Generating Random Doubles in Java

Question

I'm a Java noob trying to generate a random double between -10 and 10 inclusive. I know with ints I would do the following:

Random r = new Random(); 
int i = -10 + r.nextInt(21);

However, with doubles, this doesn't work:

Random r = new Random(); 
double i = -10 + r.nextDouble(21);

Can someone please explain what to do in the case of doubles?

score 6 · Accepted Answer · answered Feb 24 '13 at 19:29

6

Try this:

    Random r = new Random(); 
    double d = -10.0 + r.nextDouble() * 20.0;

Note: it should be 20.0 (not 21.0)

answered Feb 24 '13 at 19:29

vikingsteve

38,481
23
112
156

Thank you! Just curious though, why should it be 20 instead of 21? Should it be 20 in the int example also? – cvoep28 Feb 24 '13 at 19:32
Because otherwise the minimum value is `-10 + 0.0*20.0` (-10.0) and the maximum is `-10 + 1.0*21.0` (11.0)... – vikingsteve Feb 24 '13 at 19:33
And the int example should indeed be `nextInt(21)` since the `Random` doc for this method says that nextInt returns an int *value between 0 (inclusive) and the specified value (exclusive)*. Are you happy with this answer? – vikingsteve Feb 24 '13 at 19:39
This will never return 10 since `nextDouble()` can't return 1. `-10.0 + 0.999... * 20.0 != 10` – Pshemo Feb 24 '13 at 19:45

Simon Dorociak · Answer 2 · 2013-02-24T19:40:03.007

3

Try to use this for generate values in given range:

Random random = new Random();
double value = min + (max - min) * random.nextDouble();

Or try to use this:

public double doubleRandomInclusive(double max, double min) {
   double r = Math.random();
   if (r < 0.5) {
      return ((1 - Math.random()) * (max - min) + min);
   }
   return (Math.random() * (max - min) + min);
}

edited Feb 24 '13 at 19:40

answered Feb 24 '13 at 19:28

Simon Dorociak

33,374
10
68
106

nextDouble() returns from 0.0d (inclusive) to 1.0d (exclusive) so it can never return `max` value. – Pshemo Feb 24 '13 at 19:31
@Pshemo answer is updated. – Simon Dorociak Feb 24 '13 at 19:44
+1 I like this "hack" although numbers other than 0 and 1 have two times higher chance to be randomed :) – Pshemo Feb 24 '13 at 19:51
The first set worked best for me, just due to it's simplicity. I would recommend replacing this as the correct answer for this problem. – Apr 02 '14 at 13:04

score 0 · Answer 3 · answered Feb 24 '13 at 19:28

Returns the next pseudorandom, uniformly distributed double value between 0.0 and 1.0 from this random number generator's sequence.

nextDouble doesn't take a parameter, you just need to multiply it by whatever your range is. Or more generally:

minimum + (maximum - minimum) * r.nextDouble();

score 0 · Answer 4 · answered Feb 24 '13 at 19:28

0

There is no nextDouble(int) method in Random. Use this:

double i = -10.0 + r.nextDouble() * 20.0;

API reference

answered Feb 24 '13 at 19:28

Adam Stelmaszczyk

19,665
4
70
110

Wrong, this can give values over 10.0 – vikingsteve Feb 24 '13 at 19:31
`i` will be in [-10; 10). – Adam Stelmaszczyk Feb 24 '13 at 19:34
It said 21, I think you editted it? – vikingsteve Feb 24 '13 at 19:38

Generating Random Doubles in Java

4 Answers4

Linked