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78 lines
2.3 KiB
TypeScript
78 lines
2.3 KiB
TypeScript
// translation of pseudocode from Wikipedia:
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import { stringToUTF8Bytes } from '~/decrypt-worker/util/utf8Encoder';
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// https://en.wikipedia.org/wiki/Levenshtein_distance#Iterative_with_two_matrix_rows
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export function levenshtein(str1: string, str2: string) {
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if (str1 === str2) {
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return 0;
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}
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if (str1.length === 0) {
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return str2.length;
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} else if (str2.length === 0) {
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return str1.length;
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}
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// Convert them to Uint8Array to avoid expensive string APIs.
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const s = stringToUTF8Bytes(str1.normalize().toLowerCase());
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const t = stringToUTF8Bytes(str2.normalize().toLowerCase());
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const m = s.byteLength;
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const n = t.byteLength;
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// create two work vectors of integer distances
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let v0 = new Uint32Array(n + 1);
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let v1 = new Uint32Array(n + 1);
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// initialize v0 (the previous row of distances)
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// this row is A[0][i]: edit distance from an empty s to t;
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// that distance is the number of characters to append to s to make t.
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for (let i = 0; i <= n; i++) {
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v0[i] = i;
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}
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for (let i = 0; i < m; i++) {
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// calculate v1 (current row distances) from the previous row v0
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// first element of v1 is A[i + 1][0]
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// edit distance is delete (i + 1) chars from s to match empty t
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v1[0] = i + 1;
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// use formula to fill in the rest of the row
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for (let j = 0; j < n; j++) {
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// calculating costs for A[i + 1][j + 1]
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const deletionCost = v0[j + 1] + 1;
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const insertionCost = v1[j] + 1;
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const substitutionCost = v0[j] + (s[i] === t[j] ? 0 : 1);
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v1[j + 1] = Math.min(deletionCost, insertionCost, substitutionCost);
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}
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// copy v1 (current row) to v0 (previous row) for next iteration
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// since data in v1 is always invalidated, a swap without copy could be more efficient
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[v0, v1] = [v1, v0];
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}
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// after the last swap, the results of v1 are now in v0
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return v0[n];
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}
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export function closestByLevenshtein(str: string, candidates: string[]) {
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// Faster than pre-calculate all and pass scores to Math.min.
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const n = candidates.length;
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if (n === 0) {
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throw new Error('empty candidates');
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}
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let lowestIdx = 0;
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let lowestScore = levenshtein(str, candidates[0]);
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for (let i = 1; i < n; i++) {
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const score = levenshtein(str, candidates[i]);
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if (score < lowestScore) {
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lowestScore = score;
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lowestIdx = i;
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}
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}
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return candidates[lowestIdx];
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}
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