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
我的一便士半:
Array.prototype.concat_n_dedupe = function(other_array) {
return this
.concat(other_array) // add second
.reduce(function(uniques, item) { // dedupe all
if (uniques.indexOf(item) == -1) {
uniques.push(item);
}
return uniques;
}, []);
};
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
var result = array1.concat_n_dedupe(array2);
console.log(result);
其他回答
对于ES6,只有一行:
a = [1, 2, 3, 4]
b = [4, 5]
[...new Set(a.concat(b))] // [1, 2, 3, 4, 5]
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"]
如果要检查唯一对象,请在比较中使用JSON.stringify。
function arrayUnique(array) {
var a = array.concat();
for(var i=0; i<a.length; ++i) {
for(var j=i+1; j<a.length; ++j) {
if(JSON.stringify(a[i]) === JSON.stringify(a[j]))
a.splice(j--, 1);
}
}
return a;
}
//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
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。