js中完成对两个数组数据取差集

方法一：

例如求var arr1 = [1]; var arr2 = [1,2];的差集
Array.prototype.diff = function(a) {
    return this.filter(function(i) {return a.indexOf(i) < 0;});
};
 [1,2].diff([1]);//[2]

方法二：

例如求var arr1 = [1]; var arr2 = [1,2];的差集
1 var isNaN = Number.isNaN;
2 var difference = function(arr1, arr2) {
3     return arr1.reduce(function(previous, i) {
4         var found = arr2.findIndex(function(j) {
5             return j === i || (isNaN(i) && isNaN(j));
6         });
7         return (found < 0 && previous.push(i), previous);
8     }, []);
9 };
10 
11 
12 var arr1 = [1];
13 var arr2 = [1, 2];
14 
15 console.log(difference(arr2, arr1)); //[ 2 ]

方法三：

1 // ES6 的 Set 来处理，这是真正按照数学上的集合来进行的，不会有重复元素
2 var subSet = function(arr1, arr2) {
3     var set1 = new Set(arr1);
4     var set2 = new Set(arr2);
5 
6     var subset = [];
7 
8     for (let item of set1) {
9         if (!set2.has(item)) {
10             subset.push(item);
11         }
12     }
13 
14     return subset;
15 };
16 //普通方法
17 var subSet = function(arr1, arr2) {
18     var len = arr1.length;
19     var arr = [];
20 
21     while (len--) {
22         if (arr2.indexOf(arr1[len]) < 0) {
23             arr.push(arr1[len]);
24         }
25     }
26 
27     return arr;
28 };

方法四：

1 // 数组求差值
2 var arr1 = [1, 2, 4, 9, 0];
3 var arr2 = [2, 4, 7, 8];
4 
5 var difference = function(arr1, arr2) {
6   var diff = [];
7   var tmp = arr2;
8 
9   arr1.forEach(function(val1, i){
10     if (arr2.indexOf(val1) < 0) {
11       diff.push(val1);
12     } else {
13       tmp.splice(tmp.indexOf(val1), 1);
14     }
15   });
16 
17   console.log(diff.concat(tmp));
18 }
19 
20 // 输出 [ 1, 9, 0, 7, 8 ]
21 difference(arr1, arr2);

原文引用自：https://www.cnblogs.com/jiechen/p/5694966.html