根据之前的答案…取决于你是想要一个高效的还是“漂亮的联机”解决方案。

一般有三种方法……

"manual iterative" (using indexOf) - naive with O(n2) complexity (slow) var array_diff_naive = function(a,b){ var i, la = a.length, lb = b.length, res = []; if (!la) return b; else if (!lb) return a; for (i = 0; i < la; i++) { if (b.indexOf(a[i]) === -1) res.push(a[i]); } for (i = 0; i < lb; i++) { if (a.indexOf(b[i]) === -1) res.push(b[i]); } return res; } "abstract iterative" (using filter and concat library methods) - syntactic sugar for manual iterative (looks nicer, still sucks) var array_diff_modern = function(a1,a2){ return a1.filter(function(v) { return !a2.includes(v); } ) .concat(a2.filter(function(v) { return !a1.includes(v);})); } "using hashtable" (using object keys) - much more efficient - only O(n), but has slightly limited range of input array values var array_diff_hash = function(a1,a2){ var a = [], diff = []; for (var i = 0; i < a1.length; i++) { a[a1[i]] = true; } for (var i = 0; i < a2.length; i++) { if (a[a2[i]]) { delete a[a2[i]]; } else { a[a2[i]] = true; } } for (var k in a) { diff.push(k); } return diff; }

在jsperf上可以看到 https://jsperf.com/array-diff-algo

如果你有两个对象列表

const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]

let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))

Samuel:“对于我的代码，我也需要删除副本，但我想这并不总是可取的。 我想主要的缺点是它可能会比较许多已经被拒绝的选择。”

当比较TWO列表、数组等，且元素小于1000时，行业标准中的 3GL世界是使用冒泡排序，以避免欺骗。

代码看起来像这样……(未经测试，但应该有效)

var Array01=new Array('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P');
var Array02=new Array('X','B','F','W','Z','X','J','P','P','O','E','N','Q');
var Array03=Array01;

for(x=1; x<Array02.length; x++) {
 for(y=0; y<Array01.length-1; y++) {
  if (Array01[y]==Array02[x]) {Array03.splice(y,1);}}}

Array01=Array03;

要测试输出…

for(y=0; y<Array01.length; y++) {document.write(Array01[y])}

纯JavaScript解决方案(没有库) 与旧浏览器兼容(不使用过滤器) O (n ^ 2) 可选的fn回调参数，用于指定如何比较数组项

function diff(a, b, fn){ var max = Math.max(a.length, b.length); d = []; fn = typeof fn === 'function' ? fn : false for(var i=0; i < max; i++){ var ac = i < a.length ? a[i] : undefined bc = i < b.length ? b[i] : undefined; for(var k=0; k < max; k++){ ac = ac === undefined || (k < b.length && (fn ? fn(ac, b[k]) : ac == b[k])) ? undefined : ac; bc = bc === undefined || (k < a.length && (fn ? fn(bc, a[k]) : bc == a[k])) ? undefined : bc; if(ac == undefined && bc == undefined) break; } ac !== undefined && d.push(ac); bc !== undefined && d.push(bc); } return d; } alert( "Test 1: " + diff( [1, 2, 3, 4], [1, 4, 5, 6, 7] ).join(', ') + "\nTest 2: " + diff( [{id:'a',toString:function(){return this.id}},{id:'b',toString:function(){return this.id}},{id:'c',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}], [{id:'a',toString:function(){return this.id}},{id:'e',toString:function(){return this.id}},{id:'f',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}], function(a, b){ return a.id == b.id; } ).join(', ') );

简单地比较所有值，并返回数组与不重复的值。

var main = [9, '$', 'x', 'r', 3, 'A', '#', 0, 1];

var arr0 = ['Z', 9, 'e', '$', 'r'];
var arr1 = ['x', 'r', 3, 'A', '#'];
var arr2 = ['m', '#', 'a', 0, 'r'];
var arr3 = ['$', 1, 'n', '!', 'A'];


Array.prototype.diff = function(arrays) {
    var items = [].concat.apply(this, arguments);
    var diff = [].slice.call(items), i, l, x, pos;

    // go through all items
    for (x = 0, i = 0, l = items.length; i < l; x = 0, i++) {
        // find all positions
        while ((pos = diff.indexOf(items[i])) > -1) {
            // remove item + increase found count
            diff.splice(pos, 1) && x++;
        }
        // if item was found just once, put it back
        if (x === 1) diff.push(items[i]);
    }
    // get all not duplicated items
    return diff;
};

main.diff(arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"

[].diff(main, arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"

function diff(arr1, arr2) {
  var filteredArr1 = arr1.filter(function(ele) {
    return arr2.indexOf(ele) == -1;
  });

  var filteredArr2 = arr2.filter(function(ele) {
    return arr1.indexOf(ele) == -1;
  });
  return filteredArr1.concat(filteredArr2);
}

diff([1, "calf", 3, "piglet"], [1, "calf", 3, 4]); // Log ["piglet",4]