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))
);

新解决方案（使用Array.prototype.indexOf和Array.prototype.cocat）：

Array.prototype.uniqueMerge = function( a ) {
    for ( var nonDuplicates = [], i = 0, l = a.length; i<l; ++i ) {
        if ( this.indexOf( a[i] ) === -1 ) {
            nonDuplicates.push( a[i] );
        }
    }
    return this.concat( nonDuplicates )
};

用法：

>>> ['Vijendra', 'Singh'].uniqueMerge(['Singh', 'Shakya'])
["Vijendra", "Singh", "Shakya"]

Array.prototype.indexOf（用于internet explorer）：

Array.prototype.indexOf = Array.prototype.indexOf || function(elt)
  {
    var len = this.length >>> 0;

    var from = Number(arguments[1]) || 0;
    from = (from < 0) ? Math.ceil(from): Math.floor(from); 
    if (from < 0)from += len;

    for (; from < len; from++)
    {
      if (from in this && this[from] === elt)return from;
    }
    return -1;
  };

你可以试试这个：

常量union=（a，b）=>Array.from（新集合（[…a，…b]））；console.log（联合（[“neymar”，“mesi”]，[“ronaldo”，“neymmar”]）；

使用Lodash

我发现@GijsjanB的答案很有用，但我的数组包含具有许多属性的对象，因此我不得不使用其中一个属性来消除它们的重复。

这是我使用lodash的解决方案

userList1 = [{ id: 1 }, { id: 2 }, { id: 3 }]
userList2 = [{ id: 3 }, { id: 4 }, { id: 5 }]
// id 3 is repeated in both arrays

users = _.unionWith(userList1, userList2, function(a, b){ return a.id == b.id });

// users = [{ id: 1 }, { id: 2 }, { id: 3 }, { id: 4 }, { id: 5 }]

作为第三个参数传递的函数有两个参数（两个元素），如果它们相等，则必须返回true。

您可以合并结果并过滤重复项：

let combinedItems = [];

// items is an Array of arrays: [[1,2,3],[1,5,6],...]    
items.forEach(currItems => {
    if (currItems && currItems.length > 0) {
        combinedItems = combinedItems.concat(currItems);
    }
});

let noDuplicateItems = combinedItems.filter((item, index) => {
    return !combinedItems.includes(item, index + 1);
});

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}]