Is there a way to retrieve a SINGLE RECORD randomly in a List using Shuffle()? - C#

Question

Found this Post and it has good solution when shuffling items in List<T>.

But in my case i have a class Person which is defined as this:

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

This is my implementation and usage:

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"}
};

Now i want to shuffle all records and get 1 Sales Clerk. Here is my code and is working:

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);

But according to the given Post GUID is not good for randomizing record. And the worst is i cant use Shuffle() and call the Where and First() extensions to get the desired result.

How can i do that with Shuffle() extension?

score 3 · Accepted Answer · answered Apr 08 '16 at 03:32

3

If the question is how to get it so you can chain Shuffle() with the rest of your Linq operators, the answer is to modify the Shuffle method to return reference to the list shuffled:

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;
}

Your code then becomes:

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

answered Apr 08 '16 at 03:32

gnalck

972
6
12

1

`var worker = workers.Shuffle().Firts(x => x.Position == "Sales Clerk");` – Jacob Seleznev Apr 08 '16 at 03:33
Awesome! Great answer! – Shift 'n Tab Apr 08 '16 at 03:37
why not to use Random.Next(n) instead of this math? What advantage does it provide? – Mike Tsayper Apr 08 '16 at 04:14
@Tsayper Random.Next(n) is limited only for a random, `Shuffle()` extension is re-usable and you can call other LINQ extensions too not only for `Where()` and `First()`. – Shift 'n Tab Apr 08 '16 at 04:50

score 1 · Answer 2 · edited Apr 08 '16 at 04:28

1

Random oRandom = new Random(); 
var worker = workers[oRandom.Next(0,workers.Count)];

edited Apr 08 '16 at 04:28

Andy

49,085
60
166
233

answered Apr 08 '16 at 04:19

slnit

35
6

Is there a way to retrieve a SINGLE RECORD randomly in a List using Shuffle()? - C#

2 Answers2