let set1 = new Set(['a','b','c','d','e','f']);
let set2 = new Set(['d','e','f','g','h','i']);

//并集
let union = new Set([...set1,...set2]);
//[a,b,c,d,e,f,g,h,i]
//交集
let intersect = new Set([...set1].filter(x => set2.has(x)));
//[d,e,f]
//差集
let dfrcset = new Set([...set1].filter(x => !set2.has(x)));

最近在網上看到，不知道我的總是報錯
在進行差值的校驗
自己總結了一個：
/**

• arr1數(shù)組1
  *arr2數(shù)組2
  **/
  getDifference(arr1,arr2){
  return arr1.concat(arr2).filter((v,i ,arr)={
  return arr.indexOf(v)===arr.lastIndexOf(v)
  })
  進行調用
  this.getDifference(arr1,arr2)

let a=[1,2,3,4,5];

let b=[1,2,3,4,5,6,7,8];

一直接用filter 和contact

//交集
let c=a.filter(res=>{return b.indexOf(res)>-1})
//差集
let d=a.filter(res=>{return b.indexOf(res)==-1})
//補集
let e=a.fliter(res=>{return !(b.indexOf(res)>-1)}).contact(b.indexOf(res=>{return !(a.indexOf(res)>-1)}))
//并集
let d=a.contact(b.filter(res=>{return !(a.indexOf(res)>-1)}))

在遍歷中操作數(shù)組

let set=new Set([1,2,3]);
set =new Set([...set].map(val=>val*2)) //2,4,6
或者是
let a=new Set([1,2,3]);
a=new Set(Array.from(a, val=>val *2)) // 值是2,4,6

二.對 Array 進行擴展

//數(shù)組功能擴展
//數(shù)組迭代函數(shù)
Array.prototype.each = function(fn){
  fn = fn || Function.K;
   var a = [];
   var args = Array.prototype.slice.call(arguments, 1);
   for(var i = 0; i < this.length; i++){
       var res = fn.apply(this,[this[i],i].concat(args));
       if(res != null) a.push(res);
   }
   return a;
};
 
//數(shù)組是否包含指定元素
Array.prototype.contains = function(suArr){
  for(var i = 0; i < this.length; i ++){
      if(this[i] == suArr){
          return true;
      }
   }
   return false;
}
 
//不重復元素構成的數(shù)組
Array.prototype.uniquelize = function(){
   var ra = new Array();
   for(var i = 0; i < this.length; i ++){
      if(!ra.contains(this[i])){
          ra.push(this[i]);
      }
   }
   return ra;
};
 
//兩個數(shù)組的交集
Array.intersect = function(a, b){
   return a.uniquelize().each(function(o){return b.contains(o) ? o : null});
};
 
//兩個數(shù)組的差集
Array.minus = function(a, b){
   return a.uniquelize().each(function(o){return b.contains(o) ? null : o});
};
 
//兩個數(shù)組的補集
Array.complement = function(a, b){
   return Array.minus(Array.union(a, b),Array.intersect(a, b));
};
 
//兩個數(shù)組并集
Array.union = function(a, b){
   return a.concat(b).uniquelize();
};

樣例：

ar a = [1,2,3,4,5]
var b = [2,4,6,8,10]
console.log("數(shù)組a：", a);
console.log("數(shù)組b：", b);
console.log("a與b的交集：", Array.intersect(a, b));
console.log("a與b的差集：", Array.minus(a, b));
console.log("a與b的補集：", Array.complement(a, b));
console.log("a與b的并集：", Array.union(a, b));

三：es6

var a = [1,2,3,4,5]
var b = [2,4,6,8,10]
console.log("數(shù)組a：", a);
console.log("數(shù)組b：", b);
 
var sa = new Set(a);
var sb = new Set(b);
 
// 交集
let intersect = a.filter(x => sb.has(x));
 
// 差集
let minus = a.filter(x => !sb.has(x));
 
// 補集
let complement  = [...a.filter(x => !sb.has(x)), ...b.filter(x => !sa.has(x))];
 
// 并集
let unionSet = Array.from(new Set([...a, ...b]));
 
console.log("a與b的交集：", intersect);
console.log("a與b的差集：", minus);
console.log("a與b的補集：", complement);
console.log("a與b的并集：", unionSet);

四：利用jq

var a = [1,2,3,4,5]
var b = [2,4,6,8,10]
console.log("數(shù)組a：", a);
console.log("數(shù)組b：", b);
 
var sa = new Set(a);
var sb = new Set(b);
 
// 交集
let intersect = a.filter(x => sb.has(x));
 
// 差集
let minus = a.filter(x => !sb.has(x));
 
// 補集
let complement  = [...a.filter(x => !sb.has(x)), ...b.filter(x => !sa.has(x))];
 
// 并集
let unionSet = Array.from(new Set([...a, ...b]));
 
console.log("a與b的交集：", intersect);
console.log("a與b的差集：", minus);
console.log("a與b的補集：", complement);
console.log("a與b的并集：", unionSet);

部分轉自[hangge] (http://www.hangge.com/blog/cache/detail_1862.html)