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?
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
Array.prototype.add = function(b){
var a = this.concat(); // clone current object
if(!b.push || !b.length) return a; // if b is not an array, or empty, then return a unchanged
if(!a.length) return b.concat(); // if original is empty, return b
// go through all the elements of b
for(var i = 0; i < b.length; i++){
// if b's value is not in a, then add it
if(a.indexOf(b[i]) == -1) a.push(b[i]);
}
return a;
}
// Example:
console.log([1,2,3].add([3, 4, 5])); // will output [1, 2, 3, 4, 5]
其他回答
如果不希望复制特定属性(例如ID)
let noDuplicate = array1.filter ( i => array2.findIndex(a => i.id==a.id)==-1 );
let result = [...noDuplicate, ...array2];
这是我需要合并(或返回两个数组的并集)时使用的函数。
var union = function (a, b) {
for (var i = 0; i < b.length; i++)
if (a.indexOf(b[i]) === -1)
a.push(b[i]);
return a;
};
var a = [1, 2, 3, 'a', 'b', 'c'];
var b = [2, 3, 4, 'b', 'c', 'd'];
a = union(a, b);
//> [1, 2, 3, "a", "b", "c", 4, "d"]
var array1 = ["Vijendra", "Singh"];
var array2 = ["Singh", "Shakya"];
var array3 = union(array1, array2);
//> ["Vijendra", "Singh", "Shakya"]
只需使用Undercore.js的=>uniq即可实现:
array3 = _.uniq(array1.concat(array2))
console.log(array3)
它将印刷[“Vijendra”、“Singh”、“Shakya”]。
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"]
之前写过同样的原因(适用于任意数量的数组):
/**
* Returns with the union of the given arrays.
*
* @param Any amount of arrays to be united.
* @returns {array} The union array.
*/
function uniteArrays()
{
var union = [];
for (var argumentIndex = 0; argumentIndex < arguments.length; argumentIndex++)
{
eachArgument = arguments[argumentIndex];
if (typeof eachArgument !== 'array')
{
eachArray = eachArgument;
for (var index = 0; index < eachArray.length; index++)
{
eachValue = eachArray[index];
if (arrayHasValue(union, eachValue) == false)
union.push(eachValue);
}
}
}
return union;
}
function arrayHasValue(array, value)
{ return array.indexOf(value) != -1; }