Linq Orderby random ThreadSafe for use in ASP.NET

i’m using Asp.net MVC with Sharp Architecture.

I have this code:

return _repositoryKeyWord.FindAll(x => x.Category.Id == idCAtegory)
                .Take(50).ToList();

How can i order by random?
Note: i don’t want to order the 50 extracted items, i want order before and then extract 50 items.

thks

Answers:

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Method 1

One way to achieve efficiently is to add a column to your data Shuffle that is populated with a random int (as each record is created).

The query to access the table then becomes …

Random random = new Random();
int seed = random.Next();
result = result.OrderBy(s => (~(s.Shuffle & seed)) & (s.Shuffle | seed)); // ^ seed);

This does an XOR operation in the database and orders by the results of that XOR.

Advantages:-

1. Efficient: SQL handles the
   ordering, no need to fetch the whole
   table
2. Repeatable: (good for
   testing) – can use the same random
   seed to generate the same random
   order
3. Works on most (all?) Entity Framework supported
   databases

This is the approach used by my home automation system to randomize playlists. It picks a new seed each day giving a consistent order during the day (allowing easy pause / resume capabilities) but a fresh look at each playlist each new day.

Method 2

You can do this in T-Sql as described here. I don’t think you can do it in linq without loading the whole result set into memory and then throwing most of it away, which you do not want to do.

Method 3

Random random = new Random();
return _repositoryKeyWord.FindAll(x => x.Category.Id == idCAtegory)
                .OrderBy(x => r.Next())
                .Take(50).ToList();

Method 4

Probably best to write your own extension method to do it.

public static class Extensions
{
    static readonly Random random = new Random();

    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> items)
    {
        return Shuffle(items, random);
    }

    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> items, Random random)
    {
        // Un-optimized algorithm taken from
        // http://en.wikipedia.org/wiki/Knuth_shuffle#The_modern_algorithm
        List<T> list = new List<T>(items);
        for (int i = list.Count - 1; i >= 1; i--) 
        {
            int j = random.Next(0, i);
            T temp = list[i];
            list[i] = list[j];
            list[j] = temp;
        }
        return list;
    }
}

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0