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
如果你不反对使用图书馆，http://underscorejs.org/有一种相交方法，可以简化这一点：
var _ = require('underscore');
var target = [ 'apple', 'orange', 'banana'];
var fruit2 = [ 'apple', 'orange', 'mango'];
var fruit3 = [ 'mango', 'lemon', 'pineapple'];
var fruit4 = [ 'orange', 'lemon', 'grapes'];
console.log(_.intersection(target, fruit2)); //returns [apple, orange]
console.log(_.intersection(target, fruit3)); //returns []
console.log(_.intersection(target, fruit4)); //returns [orange]
交集函数将返回一个包含匹配项的新数组，如果不匹配，则返回空数组。
其他回答
我在node中使用下划线js提出了一个解决方案，如下所示：
var checkRole = _.intersection(['A','B'], ['A','B','C']);
if(!_.isEmpty(checkRole)) {
next();
}
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/
如果你不反对使用图书馆，http://underscorejs.org/有一种相交方法，可以简化这一点：
var _ = require('underscore');
var target = [ 'apple', 'orange', 'banana'];
var fruit2 = [ 'apple', 'orange', 'mango'];
var fruit3 = [ 'mango', 'lemon', 'pineapple'];
var fruit4 = [ 'orange', 'lemon', 'grapes'];
console.log(_.intersection(target, fruit2)); //returns [apple, orange]
console.log(_.intersection(target, fruit3)); //returns []
console.log(_.intersection(target, fruit4)); //returns [orange]
交集函数将返回一个包含匹配项的新数组，如果不匹配，则返回空数组。
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
console.log("searching Array: "+finding_array);
console.log("searching in:"+reference_array);
var check_match_counter = 0;
for (var j = finding_array.length - 1; j >= 0; j--)
{
if(reference_array.indexOf(finding_array[j]) > 0)
{
check_match_counter = check_match_counter + 1;
}
}
var match = (check_match_counter > 0) ? true : false;
console.log("Final result:"+match);
