对象数组求差集，交集，并集

数组

ES7新增了一个Array.prototype.includes的数组方法，用于返回一个数组是否包含指定元素，结合filter方法。

a = [1, 2, 3]，b = [2, 4, 5]
// 并集
let union = a.concat(b.filter(v => !a.includes(v))) // [1,2,3,4,5]
// 交集
let intersection = a.filter(v => b.includes(v)) // [2]
// 差集
let difference = a.concat(b).filter(v => !a.includes(v) || !b.includes(v)) // [1,3,4,5]

ES6中新增的一个Array.from方法，用于将类数组对象和可遍历对象转化为数组。只要类数组有length长度，基本都可以转化为数组。结合Set结构实现数学集求解。

let aSet = new Set(a)
let bSet = new Set(b)
// 并集
let union = Array.from(new Set(a.concat(b))) // [1,2,3,4,5]
// 交集
let intersection = Array.from(new Set(a.filter(v => bSet.has(v)))) // [2]
// 差集
let difference = Array.from(new Set(a.concat(b).filter(v => !aSet.has(v) || !bSet.has(v)))) // [1,3,4,5]

对象数组

求交集

var arr1 = [{name:'name1',id:1},{name:'name2',id:2},{name:'name3',id:3}];
var arr1Id = [1,2,3]
var arr2 = [{name:'name1',id:1},{name:'name2',id:2},{name:'name3',id:3},{name:'name4',id:4},{name:'name5',id:5}];
var result = arr2.filter(function(v){
      return arr1Id.indexOf(v.id)!==-1 // 利用filter方法来遍历是否有相同的元素
})

求并集
      let arr1 = [{name:'name1',id:1},{name:'name2',id:2},{name:'name3',id:3}];
      let arr2 = [{name:'name1',id:1},{name:'name4',id:4},{name:'name5',id:5}];
      let arr3 = arr1.concat(arr2);
      let result = [];
      var obj = [];
      result = arr3.reduce(function(prev, cur, index, arr) {
        console.log(prev, cur);
        obj[cur.id] ? '' : obj[cur.id] = true && prev.push(cur);
        return prev;
      }, []);

求差集　　　
   let arr1 = [{name:'name1',id:1},{name:'name2',id:2},{name:'name3',id:3}];
   let arr1Id = [1,2,3];
   let arr2 = [{name:'name1',id:1},{name:'name4',id:4},{name:'name5',id:5}];
   let arr2Id = [1,4,5];
   let arr3 = arr1.concat(arr2);
   let result = arr3.filter(function(v){
       return arr1Id.indexOf(v.id)===-1 || (arr2Id.indexOf(v.id)===-1)
   })