shuffle array in Go - arrays

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

dystroy's answer is perfectly reasonable, but it's also possible to shuffle without allocating any additional slices.
for i := range slice {
j := rand.Intn(i + 1)
slice[i], slice[j] = slice[j], slice[i]
}
See this Wikipedia article for more details on the algorithm. rand.Perm actually uses this algorithm internally as well.

Answer 2

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 3

Since 1.10 Go includes an official Fisher-Yates shuffle function.
Documentation: pkg/math/rand/#Shuffle
math/rand: add Shuffle
Shuffle uses the Fisher-Yates algorithm.
Since this is new API, it affords us the opportunity
to use a much faster Int31n implementation that mostly avoids division.
As a result, BenchmarkPerm30ViaShuffle is
about 30% faster than BenchmarkPerm30,
despite requiring a separate initialization loop
and using function calls to swap elements.
See also the original CL 51891
First, as commented by shelll:
Do not forget to seed the random, or you will always get the same order.
For example rand.Seed(time.Now().UnixNano())
Example:
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)

Answer 4

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 5

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 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

Answer 7

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 8

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.