Perhaps the two most efficient ways to find the last index:

def rindex(lst, value):
    lst.reverse()
    i = lst.index(value)
    lst.reverse()
    return len(lst) - i - 1

import operator

def rindex(lst, value):
    return len(lst) - operator.indexOf(reversed(lst), value) - 1

Both take only O(1) extra space and the two in-place reversals of the first solution are much faster than creating a reverse copy. Let's compare it with the other solutions posted previously:

def rindex(lst, value):
    return len(lst) - lst[::-1].index(value) - 1

def rindex(lst, value):
    return len(lst) - next(i for i, val in enumerate(reversed(lst)) if val == value) - 1

Benchmark results, my solutions are the red and green ones:

This is for searching a number in a list of a million numbers. The x-axis is for the location of the searched element: 0% means it's at the start of the list, 100% means it's at the end of the list. All solutions are fastest at location 100%, with the two reversed solutions taking pretty much no time for that, the double-reverse solution taking a little time, and the reverse-copy taking a lot of time.

A closer look at the right end:

At location 100%, the reverse-copy solution and the double-reverse solution spend all their time on the reversals (index() is instant), so we see that the two in-place reversals are about seven times as fast as creating the reverse copy.

The above was with lst = list(range(1_000_000, 2_000_001)), which pretty much creates the int objects sequentially in memory, which is extremely cache-friendly. Let's do it again after shuffling the list with random.shuffle(lst) (probably less realistic, but interesting):

All got a lot slower, as expected. The reverse-copy solution suffers the most, at 100% it now takes about 32 times (!) as long as the double-reverse solution. And the enumerate-solution is now second-fastest only after location 98%.

Overall I like the operator.indexOf solution best, as it's the fastest one for the last half or quarter of all locations, which are perhaps the more interesting locations if you're actually doing rindex for something. And it's only a bit slower than the double-reverse solution in earlier locations.

All benchmarks done with CPython 3.9.0 64-bit on Windows 10 Pro 1903 64-bit.

I've made a benchmark for several methods:

• argwhere
• nonzero as in the question
• .tostring() as in @Rob Reilink's answer
• python loop
• Fortran loop

The Python and Fortran code are available. I skipped the unpromising ones like converting to a list.

The results on log scale. X-axis is the position of the needle (it takes longer to find if it's further down the array); last value is a needle that's not in the array. Y-axis is the time to find it.

The array had 1 million elements and tests were run 100 times. Results still fluctuate a bit, but the qualitative trend is clear: Python and f2py quit at the first element so they scale differently. Python gets too slow if the needle is not in the first 1%, whereas f2py is fast (but you need to compile it).

To summarize, f2py is the fastest solution, especially if the needle appears fairly early.

It's not built in which is annoying, but it's really just 2 minutes of work. Add this to a file called search.f90:

subroutine find_first(needle, haystack, haystack_length, index)
    implicit none
    integer, intent(in) :: needle
    integer, intent(in) :: haystack_length
    integer, intent(in), dimension(haystack_length) :: haystack
!f2py intent(inplace) haystack
    integer, intent(out) :: index
    integer :: k
    index = -1
    do k = 1, haystack_length
        if (haystack(k)==needle) then
            index = k - 1
            exit
        endif
    enddo
end

If you're looking for something other than integer, just change the type. Then compile using:

f2py -c -m search search.f90

after which you can do (from Python):

import search
print(search.find_first.__doc__)
a = search.find_first(your_int_needle, your_int_array)

If you need the index of the first occurrence of only one value, you can use nonzero (or where, which amounts to the same thing in this case):

>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])
>>> nonzero(t == 8)
(array([6, 8, 9]),)
>>> nonzero(t == 8)[0][0]
6

If you need the first index of each of many values, you could obviously do the same as above repeatedly, but there is a trick that may be faster. The following finds the indices of the first element of each subsequence:

>>> nonzero(r_[1, diff(t)[:-1]])
(array([0, 3, 5, 6, 7, 8]),)

Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s:

[1, 1, 1, 2, 2, 3, 8, 3, 8, 8]

So it's slightly different than finding the first occurrence of each value. In your program, you may be able to work with a sorted version of t to get what you want:

>>> st = sorted(t)
>>> nonzero(r_[1, diff(st)[:-1]])
(array([0, 3, 5, 7]),)