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 using a datetime.now.ticks seed:

public Random a = new Random(DateTime.Now.Ticks.GetHashCode());
private void NewNumber()
  {
     MyNumber = a.Next(0, 10);
  }

I'm calling NewNumber() regularly, but the problem is I often get repeated numbers. Some people suggested because I was declaring the random every time I did it, it would not produce a random number, so I put the declaration outside my function. Any suggestions or better ways than using System.Random ? Thank you

Answer 1

NOTE, I dont recommend this :). Here's a "oneliner" as well:

//This code generates numbers between 1 - 100 and then takes 10 of them.
var result = Enumerable.Range(1,101).OrderBy(g => Guid.NewGuid()).Take(10).ToArray();

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

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 = Random.Range(0, numbersInOrder.Count);
        result[x] = numbersInOrder[randomIndex];
        numbersInOrder.RemoveAt(randomIndex);
    }

    return result;
}

Answer 4

randomNumber function return unqiue integer value between 0 to 100000

  bool check[] = new bool[100001];
  Random r = new Random();
  public int randomNumber() {
      int num = r.Next(0,100000);
       while(check[num] == true) {
             num = r.Next(0,100000);
     }
    check[num] = true;
   return num;
 }

Answer 5

You might try shuffling an array of possible ints if your range is only 0 through 9. This adds the benefit of avoiding any conflicts in the number generation.

var nums = Enumerable.Range(0, 10).ToArray();
var rnd = new Random();

// Shuffle the array
for (int i = 0;i < nums.Length;++i)
{
    int randomIndex = rnd.Next(nums.Length);
    int temp = nums[randomIndex];
    nums[randomIndex] = nums[i];
    nums[i] = temp;
}

// Now your array is randomized and you can simply print them in order
for (int i = 0;i < nums.Length;++i)
    Console.WriteLine(nums[i]);

Answer 6

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

Random.Next doesn't guarntee 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);
}