Python regex how do I find matching number of occurrences of a (sub)string?

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I need a construction that boils down to this: "at least 1 occurrence of (substring) a, followed by the SAME number of occurrences of (substring) b". So "ab", "aaabbb" and "aaaaabbbbb" are accepted, "aab" or "aaabbbb" are not.

I found that if the number of occurrences was fixed (say 5) I could use re.compile("a{5}b{5}") , but I don't know the number of occurrences. I just need them to be equal. I tried re.compile("a{x}b{x}") but that was wishful thinking I guess.

python regex
1 Answers

The built-in re does not support possessive quantifiers/atomic groups and does not support recursion, nor balanced groups, all those features that could help you build this pattern.

Thus, the easiest solution is to install the PyPi regex library with pip install regex and then use the How can we match a^n b^n? solution.

Otherwise, throw in some Python code and a simple (a+)(b+) regex:

import re
texts = [ "ab", "aaabbb", "aaaaabbbbb", "aab", "aaabbbb" ]
for text in texts:
    match = re.fullmatch(r'(a+)(b+)', text)
    if len(match.group(1)) == len(match.group(2)):
        print( text, '- MATCH' )
    else:
        print( text, '- NO MATCH' )

See this demo yielding

ab - MATCH
aaabbb - MATCH
aaaaabbbbb - MATCH
aab - NO MATCH
aaabbbb - NO MATCH

NOTE:

  • re.fullmatch(r'(a+)(b+)', text) matches an entire string that only contains one or more as and then one or more bs
  • if len(match.group(1)) == len(match.group(2)): is checking the lengh of as and bs, and only passes the strings where their counts are equal.
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