Randomize a List

Question

What is the best way to randomize the order of a generic list in C#? I've got a finite set of 75 numbers in a list I would like to assign a random order to, in order to draw them for a lottery type application.

Answer 1

 public Deck(IEnumerable<Card> initialCards) 
    {
    cards = new List<Card>(initialCards);
    public void Shuffle() 
     }
    {
        List<Card> NewCards = new List<Card>();
        while (cards.Count > 0) 
        {
            int CardToMove = random.Next(cards.Count);
            NewCards.Add(cards[CardToMove]);
            cards.RemoveAt(CardToMove);
        }
        cards = NewCards;
    }

public IEnumerable<string> GetCardNames() 

{
    string[] CardNames = new string[cards.Count];
    for (int i = 0; i < cards.Count; i++)
    CardNames[i] = cards[i].Name;
    return CardNames;
}

Deck deck1;
Deck deck2;
Random random = new Random();

public Form1() 
{

InitializeComponent();
ResetDeck(1);
ResetDeck(2);
RedrawDeck(1);
 RedrawDeck(2);

}



 private void ResetDeck(int deckNumber) 
    {
    if (deckNumber == 1) 
{
      int numberOfCards = random.Next(1, 11);
      deck1 = new Deck(new Card[] { });
      for (int i = 0; i < numberOfCards; i++)
           deck1.Add(new Card((Suits)random.Next(4),(Values)random.Next(1, 14)));
       deck1.Sort();
}


   else
    deck2 = new Deck();
 }

private void reset1_Click(object sender, EventArgs e) {
ResetDeck(1);
RedrawDeck(1);

}

private void shuffle1_Click(object sender, EventArgs e) 
{
    deck1.Shuffle();
    RedrawDeck(1);

}

private void moveToDeck1_Click(object sender, EventArgs e) 
{

    if (listBox2.SelectedIndex >= 0)
    if (deck2.Count > 0) {
    deck1.Add(deck2.Deal(listBox2.SelectedIndex));

}

    RedrawDeck(1);
    RedrawDeck(2);

}

Answer 2

A very simple approach to this kind of problem is to use a number of random element swap in the list.

In pseudo-code this would look like this:

do 
    r1 = randomPositionInList()
    r2 = randomPositionInList()
    swap elements at index r1 and index r2 
for a certain number of times

Answer 3

Idea is get anonimous object with item and random order and then reorder items by this order and return value:

var result = items.Select(x => new { value = x, order = rnd.Next() })
            .OrderBy(x => x.order).Select(x => x.value).ToList()

Answer 4

Here's a thread-safe way to do this:

public static class EnumerableExtension
{
    private static Random globalRng = new Random();

    [ThreadStatic]
    private static Random _rng;

    private static Random rng 
    {
        get
        {
            if (_rng == null)
            {
                int seed;
                lock (globalRng)
                {
                    seed = globalRng.Next();
                }
                _rng = new Random(seed);
             }
             return _rng;
         }
    }

    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> items)
    {
        return items.OrderBy (i => rng.Next());
    }
}

Answer 5

This is my preferred method of a shuffle when it's desirable to not modify the original. It's a variant of the Fisher–Yates "inside-out" algorithm that works on any enumerable sequence (the length of source does not need to be known from start).

public static IList<T> NextList<T>(this Random r, IEnumerable<T> source)
{
  var list = new List<T>();
  foreach (var item in source)
  {
    var i = r.Next(list.Count + 1);
    if (i == list.Count)
    {
      list.Add(item);
    }
    else
    {
      var temp = list[i];
      list[i] = item;
      list.Add(temp);
    }
  }
  return list;
}

This algorithm can also be implemented by allocating a range from 0 to length - 1 and randomly exhausting the indices by swapping the randomly chosen index with the last index until all indices have been chosen exactly once. This above code accomplishes the exact same thing but without the additional allocation. Which is pretty neat.

With regards to the Random class it's a general purpose number generator (and If I was running a lottery I'd consider using something different). It also relies on a time based seed value by default. A small alleviation of the problem is to seed the Random class with the RNGCryptoServiceProvider or you could use the RNGCryptoServiceProvider in a method similar to this (see below) to generate uniformly chosen random double floating point values but running a lottery pretty much requires understanding randomness and the nature of the randomness source.

var bytes = new byte[8];
_secureRng.GetBytes(bytes);
var v = BitConverter.ToUInt64(bytes, 0);
return (double)v / ((double)ulong.MaxValue + 1);

The point of generating a random double (between 0 and 1 exclusively) is to use to scale to an integer solution. If you need to pick something from a list based on a random double x that's always going to be 0 <= x && x < 1 is straight forward.

return list[(int)(x * list.Count)];

Enjoy!

Answer 6

If you don't mind using two Lists, then this is probably the easiest way to do it, but probably not the most efficient or unpredictable one:

List<int> xList = new List<int>() { 1, 2, 3, 4, 5 };
List<int> deck = new List<int>();

foreach (int xInt in xList)
    deck.Insert(random.Next(0, deck.Count + 1), xInt);