Question

我有一个对象和2个数组。我想根据object来更改arr2键，对象键和arr1的值是相同的，而数组中值的arr1和arr2的位置是相同的，但不相同重视自己，而我不想问如何解决它，想知道我应该如何解决这个问题，谢谢您的时间

    const object = {
        name1: "some value",
        name3: "some value",
        name2: "some value",
        etc...}
    const arr1 = ["name1", "name2", "name3"]
    const arr2 = ["Name1", "Name2", "Name3"]

预期结果

    const newObject = {
        Name1: "some value",
        Name3: "some value",
        Name2: "some value",
        etc...}

Answer 1

我假设您正在按arr1和arr2中的位置映射它们。只需运行一个循环。

 const oldObject = {
        name1: "some value",
        name3: "some value",
        name2: "some value",
};

const arr1 = ["name1", "name2", "name3"];
const arr2 = ["Name1", "Name2", "Name3"];

var newObj = {};

for (var i = 0; i < arr2.length; i++) {
  newObj[arr2[i]] = oldObject[arr1[i]];
}

console.log(newObj);

Answer 2

您可以使用reduce()

const object = {
        name1: "some value",
        name3: "some value",
        name2: "some value"
}

const arr1 = ["name1", "name2", "name3"]
const arr2 = ["Name1", "Name2", "Name3"]

const res = arr1.reduce((ac,a,i) => (ac[arr2[i]] = object[a],ac),{});
console.log(res)

Answer 3

您可以使用reduce：

const object = {
  name1: "some value",
  name3: "some value",
  name2: "some value",
}
const arr1 = ["name1", "name2", "name3"]
const arr2 = ["Name1", "Name2", "Name3"]

const newObj = arr1.reduce((acc, value, index) => {
   acc[arr2[index]] = object[value]
   return acc
}, {})

console.log(newObj)

Reduce将遍历数组1，使用第二个数组中的相同索引来分配键，然后将第一个数组中的值与对象中的键进行比较，然后将其分配给新对象。

Answer 4

您可以简单地执行以下操作：

var output={};
arr1.forEach(function(el, index){
   output[arr2[index]] = object[el];
}, this);

Answer 5

假设您要保留新对象上arr1中不存在的键名。然后，您可以先在旧名称和新名称之间创建一个Map（如果ES6不是您的选择，请在此处使用对象用于映射）。然后，您可以使用Object.entries()遍历Array.reduce()来生成新对象，其中仅将相关的键名映射到新的键名，其余的保留。

const obj = {
  name1: "some value 1",
  name3: "some value 2",
  name2: "some value 3",
  name4: "some value 4",
  name5: "some value 5"
}

const arr1 = ["name1", "name2", "name3"];
const arr2 = ["Name1", "Name2", "Name3"];

// Create a map between old names and new names.
let keysMap = new Map(arr1.map((key, idx) => [key, arr2[idx]]));

// Generate the new object, preserving non-mapped keys names.
let newObj = Object.entries(obj).reduce((acc, [k, v]) =>
{
    acc[keysMap.has(k) ? keysMap.get(k) : k] = v;
    return acc;
}, {});

console.log(newObj);

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Answer 6

您可以先通过Array.map然后通过Object.fromEntries一行：

  const object = {
    name1: "some value A",
    name3: "some value B",
    name2: "some value C"
  }
  const old = ["name1", "name2", "name3"]
  const keys = ["Name1", "Name2", "Name3"]

let result = Object.fromEntries(old.map((x,i) => [keys[i], object[x]]))

console.log(result)

根据数组更改对象键

6 个答案: