks for the input guys, but seems like there is confusion about how these methods are implemented. By copy I meant, a deep copy, not an iterator. Iterators are fine but deep copies would slow my code considerable, for several millions of data.

Joan Venge 2009-02-09 20:46:53

@Joan, you shouldn't be concerned with 'deep' versus 'shallow' copies. That only matters when you're dealing with nested lists. Even a 'shallow' copy will be expensive - it will duplicate the entire list. I believe my answer is the only one that matches your specs AND uses no extra memory.

Triptych 2009-02-09 21:30:07

Thanks Triptych, that's what I mean, not collection copy but an iterator copy.

Joan Venge 2009-02-09 23:47:44

Answer 2

+1 A:

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

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

Bill Konrad 2009-02-09 19:07:13

Answer 3

A:

the reverse function comes in handy here:

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

bchhun 2009-02-09 19:07:30

list.reverse() has no return value

Georg 2009-02-09 19:09:12

Answer 4

+10 A:

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").

mipadi 2009-02-09 19:07:38

`[::-1]` creates a shallow copy, therefore it doesn't change the array neither "permanently" nor "temporary".

J.F. Sebastian 2009-02-09 19:15:23

Answer 5

+10 A:

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

Triptych 2009-02-09 19:14:12

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.

ΤΖΩΤΖΙΟΥ 2009-02-09 20:58:30

`reversed(xrange(len(L)))` produces the same indices as `xrange(len(L)-1, -1, -1)`.

J.F. Sebastian 2009-02-10 16:52:50

@JF - yeah at the time I thought I had some reason for not doing it that way, but now I can't remember. Changing it now.

Triptych 2009-02-10 17:22:45

Answer 6

A:

Can it be done like so?:

for i in range (len(collection):0) :
    print collection[i]

Joan Venge 2009-02-09 19:16:30

Greg's solution more pythonic.

epochwolf 2009-02-09 19:23:34

Yes, but my question is would this work? Not style.

Joan Venge 2009-02-09 20:43:16

Your syntax is wrong. : is used with [], not with arguments (such as for the range function). You could do: for i in range(len(collection)-1, -1, -1): print collection[i]. If you don't understand the arguments to range() there, check the tutorial and do some experimentation.

John Fouhy 2009-02-09 23:22:03

Answer 7

+1 A:

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)

Alan Rowarth 2009-02-09 23:18:11

For versions of Python prior to 3.0, I believe xrange is preferable to range for large len(collection).

Brian M. Hunt 2009-02-09 23:26:52

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 2009-02-09 23:57:38

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Traverse a list in reverse in Python

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