Random number within range and a given granularity in Golang

Question

I've written the following code to create a random number between 0.0 and 10.0.

const minRand = 0
const maxRand = 10
v := minRand + rand.Float64()*(maxRand-minRand)

However, I would like to set the granularity to 0.05, so having all the digits as the least significant decimal should not be allowed, only 0 and 5 should be allowed, e.g.:

the value 7.73 is NOT VALID,
the values 7.7 and 7.75 ARE VALID.

How can I produce such numbers in Go?

georgeok · Accepted Answer · 2019-06-10T15:40:33.090

4

You can divide with the granularity, get a pseudo random integer and then multiply with the granularity to scale the result down.

const minRand = 8
const maxRand = 10
v := float64(rand.Intn((maxRand-minRand)/0.05))*0.05 + minRand
fmt.Printf("%.2f\n", v)

This will print:

8.05
8.35
8.35
8.95
8.05
9.90
....

If you don't want to get the same sequence every time rand.Seed(time.Now().UTC().UnixNano()).

From the docs

Seed uses the provided seed value to initialize the default Source to a deterministic state. If Seed is not called, the generator behaves as if seeded by Seed(1). Seed values that have the same remainder when divided by 2^31-1 generate the same pseudo-random sequence. Seed, unlike the Rand.Seed method, is safe for concurrent use.

edited Jun 10 '19 at 15:40

answered Jun 10 '19 at 15:01

georgeok

5,321
2
39
61

1

Good answer! Last line is wrong though - it's a PRNG whether you seed it or not. – Adrian Jun 10 '19 at 15:02
That is working, could you please add the lower bound as well? Thanks @georgeok – TPPZ Jun 10 '19 at 15:16
Why do you need the lower limit? If it's zero you don't need it. `rand.Intn` will always return a positive number. – georgeok Jun 10 '19 at 15:25
Consider `minRand` being an arbitrary number, so the solution is general. – TPPZ Jun 10 '19 at 15:28

score 1 · Answer 2 · answered Jun 10 '19 at 15:39

1

With lower bounds

const minRand = 0
const maxRand = 10
const stepRand = 0.05

v := float64(rand.Intn((maxRand-minRand)/stepRand))*stepRand + minRand
fmt.Printf("%.2f\n", v)

answered Jun 10 '19 at 15:39

Kent

568
4
12

Random number within range and a given granularity in Golang

2 Answers2