你可以:

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

我需要更多的项目，而不仅仅是一个。所以，我写了这个:

public static TList GetSelectedRandom<TList>(this TList list, int count)
       where TList : IList, new()
{
    var r = new Random();
    var rList = new TList();
    while (count > 0 && list.Count > 0)
    {
        var n = r.Next(0, list.Count);
        var e = list[n];
        rList.Add(e);
        list.RemoveAt(n);
        count--;
    }

    return rList;
}

有了这个，你可以像这样随机获取你想要的元素:

var _allItems = new List<TModel>()
{
    // ...
    // ...
    // ...
}

var randomItemList = _allItems.GetSelectedRandom(10); 

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>();

为什么不:

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

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]);

为什么不[2]:

public static T GetRandom<T>(this List<T> list)
{
     return list[(int)(DateTime.Now.Ticks%list.Count)];
}