An extension method on IEnumerable needed for shuffling

Question

I need an extension method which will shuffle an IEnumerable<T>. It can also take an int to specify the size of the returned IEnumerable. Better keeping Immutability of the IEnumerable. My current solution for IList-

public static IList<T> Shuffle<T>(this IList<T> list, int size)
{
    Random rnd = new Random();
    var res = new T[size];

    res[0] = list[0];
    for (int i = 1; i < size; i++)
    {
        int j = rnd.Next(i);
        res[i] = res[j];
        res[j] = list[i];
    }
    return res;
}

public static IList<T> Shuffle<T>(this IList<T> list)
{ return list.Shuffle(list.Count); }

Answer 1

You can use a Fisher-Yates-Durstenfeld shuffle. There's no need to explicitly pass a size argument to the method itself, you can simply tack on a call to Take if you don't need the entire sequence:

var shuffled = originalSequence.Shuffle().Take(5);

// ...

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(nameof(source));
        if (rng == null) throw new ArgumentNullException(nameof(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];
        }
    }
}