我有两个JavaScript数组:

var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];

我希望输出为:

var array3 = ["Vijendra","Singh","Shakya"];

输出数组应删除重复的单词。

如何在JavaScript中合并两个数组,以便从每个数组中只获得唯一的项目,其顺序与它们插入原始数组的顺序相同?


当前回答

如果您合并对象数组,请考虑使用lodash UnionBy函数,它允许您设置自定义谓词比较对象:

import { unionBy } from 'lodash';

const a = [{a: 1, b: 2}];
const b = [{a: 1, b: 3}];
const c = [{a: 2, b: 4}];

const result = UnionBy(a,b,c, x => x.a);

结果是:〔{a:1;b:2},{a:2;b:4}〕

结果中使用了来自数组的第一个传递匹配

其他回答

带过滤器的最简单解决方案:

var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];

var mergedArrayWithoutDuplicates = array1.concat(
  array2.filter(seccondArrayItem => !array1.includes(seccondArrayItem))
);

作为LiraNuna的一部分的单线解决方案:

let array1 = ["Vijendra","Singh"];
let array2 = ["Singh", "Shakya"];

// Merges both arrays
let array3 = array1.concat(array2); 

//REMOVE DUPLICATE
let removeDuplicate = [...new Set(array3)];
console.log(removeDuplicate);

如果您有非常大的列表,则不执行此操作,因为已经记录了许多解决方案,所以这不适合合并,但我用此解决方案解决了我的问题(因为大多数数组过滤解决方案都适用于简单数组)

const uniqueVehiclesServiced = 
  invoice.services.sort().filter(function(item, pos, ary) {
    const firstIndex = invoice.services.findIndex((el, i, arr) => el.product.vin === item.product.vin)

  return !pos || firstIndex == pos;
});

在当今时代,使用现有的图书馆提供更简单、更优雅的内容:

import {pipe, concat, distinct} from 'iter-ops';

// our inputs:
const array1 = ['Vijendra', 'Singh'];
const array2 = ['Singh', 'Shakya'];

const i = pipe(
    array1,
    concat(array2), // adding array
    distinct() // making it unique
);

console.log([...i]); //=> ['Vijendra', 'Singh', 'Shakya']

这两者都是高性能的,因为我们只迭代一次,而且代码非常容易阅读。

注:我是iter ops的作者。

以下是带有对象数组的对象的选项:

const a = [{param1: "1", param2: 1},{param1: "2", param2: 2},{param1: "4", param2: 4}]
const b = [{param1: "1", param2: 1},{param1: "4", param2: 5}]


var result = a.concat(b.filter(item =>
         !JSON.stringify(a).includes(JSON.stringify(item))
    ));

console.log(result);
//Result [{param1: "1", param2: 1},{param1: "2", param2: 2},{param1: "4", param2: 4},{param1: "4", param2: 5}]