Question

的javascript

我正在尝试在同一索引处合并2个多维数组，或者将两个数组的相同索引随机化。

var arr1 = [[a, b, c], [d, e], [f, g, h, i]]
var arr2 = [[1, 2, 3], [5, 6], [7, 8, 9, 10]]

preferredResult = [{a: 1, b: 2, c: 3}, {d: 5, e:6}, {f: 7, g: 8, h: 9, i: 10}]

我尝试过.maps，嵌套for循环，.push在不同的版本中，并且无法解决这个问题。

或者，如果我能弄清楚如何以相同的方式随机化两个数组数组，那也可以，即：arr1 [0]中的字母和arr2 [0]中的数字可以设置为相同的随机化，然后是arr1 [1]＆amp; arr2 [1]等等。

Answer 1

function merge(arr1, arr2) {
   return arr1.map(function (arr, i) {
      return mergeIntoObject(arr1[i], arr2[i])); 
   };
}

function mergeIntoObject(arr1, arr2) {
  var result = {};
  arr1.forEach((arr, i) => {
    result[arr1[i]] = arr2[i];
  });
  return result;
}

merge(arr1, arr2);

Answer 2

使用嵌入式for循环

＆＃13;

var arr1 = [['a', 'b', 'c'], ['d', 'e'], ['f', 'g', 'h', 'i']]
var arr2 = [[1, 2, 3], [5, 6], [7, 8, 9, 10]]

var result = []
for (let i = 0; i < arr1.length; i++){
    let obj = {};
    for (let j = 0; j < arr1[i].length; j++){
        obj[arr1[i][j]]=arr2[i][j]
    }
    result.push(obj);
}
console.log(result)

＆＃13;

Answer 3

您可以使用array#map和array#reduce

const arr1 = [['a', 'b', 'c'], ['d','e' ], ['f', 'g', 'h', 'i']],
      arr2 = [[1, 2, 3], [5, 6], [7, 8, 9, 10]],
      result = arr1.map(function(a, i){
        return a.reduce(function(r, v, j){
          r[v] = arr2[i][j];
          return r;
        }, {});
      });
      
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

Lodash可以提供超级帮助。

var arr1 = [['a', 'b', 'c'], ['d', 'e'], ['f', 'g', 'h', 'i']]
var arr2 = [[1, 2, 3], [5, 6], [7, 8, 9, 10]]

var out = _.zipWith(arr1, arr2, (x,y)=>_.fromPairs(_.zip(x,y)));
console.log(out)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>

zip将两个数组合并成一对，zipWith也会这样做，但是让你选择如何组合它们。

Answer 5

使用public final class ImmutableButton { public final String text; public ImmutableButton(String text) { this.text = text; } protected void onClick() { // notify listeners somehow, hoping they haven't changed } } public final class ImmutableWindow { public final ImmutableButton button; public ImmutableWindow(ImmutableButton button) { this.button = button; } protected void listenForButtonClick() { // somehow register with button and receive events, despite this object // being entirely recreated whenever a field changes } }＆amp; map方法

forEach

在同一索引处合并2个多维数组

5 个答案: