github-pages-deploy-action/node_modules/graphql/jsutils/suggestionList.mjs

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2020-03-07 11:45:40 +08:00
/**
* Given an invalid input string and a list of valid options, returns a filtered
* list of valid options sorted based on their similarity with the input.
*/
export default function suggestionList(input, options) {
var optionsByDistance = Object.create(null);
var inputThreshold = input.length / 2;
for (var _i2 = 0; _i2 < options.length; _i2++) {
var option = options[_i2];
var distance = lexicalDistance(input, option);
var threshold = Math.max(inputThreshold, option.length / 2, 1);
if (distance <= threshold) {
optionsByDistance[option] = distance;
}
}
return Object.keys(optionsByDistance).sort(function (a, b) {
var distanceDiff = optionsByDistance[a] - optionsByDistance[b];
return distanceDiff !== 0 ? distanceDiff : a.localeCompare(b);
});
}
/**
* Computes the lexical distance between strings A and B.
*
* The "distance" between two strings is given by counting the minimum number
* of edits needed to transform string A into string B. An edit can be an
* insertion, deletion, or substitution of a single character, or a swap of two
* adjacent characters.
*
* Includes a custom alteration from Damerau-Levenshtein to treat case changes
* as a single edit which helps identify mis-cased values with an edit distance
* of 1.
*
* This distance can be useful for detecting typos in input or sorting
*
* @param {string} a
* @param {string} b
* @return {int} distance in number of edits
*/
function lexicalDistance(aStr, bStr) {
if (aStr === bStr) {
return 0;
}
var d = [];
var a = aStr.toLowerCase();
var b = bStr.toLowerCase();
var aLength = a.length;
var bLength = b.length; // Any case change counts as a single edit
if (a === b) {
return 1;
}
for (var i = 0; i <= aLength; i++) {
d[i] = [i];
}
for (var j = 1; j <= bLength; j++) {
d[0][j] = j;
}
for (var _i3 = 1; _i3 <= aLength; _i3++) {
for (var _j = 1; _j <= bLength; _j++) {
var cost = a[_i3 - 1] === b[_j - 1] ? 0 : 1;
d[_i3][_j] = Math.min(d[_i3 - 1][_j] + 1, d[_i3][_j - 1] + 1, d[_i3 - 1][_j - 1] + cost);
if (_i3 > 1 && _j > 1 && a[_i3 - 1] === b[_j - 2] && a[_i3 - 2] === b[_j - 1]) {
d[_i3][_j] = Math.min(d[_i3][_j], d[_i3 - 2][_j - 2] + cost);
}
}
}
return d[aLength][bLength];
}