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;
}

为什么不:

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

你可以:

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

我通常使用这个扩展方法的小集合:

public static class EnumerableExtension
{
    public static T PickRandom<T>(this IEnumerable<T> source)
    {
        return source.PickRandom(1).Single();
    }

    public static IEnumerable<T> PickRandom<T>(this IEnumerable<T> source, int count)
    {
        return source.Shuffle().Take(count);
    }

    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source)
    {
        return source.OrderBy(x => Guid.NewGuid());
    }
}

对于强类型列表，可以这样写:

var strings = new List<string>();
var randomString = strings.PickRandom();

如果你只有一个数组列表，你可以强制转换它:

var strings = myArrayList.Cast<string>();

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]);

创建一个随机实例:

Random rnd = new Random();

获取一个随机字符串:

string s = arraylist[rnd.Next(arraylist.Count)];

记住，如果你经常这样做，你应该重用随机对象。将它作为类中的静态字段，这样它只初始化一次，然后访问它。