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JS数组去重整理合集

2023-03-15 22:47:17 时间

1.利用splice

 1 var arr = [1,2,3,4,5,6,7,8,9,9,8,7,6,5,4,3,2,1];
 2 
 3         function repeat(arr){
 4             for(var i = 0;i<arr.length;i++){
 5                 for(var j = i+1;j<arr.length;j++){
 6                     if(arr[i] == arr[j]){
 7                         arr.splice(j,1);
 8                         j--;
 9                     }
10                 }
11             }
12         return arr
13         }
14         console.log(repeat(arr));
15         //[1, 2, 3, 4, 5, 6, 7, 8, 9]

2.利用reduce

1         var arr = [1,2,3,4,5,6,7,8,9,9,8,7,6,5,4,3,2,1]
2         var newArr = arr.reduce((prev, cur) => {
3               prev.indexOf(cur) === -1 && prev.push(cur);
4                 return prev;
5             }, []);
6         console.log(newArr);
7         //[1, 2, 3, 4, 5, 6, 7, 8, 9]

3.利用Set

 1         function repeat(newarr){
 2             var s = new Set(arr);
 3             var a = [];
 4             for(var i of s){
 5                 a.push(i)
 6             }
 7         return a;
 8         }
 9         var a = repeat(arr)
10         console.log(a)
11         //[1, 2, 3, 4, 5, 6, 7, 8, 9]
12     
console.log([...new Set([1, 2, 3, 1])]);

4.利用indexOf

 1         function repeat(newarr){
 2             var array = [];
 3             for(var i=0;i < newarr.length;i++){
 4                 if(array.indexOf(newarr[i]) === -1){
 5                     array.push(newarr[i])
 6                 }
 7             }
 8             return array;
 9         }
10         console.log(repeat(arr))
11         //[1, 2, 3, 4, 5, 6, 7, 8, 9]

5.利用includes

 1         function repeat(newarr) {
 2             var array =[];
 3             for(var i = 0; i < newarr.length; i++) {
 4                 if( !array.includes( newarr[i]) ) {
 5                     array.push(newarr[i]);
 6                 }
 7             }
 8             return array;
 9         }
10         console.log(repeat(arr));
11         //[1, 2, 3, 4, 5, 6, 7, 8, 9]

6.利用filter

1         function repeat(newarr) {
2             return newarr.filter((item, index, arr)=> {
3                 return newarr.indexOf(item, 0) === index;
4             });
5         }
6         console.log(repeat(arr));
7         //[1, 2, 3, 4, 5, 6, 7, 8, 9]

 7.利用封装函数(逻辑简单)

 1  function has(arr, n) {
 2             for (var i = 0; i < arr.length; i++) {
 3                 if (arr[i] == n) {
 4                     return true;    //如果在里面则返回true;
 5                 }
 6             }
 7         return false;      //否则返回false;
 8         }
 9 
10         function norepeat(myArr) {
11             var newArr = [];            //定义一个空数组,存放符合条件的值;
12                 for (var j = 0; j < myArr.length; j++) {
13                     if (!has(newArr, myArr[j])) {  //利用封装好的函数has判断空数组中是否存在当前的值;
14                     newArr.push(myArr[j]);    //如果空数组中没有,那么则push进空数组;
15                 }
16             }
17         return newArr;
18         }
19         console.log(norepeat(arr))
20         //[1, 2, 3, 4, 5, 6, 7, 8, 9]