score:50

Accepted answer

What about:

SelectedPost = q.ElementAt(r.Next(1, Answers.Count()));

Further reading:

The comments below make good contributions to closely related questions, and I'll include them here, since as @Rouby points out, people searching for an answer to these may find this answer and it won't be correct in those cases.

Random Element Across Entire Input

To make all elements a candidate in the random selection, you need to change the input to r.Next:

SelectedPost = Answers.ElementAt(r.Next(0, Answers.Count()));

@Zidad adds a helpful extension method to get random element over all elements in the sequence:

public static T Random<T>(this IEnumerable<T> enumerable)
{
    if (enumerable == null)
    {
         throw new ArgumentNullException(nameof(enumerable));
    }

    // note: creating a Random instance each call may not be correct for you,
    // consider a thread-safe static instance
    var r = new Random();  
    var list = enumerable as IList<T> ?? enumerable.ToList(); 
    return list.Count == 0 ? default(T) : list[r.Next(0, list.Count)];
}

score:0

I have product table in database ,every time user enters one product detail I want to show 10 similar products in below of page.And in every refresh this list must be change .it must come randomly.

Linq looks like this

var products =
            DataContextFactory.GetDataContext()
                .Set<Product>()
                .Where(x =>x.Id!=id)
                .OrderBy(emp => Guid.NewGuid())
                .Take(10).ToList();

x.Id!=id 

this only for not put selected product to list .

It works perfect

score:1

Pulling all of the answers and looping them isn't the most efficient way as you're moving lots of data from the database. If you're using an integer primary key that's automatically incrementing, you should get the Max of your primary key and then find the random integer within that range. Then directly get the single answer based on the primary key derived from the random function.

score:1

I'm posting an answer because I don't have enough reputation to comment.

I like this answer:

SelectedPost = q.ElementAt(r.Next(1, Answers.Count()));

But ElementAt is zero based, surely starting at 1 and going to Answers.Count() you are going to end up potentially throwing an out of range, and you are never going to get the first entity.

Wouldn't

SelectedPost = q.ElementAt(r.Next(0, Answers.Count() - 1));

Be better?

score:4

var rand = new Random();
var selectedPost = q.Skip(rand.Next(0, q.Count())).Take(1).FirstOrDefault();

Optimally, you want to only ever make the function query for a single value, so you set up the Skip/Take to jump up to the sequence number matching the random number you're generating (bounded by dataset's itemcount, so the missing row problem bounding based on MAX(pkey) isn't an issue) and then snag the first item at that point in the sequence.

In SQL this is the same as querying for SELECT Count(*) FROM q, then SELECT * FROM q LIMIT {0}, 1 where {0} is rand.Next(0, count), which should be pretty efficient.

score:4

Generic extension method based on the accepted answer (which doesn't always skip the first, and only enumerates the enumerable once):

 public static class EnumerableExtensions
    {
        public static T Random<T>(this IEnumerable<T> enumerable)
        {
            var r = new Random();
            var list = enumerable as IList<T> ?? enumerable.ToList();
            return list.ElementAt(r.Next(0, list.Count()));
        }
    }

score:4

Late to the party but this is a high-up Google result. A succinct version could be:

var rnd = new Random();
var SelectedPost = q.OrderBy(x => rnd.Next()).Take(1);

It has the disadvantage that it'll apply a random number to all elements, but is compact and could easily be modified to take more than one random element.

score:10

Another wacky approach (not the most efficient for larger data sets):

SelectedPost = q.OrderBy(qu => Guid.NewGuid()).First();

score:10

Use a Fisher-Yates-Durstenfeld shuffle.

(You could use a helper/extension method to shuffle your IEnumerable<T> sequence. Alternatively, if you were using an IList<T> you could perform an in-place shuffle, if you prefer.)


Related Query

More Query from same tag