Using regex (Python) I want to capture a group \d-. ? that is immediately followed by another pattern \sLEFT|\sRIGHT|\sUP.

Here is my test set (from http://nflsavant.com/about.php):

(9:03) (SHOTGUN) 30-J.RICHARD LEFT GUARD PUSHED OB AT MIA 9 FOR 18 YARDS (29-BR.JONES; 21-E.ROWE).
(1:06) 69-R.HILL REPORTED IN AS ELIGIBLE.  33-D.COOK LEFT GUARD TO NO 4 FOR -3 YARDS (56-D.DAVIS; 93-D.ONYEMATA).
(3:34) (SHOTGUN) 28-R.FREEMAN LEFT TACKLE TO LAC 37 FOR 6 YARDS (56-K.MURRAY JR.).
(1:19) 22-L.PERINE UP THE MIDDLE TO CLE 43 FOR 2 YARDS (54-O.VERNON; 51-M.WILSON).

My best attempt is (\d*-. ?)(?=\sLEFT|\sRIGHT|\sUP), which works unless other characters appear between a matching capture group and my positive lookahead. In the second line of my test set this expression captures "69-R.HILL REPORTED IN AS ELIGIBLE. 33-D.COOK." instead of the desired "33-D.COOK".

My inputs are also saved on regex101, here: https://regex101.com/r/tEyuiJ/1

How can I modify (or completely rewrite) my regex to only capture the group immediately followed by my exact positive lookahead with no extra characters between?

CodePudding user response：

If you want a capture group without any lookarounds:

\b(\d -\S*)\s(?:LEFT|RIGHT|UP)\b

Explanation

• \b A word boundary to prevent a partial word match
• (\d -\S*) Capture group 1, match 1 digits - and optional non whitespace characters
• \s Match a single whitespace character
• (?:LEFT|RIGHT|UP) Match any of the alternatives
• \b A word boundary

See the capture group value on regex101.

CodePudding user response：

To prevent skipping over digits, use \D non-digit (upper is negated \d).

\b(\d -\D ?)\s(?:LEFT|RIGHT|UP)

See this demo at regex101

Further added a word boundary and changed the lookahead to a group.

CodePudding user response：

This is why you should be careful about using . to match anything and everything unless it's absolutely necessary. From the example you provided, it appears that what you're actually wanting to capture contains no spaces, thus we could utilize a negative character class [^\s] or alternatively more precisely [\w.], with either case using a * quantifier.

Your end result would look like "(\d*-[\w.]*)(?=\sLEFT|\sRIGHT|\sUP)"gm. And of course, when . is within the character class it's treated as a literal string - so it's not required to be escaped.

See it live at regex101.com

CodePudding user response：

Try this:

\b\d -[^\r \n] (?=  (?:LEFT|RIGHT|UP)\b)

\b\d -[^\r \n]

• \b word boundary to ignore things like foo30-J.RICHARD

• \d match one or more digit.

• - match a literal -.

• [^\r \n] match on or more character except \r, \n and a literal space . Excluding \r and \n helps us not to cross newlines, and that is why \s is not used(i.e., it matches \r and \n too)

(?= (?:LEFT|RIGHT|UP)\b) Using positive lookahead.

• Ensure there is one or more literal space .
• (?:LEFT|RIGHT|UP)\b using non-caputring group, ensure our previous space followed by one of these words LEFT, RIGHT or UP. \b word boundary to ignore things like RIGHTfoo or LEFTbar.

See regex demo