判断元素是否在数组中,有就返回元素的索引,没有返回-1 新建一个空的结果数组,for循环元素组,判断结果数组是否存在该元素,存在跳过,不存在就push
<script> ;(function () { function unique(){ let newArr = []; // this表示arr for(let i = 0; i<this.length; i++){ // this[i]表示数组中的每一个元素 if(newArr.indexOf(this[i]) != -1 ){ }else{ newArr.push(this[i]) } } return newArr; } Array.prototype.unique = unique; }()) let arr = [1,2,3,3,2,1,1,2,5,1,9,10,10,3]; // => [1,2,3] let newArr = arr.unique(); console.log(newArr) // => [1,2,3] </script>利用对象的属性不能重复的特点(可能有问题,不建议用)
<script> ; (function () { function unique() { let newArr = []; let o = {}; for (let i = 0; i < this.length; i++) { var t = this[i] if (o[t]) { } else { newArr.push(t); o[t] = true } } return newArr } Array.prototype.unique = unique }()) let arr = [1, 2, 3, 3, 2, 1, 1, 2, 5, 1, 9, 10, 10, 3]; let newArr = arr.unique() console.log(newArr); </script>先使用sort() 方法进行排序,然后排序数组进行遍历,比较相邻元素。
<script> ; (function () { function unique() { let arr = this.sort(); // 排序后,把相同的元素放在一起 let newArr = [arr[0]] for (let i = 1; i < arr.length; i++) { if (arr[i] !== arr[i - 1]) { newArr.push(arr[i]) } } return newArr; } Array.prototype.unique = unique; }()) let arr = [1,2,3,3,2,1,5,4,2,1,2,4,8,4]; let newArr = arr.unique(); console.log(newArr) </script>set学习:https://www.jianshu.com/p/80bf2e6139dc
<script> let arr = [1,2,3,3,2,1,1,2,5,1,9,10,10,3]; let newArr = [...new Set(arr)] console.log(newArr) </script>let newArr = new […new Set(arr)]
hasOwnProperty判断自身属性是否存在
<script> function fn(arr){ let obj = {}; for(let i =0;i<arr.length;i++){ let t = arr[i]; if(obj.hasOwnProperty(t)){ obj[t] +=1; }else{ obj[t]=1 } } return obj; } let arr = [1,2,3,4,3,2,6,7,3,9,5]; let res = fn(arr); console.log(res); </script>…展开数组,Set 去重 concat方法用于连接两个或多个数组
<script> let arr1 = [1,2,3,3] let arr2 = [4,5,6,3,2,4,9] function union(arr1,arr2){ // let res = new Set([...arr1,...arr2]) let res = new Set(arr1.concat(arr2)) return res } console.log(union(arr1,arr2)); </script>filter自带遍历 过滤器
<script> function intersection(arr1,arr2) { let s1 = new Set(arr1); let s2 = new Set(arr2); let newArr = [...s1].filter(item=>{ return s2.has(item) }) return newArr; } let arr1 = [1,2,3] let arr2 = [3,4,5] console.log(intersection(arr1,arr2)); </script>