js数组去重

判断是否为数组

//判断是否为数组
    let arr = [1,2]
    //es6
    console.log(Array.isArray(arr)) //true

    console.log(typeof arr === Array) //false
    console.log(arr instanceof Array) //true
    console.log(arr.constructor === Array) //true
    //推荐
    console.log(Object.prototype.toString.call(arr) === '[object Array]') //true
    console.log(arr.__proto__ === Array.prototype)


    let fn = function(){}
    console.log(fn.constructor === Function) //true

数组去重

let arr2 = [1,3,2,2,1,4,5,6,2,4,1,9,8]
    //console.log('++'+arr2.indexOf(1))
    //数组去重 method1
    let myset = new Set(arr2)
    console.log(myset)
    console.log(myset.has(2))
    myset.add(99)
    //set的遍历
    for(let i of myset) {
       console.log(i)
    }
    //直接使用，返回数组推荐
    let arr2_filter = [...new Set(arr2)]
    console.log(arr2_filter)
    
    //同样可以用来，去除重复字符串中的字符
    console.log([...new Set('abcadcfpk')].join(''))

    //也可以使用Arrary.from 返回数组   dedupe()
    console.log(Array.from(new Set(arr2)))
    
   
    //去重method2  includes
    function debupe(arr2){
      let res = []
      for(let i = 0; i < arr2.length; i++) {
      
      if(!res.includes(arr2[i])) {
         res.push(arr2[i])
       }
     }
      return res
    }
    console.log('原始方法')
    console.log(debupe(arr2))

    function debupe1(arr) {
      return arr.filter((item, index, arr) => {
         return arr.indexOf(item) === index
      })
    }
    console.log('filter-indexOf')
    console.log(debupe1(arr2))
    
    //去重 method3   filter
    function debupe2(arr) {
      let myset = new Set()
      return arr.filter(res => !myset.has(res)&&myset.add(res))
      //return [...myset]
    }
    console.log('filter')
    console.log(debupe2(arr2))

   function debupe3(arr) {
      let mymap = new Map()
      //return arr.filter(res => !mymap.has(res)&&mymap.set(res,1))
      arr.filter(res => !mymap.has(res)&&mymap.set(res,1))
      return mymap.keys()
    }
    console.log('filter2')
    console.log(debupe3(arr2))

     // method3：reduce传统方法
      let oldArr = [12,1,2,3,4,4,1,2,3]
      let res11 = oldArr.reduce((cur, next) => {
         if(!cur.includes(next)){
            cur.push(next)
         }
         return cur
      },[])

js两个数组交集，并集，差集

//数组求交集，并集，补集
    let arr3 = [1,2,3,4,8]
    let arr4 = [2,4,6,8]

    //并集 Union
    //let arr5 = arr3.concat(arr4)
    //es6 数组合并新写法
    let arr5 = [...arr3, ...arr4]
    console.log([...new Set(arr5)])

    //交集 intersect
    let jiao = []
    for(let i = 0; i < arr3.length; i++){
      for(let j = 0; j < arr4.length; j++) {
          if(arr3[i] === arr4[j]){
            jiao.push(arr3[i])
          }
      }
    }
   console.log(jiao)
  
   //交集 intersect es6 
   let arr6 = [...new Set(arr3)].filter(item => new Set(arr4).has(item))
   console.log(arr6)
   //补集/差集  difference  = A - A ∩ B
   //[...new Set(arr5)]
   

   //A相对于B的差集
   let difference = [...new Set(arr3)].filter(item => !new Set(arr4).has(item))

   console.log(difference)

本文由 dealdot ＜dealdot＃163.com＞ 创作, Full Stack Developer @ DeepBlue
本文最后编辑时间为: May 17, 2021 at 10:28 am
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