let aa = new Set([1,2,true],[3,6],0); aa.add(5) aa.add(20) aa.delete(9) aa.delete(1) // add 增加\delete 删除\ clear 清空 //通过new Set就能创建一个元素不重复的set数组(自动去重)。[...aa]就能将set数组转化成数组。
aa.forEach(ele => { console.log(ele) })
for(let prop of aa){ console.log(prop) }
//怎么将set的对象转换成数组,Array.from()和... // Array.from(aa) // [...aa] //Array.from('abc')将'abc'转化成['a','b','c']
let arr1 = new Set([1,21,1,3,32]); let arr2 = new Set([2,3,51,2,21,3]);
let arrsum = [...arr1,...arr2] //并集 let arrmin = [...arr1].filter(ele => arr2.has(ele)) 交集
let arrChaji1 = [...arr1].filter(ele => !arr2.has(ele)) let arrChaji2 = [...arr2].filter(ele => !arr1.has(ele)) let arrChaji = [...arrChaji1,...arrChaji2] //差集就是,a数组在b数组里面没有的,b数组在a数组里面没有的,再合集。