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

ES2015的功能方法

按照函数方法,两个数组的联合只是concat和filter的组合。为了提供最佳性能,我们使用本机Set数据类型,该类型针对属性查找进行了优化。

无论如何,结合联合函数的关键问题是如何处理重复项。以下排列是可能的:

Array A      + Array B

[unique]     + [unique]
[duplicated] + [unique]
[unique]     + [duplicated]
[duplicated] + [duplicated]

前两种排列很容易用一个函数处理。然而,后两个更复杂,因为只要依赖Set查找,就无法处理它们。由于切换到普通的旧Object属性查找会严重影响性能,因此下面的实现只会忽略第三个和第四个排列。你必须构建一个单独的union版本来支持它们。


//小型、可重复使用的辅助功能常量comp=f=>g=>x=>f(g(x));常量应用=f=>a=>f(a);常量flip=f=>b=>a=>f(a)(b);常量concat=xs=>y=>xs.contat(y);const afrom=应用(Array.from);const createSet=xs=>新集合(xs);常量过滤器=f=>xs=>xs.filter(apply(f));//重复数据消除常量重复数据删除=comp(afrom)(createSet);//实际联合函数常量并集=xs=>ys=>{const zs=创建集(xs);返回凹面(xs)(滤波器(x=>zs.has(x)? 假的:zs.add(x))(ys));}//模拟数据常量xs=[1,2,2,3,4,5];常量=[0,1,2,3,3,4,5,6,6];//我们来了console.log(“unique/unique”,union(重复数据消除(xs))(ys));console.log(“重复/唯一”,union(xs)(ys));

从这里开始,实现unionn函数变得很简单,它接受任意数量的数组(灵感来自naomik的评论):

//小型、可重复使用的辅助功能常量未修正=f=>(a,b)=>f(a)(b);常量foldl=f=>acc=>xs=>xs.reduce(uncurry(f),acc);常量应用=f=>a=>f(a);常量flip=f=>b=>a=>f(a)(b);常量concat=xs=>y=>xs.contat(y);const createSet=xs=>新集合(xs);常量过滤器=f=>xs=>xs.filter(apply(f));//工会和工会常量并集=xs=>ys=>{const zs=创建集(xs);返回凹面(xs)(滤波器(x=>zs.has(x)? 假的:zs.add(x))(ys));}常量unionn=(头,…尾)=>foldl(联合)(头)(尾);//模拟数据常量xs=[1,2,2,3,4,5];常量=[0,1,2,3,3,4,5,6,6];常数zs=[0,1,2,3,4,5,6,7,8,9];//我们来了console.log(unionn(xs,ys,zs));

事实证明unionn只是foldl(又名Array.protocol.reduce),它将union作为其缩减器。注意:由于实现没有使用额外的累加器,所以当您在没有参数的情况下应用它时,它会抛出错误。

其他回答

//Array.indexOf was introduced in javascript 1.6 (ECMA-262) 
//We need to implement it explicitly for other browsers, 
if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(elt, from)
  {
    var len = this.length >>> 0;

    for (; from < len; from++)
    {
      if (from in this &&
          this[from] === elt)
        return from;
    }
    return -1;
  };
}
//now, on to the problem

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

var merged = array1.concat(array2);
var t;
for(i = 0; i < merged.length; i++)
  if((t = merged.indexOf(i + 1, merged[i])) != -1)
  {
    merged.splice(t, 1);
    i--;//in case of multiple occurrences
  }

其他浏览器的indexOf方法的实现取自MDC

使用Undercore.js或Lo Dash,您可以执行以下操作:

console.log(_.union([1,2,3],[101,2,1,10],[2,1]));<script src=“https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js“></script>

http://underscorejs.org/#union

http://lodash.com/docs#union

这很简单,可以用jQuery在一行中完成:

var arr1 = ['Vijendra', 'Singh'], arr2 =['Singh', 'Shakya'];

$.unique(arr1.concat(arr2))//one line

["Vijendra", "Singh", "Shakya"]

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}]

使用reduce func查看的另一种方法:

function mergeDistinct(arResult, candidate){
  if (-1 == arResult.indexOf(candidate)) {
    arResult.push(candidate);
  }
  return arResult;
}

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

var arMerge = [];
arMerge = array1.reduce(mergeDistinct, arMerge);
arMerge = array2.reduce(mergeDistinct, arMerge);//["Vijendra","Singh","Shakya"];