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

您可以使用loadash unionWith-_.unionWith(〔arrays〕,〔comparator〕)

此方法类似于_.union,只是它接受被调用来比较数组元素的比较器。结果值从出现该值的第一个数组中选择。比较器由两个参数调用:(arrVal,othVal)。

var array1=[“Vijendra”,“Singh”];var array2=[“Singh”,“Shakya”];var array3=_.unionWith(array1,array2,_.isEqual);console.log(array3);<script src=“https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js“></script>

其他回答

ES 6版本

试试这个。。。这应该能解决你的问题

var array1=[“Vijendra”,“Singh”];var array2=[“Singh”,“Shakya”];

var输出=[…new Set([…array1,…array2])]

console.log(“合并数组”,输出)

合并无限数量的数组或非数组并保持其唯一性:

function flatMerge() {
    return Array.prototype.reduce.call(arguments, function (result, current) {
        if (!(current instanceof Array)) {
            if (result.indexOf(current) === -1) {
                result.push(current);
            }
        } else {
            current.forEach(function (value) {
                console.log(value);
                if (result.indexOf(value) === -1) {
                    result.push(value);
                }
            });
        }
        return result;
    }, []);
}

flatMerge([1,2,3], 4, 4, [3, 2, 1, 5], [7, 6, 8, 9], 5, [4], 2, [3, 2, 5]);
// [1, 2, 3, 4, 5, 7, 6, 8, 9]

flatMerge([1,2,3], [3, 2, 1, 5], [7, 6, 8, 9]);
// [1, 2, 3, 5, 7, 6, 8, 9]

flatMerge(1, 3, 5, 7);
// [1, 3, 5, 7]

看起来接受的答案是我测试中最慢的;

注意,我正在按Key合并2个对象数组

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width">
  <title>JS Bin</title>
</head>
<body>
<button type='button' onclick='doit()'>do it</button>
<script>
function doit(){
    var items = [];
    var items2 = [];
    var itemskeys = {};
    for(var i = 0; i < 10000; i++){
        items.push({K:i, C:"123"});
        itemskeys[i] = i;
    }

    for(var i = 9000; i < 11000; i++){
        items2.push({K:i, C:"123"});
    }

    console.time('merge');
    var res = items.slice(0);

    //method1();
    method0();
    //method2();

    console.log(res.length);
    console.timeEnd('merge');

    function method0(){
        for(var i = 0; i < items2.length; i++){
            var isok = 1;
            var k = items2[i].K;
            if(itemskeys[k] == null){
                itemskeys[i] = res.length;
                res.push(items2[i]);
            }
        }
    }

    function method1(){
        for(var i = 0; i < items2.length; i++){
            var isok = 1;
            var k = items2[i].K;

            for(var j = 0; j < items.length; j++){
                if(items[j].K == k){
                    isok = 0;
                    break;
                }
            }

            if(isok) res.push(items2[i]);
        }  
    }

    function method2(){
        res = res.concat(items2);
        for(var i = 0; i < res.length; ++i) {
            for(var j = i+1; j < res.length; ++j) {
                if(res[i].K === res[j].K)
                    res.splice(j--, 1);
            }
        }
    }
}
</script>
</body>
</html>

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"]

只需避开嵌套循环(O(n^2))和.indexOf()(+O(n))。

函数合并(a,b){var哈希={};变量i;对于(i=0;i<a.length;i++){hash[a[i]=真;}对于(i=0;i<b.length;i++){hash[b[i]]=真;}return Object.keys(哈希);}var array1=[“Vijendra”,“Singh”];var array2=[“Singh”,“Shakya”];var array3=合并(array1,array2);console.log(array3);