我已经使用这个ExtensionMethod一段时间了:

public static IEnumerable<T> GetRandom<T>(this IEnumerable<T> list, int count)
{
    if (count <= 0)
      yield break;
    var r = new Random();
    int limit = (count * 10);
    foreach (var item in list.OrderBy(x => r.Next(0, limit)).Take(count))
      yield return item;
}

或者像这样简单的扩展类:

public static class CollectionExtension
{
    private static Random rng = new Random();

    public static T RandomElement<T>(this IList<T> list)
    {
        return list[rng.Next(list.Count)];
    }

    public static T RandomElement<T>(this T[] array)
    {
        return array[rng.Next(array.Length)];
    }
}

然后只需呼叫:

myList.RandomElement();

也适用于数组。

我将避免调用OrderBy()，因为对于较大的集合来说，它的开销很大。为此，可以使用索引集合，如List<T>或数组。

Create an instance of Random class somewhere. Note that it's pretty important not to create a new instance each time you need a random number. You should reuse the old instance to achieve uniformity in the generated numbers. You can have a static field somewhere (be careful about thread safety issues): static Random rnd = new Random(); Ask the Random instance to give you a random number with the maximum of the number of items in the ArrayList: int r = rnd.Next(list.Count); Display the string: MessageBox.Show((string)list[r]);

你可以:

list.OrderBy(x => Guid.NewGuid()).FirstOrDefault()

为什么不:

public static T GetRandom<T>(this IEnumerable<T> list)
{
   return list.ElementAt(new Random(DateTime.Now.Millisecond).Next(list.Count()));
}

我已经使用这个ExtensionMethod一段时间了:

public static IEnumerable<T> GetRandom<T>(this IEnumerable<T> list, int count)
{
    if (count <= 0)
      yield break;
    var r = new Random();
    int limit = (count * 10);
    foreach (var item in list.OrderBy(x => r.Next(0, limit)).Take(count))
      yield return item;
}