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
可以使用Set完成。
var array1=[“Vijendra”,“Singh”];var array2=[“Singh”,“Shakya”];var array3=array1.concat(array2);var tempSet=新集合(array3);array3=Array.from(tempSet);//显示输出document.body.querySelector(“div”).innerHTML=JSON.stringify(array3);<div style=“width:100%;height:4rem;line height:4rem;background color:steelblue;color:#DDD;text align:center;font-family:Calibri”>临时文本</div>
其他回答
//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
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;
}
});
新解决方案(使用Array.prototype.indexOf和Array.prototype.cocat):
Array.prototype.uniqueMerge = function( a ) {
for ( var nonDuplicates = [], i = 0, l = a.length; i<l; ++i ) {
if ( this.indexOf( a[i] ) === -1 ) {
nonDuplicates.push( a[i] );
}
}
return this.concat( nonDuplicates )
};
用法:
>>> ['Vijendra', 'Singh'].uniqueMerge(['Singh', 'Shakya'])
["Vijendra", "Singh", "Shakya"]
Array.prototype.indexOf(用于internet explorer):
Array.prototype.indexOf = Array.prototype.indexOf || function(elt)
{
var len = this.length >>> 0;
var from = Number(arguments[1]) || 0;
from = (from < 0) ? Math.ceil(from): Math.floor(from);
if (from < 0)from += len;
for (; from < len; from++)
{
if (from in this && this[from] === elt)return from;
}
return -1;
};
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。
带过滤器的最简单解决方案:
var array1 = ["Vijendra","Singh"];
var array2 = ["Singh", "Shakya"];
var mergedArrayWithoutDuplicates = array1.concat(
array2.filter(seccondArrayItem => !array1.includes(seccondArrayItem))
);