Overlapping regular expression in python

I did not realize that by default python only find patterns in a non-overlapping way. For example, we want to find “AA” in “AAAA”, it will give you positions 0 and 2, eg the first two AA and the last two AA. However, the middle two AA is also a match. To get all the overlapping matches, we need “lookahead assertion”. From the python regex doc:

(?=…)
Matches if … matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match ‘Isaac ’ only if it’s followed by ‘Asimov’.

A great example is in this post: https://stackoverflow.com/questions/5616822/python-regex-find-all-overlapping-matches

So to find all the starting positions of the pattern, we need to put the pattern to (?=...):

import re
ss = "AAAA"
for m in re.finditer('(?=(AA))', ss):
    print m.start(), m.end(), m.group(1)

m.start() and m.end() are the same, because actually our pattern is an empty string “”, m.group(1) will give us the string matched the pattern “AA”.

re.findall(r'(?=(AA))', ss) and re.findall(r'(?=AA)', ss) give different results: the first one gives the matched “AA”, but the 2nd one gives the matched “”, but the m.start() is the same. I have not figured out why.