我有一个目标数组[“apple”、“banana”、“orange”]，我想检查其他数组是否包含任何一个目标阵列元素。
例如：
["apple","grape"] //returns true;
["apple","banana","pineapple"] //returns true;
["grape", "pineapple"] //returns false;
如何在JavaScript中实现？
当前回答
您正在寻找两个数组之间的交集。你有两种主要的交叉点类型：“每个”和“一些”。让我举几个好例子：
每
let brands1 = ['Ford', 'Kia', 'VW', 'Audi'];
let brands2 = ['Audi', 'Kia'];
// Find 'every' brand intersection.
// Meaning all elements inside 'brands2' must be present in 'brands1':
let intersectionEvery = brands2.every( brand => brands1.includes(brand) );
if (intersectionEvery) {
const differenceList = brands1.filter(brand => !brands2.includes(brand));
console.log('difference list:', differenceList);
const commonList = brands1.filter(brand => brands2.includes(brand));
console.log('common list:', commonList);
}
如果条件不满足（比如你在品牌中加入了“梅赛德斯”），那么“intersectionEvery”就不满足了——这将是错误的。
如果满足条件，它将把[“福特”、“大众”]列为区别，把[“起亚”、“奥迪”]列为了常见列表。
沙盒：https://jsfiddle.net/bqmg14t6/
SOME
let brands1 = ['Ford', 'Kia', 'VW', 'Audi'];
let brands2 = ['Audi', 'Kia', 'Mercedes', 'Land Rover'];
// Find 'some' brand intersection.
// Meaning some elements inside 'brands2' must be also present in 'brands1':
let intersectionSome = brands2.some( brand => brands1.includes(brand) );
if (intersectionSome) {
const differenceList = brands1.filter(brand => !brands2.includes(brand));
console.log('difference list:', differenceList);
const commonList = brands1.filter(brand => brands2.includes(brand));
console.log('common list:', commonList);
}
我们在这里寻找一些常见的品牌，但不一定全部。
它将把[“福特”、“大众”]列为不同品牌，把[“起亚”、“奥迪”]列为了共同品牌。
沙盒：https://jsfiddle.net/zkq9j3Lh/
其他回答
const areCommonElements = (arr1, arr2) => {
const arr2Set = new Set(arr2);
return arr1.some(el => arr2Set.has(el));
};
或者，如果您首先找出这两个数组中的哪一个更长，并对最长的数组进行设置，同时对最短的数组应用一些方法，您甚至可以获得更好的性能：
const areCommonElements = (arr1, arr2) => {
const [shortArr, longArr] = (arr1.length < arr2.length) ? [arr1, arr2] : [arr2, arr1];
const longArrSet = new Set(longArr);
return shortArr.some(el => longArrSet.has(el));
};
ES6溶液：
let arr1 = [1, 2, 3];
let arr2 = [2, 3];
let isFounded = arr1.some( ai => arr2.includes(ai) );
与之不同：必须包含所有值。
let allFounded = arr2.every( ai => arr1.includes(ai) );
希望，会有所帮助。
ES6（最快）
const a = ['a', 'b', 'c'];
const b = ['c', 'a', 'd'];
a.some(v=> b.indexOf(v) !== -1)
2016年
const a = ['a', 'b', 'c'];
const b = ['c', 'a', 'd'];
a.some(v => b.includes(v));
强调
const a = ['a', 'b', 'c'];
const b = ['c', 'a', 'd'];
_.intersection(a, b)
演示：https://jsfiddle.net/r257wuv5/
jsPerf（性能）：https://jsperf.com/array-contains-any-element-of-another-array
还有一个解决方案
var a1 = [1, 2, 3, 4, 5]
var a2 = [2, 4]
检查a1是否包含a2的所有元素
var result = a1.filter(e => a2.indexOf(e) !== -1).length === a2.length
console.log(result)
良好的性能解决方案：
我们应该将其中一个数组转换为对象。
const contains = (arr1, mainObj) => arr1.some(el => el in mainObj);
const includes = (arr1, mainObj) => arr1.every(el => el in mainObj);
用法：
const mainList = ["apple", "banana", "orange"];
// We make object from array, you can use your solution to make it
const main = Object.fromEntries(mainList.map(key => [key, true]));
contains(["apple","grape"], main) // => true
contains(["apple","banana","pineapple"], main) // => true
contains(["grape", "pineapple"], main) // => false
includes(["apple", "grape"], main) // => false
includes(["banana", "apple"], main) // => true
您可能会面临由in运算符检查的一些缺点（例如｛｝//=>true中的“toString”），因此您可以将解决方案更改为obj[key]检查器
