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
仅合并阵列(不删除重复项)
ES5版本使用Array.concat:
var array1=[“Vijendra”,“Singh”];var array2=[“Singh”,“Shakya”];array1=array1.concat(array2);console.log(array1);
ES6版本使用销毁
const array1 = ["Vijendra","Singh"];
const array2 = ["Singh", "Shakya"];
const array3 = [...array1, ...array2];
由于没有“内置”的方法来删除重复项(ECMA-262实际上有Array.forEach,这将非常好),因此我们必须手动执行:
Array.prototype.unique = function() {
var a = this.concat();
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);
}
}
return a;
};
然后,要使用它:
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
// Merges both arrays and gets unique items
var array3 = array1.concat(array2).unique();
这也将保持阵列的顺序(即,不需要排序)。
由于许多人对Array.prototype的原型扩充和in循环感到恼火,这里有一种不太侵入性的方法来使用它:
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(a[i] === a[j])
a.splice(j--, 1);
}
}
return a;
}
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
// Merges both arrays and gets unique items
var array3 = arrayUnique(array1.concat(array2));
对于那些有幸使用ES5可用的浏览器的人,可以像这样使用Object.defineProperty:
Object.defineProperty(Array.prototype, 'unique', {
enumerable: false,
configurable: false,
writable: false,
value: function() {
var a = this.concat();
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);
}
}
return a;
}
});
其他回答
function set(a, b) {
return a.concat(b).filter(function(x,i,c) { return c.indexOf(x) == i; });
}
首先连接两个数组,然后只过滤出唯一的项:
变量a=[1,2,3],b=[101,2,1,10]var c=交流电(b)var d=c.filter((项目,位置)=>c.indexOf(项目)===位置)console.log(d)//d为[1,2,3,101,10]
Edit
正如所建议的,一个更具性能的解决方案是在与a连接之前过滤掉b中的唯一项:
变量a=[1,2,3],b=[101,2,1,10]var c=a.oncat(b.filter((项)=>a.indexOf(项)<0))console.log(c)//c为[1,2,3,101,10]
使用reduce func查看的另一种方法:
function mergeDistinct(arResult, candidate){
if (-1 == arResult.indexOf(candidate)) {
arResult.push(candidate);
}
return arResult;
}
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
var arMerge = [];
arMerge = array1.reduce(mergeDistinct, arMerge);
arMerge = array2.reduce(mergeDistinct, arMerge);//["Vijendra","Singh","Shakya"];
//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
*/
假设原始阵列不需要重复数据消除,这应该非常快,保持原始顺序,并且不会修改原始阵列。。。
function arrayMerge(base, addendum){
var out = [].concat(base);
for(var i=0,len=addendum.length;i<len;i++){
if(base.indexOf(addendum[i])<0){
out.push(addendum[i]);
}
}
return out;
}
用法:
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
var array3 = arrayMerge(array1, array2);
console.log(array3);
//-> [ 'Vijendra', 'Singh', 'Shakya' ]