How to calculate cartesian product of iterable objects

Question

I am looking for a way to calculate cartesian product in EcmaScript 6

Example :

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

Expected result :

[["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"]]

Example :

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

Expected result :

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

What is purpose of nested for..of loops if arr2 is not and element of arr1? — guest271314
– guest271314, Commented Apr 22, 2017 at 4:08
Thought, it's more a challenge, than something really useful ! — Santa Claus
– Santa Claus, Commented Apr 22, 2017 at 4:20
Given example javascript at Question, and none of nested loops appearing to depend on property or value from preceding for..loop, not certain what expected result is? — guest271314
– guest271314, Commented Apr 22, 2017 at 4:26
The goal is to get a cartesian product = all the possibilities when you combine the arrays — Santa Claus
– Santa Claus, Commented Apr 22, 2017 at 4:30

Ry- · Accepted Answer · 2017-04-22 05:32:57Z

2

The “pretend JavaScript is Haskell” version:

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)) :
      [[]]

Use as follows (well, don’t, actually):

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

re makes a pure generator reusable.

answered Apr 22, 2017 at 5:32

Ry-♦

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1 Comment

Santa Claus Over a year ago

Thank you ! Very interesting !

Santa Claus · Accepted Answer · 2017-09-22 18:28:20Z

1

Based on Ryan's idea :

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])
    ));
}

edited Sep 22, 2017 at 18:28

answered Apr 22, 2017 at 6:57

Santa Claus

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