Question

我有两个要合并的数组。

var arrayOne = [{id:1},{id:2}];
var arrayTwo = [{age: 20}, {age: 25}, {age: 22}];

合并数组应该是这样的

[{id:1, age: 20}, {id:2, age: 25}, {age: 22}]

我从下划线库中尝试过_.zip方法，但它只合并数组而不是对象数组。

由于

Answer 1

var merged = arrayTwo.map((o, i) => Object.assign({}, o, (arrayOne[i] || {})));

Answer 2

更普遍的事情：

function merge(...arrays){
  var length = arrays.reduce((len, arr) => Math.max(len, arr.length), 0);
  var assign = i => (into, arr) => Object.assign(into, arr[i]);
  return Array.from({ length }, (v,i) => arrays.reduce(assign(i), {}));
}

var arrayOne = [{id:1},{id:2}];
var arrayTwo = [{age: 20}, {age: 25}, {age: 22}];

merge(arrayOne, arrayTwo /*or as many as you want*/ );

Answer 3

ES6方法：

int n = fscanf(binfile, "%s %s %u %f", firstName, lastName, &id, &gpa);

请注意，长度可能不同，顺序无关紧要（您可以在之前或之后传递较大的数组，结果相同）。

Answer 4

Enter the lenght of the array: 6
Enter value number 1: -4
Enter value number 2: 3
Enter value number 3: -9
Enter value number 4: 0
Enter value number 5: 4
Enter value number 6: 1
Array entered:  [-4 3 -9 0 4 1]
There are 3 positive number(s) of 6 in the array. Fraction: 0.500000
There are 2 negative number(s) of 6 in the array. Fraction: 0.333333
There are 1 zero(s) of 6 in the array. Fraction: 0.166667

或在ES6中：

function zipObjects() {
    return _.zip.apply(_, arguments).map(
        function(arr) {
            return _.extend.apply(_, [{}].concat(arr));
        }
    );
}

Answer 5

具有对象传播道具的生成器的良好用例：

function* zip(...args) {
  const len = Math.max(...args.map(a => a.length));
  for (let i = 0; i < len; i++) {
      yield Object.assign({}, ...args.map(a => a[i]));
  }
}

这将产生一个迭代器，而不是一个数组，所以你可以得到一个简单Array.from或数组扩展的数组结果，如此

var arrayOne = [{id:1},{id:2}];
var arrayTwo = [{age: 20}, {age: 25}, {age: 22}];

console.log([...zip(arrayTwo, arrayOne)]);

编辑：如果你不喜欢屈服，你可以从一个新的阵列映射到最长的长度。适用于任何数组排序和任意数量的对象数组。

const zipAr = (...args) => Array
.from({ length: Math.max(...args.map(a => a.length)) })
.map((_, i) => Object.assign({}, ...args.map(a => a[i])));

Answer 6

Object.assign（）将是合并对象的关键方法。

示例：

var arrayOne = [{id:1},{id:2}]; var arrayTwo = [{age: 20}, {age: 25}, {age: 22}]; //Lets assume that we want to update arrayTwo .. for(var i = 0 ; i < arrayOne.length; i++){ arrayTwo[i] = Object.assign(arrayTwo[i] , arrayOne[i]) } console.log(arrayTwo);

希望这会有所帮助..

Answer 7

使用array.map()＆amp; Object.assign()方法。

<强>样本

＆＃13;

var arrayOne = [{id:1},{id:2}];
var arrayTwo = [{age: 20}, {age: 25}, {age: 22}];

var res = arrayTwo.map(function(item, index) { return Object.assign(item, arrayOne[index]);
});

console.log(res);

＆＃13;

Answer 8

function compareMyArrays(arrayOne, arrayTwo){
  var newArray = [];
  var maxCounter = (arrayOne.lenght > arrayTwo.length) ? arrayOne.length : arrayTwo.length;
  for(var i = 0; i < arrayOne.length; i++) {
    var value1 = null;
    if(arrayOne.length-1 <= i){
      value1 = arrayOne[i];
    }
    var value2 = null;
      if(arrayOne.length-1 <= i){
      value1 = arrayOne[i];
    }
    if(!!value1 && !! value2){
      newArray[i] = {value1, value2}
    } else if(!!value1) {
      newArray[i] = {value1}
    } else if(!!value2) {
      newArray[i] = {value2}
    }
  }
  return newArray;
}

确切地返回您要查找的内容，但仅在两个数组具有相同长度时才起作用

编辑：现在也可以使用不同长度的数组

合并两个对象数组及其属性

8 个答案: