对象数组优化

Question

下面我有一组从服务器接收的数据。

我需要从数据中找到哪个“id”连接到哪个“apip”

我有下面的工作代码。

我的问题是......我是否以最有效的方式做到了这一点？

我的策略如下：

第 1 步 - 捕获所有的 apip 将它们保存到一个数组中

第 2 步 - 清除所有重复的 apip

第 3 步 - for 循环比较 apip 是否匹配 apip

- 如果真的捕获该 ID

-else 将该 ID 存储在 apip 2 上

（我知道我只有 2 个可能的 APIP）

第 4 步 - 清除任何重复的“id”

这是最好的方法吗？

Answer 1

要找到uniqs 中的一个（我们称之为 s oneApip），只需检查s的第一个元素data- 您不需要 Set 或迭代所有这些元素。然后迭代dataand，对于每个元素，添加到一个或另一个集合，具体取决于它是否apip匹配oneApip：var data=[{time:"1571965891.8420029",rssi:"30",id:"123456789",apip:"172.172.172.172.1"},{time:"1571971066.8283374",rssi:"30",id:"100",apip:"172.172.172.172.2"},{time:"1571965476.4821894",rssi:"30",id:"123456789",apip:"172.172.172.172.2"},{time:"1571965894.140705",rssi:"30",id:"123456789",apip:"172.172.172.172.2"},{time:"1571965893.1654441",rssi:"30",id:"123456789",apip:"172.172.172.172.1"},{time:"1571970952.7499905",rssi:"30",id:"9999999",apip:"172.172.172.172.1"},{time:"1571965888.1338017",rssi:"30",id:"123456789",apip:"172.172.172.172.2"},{time:"1571970925.342063",rssi:"30",id:"66666",apip:"172.172.172.172.1"},{time:"1571965890.158157",rssi:"30",id:"123456789",apip:"172.172.172.172.1"}];const oneApip = data[0].apip;const set1 = new Set();const set2 = new Set();for (const { apip, id } of data) {  (apip === oneApip ? set1 : set2).add(id);}const uniq1 = [...set1];const uniq2 = [...set2];console.log(uniq1);console.log(uniq2);更一般地，对于任意数量的apips，创建一个由 索引的对象apip，其值为集合：var data=[{time:"1571965891.8420029",rssi:"30",id:"123456789",apip:"172.172.172.172.1"},{time:"1571971066.8283374",rssi:"30",id:"100",apip:"172.172.172.172.2"},{time:"1571965476.4821894",rssi:"30",id:"123456789",apip:"172.172.172.172.2"},{time:"1571965894.140705",rssi:"30",id:"123456789",apip:"172.172.172.172.2"},{time:"1571965893.1654441",rssi:"30",id:"123456789",apip:"172.172.172.172.1"},{time:"1571970952.7499905",rssi:"30",id:"9999999",apip:"172.172.172.172.1"},{time:"1571965888.1338017",rssi:"30",id:"123456789",apip:"172.172.172.172.2"},{time:"1571970925.342063",rssi:"30",id:"66666",apip:"172.172.172.172.1"},{time:"1571965890.158157",rssi:"30",id:"123456789",apip:"172.172.172.172.1"}];const grouped = {};for (const { apip, id } of data) {  if (!grouped[apip]) {    grouped[apip] = new Set();  }  grouped[apip].add(id);}const arrs = Object.entries(grouped)  .map(([apip, set]) => [apip, [...set]]);console.log(arrs);

Answer 2

你可以做：const data = [{"time": "1571965891.8420029","rssi": "30","id": "123456789","apip": "172.172.172.172.1"}, {"time": "1571971066.8283374","rssi": "30","id": "100","apip": "172.172.172.172.2"}, {"time": "1571965476.4821894","rssi": "30","id": "123456789","apip": "172.172.172.172.2"}, {"time": "1571965894.140705","rssi": "30","id": "123456789","apip": "172.172.172.172.2"}, {"time": "1571965893.1654441","rssi": "30","id": "123456789","apip": "172.172.172.172.1"}, {"time": "1571970952.7499905","rssi": "30","id": "9999999","apip": "172.172.172.172.1"}, {"time": "1571965888.1338017","rssi": "30","id": "123456789","apip": "172.172.172.172.2"}, {"time": "1571970925.342063","rssi": "30","id": "66666","apip": "172.172.172.172.1"}, {"time": "1571965890.158157","rssi": "30","id": "123456789","apip": "172.172.172.172.1"}]const sets = data.reduce(    (a, { apip, id }, _, arr) => ((apip === arr[0].apip ? a.one : a.two).add(id), a),    {       one: new Set(),       two: new Set()     }  ),  result = {    uniqDataFromAP1: [...sets.one],    uniqDataFromAP2: [...sets.two]  }console.log(result).as-console-wrapper {max-height: 100% !important;top: 0;}