remove digits NOT between brackets

I am trying to write a regular expression in python to remove digits except those between brackets here is an example : "[Verse 1: Bankroll Fresh] Dope boy, '[9]5 Air Max on Came from 1952.."

I would like : "[Verse 1: Bankroll Fresh] Dope boy, '[9] Air Max on Came from .."

CodePudding user response：

Using re.sub(pattern, replacement, string)

output = re.sub(r"(?!\d \])\d ", "", input_string)

(?!...) is a negative lookahead (aka. do not find the pattern). \d \] finds digit/s followed by a ]. \d is digit/s.

For more information:

Python Regex Documentation

More details about re.sub()

Regex pattern information

CodePudding user response：

Can you try this regex r'(?<!\[)\d (?!\])'?

st = "[Verse 1: Bankroll Fresh] Dope boy, '[9]5 Air Max on Came from 1952.."
st = re.sub(r'\[\D*\d\D*\]|(?<!\[)\d (?!\])', lambda x: x.group(0) if x.group(0).startswith('[') else '', st)

# [Verse 1: Bankroll Fresh] Dope boy, '[9] Air Max on Came from ..
print(st)

(?<!\[) and (?!\]) are matches if \d NOT preceded by [ and NOT followed by ], respectively.