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pythonregexregex-groupregex-negation

How do I exact match all the words in a string includes those ends with '!,?.'"' but do not match those with any other punctuation using regex?

For example, if the pattern I want to search for is apple

The string I want to search into is apple apple@323 apple.. apple??...!! apple??%% Ilovesapple

Expected matches:

apple 
apple.. 
apple??...!!

I want to match all the exact words, the only exception is that the word ends with the punctuation in !,?.'"

Solution

  • Is this what you want ?

    \bapple[!,?.'\"]*(?=\s+|$)

    Regex Demo

    enter image description here

    Details:

    • \b word boundary for 'apple'
    • apple the pattern you want to search
    • [!,?.'\"]* zero or more occurrences of the special character(s) at the end that you want to match
    • (?=\s+|$) Positive Lookahead to ensure the matching word is followed by white space or end of line.

    Another version

    Another possibility is this if you don't allow any non-space characters before 'apple':

    (?:\s+|^)(apple[!,?.'\"]*)(?=\s+|$)

    Regex Demo Note that @apple? is not matched because it starts with @

    Refer to Group 1 for the matches

    enter image description here

    Details:

    • (?:\s+|^) non-capturing group for white space(s) or beginning of line before 'apple'

    • ( start of capturing group (captured into Group 1)

      • apple the pattern you want to search
      • [!,?.'\"]* zero or more occurrences of the special character(s) at the end that you want to match

    • ) end of capturing group (captured into Group 1)

    • (?=\s+|$) Positive Lookahead to ensure the matching word is followed by white space or end of line.