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

具有部分匹配和不区分大小写的Vanilla JS

以前的一些方法的问题是,它们需要每个单词的精确匹配。但是,如果您想提供部分匹配的结果呢?

function search(arrayToSearch, wordsToSearch) {
    arrayToSearch.filter(v => 
        wordsToSearch.every(w => 
            v.toLowerCase().split(" ").
                reduce((isIn, h) => isIn || String(h).indexOf(w) >= 0, false)
            )
        )
}
//Usage
var myArray = ["Attach tag", "Attaching tags", "Blah blah blah"];
var searchText = "Tag attach";
var searchArr = searchText.toLowerCase().split(" "); //["tag", "attach"]

var matches = search(myArray, searchArr);
//Will return
//["Attach tag", "Attaching tags"]

当您希望提供一个搜索框,用户可以在其中键入单词,结果可以以任何顺序、位置和大小写显示这些单词时,这非常有用。

其他回答

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。

这可以通过简单地遍历主数组并检查其他数组是否包含任何目标元素来完成。

试试看:

function Check(A) {
    var myarr = ["apple", "banana", "orange"];
    var i, j;
    var totalmatches = 0;
    for (i = 0; i < myarr.length; i++) {
        for (j = 0; j < A.length; ++j) {
            if (myarr[i] == A[j]) {

                totalmatches++;

            }

        }
    }
    if (totalmatches > 0) {
        return true;
    } else {
        return false;
    }
}
var fruits1 = new Array("apple", "grape");
alert(Check(fruits1));

var fruits2 = new Array("apple", "banana", "pineapple");
alert(Check(fruits2));

var fruits3 = new Array("grape", "pineapple");
alert(Check(fruits3));

JSFIDDLE演示

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

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