RegExp detect multiple single-letter instances in a row?

Question

So I'm making a program to parse twitch chat, and I'm wondering if there's a way I can use regex to parse the following into the desired result:

"f o o b a r" into "foobar"

So far, the code I have is /(?:(\w)\s){3,}/g and this works to an extent, but consider the following situation:

"FrankerZ R I O T FrankerZ" captures "T" (the last letter in "R I O T") and selects "Z R I O T"

What I would want for this is to figure out how to detect if there is a single letter with a space before and after it, and if there are at minimum 3 of those in a row (so "test a b test" isn't selected as ab, only captures if there are 3+)

Any help? Thanks!

Answer 1

Try this pattern: /(?:\b\w(?:\s|$)){3,}/g

This uses the word boundary metacharacter \b so you get a proper whole word match instead of the partial match you saw with FrankerZ. Also, the \s|$ bit addresses the last letter being lost when no space comes after it, e.g., the "T" in R I O T.

Example:

var inputs = [
  "R I",
  "R I O T",
  "FrankerZ R I O T FrankerZ",
  "f o o b a r"
];

var re = /(?:\b\w(?:\s|$)){3,}/g;

inputs.forEach(function(s) {
  var match = s.match(re);
  if (match) {
    var result = match[0].replace(/\s/g, '');
    console.log('Original: ' + s);
    console.log('Result: ' + result);
  } else {
    console.log('No match: ' + s);
  }
});

Demo: JSBin

EDIT: updated to cover 3+ single letters and example of no match.

Answer 2

Here is a good reference how to replace with matches Javascript replace with reference to matched group?

So you could do:

'string'.replace(/(\s|^)((?:\w\s){2,}\w)(\s|$)/g, function(a, b, c, d) {
     return b + c.replace(/\s/g, '') + d;
});

See demo

Answer 3

Thank you to Sam Burns for suggesting the use of \b. What works for me was:

/\b((?:\w ?\b){3,})/g

This would select the following:

H Y P E from FrankerZ H Y P E FrankerZ, and f o o b a r (doesn't end or begin with a space character, was giving me issues as well)

Specifying the literal space " " character instead of \s was also important for avoiding line breaks and other instances when I only wanted to check for just the space character in the first place.

For replacing it without spaces, I'll simply do .replace(" ","") to receive the exact result I wanted. Thanks again for everyone's help :)