Creating a number with Math.random to the tenths place

Question

I need to fill an array with random numbers but the the numbers can only go to the tenths place. Is there a way to do that with Math.random? Here is what I have for my Math.random:

double[] array = new double[size];
for (int i = 0; i < array.length; i++) {
    double num = ((Math.random() * 10000.0) + 0.0);
    array[i] = num;
}
return array;

possible duplicate of http://stackoverflow.com/questions/50532/how-do-i-format-a-number-in-java , I personally can vouch for DecimalFormat — Gordon Allocman, Mar 28 '16 at 15:31
Can you clarify what you are trying to do? Are you trying to make larger random numbers? — ostrichofevil, Mar 28 '16 at 15:31
You should be using `BigDecimal` instead of `double`s if you care about exact decimal places. — Alex Shesterov, Mar 28 '16 at 15:33
If I understand correctly, try looking at `setScale` on the `BigDecimal` class and supply a `RoundingMode` — riddle_me_this, Mar 28 '16 at 15:35

Peter Lawrey · Accepted Answer · 2016-03-28T16:20:42.717

If you want random number with one decimal place you can use

int size = 20;
double[] array = new double[size];
Random rand = new Random();
for (int i = 0; i < size; i++)
    array[i] = rand.nextInt(10_000 * 10) / 10.0;
System.out.println(Arrays.toString(array));

prints

[8773.3, 6305.2, 2758.4, 7872.7, 4648.9, 3565.2, 9546.4, 7533.1, 2369.7, 3218.0, 9960.2, 7418.0, 8149.3, 8813.4, 638.5, 8178.9, 2831.2, 4758.7, 8238.2, 6409.1]

Note: while double has a representation error, the Double.toString() is written to display the shortest decimal which would have that representation so you don't see any error (nor will you for an number for at least up to 100 trillion for one decimal place)

Note2: this is why you must do a (x * N) / (double) N as it avoid cumulated errors. e.g. if you do

double y = x * 0.1

and

double y = x / 10.0;

this does not always produce the same result. This is because 0.1 cannot be represented accurately and when you use it in an operation, you compound this error. If you use 10.0, it can be represented without error, so the error in the result will be small enough that when you call Double.toString() you will get the expected result.

Creating a number with Math.random to the tenths place

1 Answers1