C#: generate RANDOM numbers between two limits IN ORDER

Question

I want to generate random (double) numbers between two limits, let say: lim1 and lim2.

But, I want this numbers to be generated in order. E.g.: between 1 and 6 : 1.53412 1.654564 2.213123 5.13522 . Thanks!

What have you tried so far and why isn't your current solution working? Please post some code. — ZeroOne, Apr 22 '12 at 17:04

Allon Guralnek · Accepted Answer · 2012-04-22T17:20:48.570

4

public static double[] GenerateRandomOrderedNumbers(double lowerBoundInclusive, double upperBoundExclusive, int count, Random random = null)
{
    random = random ?? new Random();
    return Enumerable.Range(0, count)
        .Select(i => random.NextDouble() * (upperBoundExclusive - lowerBoundInclusive) + lowerBoundInclusive)
        .OrderBy(d => d)
        .ToArray();
}

Not perfect, but I hope this puts you in the right direction.

edited Apr 22 '12 at 17:20

answered Apr 22 '12 at 17:09

Allon Guralnek

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I'd pass in the `Random` instance as a parameter, to avoid seeding issues. – CodesInChaos Apr 22 '12 at 17:12
2

Good answer, but it doesn't generate numbers between two limits. Easy enough fix :) – Daniel Mann Apr 22 '12 at 17:12

score 1 · Answer 2 · answered Apr 22 '12 at 17:09

1

Generate the random numbers and put them on a list:

var numbers = new List<int>();
Random random = new Random();

Add your numbers:

var number = random.Next(min, max); 
numbers.Add(number);

Then sort the list:

var orderList = from n
  in numbers
  orderby n
  select n;

answered Apr 22 '12 at 17:09

Ulises

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`Random.Next` returns `int`s, the OP wants doubles. – CodesInChaos Apr 22 '12 at 17:11

score 0 · Answer 3 · edited May 23 '17 at 11:56

0

What about using this to generate a set of random numbers:

lim1 + random.Next(lim2 - lim1)

and then simply sorting them?

C#: generate RANDOM numbers between two limits IN ORDER

3 Answers3

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