解决这类问题的一个非常简单的方法是在列表中使用一些随机的元素交换。

在伪代码中是这样的:

do 
    r1 = randomPositionInList()
    r2 = randomPositionInList()
    swap elements at index r1 and index r2 
for a certain number of times

这里是Fisher-Yates shuffle的实现，允许指定返回的元素数量;因此，在获取所需数量的元素之前，没有必要首先对整个集合进行排序。

交换元素的顺序与默认值相反;从第一个元素到最后一个元素，因此检索集合的一个子集与洗牌整个集合产生相同的(部分)序列:

collection.TakeRandom(5).SequenceEqual(collection.Shuffle().Take(5)); // true

该算法基于Durstenfeld在维基百科上的(现代)Fisher-Yates shuffle。

public static IList<T> TakeRandom<T>(this IEnumerable<T> collection, int count, Random random) => shuffle(collection, count, random);
public static IList<T> Shuffle<T>(this IEnumerable<T> collection, Random random) => shuffle(collection, null, random);
private static IList<T> shuffle<T>(IEnumerable<T> collection, int? take, Random random)
{
    var a = collection.ToArray();
    var n = a.Length;
    if (take <= 0 || take > n) throw new ArgumentException("Invalid number of elements to return.");
    var end = take ?? n;
    for (int i = 0; i < end; i++)
    {
        var j = random.Next(i, n);
        (a[i], a[j]) = (a[j], a[i]);
    }

    if (take.HasValue) return new ArraySegment<T>(a, 0, take.Value);
    return a;
}

    public static List<T> Randomize<T>(List<T> list)
    {
        List<T> randomizedList = new List<T>();
        Random rnd = new Random();
        while (list.Count > 0)
        {
            int index = rnd.Next(0, list.Count); //pick a random item from the master list
            randomizedList.Add(list[index]); //place it at the end of the randomized list
            list.RemoveAt(index);
        }
        return randomizedList;
    }

使用基于Fisher-Yates Shuffle的扩展方法Shuffle任意(I)List:

private static Random rng = new Random();  

public static void Shuffle<T>(this IList<T> list)  
{  
    int n = list.Count;  
    while (n > 1) {  
        n--;  
        int k = rng.Next(n + 1);  
        T value = list[k];  
        list[k] = list[n];  
        list[n] = value;  
    }  
}

用法:

List<Product> products = GetProducts();
products.Shuffle();

上面的代码使用了备受批评的系统。选择交换候选的随机方法。它速度很快，但并不像它应该的那样随机。如果你需要更好的随机质量，可以使用System.Security.Cryptography中的随机数生成器，如下所示:

using System.Security.Cryptography;
...
public static void Shuffle<T>(this IList<T> list)
{
    RNGCryptoServiceProvider provider = new RNGCryptoServiceProvider();
    int n = list.Count;
    while (n > 1)
    {
        byte[] box = new byte[1];
        do provider.GetBytes(box);
        while (!(box[0] < n * (Byte.MaxValue / n)));
        int k = (box[0] % n);
        n--;
        T value = list[k];
        list[k] = list[n];
        list[n] = value;
    }
}

一个简单的比较可以在这个博客(WayBack Machine)上找到。

Edit: Since writing this answer a couple years back, many people have commented or written to me, to point out the big silly flaw in my comparison. They are of course right. There's nothing wrong with System.Random if it's used in the way it was intended. In my first example above, I instantiate the rng variable inside of the Shuffle method, which is asking for trouble if the method is going to be called repeatedly. Below is a fixed, full example based on a really useful comment received today from @weston here on SO.

Program.cs:

using System;
using System.Collections.Generic;
using System.Threading;

namespace SimpleLottery
{
  class Program
  {
    private static void Main(string[] args)
    {
      var numbers = new List<int>(Enumerable.Range(1, 75));
      numbers.Shuffle();
      Console.WriteLine("The winning numbers are: {0}", string.Join(",  ", numbers.GetRange(0, 5)));
    }
  }

  public static class ThreadSafeRandom
  {
      [ThreadStatic] private static Random Local;

      public static Random ThisThreadsRandom
      {
          get { return Local ?? (Local = new Random(unchecked(Environment.TickCount * 31 + Thread.CurrentThread.ManagedThreadId))); }
      }
  }

  static class MyExtensions
  {
    public static void Shuffle<T>(this IList<T> list)
    {
      int n = list.Count;
      while (n > 1)
      {
        n--;
        int k = ThreadSafeRandom.ThisThreadsRandom.Next(n + 1);
        T value = list[k];
        list[k] = list[n];
        list[n] = value;
      }
    }
  }
}

    List<T> OriginalList = new List<T>();
    List<T> TempList = new List<T>();
    Random random = new Random();
    int length = OriginalList.Count;
    int TempIndex = 0;

    while (length > 0) {
        TempIndex = random.Next(0, length);  // get random value between 0 and original length
        TempList.Add(OriginalList[TempIndex]); // add to temp list
        OriginalList.RemoveAt(TempIndex); // remove from original list
        length = OriginalList.Count;  // get new list <T> length.
    }

    OriginalList = new List<T>();
    OriginalList = TempList; // copy all items from temp list to original list.

编辑 RemoveAt是我以前版本的一个弱点。这个解决方案克服了这个问题。

public static IEnumerable<T> Shuffle<T>(
        this IEnumerable<T> source,
        Random generator = null)
{
    if (generator == null)
    {
        generator = new Random();
    }

    var elements = source.ToArray();
    for (var i = elements.Length - 1; i >= 0; i--)
    {
        var swapIndex = generator.Next(i + 1);
        yield return elements[swapIndex];
        elements[swapIndex] = elements[i];
    }
}

请注意可选的Random生成器，如果Random的基本框架实现不是线程安全的，或者加密性不够强，您可以将您的实现注入到操作中。

在这个答案中可以找到线程安全的加密强随机实现的合适实现。

这里有一个想法，以一种(希望)有效的方式扩展IList。

public static IEnumerable<T> Shuffle<T>(this IList<T> list)
{
    var choices = Enumerable.Range(0, list.Count).ToList();
    var rng = new Random();
    for(int n = choices.Count; n > 1; n--)
    {
        int k = rng.Next(n);
        yield return list[choices[k]];
        choices.RemoveAt(k);
    }

    yield return list[choices[0]];
}