shuffle array in Go

Question

I tried to translate the following Python code to Go

import random

list = [i for i in range(1, 25)]
random.shuffle(list)
print(list)

but found my Go version lengthy and awkward because there is no shuffle function and I had to implement interfaces and convert types.

What would be an idiomatic Go version of my code?

Answer 1

As your list is just the integers from 1 to 25, you can use Perm :

list := rand.Perm(25)
for i, _ := range list {
    list[i]++
}

Note that using a permutation given by rand.Perm is an effective way to shuffle any array.

dest := make([]int, len(src))
perm := rand.Perm(len(src))
for i, v := range perm {
    dest[v] = src[i]
}

Answer 2

Use Shuffle() from the math/rand library.

Here's an example:

package main

import (
    "fmt"
    "math/rand"
    "strings"
)

func main() {
    words := strings.Fields("ink runs from the corners of my mouth")
    rand.Shuffle(len(words), func(i, j int) {
        words[i], words[j] = words[j], words[i]
    })
    fmt.Println(words)
}

Since it comes from the math/rand library it needs to be seeded. See here for more details.

Answer 3

Raed's approach is very inflexible because of []interface{} as input. Here is more convenient version for go>=1.8:

func Shuffle(slice interface{}) {
    rv := reflect.ValueOf(slice)
    swap := reflect.Swapper(slice)
    length := rv.Len()
    for i := length - 1; i > 0; i-- {
            j := rand.Intn(i + 1)
            swap(i, j)
    }
}

Example usage:

    rand.Seed(time.Now().UnixNano()) // do it once during app initialization
    s := []int{1, 2, 3, 4, 5}
    Shuffle(s)
    fmt.Println(s) // Example output: [4 3 2 1 5]

And also, don't forget that a little copying is better than a little dependency

Answer 4

Maybe you can also use the following function:

func main() {
   slice := []int{10, 12, 14, 16, 18, 20}
   Shuffle(slice)
   fmt.Println(slice)
}

func Shuffle(slice []int) {
   r := rand.New(rand.NewSource(time.Now().Unix()))
   for n := len(slice); n > 0; n-- {
      randIndex := r.Intn(n)
      slice[n-1], slice[randIndex] = slice[randIndex], slice[n-1]
   }
}

Answer 5

Answer by Evan Shaw has a minor bug. If we iterate through the slice from lowest index to highest, to get a uniformly (pseudo) random shuffle, according to the same article, we must choose a random integer from interval [i,n) as opposed to [0,n+1).

That implementation will do what you need for larger inputs, but for smaller slices, it will perform a non-uniform shuffle.

To utilize rand.Intn(), we can do:

    for i := len(slice) - 1; i > 0; i-- {
        j := rand.Intn(i + 1)
        slice[i], slice[j] = slice[j], slice[i]
    }

following the same algorithm from Wikipedia article.

Answer 6

When using the math/rand package, do not forget to set a source

// Random numbers are generated by a Source. Top-level functions, such as
// Float64 and Int, use a default shared Source that produces a deterministic
// sequence of values each time a program is run. Use the Seed function to
// initialize the default Source if different behavior is required for each run.

So I wrote a Shuffle function that takes this into consideration:

import (
    "math/rand"
)

func Shuffle(array []interface{}, source rand.Source) {
    random := rand.New(source)
    for i := len(array) - 1; i > 0; i-- {
        j := random.Intn(i + 1)
        array[i], array[j] = array[j], array[i]
    }
}

And to use it:

source := rand.NewSource(time.Now().UnixNano())
array := []interface{}{"a", "b", "c"}

Shuffle(array, source) // [c b a]

If you would like to use it, you can find it here https://github.com/shomali11/util