return index of last non-zero element in list

Question

I get data in the following format:

[-2, -2, 0, 0, 0, 0, 0]
[-2, 20, -1, 0, 3, 0, 0]

with each line being a different input. The lists could be longer than 7 elements. I need to return the index position of the last non-zero element, so:

[-2, -2, 0, 0, 0, 0, 0]
>>> 1
[-2, 20, -1, 0, 3, 0, 0]
>>> 4

The following code does this most of the time:

def getIndex(list):
    for element in reversed(list):
        if element != 0:
            return list.index(element)

However, it doesn't work when there are two of the same numbers, as in the first example above, which returns 0 because -2 is in both the 0th and 1st position of the list.

So how to get the index of the last non-zero element of a list, even when there are elements with the same value?

Answer 1

Your solution changes the list constantly, I think even the naive for loop is better:

last_nonzero_index = None
for idx, item in enumerate(originalList):
    if item != 0:
        last_nonzero_index = idx

Answer 2

Here is a pythonic and optimized approach using itertools.takewhile():

from itertools import takewhile
from operator import not_
len(lst) - sum(1 for _ in takewhile(not_, reversed(lst))) - 1

Demo:

In [115]: lst = [-2, -2, 0, 0, 0, 0, 0]
In [118]: len(lst) - sum(1 for _ in takewhile(lambda x : not bool(x), lst[::-1])) - 1
Out[118]: 1

In [119]: lst = [-2, 20, -1, 0, 3, 0, 0]

In [120]: len(lst) - sum(1 for _ in takewhile(lambda x : not bool(x), lst[::-1])) - 1
Out[120]: 4

Here is a benchmarck with another answers on a longer list:

In [141]: lst = lst * 1000

In [142]: %timeit len(lst) - len(list(takewhile(not_, reversed(lst)))) - 1
1000000 loops, best of 3: 758 ns per loop

In [143]: %timeit get_last_non_zero_index(lst)
1000000 loops, best of 3: 949 ns per loop

In [146]: %timeit index_of_last_nonzero(lst)
1000000 loops, best of 3: 590 ns per loop