如何在JavaScript中合并两个数组并消除重复项
我有两个JavaScript数组:如何在JavaScript中合并两个数组并消除重复项,javascript,arrays,merge,Javascript,Arrays,Merge,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字 如何在JavaScript中合并两个数组,以便按照插入原始数组的相同顺序仅从每个数组中获取唯一项?新解决方案(使用array.prototype.indexOf和array.prototype.
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字
如何在JavaScript中合并两个数组,以便按照插入原始数组的相同顺序仅从每个数组中获取唯一项?新解决方案(使用array.prototype.indexOf
和array.prototype.concat
):
Array.prototype.indexOf(用于internet explorer):
Array.prototype.indexOf=Array.prototype.indexOf | |函数(elt)
{
var len=this.length>>>0;
var from=Number(参数[1])| | 0;
from=(from<0)?Math.ceil(from):Math.floor(from);
如果(from<0)from+=len;
for(;from
仅合并阵列(不删除重复项)
ES5版本使用数组。concat
:
var array1=[“Vijendra”,“Singh”];
var array2=[“辛格”、“沙基亚”];
console.log(array1.concat(array2))代码>//Array.indexOf是在javascript 1.6(ECMA-262)中引入的
//我们需要为其他浏览器显式地实现它,
if(!Array.prototype.indexOf)
{
Array.prototype.indexOf=函数(elt,from)
{
var len=this.length>>>0;
for(;from
其他浏览器的indexOf
方法的实现取自为什么不使用对象?看起来您正在尝试为一组对象建模。然而,这不会维持秩序
var set1 = {"Vijendra":true, "Singh":true}
var set2 = {"Singh":true, "Shakya":true}
// Merge second object into first
function merge(set1, set2){
for (var key in set2){
if (set2.hasOwnProperty(key))
set1[key] = set2[key]
}
return set1
}
merge(set1, set2)
// Create set from array
function setify(array){
var result = {}
for (var item in array){
if (array.hasOwnProperty(item))
result[array[item]] = true
}
return result
}
Array.prototype.merge=函数(/*数组的可变数目*/){
for(var i=0;i
一个更好的数组合并函数 使用下划线.js或Lo Dash,您可以执行以下操作:
console.log(u.union([1,2,3],[101,2,1,10],[2,1])代码>
我只要把两分钱投进去就行了
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;
}
函数合并字符串数组(a,b){
var hash={};
var-ret=[];
对于(变量i=0;i
这是我经常使用的一种方法,它使用一个对象作为hashlookup表来执行重复检查。假设散列是O(1),那么它在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运算的,但它在数千个数据点上表现良好。Array.prototype.add=function(b){
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]
var a=this.concat();//克隆当前对象
如果(!b.push | | |!b.length)返回a;//如果b不是数组或空,则返回a不变
如果(!a.length)返回b.concat();//如果original为空,则返回b
//检查b的所有元素
对于(变量i=0;i
首先连接两个数组,然后仅过滤出唯一项:
var a=[1,2,3],b=[101,2,1,10]
var c=a.concat(b)
变量d=c.filter((项目,位置)=>c.indexOf(项目)==pos)
log(d)//d是[1,2,3,101,10]
这里是对循环的一个稍微不同的理解。在最新版本的Chrome中进行了一些优化,这是解决两个阵列合并的最快方法(Chrome38.0.2111)
while循环:~589k次操作/s
过滤器:~445k操作/秒
洛达斯:308k运营/服务
环路:225k运行/秒
一条评论指出,我的一个设置变量导致我的循环领先于其他变量,因为它不必初始化要写入的空数组。我同意这一点,所以我重新编写了测试,以平衡竞争环境,并包括一个更快的选项
在这个替代解决方案中,我结合了一个答案的关联数组解决方案,以消除循环中的.indexOf()
调用,该调用通过第二个循环大大降低了速度,并包括了其他用户在其答案中建议的一些其他优化
在这里,每个值(i-1)上都有双循环的最高答案仍然要慢得多。lodash仍然做得很好,我仍然会向任何不介意在他们的项目中添加库的人推荐它。对于那些不想这样做的人来说,我的while循环仍然是一个很好的答案,而过滤器的答案在这里有很强的表现力,在我的测试中,我用最新的Canary Chrome(44.0.2360)击败了所有测试
看看吧,如果你想把它的速度提高一个档次。在浏览了几乎所有可行的答案之后,这些是迄今为止最快的结果。这是一个使用和数组泛型的ECMAScript 6解决方案
目前它只适用于Firefox,也可能适用于InternetExplorer技术预览版
但是如果你使用,你现在就可以拥有它
const输入=[
[1, 2, 3],
[101, 2, 1, 10],
[2, 1]
];
常量合并重复数据消除=(arr)=>{
返回[…新集合([].concat(…arr))];
}
console.log('output',merg
//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
}
var set1 = {"Vijendra":true, "Singh":true}
var set2 = {"Singh":true, "Shakya":true}
// Merge second object into first
function merge(set1, set2){
for (var key in set2){
if (set2.hasOwnProperty(key))
set1[key] = set2[key]
}
return set1
}
merge(set1, set2)
// Create set from array
function setify(array){
var result = {}
for (var item in array){
if (array.hasOwnProperty(item))
result[array[item]] = true
}
return result
}
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;
};
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;
}
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]
var array1 = ["Vijendra", "Singh"];
var array2 = ["Singh", "Shakya"];
var array3 = [];
var arr = array1.concat(array2),
len = arr.length;
while (len--) {
var itm = arr[len];
if (array3.indexOf(itm) === -1) {
array3.unshift(itm);
}
}
let whileLoopAlt = function (array1, array2) {
const array3 = array1.slice(0);
let len1 = array1.length;
let len2 = array2.length;
const assoc = {};
while (len1--) {
assoc[array1[len1]] = null;
}
while (len2--) {
let itm = array2[len2];
if (assoc[itm] === undefined) { // Eliminate the indexOf call
array3.push(itm);
assoc[itm] = null;
}
}
return array3;
};
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);
a = [1, 2, 3];
b = [3, 2, 1, "prince"];
a.concat(b.filter(function(el) {
return a.indexOf(el) === -1;
}));
["prince", "asish", 5].concat(["ravi", 4])
[1, 2, 3].concat([3, 2, 1, "prince"].filter(function(el) {
return [1, 2, 3].indexOf(el) === -1;
}));
f12 > console
["prince", "asish", 5, "ravi", 4]
[1, 2, 3, "prince"]
var array1 = ["Vijendra", "Singh"];
var array2 = ["Singh", "Shakya"];
var array3 = [...new Set([...array1 ,...array2])];
console.log(array3); // ["Vijendra", "Singh", "Shakya"];
[...array1,...array2] // => don't remove duplication
[...new Set([...array1 ,...array2])]; // => remove duplication
Array A + Array B
[unique] + [unique]
[duplicated] + [unique]
[unique] + [duplicated]
[duplicated] + [duplicated]
array3 = _.uniq(array1.concat(array2))
console.log(array3)
function mergeUnique(arr1, arr2){
return arr1.concat(arr2.filter(function (item) {
return arr1.indexOf(item) === -1;
}));
}
/**
* This function merging only arrays unique values. It does not merges arrays in to array with duplicate values at any stage.
*
* @params ...args Function accept multiple array input (merges them to single array with no duplicates)
* it also can be used to filter duplicates in single array
*/
function arrayDeDuplicate(...args){
let set = new Set(); // init Set object (available as of ES6)
for(let arr of args){ // for of loops through values
arr.map((value) => { // map adds each value to Set object
set.add(value); // set.add method adds only unique values
});
}
return [...set]; // destructuring set object back to array object
// alternativly we culd use: return Array.from(set);
}
// SCENARIO
let a = [1,2,3,4,5,6];
let b = [4,5,6,7,8,9,10,10,10];
let c = [43,23,1,2,3];
let d = ['a','b','c','d'];
let e = ['b','c','d','e'];
// USEAGE
let uniqueArrayAll = arrayDeDuplicate(a, b, c, d, e);
let uniqueArraySingle = arrayDeDuplicate(b);
// OUTPUT
console.log(uniqueArrayAll); // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 43, 23, "a", "b", "c", "d", "e"]
console.log(uniqueArraySingle); // [4, 5, 6, 7, 8, 9, 10]
array1.concat(array2).filter((value, pos, arr)=>arr.indexOf(value)===pos)
someSource()
.reduce(...)
.filter(...)
.map(...)
// and now you want to concat array2 and deduplicate:
.concat(array2).filter((value, pos, arr)=>arr.indexOf(value)===pos)
// and keep chaining stuff
.map(...)
.find(...)
// etc
a = [1, 2, 3, 4]
b = [4, 5]
[...new Set(a.concat(b))] // [1, 2, 3, 4, 5]
// mutable operation(array1 is the combined array)
array1.push(...array2);
array1.unshift(...array2);
// immutable operation
const combined = array1.concat(array2);
const combined = [...array1, ...array2]; // ES6
// a little bit tricky
const merged = combined.filter((item, index) => combined.indexOf(item) === index);
const merged = [...new Set(combined)];
const merged = array1.concat(array2.filter(secItem => !array1.includes(secItem)));
var arr1 = [ {a: 1}, {a: 2}, {a: 3} ];
var arr2 = [ {a: 1}, {a: 2}, {a: 4} ];
var arr3 = arr1.concat(arr2.filter( ({a}) => !arr1.find(f => f.a == a) ));
// [ {a: 1}, {a: 2}, {a: 3}, {a: 4} ]
for (var i = 0; i < array2.length; i++)
if (array1.indexOf(array2[i]) === -1)
array1.push(array2[i]);
array2.forEach(v => array1.includes(v) ? null : array1.push(v));
var a = [...new Set([...array1 ,...array2])];
var a = array1.concat(array2);
for (var i = 0; i < a.length; ++i) {
for (var j = i + 1; j < a.length; ++j) {
if (a[i] === a[j])
a.splice(j--, 1);
}
}