How to return that last index of a list Python

Question

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Finding first and last index of some value in a list in Python

Hi I was wondering if someone could help me with Python. I am trying to create a code that returns the last index of the last occurrence of an item in a list in a recursive way. So in a list [1,2,3,4,5,2] the last it should return 4. It only takes in 2 variables which are the list and the item that it is searching for. If it does not find any matching variable then it returns -1.

So far I have this:

def lstIndex(lst, item):
    if len(lst) == 0:
        return -1
    place = lst[0]
    if place == item:
        print(place)
        #just return the index
        return lst.index(place)
    else:
        return lstIndex(lst[1:],item)

Answer 1

If recursion isn't necessary, you can use this:

def find_last(lst,item):
    try:
       return len(lst) - next(i for i,elem in enumerate(reversed(lst),1) if elem == item)
    except StopIteration:
       return -1


a = [1,2,3,4,5,4,3]
idx = find_last(a,4)
print a[idx]
print find_last(a,6)

Answer 2

Short iterative solution:

try:
    return (len(lst)-1) - lst[::-1].index(item)
except ValueError:
    return -1

But, since you are explicitly looking for a recursive solution, I'll show you how it can be done recursively. However, it will not be efficient; if you want a nice, efficient, Pythonic solution you should use an iterative solution like the others have shown (or the one above).

There's actually a few ways you can do this. You can use a helper function, which takes an extra argument specifying the last index at which the value was found:

def list_rfind(lst, item):
    def list_rfind_helper(i, item, last=-1):
        if i >= len(lst): return last
        if lst[i] == item: last = i
        return list_rfind_helper(i+1, item, last)

    return list_rfind_helper(0, item)

You can do it without a helper function:

def list_rfind(lst, item):
    if not lst:
        return -1

    res = list_rfind(lst[1:], item)
    if res >= 0:
        return res+1
    elif lst[0] == item:
        return 0
    else:
        return -1

Answer 3

lst = [1, 2, 3, 4, 3, 4]

findLast(lst, 4)

def findLast(lst, item):
    for i, val in enumerate(reversed(lst)):
        if val == item:
            return len(lst) - (i + 1)  # Return index of matched item

    return -1

Answer 4

Just for completeness:

def list_rfind(lst, item):
    return (len(lst)-1) - sum(1 for _ in iter(reversed(lst).next, item))

Answer 5

I'm not quite 100% sure I know what you want. Your statement that "... in a list of [1,2,3,4,5,2] the last it should return 4 ..." has me a bit confused; I think you want to return the index of the last appearance of your specified item. So, for 4 to be the result in the list specified, item must be 5.

As noted elsewhere, a recursive function would not be the most efficient or Pythonic solution here. I'd prefer a solution like the first one in nneonneo's answer.

However, if it must be recursive, I believe the code below gets what you want. Instead of stepping through the list from the front (by using [1:]), you need to step backwards by using [:-1] as the index range when passing the list in the recursive call:

def lstIndex(lst, item):
    if len(lst) == 0:
        return -1
    elif lst[-1] == item:
        return len(lst) - 1
    else:
        return lstIndex(lst[0:-1], item)

I tested with the following:

the_list = [1,2,3,4,5,2]
print lstIndex(the_list, 2)
print lstIndex(the_list, 1)
print lstIndex(the_list, 3)
print lstIndex(the_list, 4)
print lstIndex(the_list, 5)
print lstIndex(the_list, 6)
print lstIndex(the_list, 0)

With the following output:

5
0
2
3
4
-1
-1