Implement Ruby style Cartesian product in Go

Question

I want to get the Cartesian product of a, b, c, d:

a = ['a1']
b = ['b1', 'b2']
c = ['c1', 'c2', 'c3']
d = ['d1']

Here is code in Ruby:

e = [b, c, d]
print a.product(*e)

Output is:

[
  ["a1", "b1", "c1", "d1"],
  ["a1", "b1", "c2", "d1"],
  ["a1", "b1", "c3", "d1"],
  ["a1", "b2", "c1", "d1"],
  ["a1", "b2", "c2", "d1"],
  ["a1", "b2", "c3", "d1"]
]

Is there a similar package or function that could do product in Golang? This is just simplified version, in fact, the input data is like [['a1'], ['b1','b2'], ['c1','c2','c3],['d1'],['e1',...],...].

Answer 1

If you need an unknown-at-compile-time set of nested index loops, you can use code like this.

package main

import "fmt"

// NextIndex sets ix to the lexicographically next value,
// such that for each i>0, 0 <= ix[i] < lens(i).
func NextIndex(ix []int, lens func(i int) int) {
    for j := len(ix) - 1; j >= 0; j-- {
        ix[j]++
        if j == 0 || ix[j] < lens(j) {
            return
        }
        ix[j] = 0
    }
}

func main() {
    e := [][]string{
        {"a1"},
        {"b1", "b2"},
        {"c1", "c2", "c3"},
        {"d1"},
    }
    lens := func(i int) int { return len(e[i]) }

    for ix := make([]int, len(e)); ix[0] < lens(0); NextIndex(ix, lens) {
        var r []string
        for j, k := range ix {
            r = append(r, e[j][k])
        }
        fmt.Println(r)
    }
}

The output is:

[a1 b1 c1 d1]
[a1 b1 c2 d1]
[a1 b1 c3 d1]
[a1 b2 c1 d1]
[a1 b2 c2 d1]
[a1 b2 c3 d1]