Question

How can I obtain the number of overlapping regex matches using Python?

I’ve read and tried the suggestions from this, that and a few other questions, but found none that would work for my scenario. Here it is:

• input example string: akka
• search pattern: a.*k

A proper function should yield 2 as the number of matches, since there are two possible end positions (k letters).

The pattern might also be more complicated, for example a.*k.*a should also be matched twice in akka (since there are two k‘s in the middle).

Answer 1

Yes, it is ugly and unoptimized but it seems to be working. This is a simple try of all possible but unique variants

def myregex(pattern,text,dir=0):
    import re
    m = re.search(pattern, text)
    if m:
        yield m.group(0)
        if len(m.group('suffix')):
            for r in myregex(pattern, "%s%s%s" % (m.group('prefix'),m.group('suffix')[1:],m.group('end')),1):
                yield r
            if dir<1 :
                for r in myregex(pattern, "%s%s%s" % (m.group('prefix'),m.group('suffix')[:-1],m.group('end')),-1):
                    yield r


def myprocess(pattern, text):    
    parts = pattern.split("*")    
    for i in range(0, len(parts)-1 ):
        res=""
        for j in range(0, len(parts) ):
            if j==0:
                res+="(?P<prefix>"
            if j==i:
                res+=")(?P<suffix>"
            res+=parts[j]
            if j==i+1:
                res+=")(?P<end>"
            if j<len(parts)-1:
                if j==i:
                    res+=".*"
                else:
                    res+=".*?"
            else:
                res+=")"
        for r in myregex(res,text):
            yield r

def mycount(pattern, text):
    return set(myprocess(pattern, text))

test:

>>> mycount('a*b*c','abc')
set(['abc'])
>>> mycount('a*k','akka')
set(['akk', 'ak'])
>>> mycount('b*o','bboo')
set(['bbo', 'bboo', 'bo', 'boo'])
>>> mycount('b*o','bb123oo')
set(['b123o', 'bb123oo', 'bb123o', 'b123oo'])
>>> mycount('b*o','ffbfbfffofoff')
set(['bfbfffofo', 'bfbfffo', 'bfffofo', 'bfffo'])