Question

I use python regular expressions (re module) in my code and noticed different behaviour in theese cases:

re.findall(r'\s*(?:[a-z]\))?[^.)]+', 'a) xyz. b) abc.') # non-capturing group
# results in ['a) xyz', ' b) abc']

and

re.findall(r'\s*(?<=[a-z]\))?[^.)]+', 'a) xyz. b) abc.') # lookbehind
# results in ['a', ' xyz', ' b', ' abc']

What I need to get is just ['xyz', 'abc']. Why are the examples behave differently and how t get the desired result?

Answer 1

The reason a and b are included in the second case is because (?<=[a-z]\)) would first find a) and since lookaround’s don’t consume any character you are back at the start of string.Now [^.)]+ matches a

Now you are at ).Since you have made (?<=[a-z]\)) optional [^.)]+ matches xyz

This same thing is repeated with b) abc

remove ? from the second case and you would get the expected result i.e ['xyz', 'abc']