Regex. Python. Keep one group and replace others in string

Question

Here is my code:

In [1]: import re

In [2]: p = 'zxc(.*?)!(.*?)zxc'

In [3]: s = 'zxc wololo ! ololo zxc'

In [4]: re.sub(pattern=p, repl=r"", string=s)
Out[4]: ''

In [5]: re.sub(pattern=p, repl=r"\1", string=s)
Out[5]: ' wololo '

Expected

zxc wololo !zxc

Question

How to achieve the expected output?

I need to keep 1-st group with "prefix" and "suffix" of pattern. And let's assume that there are more then 2 groups.

Which keyword I should use in repl to achieve the expected result?

repl=r"zxc \1 !zxc"? See ideone.com/QbJ74Q and regex101.com/r/tzlx00/1 — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 27, 2019 at 10:43

anubhava · Accepted Answer · 2019-02-27 10:43:08Z

4

You may use a zero width lookaround based regex:

>>> s = 'zxc wololo ! ololo zxc'
>>> print re.sub(r'(?<=zxc)([^!]*!).*?(?=zxc)', r'\1', s)
zxc wololo !zxc

Here:

(?<=zxc) is a lookbehind assertion
(?=zxc) is a lookahead assertion
([^!]*!) matches and captures substring until ! and following ! in group #1

answered Feb 27, 2019 at 10:43

anubhava

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