c# - 有没有办法使用 Shuffle() 在列表中随机检索单个记录？ - C#

Question

找到这个 Post在 List<T> 中洗牌时有很好的解决方案.

但在我的例子中，我有一个类 Person定义如下:

class Person
{
   public int Id { get; set; }
   public string Name { get; set; }
   public string Position { get; set; }
}

---

这是我的实现和用法:

List<Person> workers = new List<Person>()
{
    new Person { Id = 1, Name = "Emp 1", Position = "Cashier"},
    new Person { Id = 2, Name = "Emp 2", Position = "Sales Clerk"},
    new Person { Id = 3, Name = "Emp 3", Position = "Cashier"},
    new Person { Id = 4, Name = "Emp 4", Position = "Sales Clerk"},
    new Person { Id = 5, Name = "Emp 5", Position = "Sales Clerk"},
    new Person { Id = 6, Name = "Emp 6", Position = "Cashier"},
    new Person { Id = 7, Name = "Emp 7", Position = "Sales Clerk"}
};

现在我想洗牌所有记录并获得 1 名销售员。这是我的代码并且正在运行:

var worker = workers.OrderBy(x => Guid.NewGuid()).Where(x => x.Position == "Sales Clerk").First();

// This can yield 1 of this random result (Emp 2, Emp 4, Emp 5 and Emp 7).
Console.WriteLine(worker.Name); 

但是根据给定的Post GUID 不利于随机化记录。最糟糕的是我不能使用 Shuffle()并调用Where和 First()扩展以获得所需的结果。

我如何使用 Shuffle() 做到这一点？扩展名？

Answer 1

如果问题是如何获取它以便您可以将 Shuffle() 与其余的 Linq 运算符链接起来，答案是修改 Shuffle 方法以返回对已洗牌列表的引用:

public static IEnumerable<T> 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;
    }

    return list;
}

您的代码将变为:

var worker = workers.Shuffle().Where(x => x.Position == "Sales Clerk").First();