Equivelant to rindex for lists in Python [duplicate]

Answer 1

This should work:

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

Answer 2

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.

Answer 3

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

Answer 4

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