我想根据其ID对两个数组的对象进行联合 Download


假设我有两个数组

let array1 = [{id: 1,name: "a"},{id: 2,name: "b"}]
let array2 = [{id: 1,name: 'c'},{id: 3,name: 'd'}]

我希望得到的数组像这样

let result = [{id: 1,name: 'c'},{id: 2,name: 'b'},{id: 3,name: 'd'}]

因此，如果第二个数组有一个对象，而第一个数组也有一个具有相同ID的对象，则将第一个对象替换为第二个数组对象。 目前，我尝试使用以下代码，但是它会检查每个值是否匹配，并且我只想基于id进行匹配

const uniqueArray = this.servicesdata.filter((item, index) => {
        const _item = JSON.stringify(item);
        return (
          index ===
          this.servicesdata.findIndex(obj => {
            return JSON.stringify(obj) === _item;
          })
        );
      });
      console.log(uniqueArray);

2 个答案:

答案 0 :(得分：0)

这是一种O(NlogN)的方法。

let array1 = [{id: 1,name: "a"},{id: 2,name: "b"}] // Let's say its size is N1
let array2 = [{id: 1,name: 'c'},{id: 3,name: 'd'}] // Let's say its size is N2
let duplicate  = new Set() // A set to check for duplicate elements
let result = array2
array2.forEach((item)=>{ // O(N2logN2)
    duplicate.add(item.id);
})
array1.forEach((item)=>{ // O(N1logN2)
    // If `duplicate` does NOT have `item.id`, that means it's unique in array1, so add it to result
    // Otherwise, skip it because for duplicates, we want value from array2
    if(!duplicate.has(item.id))
        result.push(item);
})
// Overall complexity of approach - O(N2logN2) + O(N1logN2) ==> O(NlogN)
console.log(result);

输出：

[ { id: 1, name: 'c' },
  { id: 3, name: 'd' },
  { id: 2, name: 'b' } ]

答案 1 :(得分：0)

使用lodash，您可以使用unionBy：

let array1 = [{id: 1,name: "a"},{id: 2,name: "b"}]
let array2 = [{id: 1,name: 'c'},{id: 3,name: 'd'}]
console.log(_.unionBy(array2, array1, 'id'))
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.min.js"></script>



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