Equivelant to rindex for lists in Python

Question

Is there an efficient way to find the last matching item in a list? When working with strings, you can find the last item with rindex:

    >>> a="GEORGE"
    >>> a.rindex("G")
    4

...But this method doesn't exist for lists:

    >>> a=[ "hello", "hello", "Hi." ]
    >>> a.rindex("hello")
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
    AttributeError: 'list' object has no attribute 'rindex'

Is there a way to get this without having to construct a big loop? I'd prefer not to use the reverse method if it can be avoided, as the order is important and I'd also have to do a bit of extra math to find out where the object /would/ have been. This seems wasteful.

Edit:

To clarify, I need the index number of this item.

http://stackoverflow.com/questions/6890170/python-how-to-find-last-occurrence-in-a-list-in-python — Martin Flucka, Mar 23 '12 at 09:10
Use `reversed(a)`, it create a reverse iterator and do not modify the list. — Kien Truong, Mar 23 '12 at 09:15
Dikei, could you give me an example as an answer? I'll gladly select it if it works. — Kelketek, Mar 23 '12 at 09:17

EwyynTomato · Accepted Answer · 2012-03-23T09:46:49.133

19

How about:

len(a) - a[-1::-1].index("hello") - 1

Edit (put in function as suggested):

def listRightIndex(alist, value):
    return len(alist) - alist[-1::-1].index(value) -1

edited Mar 23 '12 at 09:46

answered Mar 23 '12 at 09:34

EwyynTomato

3,879
1
29
39

5

This isn't a good solution. It works, but creates a copy of the whole list. Not reasonable for the use-case. – Guy Oct 10 '18 at 08:45

Kien Truong · Answer 2 · 2012-03-23T09:40:10.620

6

This should work:

for index, item in enumerate(reversed(a)):
    if item == "hello":
        print len(a) - index - 1
        break

edited Mar 23 '12 at 09:40

answered Mar 23 '12 at 09:24

Kien Truong

10,866
2
28
34

steveha · Answer 3 · 2022-05-20T03:42:19.703

4

I wrote a straightforward Python function, and here it is:

def list_rindex(lst, item):
    """
    Find first place item occurs in list, but starting at end of list.
    Return index of item in list, or -1 if item not found in the list.
    """
    i_max = len(lst)
    i_limit = -i_max
    i = -1
    while i >= i_limit:
        if lst[i] == item:
            return i_max + i
        i -= 1
    return -1

But while I was testing it, EwyynTomato posted a better answer. Use the "slicing" machinery to reverse the list and use the .index() method.

edited May 20 '22 at 03:42

answered Mar 23 '12 at 09:40

steveha

71,436
20
89
116

Here should be `while i >= i_limit`. Otherwise it will never find element at index 0. – Pavel Chernov May 08 '22 at 18:22
@PavelChernov You are correct. Thanks for that. I have edited the code snippet to incorporate your correction. – steveha May 20 '22 at 03:43

score 0 · Answer 4 · answered Jul 09 '12 at 01:14

0

Supports start:

def rindex(lst, val, start=None):
    if start is None:
        start = len(lst)-1
    for i in xrange(start,-1,-1):
        if lst[i] == val:
            return i

answered Jul 09 '12 at 01:14

mpen

256,080
255
805
1,172

Equivelant to rindex for lists in Python

4 Answers4

Linked