是否有一种方法可以在JavaScript中返回两个数组之间的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
当前回答
这是目前为止最简单的方法来得到你正在寻找的结果，使用jQuery:
var diff = $(old_array).not(new_array).get();
Diff现在包含了old_array中不在new_array中的内容
其他回答
Samuel:“对于我的代码，我也需要删除副本，但我想这并不总是可取的。 我想主要的缺点是它可能会比较许多已经被拒绝的选择。”
当比较TWO列表、数组等，且元素小于1000时，行业标准中的 3GL世界是使用冒泡排序，以避免欺骗。
代码看起来像这样……(未经测试，但应该有效)
var Array01=new Array('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P');
var Array02=new Array('X','B','F','W','Z','X','J','P','P','O','E','N','Q');
var Array03=Array01;
for(x=1; x<Array02.length; x++) {
for(y=0; y<Array01.length-1; y++) {
if (Array01[y]==Array02[x]) {Array03.splice(y,1);}}}
Array01=Array03;
要测试输出…
for(y=0; y<Array01.length; y++) {document.write(Array01[y])}
这个怎么样:
Array.prototype.contains = function(needle){
for (var i=0; i<this.length; i++)
if (this[i] == needle) return true;
return false;
}
Array.prototype.diff = function(compare) {
return this.filter(function(elem) {return !compare.contains(elem);})
}
var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));
这样你就可以用array1。diff(array2)来得到它们的区别算法的时间复杂度很可怕- O(array1。长度(我相信)
随着ES6的到来，带有sets和splat操作符(当时只在Firefox中工作，请检查兼容性表)，你可以编写以下一行代码:
var a = ['a', 'b', 'c', 'd'];
var b = ['a', 'b'];
var b1 = new Set(b);
var difference = [...new Set(a.filter(x => !b1.has(x)))];
结果是["c"， "d"]。
简单地比较所有值，并返回数组与不重复的值。
var main = [9, '$', '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!"
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(', ') );
