Random Number with NextDouble (C#)

Question

I'm programming a perceptron and really need to get the range from the normal NextDouble (0, 1) to (-0.5, 0.5). Problem is, I'm using an array and I'm not sure whether it is possible. Hopefully that's enough information.

Random rdm = new Random();

double[] weights = {rdm.NextDouble(), rdm.NextDouble(), rdm.NextDouble()};

Do you care if -0.5 is a possibility? – Douglas Zare Mar 28 '16 at 21:59 — Douglas Zare, Mar 28 '16 at 21:59
Could you solve this? Have you seen all answers? – Andrew Mar 30 '16 at 14:14 — Andrew, Mar 30 '16 at 14:14

score 2 · Accepted Answer · edited Mar 14 '17 at 12:41

2

Simply subtract 0.5 from your random number:

double[] weights = {
                     rdm.NextDouble() - 0.5, 
                     rdm.NextDouble() - 0.5, 
                     rdm.NextDouble() - 0.5 
                   };

edited Mar 14 '17 at 12:41

Ajay

18,086
12
59
105

answered Mar 28 '16 at 21:58

Mateusz Dryzek

651
4
18

See also: http://stackoverflow.com/questions/3975290/produce-a-random-number-in-a-range-using-c-sharp – Rana Mar 28 '16 at 22:02
Just a heads up: this way you will get values between -0.5 and 0.49999 (aprox.). – Andrew Mar 28 '16 at 22:21

Andrew · Answer 2 · 2016-03-28T22:33:11.273

If you need a only one decimal value (my wild guess from what I have seen in Wikipedia) and to include both limits, I wouldn't use a double but just a decimal value and then do the math:

(rdm.Next(11) - 5) / 10M;

That will return any of the 11 different possible values from -0.5 to 0.5.

Or you could go the double way but with a rounding, so you can actually hit the upper limit (0.5):

Math.Round(rdm.NextDouble() - 0.5, 1);

This way is probably a tiny bit slower than my first suggestion.

Random Number with NextDouble (C#)

2 Answers2