查找列表中项目的最后一次出现，不包括最后一个元素

Question

For some reason, this doesn't work when trying to find last occurrence excluding last element

len(list) - 1 - list[::-1].index(1)

list = [3, 1, 1, 4, 1, 3, 1, 3, 1]

It needs to find index = 6

Correct answer provided for a single list iteration.

Edited:

How to find the index of the last occurrence of number 1 in both lists with the same index?

list1 = [3, 1, 1, 4, 1, 3, 1, 3, 1]
list2 = [3, 1, 1, 4, 1, 3, 2, 3, 1]

The answer should be index = 4

Thank you

Answer 1

I think you need:

If you want to ignore last element

lis = [3, 1, 1, 4, 1, 3, 1, 3, 1]

print([i for i,v in enumerate(lis[:-1]) if v==1][-1])
# 6

EDIT

l1 = [3, 1, 1, 4, 1, 3, 1, 3, 1]
l2 = [3, 1, 1, 4, 1, 3, 2, 3, 1]

ls1 = [i for i,v in enumerate(l1[:-1]) if v==1]
ls2 = [i for i,v in enumerate(l2[:-1]) if v==1]

val = None
# select the list with higer length here
for i in ls1[::-1]:
    if i in ls2:
        val = i
        break
print(val)
# 4

Answer 2

The extra arguments to index might help you simplify the indexing:

len(lst) - 1 - lst[::-1].index(1, 1)

Since a list is reversible, you can avoid reallocations by using an iterator:

it = reversed(lst)
next(it) # discard last element
len(lst) - 2 - next(ind for ind, elem in it if elem == 1)