您好我有一个很长的JSON键键值对的列表：值，键：值等

Hi I have a very long list of key value pairs in json key:value, key:value and so on

car <--> wheel
wheel <--> tyre
bed <--> sheets
guitar <--> strings
guitar <--> pickup
tyre <--> rubber

我要的是把所有的关系到阵列无论多么遥远像这样

What I want is to group all relations into arrays no matter how distant like this

[car, wheel, tyre, rubber]
[guitar, strings, pickup]
[bed, sheets]

什么是有效的方式使用Javascript做到这一点？

What is an efficient way to do this with Javascript?

推荐答案

首先，我将存储的关系作为数组，这样你可以有重复的钥匙。主要方法：初始字典，包括所有相关的每一个单词字;使用地图和减少递归链的扩展;基于等效过滤链。

First of all, I would store the relationships as arrays so that you can have duplicate "keys." Key methods: an initial dictionary including every word related to each individual word; a recursive chain expander using map and reduce; filtering chains based on equivalency.

Array.prototype.getUnique = function(){
   var u = {}, a = [];
   for(var i = 0, l = this.length; i < l; ++i){
      if(u.hasOwnProperty(this[i])) {
         continue;
      }
      a.push(this[i]);
      u[this[i]] = 1;
   }
   return a;
}
var links = {};
var pairs = [
    ["car", "wheel"],
    ["wheel", "tyre"],
    ["bed", "sheets"],
    ["guitar", "strings"],
    ["guitar", "pickup"],
    ["rubber", "tyre"],
    ["truck", "wheel"],
    ["pickup", "car"]
];
pairs.map(function(pair) {
    links[pair[0]] = links[pair[0]] || [];
    links[pair[1]] = links[pair[1]] || [];

    links[pair[0]].push(pair[1]);
    links[pair[1]].push(pair[0]);
});
var append = function(list) {
    var related = list.map(function(item) {
        return links[item];
    }).reduce(function(listA, listB) {
        return listA.concat(listB);
    }).filter(function(item) {
        // make sure related only includes new links
        return list.indexOf(item) == -1
    }).getUnique();

    return related.length ? append(list.concat(related)) : list.concat(related);
};
var branches = [];
for( var word in links ) {
    branches.push(append(links[word].concat(word)));
}
var compareArrays = function(listA, listB) {
    if( listA.length != listB.length ) return false;
    return listA.map(function(element) {
        if( listB.indexOf(element) == -1 ) return 0;
        return 1;
    }).filter(function(el) {
        return el == 1;
    }).length == listA.length;
};
var _branches = branches;
var chains = branches.filter(function(branch1, i) {     
    var isUnique = _branches.filter(function(branch2) {
        // are they equivalent
        return compareArrays(branch1, branch2);
    }).length == 1; 
    delete _branches[i];
    return isUnique;
});