How to get index of the last item with a given value

Question

In a list of integer values

a = [4, 8, 2, 3, 8, 5, 8, 8, 1, 4, 8, 2, 1, 3]

I have to find the index of the last item with value 8. Is there more elegant way to do that rather than mine:

a = [4, 8, 2, 3, 8, 5, 8, 8, 1, 4, 8, 2, 1, 3]

for i in range(len(a)-1, -1, -1):
    if a[i] == 8:
        print(i)
        break

Possible duplicate of How to find the last occurrence of an item in a Python list — Aniket Navlur
– Aniket Navlur, Commented Oct 21, 2018 at 9:35

Rohit-Pandey · Accepted Answer · 2018-10-21 09:19:33Z

4

Try This:

a = [4, 8, 2, 3, 8, 5, 8, 8, 1, 4, 8, 2, 1, 3]
index = len(a) - 1 - a[::-1].index(8)

Just reverse the array and find first index of element and then subtract it from length of array.

answered Oct 21, 2018 at 9:19

Rohit-Pandey

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Comments

Aran-Fey · Accepted Answer · 2018-10-21 09:25:34Z

1

>>> lst = [4, 8, 2, 3, 8, 5, 8, 8, 1, 4, 8, 2, 1, 3]
>>> next(i for i in range(len(lst)-1, -1, -1) if lst[i] == 8)
10

This throws StopIteration if the list doesn't contain the search value.

answered Oct 21, 2018 at 9:25

Aran-Fey

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vash_the_stampede · Accepted Answer · 2018-10-21 17:02:19Z

0

Use sorted on all indexes of items with a value of 8 and take the last value, or using reverse = True take the first value

x = sorted(i for i, v in enumerate(a) if v == 8)[-1]

print(x) # => 10

answered Oct 21, 2018 at 17:02

vash_the_stampede

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