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
所选的答案只对了一半。您必须比较数组的两种方式才能得到完整的答案。
const ids_exist = [
'1234',
'5678',
'abcd',
]
const ids_new = [
'1234',
'5678',
'efjk',
'9999',
]
function __uniq_Filter (__array_1, __array_2) {
const one_not_in_two = __array_1.filter(function (obj) {
return __array_2.indexOf(obj) == -1
})
const two_not_in_one = __array_2.filter(function (obj) {
return __array_1.indexOf(obj) == -1
})
return one_not_in_two.concat(two_not_in_one)
}
let uniq_filter = __uniq_Filter(ids_exist, ids_new)
console.log('uniq_filter', uniq_filter) // => [ 'abcd', 'efjk', '9999' ]
其他回答
function array_diff(array1, array2) {
let returnArray = [];
$.each(array1, function(index, value) {
let findStatus = false;
if (Array.isArray(array2)) {
$.each(array2, function(index2, value2) {
if (value == value2) findStatus = true;
});
} else {
if (value == array2) {
findStatus = true;
}
}
if (findStatus == false) {
returnArray.push(value);
}
});
return returnArray;
}
你可以使用underscore.js: http://underscorejs.org/#intersection
你需要数组的方法:
_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]
_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2]
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(', ') );
这个答案是2009年写的，所以有点过时了，但是对于理解这个问题还是很有教育意义的。我今天最好的解决办法是
let difference = arr1.filter(x => !arr2.includes(x));
(此处致谢给其他作者)
我假设你比较的是一个普通数组。如果不是，你需要将for循环改为for ..在循环。
函数arr_diff (a1, a2) { Var a = []， diff = []; For (var I = 0;I < a1.length;我+ +){ A [a1[i]] = true; ｝ For (var I = 0;I < a2.length;我+ +){ If (a[a2[i]]) { 删除一个[a2[我]]; }其他{ A [a2[i]] = true; ｝ ｝ 对于(var k in a) { diff.push (k); ｝ 返回差异; ｝ console.log (arr_diff ([a, b], [a, b, c, d '))); console.log (arr_diff(“abcd”、"中的")); console.log (arr_diff(“必杀技”,“必杀技”));
function diffArray(newArr, oldArr) {
var newSet = new Set(newArr)
var diff = []
oldArr.forEach((a) => {
if(!newSet.delete(a))diff.push(a)
})
return diff.concat(Array.from(newSet))
}
