Select single random item from list of items

Question

I have a query that returns list of results. I want a single random item from that list.

var query = (from...
            where...
            select q).Random(1); //Something like this?

How to do this? Suppose there is 1 item i want that one to be selected. If there are more then i want one of those to be selected in a random order.

score 4 · Accepted Answer · answered Jul 24 '12 at 16:40

4

var query = (from...
            where...
            orderby Guid.NewGuid()
            select q).First();

answered Jul 24 '12 at 16:40

SLaks

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score 0 · Answer 2 · edited May 23 '17 at 12:34

You could utilise the Shuffle example posted in Randomize a List in C# in conjunction with the LINQ Take method to create an extension method on an IList<T>, you'd need to ToList your selection prior to calling Random, unless you moved that inside the extension.

public static List<T> Random<T>(this IList<T> list, int takeNumber)
{
    return list.Shuffle().Take(takeNumber);
}

public static List<T> Shuffle<T>(this IList<T> list)  
{  
    Random rng = new Random();  
    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;  
    }

    return list;
}

This will execute on the client rather than in the SQL. – SLaks Jul 24 '12 at 18:11 — SLaks, Jul 24 '12 at 18:11

shtse8 · Answer 3 · 2018-08-04T20:11:44.900

This Shuffle Method is better than others answers. it will not cost any performance issue when chaining and it's still LINQ.

public static IEnumerable<T> TakeRandom<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());
}

Usage:

var query = (from...
        where...
        select q).TakeRandom(1);

Select single random item from list of items

3 Answers3