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
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"}]
其他回答
ES2015的函数方法
计算两个数组之间的差值是Set操作之一。这个术语已经表明应该使用本机Set类型,以便提高查找速度。不管怎样,当你计算两个集合之间的差值时,有三种排列:
[+left difference] [-intersection] [-right difference]
[-left difference] [-intersection] [+right difference]
[+left difference] [-intersection] [+right difference]
下面是反映这些排列的功能性解决方案。
离开的区别:
// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( differencel(xs) (ys) );
正确的区别:
差异是微不足道的。这与翻转的参数不同。为了方便,你可以写一个函数:const differencer = flip(difference)。这是所有!
对称的区别:
现在我们有了左边和右边,实现对称的差异也变得微不足道:
// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const concat = y => xs => xs.concat(y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // symmetric difference const difference = ys => xs => concat(differencel(xs) (ys)) (flip(differencel) (xs) (ys)); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( difference(xs) (ys) );
我想这个例子是一个很好的起点,可以让你了解函数式编程的含义:
使用可以以许多不同方式组合在一起的构建块进行编程。
Array.prototype.diff =函数(a) { return this.filter(function(i) {return a.f index_of (i) < 0;}); }; ////////////// //示例// ////////////// Const dif1 =[1,2,3,4,5,6]。Diff ([3,4,5]); console.log (dif1);// => [1,2,6] const dif2 =[“test1”、“test2”,“test3”,“test4”,“test5”,“test6”].diff([“test1”、“test2”,“test3”、“test4”)); console.log (dif2);// => ["test5", "test6"]
注意:. indexof()和.filter()在IE9之前是不可用的。
使用ES7有一个更好的方法:
十字路口
let intersection = arr1.filter(x => arr2.includes(x));
对于[1,2,3][2,3],它将生成[2,3]。另一方面,for[1,2,3][2,3,5]将返回相同的结果。
区别
let difference = arr1.filter(x => !arr2.includes(x));
对于[1,2,3][2,3],它将生成[1]。另一方面,for[1,2,3][2,3,5]将返回相同的结果。
对于对称差分,你可以这样做:
let difference = arr1
.filter(x => !arr2.includes(x))
.concat(arr2.filter(x => !arr1.includes(x)));
通过这种方式,您将得到一个包含arr1中所有不在arr2中的元素的数组,反之亦然
正如@Joshaven Potter在他的回答中指出的,你可以把这个添加到Array中。原型,所以它可以这样使用:
Array.prototype.diff = function(arr2) { return this.filter(x => !arr2.includes(x)); }
[1, 2, 3].diff([2, 3])
在这种情况下,您可以使用Set。它针对这种操作(并、交、差)进行了优化。
确保它适用于你的案例,一旦它不允许重复。
var a = new JS.Set([1,2,3,4,5,6,7,8,9]);
var b = new JS.Set([2,4,6,8]);
a.difference(b)
// -> Set{1,3,5,7,9}
只是修整字符串以确保....空格不会影响差异
function arr_diff(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].trim()]) delete a[a2[i].trim()];
else a[a2[i].trim()]=true;
for(var k in a)
diff.push(k);
return diff;
}