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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responses键访问的数组，以便每个对象都键入它的迭代器。

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，但想在真正深入研究之前看看是否有简单的 es6 方法。

4

4 回答 4

3

您可以使用forEach遍历responses数组并在您的对象上为entries每个对象分配一个新键。之后只是delete原始responses数组：

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)

于 2018-05-09T21:11:04.910 回答
3

在这里使用数组.reduce方法可能会很好。由于reduce回调将接收当前索引作为第三个参数，因此您可以使用它来创建您要查找的键。

例子：

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);

于 2018-05-09T21:09:11.673 回答
2

一种方法是使用一些较新的好东西，包括、、、和spread最重要的。destructuring assignmenttemplate literalsentriesreduce

主要要点是使用 reducer 将responses数组转换为新对象，每个元素对象使用辅助 reducer 使用迭代器计数修改键，并使用新键将值分配给外部对象。

这种方法的一个主要好处是原始对象（包括其子对象）不会被修改（阅读：没有副作用）。

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});

于 2018-05-09T21:11:47.167 回答
1

您可以解构您的响应和其他数据，然后使用索引键将每个响应减少到一个对象，然后通过对象扩展运算符组装结果：

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 }
于 2018-05-09T21:18:51.063 回答