python - How do I find the shortest overlapping match using regular expressions?

Question

I'm still relatively new to regex. I'm trying to find the shortest string of text that matches a particular pattern, but am having trouble if the shortest pattern is a substring of a larger match. For example:

import re
string = "A|B|A|B|C|D|E|F|G"
my_pattern = 'a.*?b.*?c'

my_regex = re.compile(my_pattern, re.DOTALL|re.IGNORECASE)
matches = my_regex.findall(string)

for match in matches:
    print match

prints:

A|B|A|B|C

but I'd want it to return:

A|B|C

Is there a way to do this without having to loop over each match to see if it contains a substring that matches?

Answer 1

Contrary to most other answers here, this can be done in a single regex using a positive lookahead assertion with a capturing group:

>>> my_pattern = '(?=(a.*?b.*?c))'
>>> my_regex = re.compile(my_pattern, re.DOTALL|re.IGNORECASE)
>>> matches = my_regex.findall(string)
>>> print min(matches, key=len)
A|B|C

findall() will return all possible matches, so you need min() to get the shortest one.

How this works:

• We're not matching any text in this regex, just positions in the string (which the regex engine steps through during a match attempt).
• At each position, the regex engine looks ahead to see whether your regex would match at this position.
• If so, it will be captured by the capturing group.
• If not, it won't.
• In either case, the regex engine then steps ahead one character and repeats the process until the end of the string.
• Since the lookahead assertion doesn't consume any characters, all overlapping matches will be found.