var s = new Set();
[2,3,5,4,5,2,2].map(x => s.add(x))
console.log(s); //Set { 2, 3, 5, 4 }

var s = new Set();
s.add(1).add(2).add(2);
// 注意2被加入了两次
console.log(s.size) // 2
console.log(s.has(1)) // true
console.log(s.has(2)) // true
console.log(s.has(3)) // false
console.log(s.delete(2));
console.log(s.has(2)) // false

var items = new Set([1, 2, 3, 4, 5]);
var arr = Array.from(items);
//运用: 去除数组中重复元素的方法
var array = [1,2,3,2,3,4];
function fun(array) {
  return Array.from(new Set(array));
}
console.log(fun(array));//[ 1, 2, 3, 4 ]

var s = new Set([1, 2, 3, 4, 5]);
console.log(s.values());//SetIterator { 1, 2, 3, 4, 5 }
//Set结构的默认遍历器就是它的values方法
console.log(Set.prototype[Symbol.iterator] === Set.prototype.values)//true
//所以遍历可以直接使用 for...of
for (let x of s) {
  console.log(x);
}
//由于扩展运算符（...）内部使用for...of循环，将Set转化为数组。
var arr = [...s];
console.log(arr);//[ 1, 2, 3, 4, 5 ]

var set = new Set([1, 2, 3]);
set.forEach(function(value,key){
  console.log(value);
});

let set = new Set(['red', 'green', 'blue']);
for ( let item of set.keys() ){
  console.log(item);
}
// red
// green
// blue
for ( let [key, value] of set.entries() ){
  console.log(key, value);
}
// red, red
// green, green
// blue, blue

var set = new Set([1, 2, 3]);
set = new Set([...set].map(x => x * 2));
console.log(set);//返回Set结构：{2, 4, 6}
var set = new Set([1, 2, 3, 4, 5]);
set = new Set([...set].filter(x => (x % 2) == 0));
console.log(set);// 返回Set结构：{2, 4}

var map = new Map();
var map1 = new Map([["name", "张三"], ["title", "Author"]]);
console.log(map1.size)//2

var map = new Map();
map.set(['a'], 555);
console.log(map.get(['a'])) // undefined
var k1 = ['a'];
//虽然值相同，但这是一个新的值，内存地址不一样,new 的
var k2 = ['a'];
map.set(k1, 111);
map.set(k2, 222);

console.log(Map[Symbol.iterator] === Map.entries)//true
let map = new Map([[1, 'one'], [2, 'two'], [3, 'three']]);
console.log([...map.keys()]);//[ 1, 2, 3 ]
console.log([...map.values()]);//[ 'one', 'two', 'three' ]
console.log([...map.entries()]);//[ [ 1, 'one' ], [ 2, 'two' ], [ 3, 'three' ] ]
console.log([...map]);[ [ 1, 'one' ], [ 2, 'two' ], [ 3, 'three' ] ]