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"]
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
如果不使用hasOwnProperty,那么我们有不正确的元素。例如:
[1,2,3].diff([1,2]); //Return ["3", "remove", "diff"] This is the wrong version
我的版本:
Array.prototype.diff = function(array2)
{
var a = [],
diff = [],
array1 = this || [];
for (var i = 0; i < array1.length; i++) {
a[array1[i]] = true;
}
for (var i = 0; i < array2.length; i++) {
if (a[array2[i]]) {
delete a[array2[i]];
} else {
a[array2[i]] = true;
}
}
for (var k in a) {
if (!a.hasOwnProperty(k)){
continue;
}
diff.push(k);
}
return diff;
}
其他回答
以下是我使用的方法:
var newArr = a1.filter(function(elem) {
return a2.indexOf(elem) === -1;
}).concat( a2.filter(function(elem) {
return a1.indexOf(elem) === -1;
}));
console.log(newArr);
或者这个
var newArr = a1.concat(a2);
function check(item) {
if (a1.indexOf(item) === -1 || a2.indexOf(item) === -1) {
return item;
}
}
return newArr.filter(check);
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))
}
数据:
var new_storage = JSON.parse('[{"id_order":"0003"},{"id_order":"0004"},{"id_order":"0006"}]');
var old_storage = JSON.parse('[{"id_order":"0001"},{"id_order":"0002"},{"id_order":"0003"},{"id_order":"0004"},{"id_order":"0005"}]');
使用过滤器:
var diff = new_storage
.filter(x => {if(!(old_storage.filter(y => y.id_order==x.id_order)).length){return x}})
.concat(old_storage
.filter(x => {if(!(new_storage.filter(y => y.id_order==x.id_order)).length){return x}})
)
console.log(JSON.stringify(diff))
两个数组的结果不同
[{"id_order":"0006"},{"id_order":"0001"},{"id_order":"0002"},{"id_order":"0005"}]
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)
您可以使用一个公共对象并计算第一个数组中每个值的频率。对于第二个数组,减少公共对象中的值。然后遍历所有键并添加所有值大于1的键。
常量差值= (a1, a2) => { Var obj = {}; a1。forEach(obj[v] = (obj[v] || 0) + 1); a2。forEach(v => obj[v] = (obj[v] || 0) - 1); 返回对象 . keys (obj) .reduce((r,k) => { If (obj[k] > 0) r = r.concat (Array.from({长度:obj [k]}) .fill (k)); 返回r; }, []); }; const =结果不同([' a ', ' ', ' b ', ' c ', ' d '], [a, b]); console.log(结果);