Uniform random numbers to simulate with ten decimal places

Question

I'm tring to run an specific simulation, i have some code implemented but the result is not the one i'm expecting, i need to generate random uniform(ranged from 0 to 1) numbers with 10 decimal places (in order to convert that in other distribution). Example: 0,0345637897, 0,3445627876, 0,9776428487

this is the code:

double next = r.Next(1000000000, 9999999999);
return double.Parse(String.Format("0.{0}", next));

Please post your existing code so we can see what you already tried — Blachshma, May 09 '13 at 20:42
What are your results? (Aside from unexpected). What's wrong with them? — H H, May 09 '13 at 20:45

H H · Accepted Answer · 2013-05-09T22:11:40.843

3

This should be enough:

double v = Math.Round(myRandom.NextDouble(), 10);

The difference between 0.123 and 0.1230000000 is a matter of formatting, see the answer from @SamIAm.

After the Edit:

double next = r.Next(1000000000, 9999999999);
return double.Parse(String.Format("0.{0}", next));

this is getting an integer between 1000000000 and 9999999999 and then uses the default culture to convert it to a double (in the 0 ... 1.0 range).

Since you seem to use the comma (,) as a decimal separator, at least use

return double.Parse(String.Format("0.{0}", next), CultureInfo.Invariant);

edited May 09 '13 at 22:11

answered May 09 '13 at 20:51

H H

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If you dont care so much about the random quality, then yes – Half_Baked May 09 '13 at 21:11
Again, what is 'quality' in this context? – H H May 09 '13 at 22:06
Just read the post in the link I posted. It explains it – Half_Baked May 13 '13 at 18:54

Sam I am says Reinstate Monica · Answer 2 · 2013-05-09T20:53:09.290

0

Are you looking to pad your output? because you can do that with String.Format()

string s = String.Format({0:0000000000}, i);

you can also do it with a double or float

string s = String.Format("{0:0.0000000000}", d);

if you need a random double, just make one, and pad it with 10 decimal places when you print.

Random random = new Random();
double d = random.NextDouble()

edited May 09 '13 at 20:53

answered May 09 '13 at 20:44

Sam I am says Reinstate Monica

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I really don't think that's what the question is asking. – svick May 09 '13 at 20:48
That `random.NextDouble() * 9999999999` is totally wrong, the LongRandom is not what is being asked. – H H May 09 '13 at 20:56
@HenkHolterman sorry, It was the comma used as a decimal point that got me confused – Sam I am says Reinstate Monica May 09 '13 at 20:57

score -1 · Answer 3 · edited May 23 '17 at 10:32

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I suggest using RNGCryptoServiceProvider. It will produces high-quality random numbers

See this post: Why use the C# class System.Random at all instead of System.Security.Cryptography.RandomNumberGenerator?

edited May 23 '17 at 10:32

Community

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answered May 09 '13 at 20:46

Half_Baked

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Could you explain how exactly would you use that? – svick May 09 '13 at 20:47
What determines the 'quality' of a random number? – H H May 09 '13 at 20:48
@HenkHolterman How random the number is. Random() uses the computers clock or whatever, creating repeating sequences. Wich is bad! True random needs something like reading static space disturbence or something... – Half_Baked May 09 '13 at 21:09
I know the difference, see my answer to the linked question. The OP only wants to simulate something. Statistical arguments apply, security does not. – H H May 09 '13 at 22:13

Uniform random numbers to simulate with ten decimal places

3 Answers3