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?
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
使用一些/findIndex和indexOf的组合怎么样?
所以类似这样:
var array1 = ["apple","banana","orange"];
var array2 = ["grape", "pineapple"];
var found = array1.some(function(v) { return array2.indexOf(v) != -1; });
为了使其更可读,可以将此功能添加到Array对象本身。
Array.prototype.indexOfAny = function (array) {
return this.findIndex(function(v) { return array.indexOf(v) != -1; });
}
Array.prototype.containsAny = function (array) {
return this.indexOfAny(array) != -1;
}
注意:如果您想对谓词执行某些操作,可以用另一个findIndex和谓词替换内部indexOf
一个简短的写法:找到常量=arr1.some(arr2.includes)
就我个人而言,我会使用以下功能:
var arrayContains = function(array, toMatch) {
var arrayAsString = array.toString();
return (arrayAsString.indexOf(','+toMatch+',') >-1);
}
“toString()”方法将始终使用逗号分隔值。仅适用于基本类型。
香草js
/**
* @description determine if an array contains one or more items from another array.
* @param {array} haystack the array to search.
* @param {array} arr the array providing items to check for in the haystack.
* @return {boolean} true|false if haystack contains at least one item from arr.
*/
var findOne = function (haystack, arr) {
return arr.some(function (v) {
return haystack.indexOf(v) >= 0;
});
};
正如@loganofsmyth所指出的,您可以在ES2016中将其缩短为
/**
* @description determine if an array contains one or more items from another array.
* @param {array} haystack the array to search.
* @param {array} arr the array providing items to check for in the haystack.
* @return {boolean} true|false if haystack contains at least one item from arr.
*/
const findOne = (haystack, arr) => {
return arr.some(v => haystack.includes(v));
};
或者简单地称为arr.some(v=>haystalk.includes(v));
如果要确定数组是否包含其他数组中的所有项,请将some()替换为every()或作为arr.every(v=>haystalk.includes(v));
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);