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
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"]
]
其他回答
//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
*/
我简化了这个答案的最佳部分,并将其转化为一个很好的函数:
function mergeUnique(arr1, arr2){
return arr1.concat(arr2.filter(function (item) {
return arr1.indexOf(item) === -1;
}));
}
使用Lodash
我发现@GijsjanB的答案很有用,但我的数组包含具有许多属性的对象,因此我不得不使用其中一个属性来消除它们的重复。
这是我使用lodash的解决方案
userList1 = [{ id: 1 }, { id: 2 }, { id: 3 }]
userList2 = [{ id: 3 }, { id: 4 }, { id: 5 }]
// id 3 is repeated in both arrays
users = _.unionWith(userList1, userList2, function(a, b){ return a.id == b.id });
// users = [{ id: 1 }, { id: 2 }, { id: 3 }, { id: 4 }, { id: 5 }]
作为第三个参数传递的函数有两个参数(两个元素),如果它们相等,则必须返回true。
ES2019年
可以像union(array1,array2,array3,…)一样使用它
/**
* Merges two or more arrays keeping unique items. This method does
* not change the existing arrays, but instead returns a new array.
*/
function union<T>(...arrays: T[]) {
return [...new Set([...arrays].flat())];
}
这是ES2019,因为flat()函数,但您可以使用core js将其作为polyfill获取。这里的T是TypeScript泛型类型,如果不使用TypeScript,则可以删除它。如果您使用的是TypeScript,请确保在tsconfig.json中的编译器选项中添加“lib”:[“es2019.array”]。
或
只需使用lodash。union
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;
};