String count with overlapping occurrences
>>> import re>>> text = '1011101111'>>> len(re.findall('(?=11)', text))5If you didn't want to load the whole list of matches into memory, which would never be a problem! you could do this if you really wanted:
>>> sum(1 for _ in re.finditer('(?=11)', text))5As a function (re.escape makes sure the substring doesn't interfere with the regex):
>>> def occurrences(text, sub): return len(re.findall('(?={0})'.format(re.escape(sub)), text))>>> occurrences(text, '11')5
You can also try using the new Python regex module, which supports overlapping matches.
import regex as redef count_overlapping(text, search_for): return len(re.findall(search_for, text, overlapped=True))count_overlapping('1011101111','11') # 5