Find the nth occurrence of substring in a string

Here’s a more Pythonic version of the straightforward iterative solution:

def find_nth(haystack, needle, n):
    start = haystack.find(needle)
    while start >= 0 and n > 1:
        start = haystack.find(needle, start+len(needle))
        n -= 1
    return start

Example:

>>> find_nth("foofoofoofoo", "foofoo", 2)
6

If you want to find the nth overlapping occurrence of needle, you can increment by 1 instead of len(needle), like this:

def find_nth_overlapping(haystack, needle, n):
    start = haystack.find(needle)
    while start >= 0 and n > 1:
        start = haystack.find(needle, start+1)
        n -= 1
    return start

Example:

>>> find_nth_overlapping("foofoofoofoo", "foofoo", 2)
3

This is easier to read than Mark’s version, and it doesn’t require the extra memory of the splitting version or importing regular expression module. It also adheres to a few of the rules in the Zen of python, unlike the various re approaches:

1. Simple is better than complex.
2. Flat is better than nested.
3. Readability counts.