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pythonregexwhitespaceregex-lookaroundslookbehind

Regular expression: matching words between white space


Im trying to do something fairly simple with regular expression in python... thats what i thought at least.

What i want to do is matching words from a string if its preceded and followed by a whitespace. If its at the beginning of the string there is no whitespace required before - if its at the end, dont't search for whitespace either.

Example:

"WordA WordB WordC-WordD WordE"

I want to match WordA WordB WordE.

I only came up with overcomplicated way of doing this...

(?<=(?<=^)|(?<=\s))\w+(?=(?=\s)|(?=$))

It seems to me there has to be a simple way for such a simple problem.... I figured i can just start with (?<=\s|^) but that doesnt seem possible because "look-behind requires fixed-width pattern".


Solution

  • You seem to work in Python as (?<=^|\s) is perfectly valid in PCRE, Java and Ruby (and .NET regex supports infinite width lookbehind patterns).

    Use

    (?<!\S)\w+(?!\S)
    

    It will match 1 or more word chars that are enclosed with whitespace or start/end of string.

    See the regex demo.

    Pattern details:

    • (?<!\S) - a negative lookbehind that fails the match once the engine finds a non-whitespace char immediately to the left of the current location
    • \w+ - 1 or more word chars
    • (?!\S) - a negative lookahead that fails the match once the engine finds a non-whitespace char immediately to the right of the current location.