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

一个衬垫

const unique = (a) => [...new Set(a)]; const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{})); const intersection = (a, b) => a.filter((v) => b.includes(v)); const diff = (a, b) => a.filter((v) => !b.includes(v)); const symDiff = (a, b) => diff(a, b).concat(diff(b, a)); const union = (a, b) => diff(a, b).concat(b); const a = unique([1, 2, 3, 4, 5, 5]); console.log(a); const b = [4, 5, 6, 7, 8]; console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b)); console.log(uniqueBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, { id: 1, name: "abc" }, ], (v) => v.id )); const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u))); console.log(intersectionBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, ], [ { id: 1, name: "abc" }, { id: 3, name: "pqr" }, ], (v, u) => v.id === u.id )); const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u))); console.log(diffBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, ], [ { id: 1, name: "abc" }, { id: 3, name: "pqr" }, ], (v, u) => v.id === u.id ));

打印稿

操场上的链接

const unique = <T>(array: T[]) => [...new Set(array)];


const intersection = <T>(array1: T[], array2: T[]) =>
  array1.filter((v) => array2.includes(v));


const diff = <T>(array1: T[], array2: T[]) =>
  array1.filter((v) => !array2.includes(v));


const symDiff = <T>(array1: T[], array2: T[]) =>
  diff(array1, array2).concat(diff(array2, array1));


const union = <T>(array1: T[], array2: T[]) =>
  diff(array1, array2).concat(array2);


const intersectionBy = <T>(
  array1: T[],
  array2: T[],
  predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));


const diffBy = <T>(
  array1: T[],
  array2: T[],
  predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));


const uniqueBy = <T>(
  array: T[],
  predicate: (v: T, i: number, a: T[]) => string
) =>
  Object.values(
    array.reduce((acc, value, index) => {
      acc[predicate(value, index, array)] = value;
      return acc;
    }, {} as { [key: string]: T })
  );

其他回答

对于想要从一个数组中减去另一个数组的人的回应…

如果不超过1000个元素试试这个…

设置一个新变量来复制Array01,并将其命名为Array03。

现在,使用冒泡排序算法比较Array01和Array02的元素,并在找到匹配时进行比较 对Array03执行以下操作…

 if (Array01[x]==Array02[y]) {Array03.splice(x,1);}

注意:我们正在修改Array03而不是Array01,这样就不会搞砸冒泡排序的嵌套循环!

最后,用一个简单的赋值操作将Array03的内容复制到Array01,就完成了。

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}

使用indexOf()的解决方案对于小型数组是可以的,但是随着长度的增长,算法的性能将接近O(n^2)。这里有一个解决方案,将执行非常大的数组使用对象作为关联数组存储数组项作为键;它还自动消除重复项,但只适用于字符串值(或可以安全地存储为字符串的值):

function arrayDiff(a1, a2) {
  var o1={}, o2={}, diff=[], i, len, k;
  for (i=0, len=a1.length; i<len; i++) { o1[a1[i]] = true; }
  for (i=0, len=a2.length; i<len; i++) { o2[a2[i]] = true; }
  for (k in o1) { if (!(k in o2)) { diff.push(k); } }
  for (k in o2) { if (!(k in o1)) { diff.push(k); } }
  return diff;
}

var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
arrayDiff(a1, a2); // => ['c', 'd']
arrayDiff(a2, a1); // => ['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!"
Array.prototype.difference = function(e) {
    return this.filter(function(i) {return e.indexOf(i) < 0;});
};

eg:- 

[1,2,3,4,5,6,7].difference( [3,4,5] );  
 => [1, 2, 6 , 7]