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

这是工作的:基本上合并两个数组,寻找重复的,并将没有重复的推入一个新数组,这是差异。

函数diff(arr1, arr2) { var newArr = []; Var arr = ar1 .concat(arr2); For (var I in arr){ Var f = arr[i]; Var t = 0; (j = 0;j < arr.length;j + +) { If (arr[j] === f){ t + +; } } 如果(t == 1){ newArr.push (f); } } 返回newArr; }

其他回答

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]

下划线中的差分方法(或它的替换,Lo-Dash)也可以做到这一点:

(R)eturns the values from array that are not present in the other arrays

_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]

与任何下划线函数一样,你也可以以更面向对象的风格使用它:

_([1, 2, 3, 4, 5]).difference([5, 2, 10]);

CoffeeScript版本:

diff = (val for val in array1 when val not in array2)

如果你有两个对象列表

const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]

let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))

转换为字符串对象类型:

[1, 1].toString() === [1, 1].toString(); // true