How to replace a string with parts of it between markers (regex)?

user_16117441 0
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I have a text, let's say:

Lorem ipsum dolor sit [[amet]], consectetur adipiscing elit, 
sed do eiusmod [[time (sample)|tempor]]  incididunt ut [[labore]] et dolore magna aliqua.

I want to replace the [[time (sample)|tempor]] with tempor. The structure is always the same: [[string to remove|string to extract]] and can appear several times in the text.

I tried regular expressions in regex, but I wasn't successful without cutting off half of the text: re.sub(r'\[.*?\|', '', text)

How can I replace the string?

python regex
2 Answers

You may use the following regex to collect only the relevant field

r'\[\[[\w\s\(\)]+?\|(.+?)\]\]'

import re
regex = r'\[\[[\w\s\(\)]+?\|(.+?)\]\]'

text = '''
Lorem ipsum dolor sit [[amet]], consectetur adipiscing elit,
sed do eiusmod [[time (sample)|tempor]]  incididunt ut [[labore]] et dolore magna aliqua.

Lorem ipsum dolor sit [[amet]], consectetur adipiscing elit, sed do eiusmod [[time (sample)|tempor]]  incididunt ut [[labore]] et dolore magna aliqua.
'''

txt = re.sub(regex, '[[\g<1>]]', text)
print(txt)

Lorem ipsum dolor sit [[amet]], consectetur adipiscing elit,
sed do eiusmod [[tempor]]  incididunt ut [[labore]] et dolore magna aliqua.

Lorem ipsum dolor sit [[amet]], consectetur adipiscing elit, sed do eiusmod [[tempor]]  incididunt ut [[labore]] et dolore magna aliqua.

Regex101 sample here

user_131441430

Use

\[\[(?:(?!\[\[)[^|])*\|(.*?)]]

Replace with [[\1]] or \1, depending on the requirements.

See proof.

EXPLANATION

--------------------------------------------------------------------------------
  \[                       '['
--------------------------------------------------------------------------------
  \[                       '['
--------------------------------------------------------------------------------
  (?:                      group, but do not capture (0 or more times
                           (matching the most amount possible)):
--------------------------------------------------------------------------------
    (?!                      look ahead to see if there is not:
--------------------------------------------------------------------------------
      \[                       '['
--------------------------------------------------------------------------------
      \[                       '['
--------------------------------------------------------------------------------
    )                        end of look-ahead
--------------------------------------------------------------------------------
    [^|]                     any character except: '|'
--------------------------------------------------------------------------------
  )*                       end of grouping
--------------------------------------------------------------------------------
  \|                       '|'
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    .*?                      any character except \n (0 or more times
                             (matching the least amount possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  ]]                       ']]'
user_113298900
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