How to keep count in a recursive function?

Question

I wrote a recursive function to find the number of instances of a substring in the parent string.

The way I am keeping count is by declaring/initialising count

Answer 1

Just a side note: all solutions presented (from the original Q to all the As) are solving a problem that's different than the specifically stated one (I imagine that's a bug in the specific problem statement, but, worth fixing if so;-). Consider:

>>> 'banana'.count('ana')
1
>>> sum('banana'[x:x+3]=='ana' for x in range(len('banana')))
2

the first expression is counting the non-overlapping occurrences of 'ana' in 'banana'; the second one is counting all occurrences -- there are two occurrences in all, at indices 1 and 3 in 'banana', and they overlap. So given the problem statement, and I quote:

find the no. of instances of a substring in the parent string.

without any mention of "non-overlapping", it seems that overlapping occurrences should be counted. Of course, that's easy to fix, once noticed -- you just have to advance by 1 each time, instead of advancing by len(key) which leads you to skip overlapping occurrences.

So, for example:

import string

def countit(target, key, startfrom=0):
    where = string.find(target, key, startfrom)
    if where < 0: return 0
    return 1 + countit(target, key, where+1)

print countit('banana', 'ana')

prints 2, counting both (overlapping) occurrences.