Generating random, unique values C#

Question

I\'ve searched for a while and been struggling to find this, I\'m trying to generate several random, unique numbers is C#. I\'m using System.Random, and I\'m us

Answer 1

I'm calling NewNumber() regularly, but the problem is I often get repeated numbers.

Random.Next doesn't guarantee the number to be unique. Also your range is from 0 to 10 and chances are you will get duplicate values. May be you can setup a list of int and insert random numbers in the list after checking if it doesn't contain the duplicate. Something like:

public Random a = new Random(); // replace from new Random(DateTime.Now.Ticks.GetHashCode());
                                // Since similar code is done in default constructor internally
public List<int> randomList = new List<int>();
int MyNumber = 0;
private void NewNumber()
{
    MyNumber = a.Next(0, 10);
    if (!randomList.Contains(MyNumber))
        randomList.Add(MyNumber);
}

Answer 2

You could also use a dataTable storing each random value, then simply perform the random method while != values in the dataColumn

Answer 3

You can use basic Random Functions of C#

Random ran = new Random();
int randomno = ran.Next(0,100);

you can now use the value in the randomno in anything you want but keep in mind that this will generate a random number between 0 and 100 Only and you can extend that to any figure.

Answer 4

I noted that the accepted answer keeps adding int to the list and keeps checking them with if (!randomList.Contains(MyNumber)) and I think this doesn't scale well, especially if you keep asking for new numbers.

I would do the opposite.

1. Generate the list at startup, linearly
2. Get a random index from the list
3. Remove the found int from the list

This would require a slightly bit more time at startup, but will scale much much better.

public class RandomIntGenerator
{
    public Random a = new Random();
    private List<int> _validNumbers;

    private RandomIntGenerator(int desiredAmount, int start = 0)
    {
        _validNumbers = new List<int>();
        for (int i = 0; i < desiredAmount; i++)
            _validNumbers.Add(i + start);
    }

    private int GetRandomInt()
    {
        if (_validNumbers.Count == 0)
        {
            //you could throw an exception here
            return -1;
        }
        else
        {
            var nextIndex = a.Next(0, _validNumbers.Count - 1);
            var number    = _validNumbers[nextIndex];
            _validNumbers.RemoveAt(nextIndex);
            return number;
        }
    }
}

Answer 5

Try this:

private void NewNumber()
  {
     Random a = new Random(Guid.newGuid().GetHashCode());
     MyNumber = a.Next(0, 10);
  }

Some Explnations:

Guid : base on here : Represents a globally unique identifier (GUID)

Guid.newGuid() produces a unique identifier like "936DA01F-9ABD-4d9d-80C7-02AF85C822A8"

and it will be unique in all over the universe base on here

Hash code here produce a unique integer from our unique identifier

so Guid.newGuid().GetHashCode() gives us a unique number and the random class will produce real random numbers throw this

Sample: https://rextester.com/ODOXS63244

generated ten random numbers with this approach with result of:

-1541116401
7
-1936409663
3
-804754459
8
1403945863
3
1287118327
1
2112146189
1
1461188435
9
-752742620
4
-175247185
4
1666734552
7

we got two 1s next to each other, but the hash codes do not same.

Answer 6

This is a unity only answer:

Check this ready-to-use method: Give in a range & count of number you want to get.

public static int[] getUniqueRandomArray(int min, int max, int count) {
    int[] result = new int[count];
    List<int> numbersInOrder = new List<int>();
    for (var x = min; x < max; x++) {
        numbersInOrder.Add(x);
    }
    for (var x = 0; x < count; x++) {
        var randomIndex = UnityEngine.Random.Range(0, numbersInOrder.Count);
        result[x] = numbersInOrder[randomIndex];
        numbersInOrder.RemoveAt(randomIndex);
    }

    return result;
}