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
//1.merge two array into one array
var arr1 = [0, 1, 2, 4];
var arr2 = [4, 5, 6];
//for merge array we use "Array.concat"
let combineArray = arr1.concat(arr2); //output
alert(combineArray); //now out put is 0,1,2,4,4,5,6 but 4 reapeat
//2.same thing with "Spread Syntex"
let spreadArray = [...arr1, ...arr2];
alert(spreadArray); //now out put is 0,1,2,4,4,5,6 but 4 reapete
/*
if we need remove duplicate element method use are
1.Using set
2.using .filter
3.using .reduce
*/
其他回答
我学会了一种用扩展运算符连接两个数组的小方法:
var array1 = ['tom', 'dick', 'harry'];
var array2 = ['martin', 'ricky'];
array1.push(...array2);
“…”扩展运算符将以下数组拆分为单个项,然后push可以将它们作为单独的参数处理。
这是我的解决方案https://gist.github.com/4692150深度相等且易于使用的结果:
function merge_arrays(arr1,arr2)
{
...
return {first:firstPart,common:commonString,second:secondPart,full:finalString};
}
console.log(merge_arrays(
[
[1,"10:55"] ,
[2,"10:55"] ,
[3,"10:55"]
],[
[3,"10:55"] ,
[4,"10:55"] ,
[5,"10:55"]
]).second);
result:
[
[4,"10:55"] ,
[5,"10:55"]
]
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);
只是把我的两分钱扔进去。
function mergeStringArrays(a, b){
var hash = {};
var ret = [];
for(var i=0; i < a.length; i++){
var e = a[i];
if (!hash[e]){
hash[e] = true;
ret.push(e);
}
}
for(var i=0; i < b.length; i++){
var e = b[i];
if (!hash[e]){
hash[e] = true;
ret.push(e);
}
}
return ret;
}
这是我经常使用的方法,它使用一个对象作为哈希查找表来执行重复检查。假设哈希值是O(1),那么这将在O(n)中运行,其中n是a.length+b.length。老实说,我不知道浏览器是如何进行哈希的,但它在数千个数据点上表现良好。
const merge(…args)=>(新集合([].contat(…arg)))