字符串替换正则表达式匹配的捕获组，除非不同的正则表达式匹配 JS 中的相同捕获组

Question

This is about the simplest example I can come up with (and actually not a far cry from the actual use case) that I anticipate people might not tell me to simply change the regex itself, but here goes.

Say I have these lovely nonsensical input words: sent seimk semek t͡ʃeno eint͡ɬi em t͡ʃeʃeimp t͡ɬent͡ɬien keinsen

And I want to search replace all instances of ei OR e with a before n or m , which (ei|e)(?:n|m) matches, UNLESS

1. it is directly preceded (ie no negative lookbehinds, since they'll cause a false negative) by t͡ɬ , t͡ʃ , or d͡ʒ , so ie (?:t͡ɬ|t͡ʃ|d͡ʒ)(ei|e) , or

2. if the n/m is directly succeeded (ie no negative lookaheads, for the same reason) by p , t , or k , ie if (ei|e)(?:n|m)(?:p|t|k) matches

The desired output is therefore sent seimk samek t͡ʃeno ant͡ɬi am t͡ʃeʃeimp t͡ɬent͡ɬian kansan .

So if the "find and replace" function in JS is String.replace(RegExp pattern, String replacement) , then then you would have to compress all 3 regexes into 1 for the first argument, which 1) I don't think is possible? It would require negative non-capturing groups which... aren't a thing, right? and 2) in the actual use case, the regex patterns are generated by a text parser, not by hand, and I don't have enough faith in my ability to write a parser smart enough to optimize.

The other way I thought about doing this was to simply cache all the matches to the first regex in a dictionary, and then remove everything that should be ruled out by the other two, but when you look at the output:

 let sEnv = "(ei|e)(?:n|m)"; let reEnv = new RegExp(sEnv, "g"); let sExc1 = "(?:t͡ɬ|t͡ʃ|d͡ʒ)(ei|e)"; let reExc1 = new RegExp(sExc1, "g"); let sExc2 = "(ei|e)(?:n|m)(?:p|t|k)"; let reExc2 = new RegExp(sExc2, "g"); let sWords = "sent seimk semek t͡ʃeno eint͡ɬi em t͡ʃeʃeimp t͡ɬent͡ɬien keinsen"; let tWords = sWords.split(" "); for (i = 0; i < tWords.length; i++){ let sCurrentWord = tWords[i]; let result; let tMatches = {} // first cache all the matches (ignoring the exceptions) while (result = reEnv.exec(sCurrentWord)) { tMatches[result.index] = result[0].length; } // then remove the exceptions while (result = reExc1.exec(sCurrentWord)) { delete tMatches[result.index]; } while (result = reExc2.exec(sCurrentWord)) { delete tMatches[result.index]; } // then apply all remaining matches let sOutput = sCurrentWord; for (var index in tMatches){ console.log(sCurrentWord+": starting at "+index+", "+tMatches[index]+" chars long"); } }

It's... a bit confused. Not only is it apparently capturing the n despite being in a non-capturing group (result length printed keeps being 2 when the thing I want to capture is only 1 char long), but it also filtered out eint͡ɬi but kept t͡ʃeno as matches, which is the opposite of what it's supposed to do - and as the length of the input list grows this method must get quite slow.

How else can I go about this find-and-replace?

Answer 1

You actually have more hidden requirement here, namely the p , t , k cannot be followed with a diacritic mark.

Your main mistake is that you think that lookarounds cannot be used to match locations immediately preceded/followed with some pattern. In fact, lookarounds DO and ALWAYS DO match locations that are immediately preceded/followed with some patterns .

In your case, you can use (assuming you are using the ECMAScript 2018 compliant RegExp ):

text = text.replace(/(?<!t͡ɬ|t͡ʃ|d͡ʒ)(ei?)(?=[nm](?![ptk](?!\p{M})))/gu, 'a')

See the regex demo . Details:

• (?<!t͡ɬ|t͡ʃ|d͡ʒ) - a negative lookbehind that fails the match if there is t͡ɬ , t͡ʃ or d͡ʒ immediately to the left of the current location
• (ei?) - ei or e
• (?=[nm](??[ptk](?!\p{M}))) - a positive lookahead that matches a location that is immediately followed with n or m that are not immediately followed with p , t nor k that are not immediately followed with any diacrtic mark ( \p{M} ).

See the JavaScript demo:

 const regex = /(?<?t͡ɬ|t͡ʃ|d͡ʒ)(ei?)(?=[nm](?;[ptk](;.\p{M})))/gu. const text = 'sent seimk semek t͡ʃeno eint͡ɬi em t͡ʃeʃeimp t͡ɬent͡ɬien keinsen', console;log(text.replace(regex, 'a'));