[Python] Is there a command that returns the number of substrings in a string?

2
S.count(sub[, start[, end]]) -> int

Return the number of non-overlapping occurrences of substring sub in
string S[start:end]. Optional arguments start and end are interpreted
as in slice notation.

HTH,

--S

The .count string method.

re.findall?
2

For groups of non-overlapping substrings, tested only as far as you see:

8<----------------------------------------------------------------------

import re
from collections import defaultdict

def count(text, *args):
"""
2
2
3
2
2
"""
args = map(re.escape, args)
args.sort()
args.reverse()
pattern = re.compile('|'.join(args))
result = defaultdict(int)
def callback(match):
matched = match.group(0)
result[matched] += 1
return matched
pattern.sub(callback, text)
return result


if __name__ == '__main__':
import doctest
doctest.testmod()
8<----------------------------------------------------------------------

Other than the ability to handle multiple substrings, you do realise
you've effectively duplicated str.count()?

I realise that calling this count function with a single argument would
be functionally identical to calling str.count(), yes. But I can imagine
the situation of wanting to find multiple (disjoint) substrings. Is
there a reason for preferring multiple calls to str.count() in such a
case? Or is there a more obvious approach?
Forget that, that was stupid.

En Tue, 27 Oct 2009 04:31:22 -0300, Gerard Flanagan <grflanagan at gmail.com>
escribi?:
There is a more efficient algorithm (Aho-Corasick) which computes all
occurences of a set of substrings inside a given string at once. It
*should* be faster than repeatly calling str.find for every substring, and
faster than using regular expressions too. (note that you get not only the
count of occurences, but their positions too).
I've seen a couple implementations for Python; if they're *actually*
faster is to be determined...

No, I totally missed that it was counting disjoint substrings :)