Depending on the regular expression implementation you can use so called non-capturing groups with the syntax (?:…): ((?:a|b)c) Here (?:a|b) is a group but you cannot reference its match. So you can only reference the match of ((?:a|b)c) that is either ac or bc.
Have a look at re.sub: result = re.sub(r”(\d.*?)\s(\d.*?)”, r”\1 \2″, string1) This is Python’s regex substitution (replace) function. The replacement string can be filled with so-called backreferences (backslash, group number) which are replaced with what was matched by the groups. Groups are counted the same as by the group(…) function, i.e. starting from 1, from … Read more
One way to fix this is by ensuring the pattern is enclosed by escaped parentheses: :%s/\(\w\)\(\w\w\)/\1y\2/g Slightly shorter (and more magic-al) is to use \v, meaning that in the pattern after it all ASCII characters except ‘0’-‘9’, ‘a’-‘z’, ‘A’-‘Z’ and ‘_’ have a special meaning: :%s/\v(\w)(\w\w)/\1y\2/g See: :help \( :help \v
Continue calling re.exec(s) in a loop to obtain all the matches: var re = /\s*([^[:]+):\”([^”]+)”/g; var s=”[description:”aoeu” uuid:”123sth”]”; var m; do { m = re.exec(s); if (m) { console.log(m[1], m[2]); } } while (m); Try it with this JSFiddle: https://jsfiddle.net/7yS2V/
The answer is: re.sub(r'(foo)’, r’\g<1>123′, ‘foobar’) Relevant excerpt from the docs: In addition to character escapes and backreferences as described above, \g will use the substring matched by the group named name, as defined by the (?P…) syntax. \g uses the corresponding group number; \g<2> is therefore equivalent to \2, but isn’t ambiguous in a … Read more
Since we’re all guessing, I might as well give mine: I’ve always thought it stood for Python. That may sound pretty stupid — what, P for Python?! — but in my defense, I vaguely remembered this thread [emphasis mine]: Subject: Claiming (?P…) regex syntax extensions From: Guido van Rossum (gui…@CNRI.Reston.Va.US) Date: Dec 10, 1997 3:36:19 … Read more
Let me try to explain this with an example. Consider the following text: http://stackoverflow.com/ https://stackoverflow.com/questions/tagged/regex Now, if I apply the regex below over it… (https?|ftp)://([^/\r\n]+)(/[^\r\n]*)? … I would get the following result: Match “http://stackoverflow.com/” Group 1: “http” Group 2: “stackoverflow.com” Group 3: “https://stackoverflow.com/” Match “https://stackoverflow.com/questions/tagged/regex” Group 1: “https” Group 2: “stackoverflow.com” Group 3: “https://stackoverflow.com/questions/tagged/regex” But … Read more