An extension method on IEnumerable needed for shuffling [duplicate]
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); }
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];
}
}
}
With some LINQ love:
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> list, int size)
{
var r = new Random();
var shuffledList =
list.
Select(x => new { Number = r.Next(), Item = x }).
OrderBy(x => x.Number).
Select(x => x.Item).
Take(size); // Assume first @size items is fine
return shuffledList.ToList();
}