如何在golang中获取随机数样本？ [关闭]

Question

Or I have to use straightforward way like:

var arr []int

for i := 0; i < 5; i++ {
    arr = append(arr, rand.Intn(100))
}

Answer 1

What you did is clean and fast enough. What you could improve on it is to pre-allocate the slice and fill it using a for.. range loop like this:

s := make([]int, 5)
for i := range s {
    s[i] = rand.Intn(100)
}

The math/rand package also has a rand.Read() function which fills a slice with random bytes. So if you want to fill a []byte slice with random data, this is all it takes:

s := make([]byte, 100)
rand.Read(s) // This never returns an error

Another interesting way would be to take advantage of rand.Rand being an io.Reader. Which means it has a Read() method which fills a []byte with random data.

This combined with the encoding/binary package, you can "fill" variables with random data. Create and pass a rand.Rand to the binary.Read() function as the source, and that's it.

This is how it would look like:

r := rand.New(rand.NewSource(time.Now().UnixNano()))

s := make([]int32, 5)
err := binary.Read(r, binary.BigEndian, &s)
if err != nil {
    panic(err)
}
fmt.Println(s)

Output:

[203443513 1611652563 -235795288 8294855 -802604260]

This is "cool" enough to even fill structs for example:

var point struct{ X, Y int16 }
err = binary.Read(r, binary.BigEndian, &point)
if err != nil {
    panic(err)
}
fmt.Printf("%+v", point)

Output:

{X:-15471 Y:2619}

Try these examples on the Go Playground.

One handicap of using binary.Read() is that–understandably–it can only fill values of fixed-size types, and the most famous exception is the common int type, whose size is not fixed (architecture dependent). So you can't fill an []int slice or a struct with a field of int type. That's why I used int32 and int16 types in the above examples.

Of course in these solutions you could not limit the range of random numbers that are used to fill your variables. For that, the initial loop is still easier.