用另一个对象的索引填充一个对象答案

【问题标题】：Fill an object with indexes from another object用另一个对象的索引填充一个对象
【发布时间】：2017-08-02 11:34:05
【问题描述】：

我有两个对象。一个是单词数组，另一个给出第一个数组中两个单词之间的关系：

nodes = ["apple","cherry","pear","strawberry"]

data = [{word1 : "apple", word2 : "cherry" , weight : "0.1"},
        {word1 : "apple", word2 : "strawberry" , weight : "0.2"},
        {word1 : "cherry", word2 : "pear" , weight : "0.3"},
        {word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
        {word1 : "strawberry", word2 : "pear" , weight : "0.5"}]

我想创建一个具有数据结构的新对象，而不是单词，而是它们的索引。像这样：

links =  [{source : "0", target : "1" , value : "0.1"},
          {source : "0", target : "3" , value : "0.2"},
          {source : "1", target : "2" , value : "0.3"},
          {source : "1", target : "3" , value : "0.4"},
          {source : "3", target : "2" , value : "0.5"}]

到目前为止，我有这个：

for (var i = 0 ; i < nodes.length; i++) {
     test[i] = nodes.findIndex(d => d == nodes[i]);
}

这给了我这个新数组，对应于来自nodes 的索引：

test = [0,1,2,3]

那么我现在该如何将这些索引分配给links 中的属性？

【问题讨论】：

标签： javascript arrays d3.js

【解决方案1】：

尝试：

var nodes = ["apple","cherry","pear","strawberry"];

var data = [{word1 : "apple", word2 : "cherry" , weight : "0.1"},
        {word1 : "apple", word2 : "strawberry" , weight : "0.2"},
        {word1 : "cherry", word2 : "pear" , weight : "0.3"},
        {word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
        {word1 : "strawberry", word2 : "pear" , weight : "0.5"}];

var links = data.map((item) => {
    return {
        source: nodes.indexOf(item.word1),
        target: nodes.indexOf(item.word2),
        value: item.weight
        
    };
});

console.log(links);
console.log(data);

【讨论】：

【解决方案2】：

您可以在第二个数组（而不是第一个）上使用Array#map 和Object.assign 以避免原始数据的突变。

为了降低时间复杂度，可以创建一个Map，它允许在恒定时间内获取索引。该 Map 可以将this 传递给 map 方法。通过检查每个属性的匹配，您可以使其具有通用性，因此算法本身没有硬编码word1 或word2：

const nodes = ["apple","cherry","pear","strawberry"]

const data = [
    {word1 : "apple", word2 : "cherry" , weight : "0.1"},
    {word1 : "apple", word2 : "strawberry" , weight : "0.2"},
    {word1 : "cherry", word2 : "pear" , weight : "0.3"},
    {word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
    {word1 : "strawberry", word2 : "pear" , weight : "0.5"}
];
   
const result = data.map(function(o) {
    return Object.assign(...Object.keys(o).map(key => ({
        [key]: this.has(o[key]) ? this.get(o[key]) : o[key]
    })));
}, new Map(nodes.map( (s,i) => [s,i] ))); 

console.log(result);

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【讨论】：

【解决方案3】：

这可行：

nodes = ["apple", "cherry", "pear", "strawberry"]

data = [{
    word1: "apple",
    word2: "cherry",
    weight: "0.1"
  },
  {
    word1: "apple",
    word2: "strawberry",
    weight: "0.2"
  },
  {
    word1: "cherry",
    word2: "pear",
    weight: "0.3"
  },
  {
    word1: "cherry",
    word2: "strawberry",
    weight: "0.4"
  },
  {
    word1: "strawberry",
    word2: "pear",
    weight: "0.5"
  }
]

var links = [];

for (var i = 0; i < data.length; i++) {
  links.push({
    source: nodes.indexOf(data[i].word1),
    target: nodes.indexOf(data[i].word2),
    value: data[i].weight
  })
}

console.log(links);

【讨论】：

【解决方案4】：

使用Array#map() 创建一个新数组。获取indexOf(word1)和indexOf(word2)替换当前值

const nodes = ["apple","cherry","pear","strawberry"];

const data = [{word1 : "apple", word2 : "cherry" , weight : "0.1"},
        {word1 : "apple", word2 : "strawberry" , weight : "0.2"},
        {word1 : "cherry", word2 : "pear" , weight : "0.3"},
        {word1 : "cherry", word2 : "strawberry" , weight : "0.4"},
        {word1 : "strawberry", word2 : "pear" , weight : "0.5"}];
        
const updatedData = data.map((item) => ({
  source: nodes.indexOf(item.word1),
  target: nodes.indexOf(item.word2),
  weight: item.weight
}));

console.log(updatedData);

【讨论】：