Optimal LINQ query to get a random sub collection - Shuffle

Question

Please suggest an easiest way to get a random shuffled collection of count 'n' from a collection having 'N' items. where n <= N

Answer 1

Another option is to use OrderBy and to sort on a GUID value, which you can do so using:

var result = sequence.OrderBy(elem => Guid.NewGuid());

I did some empirical tests to convince myself that the above actually generates a random distribution (which it appears to do). You can see my results at Techniques for Randomly Reordering an Array.

Answer 2

Further to mquander's answer and Dan Blanchard's comment, here's a LINQ-friendly extension method that performs a Fisher-Yates-Durstenfeld shuffle:

// take n random items from yourCollection
var randomItems = yourCollection.Shuffle().Take(n);

// ...

public static class EnumerableExtensions
{
    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source)
    {
        return source.Shuffle(new Random());
    }

    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng)
    {
        if (source == null) throw new ArgumentNullException("source");
        if (rng == null) throw new ArgumentNullException("rng");

        return source.ShuffleIterator(rng);
    }

    private static IEnumerable<T> ShuffleIterator<T>(
        this IEnumerable<T> source, Random rng)
    {
        var buffer = source.ToList();
        for (int i = 0; i < buffer.Count; i++)
        {
            int j = rng.Next(i, buffer.Count);
            yield return buffer[j];

            buffer[j] = buffer[i];
        }
    }
}

Answer 3

This has some issues with "random bias" and I am sure it's not optimal, this is another possibility:

var r = new Random();
l.OrderBy(x => r.NextDouble()).Take(n);

Answer 4

Shuffle the collection into a random order and take the first n items from the result.

Answer 5

A bit less random, but efficient:

var rnd = new Random();
var toSkip = list.Count()-n;

if (toSkip > 0)
    toSkip = rnd.Next(toSkip);
else
    toSkip=0;

var randomlySelectedSequence = list.Skip(toSkip).Take(n);

Answer 6

I write this overrides method:

public static IEnumerable<T> Randomize<T>(this IEnumerable<T> items) where T : class
{
     int max = items.Count();
     var secuencia = Enumerable.Range(1, max).OrderBy(n => n * n * (new Random()).Next());

     return ListOrder<T>(items, secuencia.ToArray());
}

private static IEnumerable<T> ListOrder<T>(IEnumerable<T> items, int[] secuencia) where T : class
        {
            List<T> newList = new List<T>();
            int count = 0;
            foreach (var seed in count > 0 ? secuencia.Skip(1) : secuencia.Skip(0))
            {
                newList.Add(items.ElementAt(seed - 1));
                count++;
            }
            return newList.AsEnumerable<T>();
        }

Then, I have my source list (all items)

var listSource = p.Session.QueryOver<Listado>(() => pl)
                        .Where(...);

Finally, I call "Randomize" and I get a random sub-collection of items, in my case, 5 items:

var SubCollection = Randomize(listSource.List()).Take(5).ToList();

Answer 7

Sorry for ugly code :-), but

var result =yourCollection.OrderBy(p => (p.GetHashCode().ToString() + Guid.NewGuid().ToString()).GetHashCode()).Take(n);