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
const dbData = [{name:'ally'},
{name:'James'}]
const localData = [{name:'James'}]
const diff = dbData.filter(a =>!localData.some(b => { return a.name === b.name}))
其他回答
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"}]
要从另一个数组中减去一个数组,只需使用下面的代码片段:
var a1 = ['1','2','3','4','6'];
var a2 = ['3','4','5'];
var items = new Array();
items = jQuery.grep(a1,function (item) {
return jQuery.inArray(item, a2) < 0;
});
它将返回['1,'2','6'],这些是第一个数组中不存在的项。
因此,根据您的问题示例,以下代码是准确的解决方案:
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var _array = new Array();
_array = jQuery.grep(array2, function (item) {
return jQuery.inArray(item, array1) < 0;
});
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;
}
纯JavaScript
对于“差异”有两种可能的解释。我让你选你想要的。假设你有:
var a1 = ['a', 'b' ];
var a2 = [ 'b', 'c'];
If you want to get ['a'], use this function: function difference(a1, a2) { var result = []; for (var i = 0; i < a1.length; i++) { if (a2.indexOf(a1[i]) === -1) { result.push(a1[i]); } } return result; } If you want to get ['a', 'c'] (all elements contained in either a1 or a2, but not both -- the so-called symmetric difference), use this function: function symmetricDifference(a1, a2) { var result = []; for (var i = 0; i < a1.length; i++) { if (a2.indexOf(a1[i]) === -1) { result.push(a1[i]); } } for (i = 0; i < a2.length; i++) { if (a1.indexOf(a2[i]) === -1) { result.push(a2[i]); } } return result; }
斜线/下划线
如果你正在使用lodash,你可以使用_。差异(a1, a2)(上述情况1)或_。Xor (a1, a2)(情形2)
如果你使用的是Underscore.js,你可以使用_。情况1的差分(a1, a2)函数。
ES6 Set,用于非常大的数组
上面的代码适用于所有浏览器。然而,对于超过10,000个项目的大型数组,它变得相当慢,因为它有O(n²)的复杂度。在许多现代浏览器中,我们可以利用ES6 Set对象来加快速度。Lodash在可用时自动使用Set。如果你不使用lodash,使用下面的实现,灵感来自Axel Rauschmayer的博客文章:
function difference(a1, a2) {
var a2Set = new Set(a2);
return a1.filter(function(x) { return !a2Set.has(x); });
}
function symmetricDifference(a1, a2) {
return difference(a1, a2).concat(difference(a2, a1));
}
笔记
如果您关心-0、+0、NaN或稀疏数组,那么所有示例的行为都可能令人惊讶或不明显。(对于大多数用途来说,这并不重要。)
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)