Regex to match everything after a pattern occurrence until the next pattern occurs and so on

user_6787880 0
Viewed 78

I'd like to extract everything that follows a "line break and integer" until the next "line break and integer", where i'd like to capture everything that follows that and so on. For example for the following string:

"\na \n1 b\nc \n2 b\nc \n3 b\nc"

I'd like to capture the following groups:

["\n1 b\nc ", "\n2 b\nc ", "\n3 b\nc"]

This is what i've tried

re.findall("\n\d[\s\S]*(?=\n\d)*","\na \n1 b\nc \n2 b\nc \n3 b\nc")

But it's not splitting the matches, I think i need to make it "non-greedy" but i'm not sure how.

['\n1 b\nc \n2 b\nc \n3 b\nc']
python regex
2 Answers

You may use this regex in DOTALL or single line mode:

(?s)\n\d.*?(?=\n\d|\Z)

RegEx Demo

RegEx Details:

  • (?s): Enable single line mode to allow dot to match line break
  • \n: Match a line break
  • \d: Match a digit
  • .*?: Match 0 or more of any characters (lazy)
  • (?=\n\d|\Z): Lookahead to assert that we have either another line break and digit or end of input

Code:

>>> import re
>>> s = "\na \n1 b\nc \n2 b\nc \n3 b\nc"
>>> re.findall(r'(?s)\n\d.*?(?=\n\d|\Z)', s)
['\n1 b\nc ', '\n2 b\nc ', '\n3 b\nc']
user_5482250

Probably another way is to match from beginning of line with digits
to next beginning of line with digits.
This will allow for the beginning and and of string.
Group 1 contains the part you're interested in.

(?m)^\d+([\S\s]+?)(?=^\d|\Z)

https://regex101.com/r/ctC5KO/1

(?m)
^ \d+ 
( [\S\s]+? )     # (1)
(?=
  ^ \d 
| \Z 
)
user_155776650
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