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
下面是使用Set的另一个简单解决方案:
常量o1={a:1};const arr1=['!@#$%^&*()','gh',123,o1,1,true,undefined,null];常量arr2=['!@#$%^&*()',123,'abc',o1,0x001,true,void 0,0];const mergeUnique=(…args)=>[…新集合([].contat(…arg))];console.log(mergeUnique(arr1,arr2));
其他回答
var MergeArrays=function(arrayOne, arrayTwo, equalityField) {
var mergeDictionary = {};
for (var i = 0; i < arrayOne.length; i++) {
mergeDictionary[arrayOne[i][equalityField]] = arrayOne[i];
}
for (var i = 0; i < arrayTwo.length; i++) {
mergeDictionary[arrayTwo[i][equalityField]] = arrayTwo[i];
}
return $.map(mergeDictionary, function (value, key) { return value });
}
利用字典和Jquery,您可以合并这两个数组,而不会得到重复项。在我的示例中,我在对象上使用给定的字段,但可能只是对象本身。
这很快,可以整理任意数量的数组,并且可以处理数字和字符串。
function collate(a){ // Pass an array of arrays to collate into one array
var h = { n: {}, s: {} };
for (var i=0; i < a.length; i++) for (var j=0; j < a[i].length; j++)
(typeof a[i][j] === "number" ? h.n[a[i][j]] = true : h.s[a[i][j]] = true);
var b = Object.keys(h.n);
for (var i=0; i< b.length; i++)
b[i]=Number(b[i]);
return b.concat(Object.keys(h.s));
}
> a = [ [1,2,3], [3,4,5], [1,5,6], ["spoon", "fork", "5"] ]
> collate( a )
[1, 2, 3, 4, 5, 6, "5", "spoon", "fork"]
如果你不需要区分5和“5”,那么
function collate(a){
var h = {};
for (i=0; i < a.length; i++) for (var j=0; j < a[i].length; j++)
h[a[i][j]] = typeof a[i][j] === "number";
for (i=0, b=Object.keys(h); i< b.length; i++)
if (h[b[i]])
b[i]=Number(b[i]);
return b;
}
[1, 2, 3, 4, "5", 6, "spoon", "fork"]
可以。
如果你不介意(或者更愿意)所有值都以字符串结尾,那么就这样:
function collate(a){
var h = {};
for (var i=0; i < a.length; i++)
for (var j=0; j < a[i].length; j++)
h[a[i][j]] = true;
return Object.keys(h)
}
["1", "2", "3", "4", "5", "6", "spoon", "fork"]
如果您实际上不需要数组,但只想收集唯一值并对其进行迭代,那么(在大多数浏览器(和node.js)中):
h = new Map();
for (i=0; i < a.length; i++)
for (var j=0; j < a[i].length; j++)
h.set(a[i][j]);
这可能更好。
const merge(…args)=>(新集合([].contat(…arg)))
Array.prototype.merge = function(/* variable number of arrays */){
for(var i = 0; i < arguments.length; i++){
var array = arguments[i];
for(var j = 0; j < array.length; j++){
if(this.indexOf(array[j]) === -1) {
this.push(array[j]);
}
}
}
return this;
};
一个更好的数组合并函数。
//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