与Array.prototype.reduce的数组交集(理论集) [英] Array intersection (set-theoretic) with Array.prototype.reduce

问题描述

我有两个数组 [a，b，c，d] 和 [b，d，f，h] .

我想找回带有公共元素 [b，d] 的数组.

I want to get an array back with the common elements [b,d].

我可以结合使用 filter 和 indexOf :

[a,b,c,d].filter(el => [b,d,f,h].indexOf(el) !== -1)

但是我想知道是否可以以及如何使用 reduce 进行同样的操作.

but I was wondering if and how I can do the same with reduce.

我承认，尽管查看了许多示例，但 reduce 对我来说仍然是最晦涩的JS方法之一，因此，我非常感谢一些建议.

I admit that, despite looking at many examples, reduce still is to me one of the most obscure JS methods, so I'd really appreciate some advice.

推荐答案

ES6，与

ES6, a propoosal with Array#includes

includes() 方法确定数组是否包含某个元素，并返回 true 或 false 作为合适的.

在 aa 的每个循环中，如果在测试数组 bb 中找到该值，则reduce将元素添加到结果数组中.如果找不到，则返回先前的结果.

On every loop of aa, reduce adds the element to the result array if the value is found in the test array bb. If not found, the former result is returned.

var aa = ['a','b','c','d'],
    bb = ['b','d','f','h'],
    cc = aa.reduce((r, a) => bb.includes(a) && r.concat(a) || r, []);

console.log(cc);

使用包含所有数组的单个数组只是一种更聪明的方法.

Just a smarter approach with using a single array which contains all arrays.

var aa = ['a','b','c','d'],
    bb = ['b','d','f','h'],
    result = [aa, bb].reduce((a, b) => a.filter(c => b.includes(c)));

console.log(result);

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