Mark the shortest overlapping match using regular expressions

Question

This post shows how to find the shortest overlapping match using regex. One of the answers shows how to get the shortest match, but I am struggling with how to locate the shortest match and mark its position, or substitute it with another string.

So in the given pattern,

A|B|A|F|B|C|D|E|F|G

and the pattern I want to locate is:

my_pattern = 'A.*?B.*?C'

How can I identify the shortest match and mark it in the original given pattern like below?

A|B|[A|F|B|C]|D|E|F|G

or substitute:

A|B|AAA|F|BBB|CCC|D|E|F|G

Answer 1

I suggest to use Tim Pietzcker's answer with re.sub :

>>> p=re.findall(r'(?=(A.*?B.*?C))',s)
>>> re.sub(r'({})'.format(re.escape(min(p, key=len))),r'[\1]',s,re.DOTALL)
'A|B|[A|F|B|C]|D|E|F|G'

Answer 2

One way is to use lookahead between A and B and then B and C like this:

import re
p = re.compile(ur'A(?:(?![AC]).)*B(?:(?![AB]).)*C')
test_str = u"A|B|A|F|B|C|D|E|F|G"
result = re.sub(p, u"[$0]", test_str)
# A|B|[A|F|B|C]|D|E|F|G

test_str = u"A|B|C|F|B|C|D|E|F|G"
result = re.sub(p, u"[$0]", test_str)
# [A|B|C]|F|B|C|D|E|F|G

RegEx Demo

Answer 3

(A[^A]*?B[^B]*?C)

You can use this simple regex.Replace by [\1].

See Demo

x="A|B|A|F|B|C|D|A|B|C" print re.sub("("+re.escape(min(re.findall(r"(A[^A]*?B[^B]*?C)",x),key=len))+")",r"[\1]",x)