合并两个对象并将相同键的值相加的最佳方法是什么？

Question

我有两个与此类似的对象：

let data = [[
  {
    10: {key1: 1, key2: 2, key3: 3},
    11: {key1: 1, key2: 2, key3: 3},
    12: {key1: 1, key2: 2, key3: 3},
    },
  {},
  {}
],
[
  {
    10: {key1: 1, key2: 2, key3: 3},
    11: {key1: 1, key2: 2, key3: 3},
    12: {key1: 1, key2: 2, key3: 3},
    },
  {},
  {}
]]

我想创建一个新对象并对每个键的每个值求和。 像这样：

let newData = [
      {
        10: {key1: 2, key2: 4, key3: 6},
        11: {key1: 2, key2: 4, key3: 6},
        12: {key1: 2, key2: 4, key3: 6},
        },
      {},
      {}
    ]

每个对象都有三个对象。 在这三个对象中，有45个键，每个键都有一个具有三个键/值的对象。

我找不到任何好的解决方案来总结每一个价值。

我目前的解决方案是我首先循环data ：

let count = 0;
let count2 = 0;
for(let ob of data){
     for(let child in ob[0]){
          keyOne = ob[0][child].keyOne;
          keyTwo = ob[0][child].keyTwo;
          keyThree = ob[0][child].keyThree;
          keyFour = ob[0][child].keyFour;
          if(count < 45) {
               newData[0][count].keyOne += keyOne;
               newData[0][count].keyTwo += gkeyTwop;
               newData[0][count].keyThree += keyThree;
               newData[0][count].keyFour += keyFour;
          } else {
               newData[0][count2].keyOne += keyOne;
               newData[0][count2].keyTwo += keyTwo;
               newData[0][count2].keyThree += keyThree;
               newData[0][count2].keyFour += keyFour;
               count2++;

          }
          count++;
     }

在最坏的情况下，数据有三个对象，每个对象有45个密钥。 然后我必须这样做三次。 这看起来真的很糟糕。 肯定有更好的办法。 提示？

Answer 1

您可以使用Array#reduce并在添加值后映射内部数组的值。

 var data = [[{ 10: { key1: 1, key2: 2, key3: 3 }, 11: { key1: 1, key2: 2, key3: 3 }, 12: { key1: 1, key2: 2, key3: 3 }, }, {}, {}], [{ 10: { key1: 1, key2: 2, key3: 3 }, 11: { key1: 1, key2: 2, key3: 3 }, 12: { key1: 1, key2: 2, key3: 3 }, }, {}, {}]], result = data.reduce((a, b) => a.map((v, i) => (Object.keys(b[i]).forEach(k => Object.keys(b[i][k]).forEach(l => v[k][l] = (v[k][l] || 0) + b[i][k][l])) , v) ) ); console.log(result); 

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

Answer 2

这是使用reduce()和少量forEach()循环的解决方案，因为您还需要循环嵌套对象。

 var data = [[{"10":{"key1":1,"key2":2,"key3":3},"11":{"key1":1,"key2":2,"key3":3},"12":{"key1":1,"key2":2,"key3":3}},{},{}],[{"10":{"key1":1,"key2":2,"key3":3},"11":{"key1":1,"key2":2,"key3":3},"12":{"key1":1,"key2":2,"key3":3}},{},{}]] var result = data.reduce(function(r, e, i) { if (i == 0) r = r.concat(e) else { e.forEach(function(a, j) { Object.keys(a).forEach(function(key) { if (!r[j][key]) r[j][key] = a[key] else { Object.keys(a[key]).forEach(function(k) { r[j][key][k] += a[key][k] }) } }) }) } return r; }, []) console.log(result)