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node_modules/graphql/jsutils/suggestionList.js
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85
node_modules/graphql/jsutils/suggestionList.js
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"use strict";
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Object.defineProperty(exports, "__esModule", {
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value: true
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});
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exports.default = suggestionList;
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/**
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* Given an invalid input string and a list of valid options, returns a filtered
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* list of valid options sorted based on their similarity with the input.
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*/
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function suggestionList(input, options) {
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var optionsByDistance = Object.create(null);
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var inputThreshold = input.length / 2;
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for (var _i2 = 0; _i2 < options.length; _i2++) {
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var option = options[_i2];
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var distance = lexicalDistance(input, option);
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var threshold = Math.max(inputThreshold, option.length / 2, 1);
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if (distance <= threshold) {
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optionsByDistance[option] = distance;
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}
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}
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return Object.keys(optionsByDistance).sort(function (a, b) {
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return optionsByDistance[a] - optionsByDistance[b];
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});
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}
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/**
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* Computes the lexical distance between strings A and B.
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*
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* The "distance" between two strings is given by counting the minimum number
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* of edits needed to transform string A into string B. An edit can be an
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* insertion, deletion, or substitution of a single character, or a swap of two
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* adjacent characters.
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*
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* Includes a custom alteration from Damerau-Levenshtein to treat case changes
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* as a single edit which helps identify mis-cased values with an edit distance
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* of 1.
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*
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* This distance can be useful for detecting typos in input or sorting
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*
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* @param {string} a
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* @param {string} b
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* @return {int} distance in number of edits
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*/
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function lexicalDistance(aStr, bStr) {
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if (aStr === bStr) {
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return 0;
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}
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var d = [];
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var a = aStr.toLowerCase();
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var b = bStr.toLowerCase();
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var aLength = a.length;
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var bLength = b.length; // Any case change counts as a single edit
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if (a === b) {
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return 1;
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}
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for (var i = 0; i <= aLength; i++) {
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d[i] = [i];
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}
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for (var j = 1; j <= bLength; j++) {
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d[0][j] = j;
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}
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for (var _i3 = 1; _i3 <= aLength; _i3++) {
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for (var _j = 1; _j <= bLength; _j++) {
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var cost = a[_i3 - 1] === b[_j - 1] ? 0 : 1;
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d[_i3][_j] = Math.min(d[_i3 - 1][_j] + 1, d[_i3][_j - 1] + 1, d[_i3 - 1][_j - 1] + cost);
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if (_i3 > 1 && _j > 1 && a[_i3 - 1] === b[_j - 2] && a[_i3 - 2] === b[_j - 1]) {
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d[_i3][_j] = Math.min(d[_i3][_j], d[_i3 - 2][_j - 2] + cost);
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}
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}
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}
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return d[aLength][bLength];
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}
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