展平嵌套的对象数组，将键重命名为迭代器

Question

我有一个对象数组，每个对象如下所示（轮询响应）：

{
"slug": "18-AZ-Gov-GE-DvF",
"name": "2018 Arizona Gubernatorial GE",
"tags": [],
"charts": [],
"election_date": "2018-11-06",
"n_polls": 1,
"created_at": "2017-06-13T13:32:26.000Z",
"responses": [
 {
 "label": "Ducey",
 "name": "Doug Ducey",
 "party": "Republican",
 "incumbent": true
 },
 {
 "label": "Farley",
 "name": "Steve Farley",
 "party": "Democrat",
 "incumbent": false
 },
 {
 "label": "Other",
 "name": "Other",
 "party": null,
 "incumbent": false
 },
 {
 "label": "Undecided",
 "name": "Undecided",
 "party": null,
  "incumbent": false
 }
]
},

我需要展平由responses键访问的数组，以便将每个对象键接到其迭代器。

最终对象应如下所示：

{
"slug": "18-AZ-Gov-GE-DvF",
"name": "2018 Arizona Gubernatorial GE",
"tags": [],
"charts": [],
"election_date": "2018-11-06",
"n_polls": 1,
"created_at": "2017-06-13T13:32:26.000Z",
 "label1": "Ducey",
 "name1": "Doug Ducey",
 "party1": "Republican",
 "incumbent1": true
 "label2": "Farley",
 "name2": "Steve Farley",
 "party2": "Democrat",
 "incumbent2": false
 "label3": "Other",
 "name3": "Other",
 "party3": null,
 "incumbent3": false
 "label4": "Undecided",
 "name4": "Undecided",
 "party4": null,
  "incumbent4": false
},

我看到的答案在展平或在集合上执行时不会重命名对象键。

我尝试了几种解决方案，但想在真正开始之前先看看是否有一种简便的es6方法。

Answer 1

在这里使用array .reduce方法可能效果很好。 由于reduce回调将接收当前索引作为第三个参数，因此您可以使用该索引创建所需的键。

例：

 const myArray = [ { "label": "Ducey", "name": "Doug Ducey", "party": "Republican", "incumbent": true }, { "label": "Farley", "name": "Steve Farley", "party": "Democrat", "incumbent": false }, { "label": "Other", "name": "Other", "party": null, "incumbent": false }, { "label": "Undecided", "name": "Undecided", "party": null, "incumbent": false } ] const flattened = myArray.reduce((flat, item, index) => ({ ...flat, ...Object.keys(item).reduce((numbered, key) => ({ ...numbered, [key + (index+1)]: item[key], }), {}), }), {}); console.log(flattened); 

Answer 2

您可以使用forEach遍历responses数组，并在对象上为每个对象中的entries分配一个新键。 之后，只需delete原始responses数组即可：

 let obj = {"slug": "18-AZ-Gov-GE-DvF","name": "2018 Arizona Gubernatorial GE","tags": [],"charts": [],"election_date": "2018-11-06","n_polls": 1,"created_at": "2017-06-13T13:32:26.000Z","responses": [ { "label": "Ducey", "name": "Doug Ducey", "party": "Republican", "incumbent": true }, { "label": "Farley", "name": "Steve Farley", "party": "Democrat", "incumbent": false }, { "label": "Other", "name": "Other", "party": null, "incumbent": false }, { "label": "Undecided", "name": "Undecided", "party": null, "incumbent": false }]} obj.responses.forEach((item, i) => { // item is one object from responses // i is the index starting at 0. Concat that on the key Object.entries(item).forEach(([k, v]) => obj[k+(i+1)] = v) }) // no need for obj.responses any more delete obj.responses console.log(obj) 

Answer 3

一种方法是使用一些较新的功能，包括spread ， destructuring assignment ， template literals ， entries以及最重要的是reduce 。

主要要点是使用化简器将responses数组转换为每个元素对象的新对象，并使用辅助化简器用迭代器计数修改键，并使用新键将值分配给外部对象。

这种方法的主要好处是原始对象（包括其子对象）没有被修改（请阅读：无副作用）。

const flattened = responses.reduce((o, g, i) => {
    Object.entries(g).reduce((t, [k, v]) => {
      t[`${k}${i + 1}`] = v;
      return t;
    }, o);

    return o;
  },
  {});

完整的工作示例：

 const orig = { "slug": "18-AZ-Gov-GE-DvF", "name": "2018 Arizona Gubernatorial GE", "tags": [], "charts": [], "election_date": "2018-11-06", "n_polls": 1, "created_at": "2017-06-13T13:32:26.000Z", "responses": [{ "label": "Ducey", "name": "Doug Ducey", "party": "Republican", "incumbent": true }, { "label": "Farley", "name": "Steve Farley", "party": "Democrat", "incumbent": false }, { "label": "Other", "name": "Other", "party": null, "incumbent": false }, { "label": "Undecided", "name": "Undecided", "party": null, "incumbent": false } ] }; const {responses, ...foo} = orig; const flattened = responses.reduce((o, g, i) => { Object.entries(g).reduce((t, [k, v]) => { t[`${k}${i + 1}`] = v; return t; }, o); return o; }, {}); console.log({flattened: {...foo, ...flattened}}); console.log({orig}); 

Answer 4

您可以分解响应和其他数据，然后使用索引键将每个响应减少到一个对象，然后通过对象散布运算符组合结果：

const { responses, ...other } = data
const indexedResponses = responses.reduce((acc, r, i) => {
  Object.entries(r).forEach(([key, value]) => {
    acc[`${key}${i + 1}`] = value
  })
  return acc
}, {})
const result = { ...other, ...indexedResponses }