How To Match Overlapping Keywords With Regex

This example finds only sam. How to make it find both sam and samwise? var regex = /sam|samwise|merry|pippin/g; var string = 'samwise gamgee'; var match = string.match(regex); cons

Solution 1:

You can use lookahead regex with capturing group for this overlapping match:

var regex = /(?=(sam))(?=(samwise))/;
varstring = 'samwise';
var match = string.match( regex ).filter(Boolean);
//=> ["sam", "samwise"]

• It is important to not to use g (global) flag in the regex.
• filter(Boolean) is used to remove first empty result from matched array.

Solution 2:

Why not just mapindexOf() on array substr:

varstring = 'samwise gamgee';
var substr = ['sam', 'samwise', 'merry', 'pippin'];

var matches = substr.map(function(m) {
  return (string.indexOf(m) < 0 ? false : m);
}).filter(Boolean);

See fiddleconsole.log(matches);

Array [ "sam", "samwise" ]

Probably of better performance than using regex. But if you need the regex functionality e.g. for caseless matching, word boundaries, returned matches... use with exec method:

var matches = substr.map(function(v) {
  var re = newRegExp("\\b" + v, "i"); var m = re.exec(string); 
  return (m !== null ? m[0] : false);
}).filter(Boolean);

This one with i-flag (ignore case) returns each first match with initial \bword boundary.

Solution 3:

I can't think of a simple and elegant solution, but I've got something that uses a single regex:

functionquotemeta(s) {
    return s.replace(/\W/g, '\\$&');
}

let keywords = ['samwise', 'sam'];

let subsumed_by = {};
keywords.sort();
for (let i = keywords.length; i--; ) {
    let k = keywords[i];
    for (let j = i - 1; j >= 0 && k.startsWith(keywords[j]); j--) {
        (subsumed_by[k] = subsumed_by[k] || []).push(keywords[j]);
    }
}

keywords.sort(function (a, b) b.length - a.length);
let re = newRegExp('(?=(' + keywords.map(quotemeta).join('|') + '))[\\s\\S]', 'g');

let string = 'samwise samgee';

let result = [];
let m;
while (m = re.exec(string)) {
    result.push(m[1]);
    result.push.apply(result, subsumed_by[m[1]] || []);
}

console.log(result);

Solution 4:

How about:

var re = /((sam)(?:wise)?)/;
var m = 'samwise'.match(re); // gives ["samwise", "samwise", "sam"]
var m = 'sam'.match(re);     // gives ["sam", "sam", "sam"]

You can use Unique values in an array to remove dupplicates.

Solution 5:

If you don't want to create special cases, and if order doesn't matter, why not first match only full names with:

\b(sam|samwise|merry|pippin)\b

and then, filter if some of these doesn't contain shorter one? for example with:

(sam|samwise|merry|pippin)(?=\w+\b)

It is not one elegant regex, but I suppose it is simpler than iterating through all matches.