How to find the last occurrence of an item in a Python list

>>> (x for x in reversed(list(enumerate(li))) if x[1] == 'a').next()[0]
6

>>> len(li) - (x for x in enumerate(li[::-1]) if x[1] == 'a').next()[0] - 1
6

If you are actually using just single letters like shown in your example, then str.rindex would work handily. This raises a ValueError if there is no such item, the same error class as list.index would raise. Demo:

>>> li = ["a", "b", "a", "c", "x", "d", "a", "6"]
>>> ''.join(li).rindex('a')
6

For the more general case you could use list.index on the reversed list:

>>> len(li) - 1 - li[::-1].index('a')
6

The slicing here creates a copy of the entire list. That's fine for short lists, but for the case where li is very large, efficiency can be better with a lazy approach:

def list_rindex(li, x):
    for i in reversed(range(len(li))):
        if li[i] == x:
            return i
    raise ValueError("{} is not in list".format(x))

One-liner version:

next(i for i in reversed(range(len(li))) if li[i] == 'a')

A one-liner that's like Ignacio's except a little simpler/clearer would be

max(loc for loc, val in enumerate(li) if val == 'a')

It seems very clear and Pythonic to me: you're looking for the highest index that contains a matching value. No nexts, lambdas, reverseds or itertools required.

Many of the other solutions require iterating over the entire list. This does not.

def find_last(lst, elm):
  gen = (len(lst) - 1 - i for i, v in enumerate(reversed(lst)) if v == elm)
  return next(gen, None)

Edit: In hindsight this seems like unnecessary wizardry. I'd do something like this instead:

def find_last(lst, sought_elt):
    for r_idx, elt in enumerate(reversed(lst)):
        if elt == sought_elt:
            return len(lst) - 1 - r_idx