从数组中删除所有重复的元素

Question

请告诉我如何删除所有重复的数组。示例

let arr1=['a', 'b', 'c', 'd']
let arr2=['a', 'e', 'd', 'f', 'p']

我想要这样的结果

let arr3=['b', 'c', 'f', 'p']

请在 javascript 或汇总 MongoDB

中编写脚本

Answer 1

你可以通过这种简单的方式实现它

let arr1=['a', 'b', 'c', 'd'];
let arr2=['a', 'e', 'd', 'f', 'p'];

var getInA_Not_InB = (a, b) => { return a.filter(v => b.indexOf(v) == -1)};
var result = getInA_Not_InB(arr1, arr2)
                                       .concat(getInA_Not_InB(arr2, arr1));
console.log(result);

Answer 2

你可以试试：

let arr1 = ['a', 'b', 'c', 'd'];
let arr2 = ['a', 'e', 'd', 'f', 'p'];

let arr3 = [...findDifference(arr1, arr2), ...findDifference(arr2, arr1)];

console.log(arr3);

/* Find all elements in array1 that are not in array2 */
function findDifference(array1, array2) {
    return array1.filter(element => !array2.includes(element));
}

Answer 3

另一种方法是合并2个数组并计算频率并过滤频率为1的元素，您无需担心分别找到2个数组之间的差异并连接它们

const arr1=['a', 'b', 'c', 'd']
const arr2=['a', 'e', 'd', 'f', 'p']
let merged=[...arr1,...arr2]
const mapped = [...new Set(merged)].filter(a => merged.filter(a1 => a1 === a).length ===1 );
console.log(mapped);

从数组中删除所有重复的元素

Remove all elements from array that duplicate

javascript

aggregation

mongodb