Levenshtein distance between two given strings implemented in JavaScript and usable as a Node.js module

levenshtein.js

/*
Copyright (c) 2011 Andrei Mackenzie

Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
*/

// Compute the edit distance between the two given strings
exports.getEditDistance = function(a, b){
  if(a.length == 0) return b.length; 
  if(b.length == 0) return a.length; 

  var matrix = [];

  // increment along the first column of each row
  var i;
  for(i = 0; i <= b.length; i++){
    matrix[i] = [i];
  }

  // increment each column in the first row
  var j;
  for(j = 0; j <= a.length; j++){
    matrix[0][j] = j;
  }

  // Fill in the rest of the matrix
  for(i = 1; i <= b.length; i++){
    for(j = 1; j <= a.length; j++){
      if(b.charAt(i-1) == a.charAt(j-1)){
        matrix[i][j] = matrix[i-1][j-1];
      } else {
        matrix[i][j] = Math.min(matrix[i-1][j-1] + 1, // substitution
                                Math.min(matrix[i][j-1] + 1, // insertion
                                         matrix[i-1][j] + 1)); // deletion
      }
    }
  }

  return matrix[b.length][a.length];
};

@DerekZiemba

I'm afraid there is a bug in your code. For my example ("badbadnotgood", "s") it returned 2 instead of 13.

Have you seen this git?
JS-Levenshtein

$ npm run bench
         655,992 op/s » js-levenshtein
         542,796 op/s » leven
         497,966 op/s » talisman
         386,839 op/s » levenshtein-edit-distance
         254,941 op/s » fast-levenshtein
          69,857 op/s » levenshtein-component
          21,688 op/s » levdist
          24,631 op/s » ld
          21,834 op/s » natural
          13,984 op/s » levenshtein

Anyone know how to implement this in Javascript?
String Matching and Clustering

I'm thinking of the clustering part. So calculate the score I'm looking at this package. a javascript port of the popular Python package fuzzywuzzy

function fuzzyMatched (comparer, comparitor, matchCount) {
var isMatched = false;
a = comparer.trim().toLowerCase();
b = comparitor.trim().toLowerCase();
if(a.length == 0) return b.length;
if(b.length == 0) return a.length;
var matrix = [];

// increment along the first column of each row
var i;
for(i = 0; i <= b.length; i++){
    matrix[i] = [i];
}

// increment each column in the first row
var j;
for(j = 0; j <= a.length; j++){
    matrix[0][j] = j;
}

// Fill in the rest of the matrix
for(i = 1; i <= b.length; i++){
    for(j = 1; j <= a.length; j++){
        if(b.charAt(i-1) == a.charAt(j-1)){
            matrix[i][j] = matrix[i-1][j-1];
        } else {
            matrix[i][j] = Math.min(matrix[i-1][j-1] + 1, // substitution
                                    Math.min(matrix[i][j-1] + 1, // insertion
                                             matrix[i-1][j] + 1)); // deletion
        }
    }
}
var fuzzyDistance = matrix[b.length][a.length];
var cLength = Math.max(a.length, b.length);
var score = 1.0 - (fuzzyDistance / cLength);
if (score > matchCount)
    isMatched = true;

return isMatched;

}

In the above method youll have to pass two parameters and the maximum fuzzy match count, for example if you are passing 0.9 then the it will check match count 90% and gives you the result true or false.

fuzzyMatched("abc","abc1",0.7)
true
fuzzyMatched("abc","abc1",0.8)
false

String.prototype.levenstein = function(string) {
    var a = this, b = string + "", m = [], i, j, min = Math.min;

    if (!(a && b)) return (b || a).length;

    for (i = 0; i <= b.length; m[i] = [i++]);
    for (j = 0; j <= a.length; m[0][j] = j++);

    for (i = 1; i <= b.length; i++) {
        for (j = 1; j <= a.length; j++) {
            m[i][j] = b.charAt(i - 1) == a.charAt(j - 1)
                ? m[i - 1][j - 1]
                : m[i][j] = min(
                    m[i - 1][j - 1] + 1, 
                    min(m[i][j - 1] + 1, m[i - 1 ][j]))
        }
    }

    return m[b.length][a.length];
}

function findNearestMatchingSymbol(symbol, coins){
  let combinations = []
  for(let i = 0; i < symbol.length; i++){
    for(let j = i + 1; j <= symbol.length; j++){
      combinations.push(symbol.substring(i,j))
    }
  }

  let scores = []
  coins.forEach(coin=>{
    for(let current of combinations){
      if(coin.includes(current)){
        scores.push({
          symbol: current,
          coin: coin,
          score: current.levenstein(coin)
        }) 
      }
    }
  })
  scores.sort((a,b)=>a.score - b.score)
  console.log(scores)
}

findNearestMatchingSymbol('BTR', coins.map(i=>i.symbol))

This gets all the close symbols but takes too many iterations

Providing that string is shorter than 128 chars, this version based on kigiri's is ~25% faster (chrome 63):

const levenshtein = (a,b)=>{
  let alen = a.length;
  let blen = b.length;
  if (alen === 0) return blen;
  if (blen=== 0) return alen;
  let tmp, i, j, prev, val, row, ma, mb, mc, md, bprev;

  if (alen> blen) {
    tmp = a;
    a = b;
    b = tmp;
  }

  row = new Int8Array(alen+1);
  // init the row
  for (i = 0; i <= alen; i++) {
    row[i] = i;
  }

  // fill in the rest
  for (i = 1; i <= blen; i++) {
    prev = i;
    bprev = b[i - 1]
    for (j = 1; j <= alen; j++) {
      if (bprev === a[j - 1]) {
        val = row[j-1];
      } else {
          ma = prev+1;
          mb = row[j]+1;
          mc = ma - ((ma - mb) & ((mb - ma) >> 7));
          md = row[j-1]+1;
          val = mc - ((mc - md) & ((md - mc) >> 7));
      }
      row[j - 1] = prev;
      prev = val;
    }
    row[alen] = prev;
  }
  return row[alen];
}

https://jsperf.com/levenshteinteleranek/

@teleranek, your solution is broken as well.

levenshtein('helli', 'elli') returns 3, while levenshtein('elli', 'helli') returns 1

👏 Awesome!🤟

@DerekZiemba
Found the error in your algorithm as pointed out by @jonha892
It was missing the last write of the result to row so that value wasn't available the next iteration.

function dziemba_levenshtein(a, b){
	var tmp;
	if (a.length === 0) { return b.length; }
	if (b.length === 0) { return a.length; }
	if (a.length > b.length) { tmp = a; a = b; b = tmp; }

	var i, j, res, alen = a.length, blen = b.length, row = Array(alen);
	for (i = 0; i <= alen; i++) { row[i] = i; }

	for (i = 1; i <= blen; i++) {
		res = i;
		for (j = 1; j <= alen; j++) {
			tmp = row[j - 1];
			row[j - 1] = res;
			res = Math.min(tmp + (b[i - 1] !== a[j - 1]), res + 1, row[j] + 1);
		}
		row[j - 1] = res; // This was the missing line
	}
	return res;
}

@milot-mirdita

Hey i wanna ask, if it is okay for you if I use your code in an own npm package, where i calculate the character-error-rate for ASR.
I can also give you credits on the code

@henning410 Sure! its MIT licensed, you don't have to ask for permission as long as you stick to the license terms.

boa noite consigo rodar bot mais nao consigo fazer nenhuma da funçoes rodar

Here's a fresh TypeScript version of it: https://gist.github.com/felixeichler/53c48ec5c398e6a01d675c42d430e7c6