Regex - match a pattern (from capture) before a capture group

Question

I have to match text with this scheme:

• capture [\w-/]* which are not ['] (\1)
• capture [']+ which follow (\2)
• and replace \2\1\2 with \1

Example:

my input text is: l''''text'''
the right output is: l'text

I've tried with:

re.sub(r"(\5)(?=((([\w\-\/](?<!'))+)('+)))", r"\2", text)

Answer 1

You could match the previously matched quotes after the string:

('+)([\w/-]+)\1

The \1 matches the exact same text group 1 matched.

Online demo at https://regex101.com/r/zQ0hM2/2.

Python session demo:

>>> import re
>>> text = "l''''text'''"
>>> re.sub(r'''('+)([\w/-]+)\1''', r'\2', text)
"l'text"