如何计算乘积的迭代对象

Question

我正在寻找一种方法来计算笛卡尔乘积中的EcmaScript 6

---

例子：

product([["I", "They"], ["watch", "drink"], ["the sea", "the juice"]]) 

预期结果：

[["I", "watch", "the sea"], ["I", "watch", "the juice"], ["I", "drink", "the sea"], ["I", "drink", "the juice"], ["They", "watch", "the sea"], ["They", "watch", "the juice"], ["They", "drink", "the sea"], ["They", "drink", "the juice"]] 

---

例如：

product([[-1, -2], [10, 20]]) 

预期结果：

[[-1, 10], [-1, 20], [-2, 10], [-2, 20]] 

Answer 1

的“假装JavaScript是哈斯克尔”版本：

const re = g => a => ({ 
    [Symbol.iterator]:() => g(a) 
}) 

const cons = x => re(function* (xs) { 
    yield x 
    yield* xs 
}) 

const map = f => re(function* (xs) { 
    for (const x of xs) 
    yield f(x) 
}) 

const ap = fs => re(function* (xs) { 
    for (const f of fs) 
    for (const x of xs) 
     yield f(x) 
}) 

const product = ([x, ...xs]) => 
    x ? ap(map(cons)(x))(product(xs)) : 
     [[]] 

使用方法如下（当然，不这样做，实际上）：

for (const l of product([arr1, arr2, arr3])) 
    callback(...l) 

re使一个纯生成器可重用。

Answer 2

基于Ryan的想法：

let flatten = arr => [].concat(...arr); 

function product([x, ...xs]) { 
    if(!x) return [[]]; 

    let next = product(xs); 

    return flatten(x.map(a => 
     next.map(b => [a, ...b]) 
    )); 
}