Set accuracy of random numbers in java?

Question

I was wondering if I am able to set the accuracy of the random double numbers that I generate.

Random generator = new Random();
double randomIndex = generator.nextDouble()*10000;

That produces the random numbers within 10000. How am I able to set the accuracy to 4?

What do you mean by "accuracy" of a random number? What do you want as output? — JB Nizet, Nov 17 '13 at 22:25
Please add details about *why* you want to do this, that will help you get a more accurate answer. As it stands, your question isn't quite specific enough for us to know what you are after. — Jason C, Nov 17 '13 at 22:34

score 2 · Answer 1 · answered Nov 17 '13 at 22:32

2

         Random generator = new Random();
         double randomIndex = generator.nextDouble()*10000;            
         randomIndex=Math.floor(randomIndex * 10000) / 10000;//this is the trick

answered Nov 17 '13 at 22:32

Trying

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@Ingo It *does* work, it's just that your original question was unclear. – Jason C Nov 17 '13 at 22:34
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All solutions, including my own, have the error that a double simply does not have "decimal positions", hence you can end up with numbers that print with more (or less) decimals. – Ingo Nov 17 '13 at 22:46
@Ingo You *did* cover that by explicitly mentioning it in your answer, which is why yours is a good one in that respect. :) I also mentioned storing as integers, and converting on display, for exact precision in mine. – Jason C Nov 17 '13 at 22:51

Jason C · Accepted Answer · 2013-11-17T22:44:02.470

A couple of things. First, you mean "precision", not "accuracy". Second, you need to clarify why you want to do this, because a correct answer will depend on it.

If all you want to do is display the numbers with that precision, it is a formatting issue. You can use, e.g. System.out.printf("%.4f", value) or String.format().

If you are trying to to generate numbers with that precision, you could approximate by doing something like (rounding left out for simplicity):

double value = (int)(generateor.nextDouble() * 10000.0) / 10000.0;

Or if you want your range to be 0-10000 instead of 0-1:

double value = (int)(generateor.nextDouble() * 100000000.0) / 10000.0;

Due to the way floating-point numbers are stored, that will not be exact, but perhaps it is close enough for your purposes. If you need exact, you would want to store as integers, e.g.:

int value = (int)(generator.nextDouble() * 10000.0);

Then you can operate on that internally, and display as:

System.out.printf("%.4f", value / 10000.0);

Adjust multiplication factor above if you meant you wanted your range to be 0-10000.

If you are merely trying to generate a number in [0, 10000), you can use Random.nextInt(int) with a range specified, or simply cast the value to an int as above (optionally rounding).

score 1 · Answer 3 · answered Nov 17 '13 at 22:32

1

If you want 4 digits after the decimal mark you can simply do the following:

Random generator = new Random();
double randomIndex = Math.floor(generator.nextDouble()*10000 * 10000) /10000;

answered Nov 17 '13 at 22:32

Basim Khajwal

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Do I need to import anything by adding the "Math"? – jsh6303 Nov 17 '13 at 22:35
@JiajuShen You could a) try it, or b) http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html. – Jason C Nov 17 '13 at 22:38

score 1 · Answer 4 · answered Nov 17 '13 at 22:34

1

Random generator = new Random();
double randomIndex = Double.parseDouble(new DecimalFormat("##.####")
                .format(generator.nextDouble() * 10000));

answered Nov 17 '13 at 22:34

Sandhu Santhakumar

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That is a rather roundabout way of doing `(int)(value * 10000.0 + 0.5) / 10000.0`, I think. – Jason C Nov 17 '13 at 22:35

Ingo · Answer 5 · 2013-11-17T22:50:01.167

1

Simply

double result = generator.nextLong() / 10000.0;

Note, hoewever, that you can never be sure that the number has exactly 4 decimals, whenever you hit a number that is not representable in a double.

Anyway, the requirement is silly, because a double simply does not have decimal positions. Hence, to request 4 of them makes no sense.

edited Nov 17 '13 at 22:50

answered Nov 17 '13 at 22:37

Ingo

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It is worth noting that the theoretical range of this is -922337203685477.5808 to 922337203685477.5807; actual range and distribution susceptible to limits of `nextLong()` and internal representation of a `double`. – Jason C Nov 17 '13 at 22:41

Set accuracy of random numbers in java?

5 Answers5