C#: Elegant code for getting a random value from an IEnumerable

Question

In Python, I can do this:

>>> import random
>>> ints = [1,2,3]
>>> random.choice(ints)
3

In C# the first thing I did was:

var randgen = new Random();
var ints = new int[] { 1, 2, 3 };
ints[randgen.Next(ints.Length)];

But this requires indexing, also the duplication of ints bothers me. So, I came up with this:

var randgen = new Random();
var ints = new int[] { 1, 2, 3 };
ints.OrderBy(x=> randgen.Next()).First();

Still not very nice and efficient. Is there a more elegant way of getting a random value from an IEnumberable?

Answer 1

Here's a couple extension methods for you:

public static T RandomElement<T>(this IEnumerable<T> enumerable)
{
    return enumerable.RandomElementUsing<T>(new Random());
}

public static T RandomElementUsing<T>(this IEnumerable<T> enumerable, Random rand)
{
    int index = rand.Next(0, enumerable.Count());
    return enumerable.ElementAt(index);
}

// Usage:
var ints = new int[] { 1, 2, 3 };
int randomInt = ints.RandomElement();

// If you have a preexisting `Random` instance, rand, use it:
// this is important e.g. if you are in a loop, because otherwise you will create new
// `Random` instances every time around, with nearly the same seed every time.
int anotherRandomInt = ints.RandomElementUsing(rand);

For a general IEnumerable<T>, this will be O(n), since that is the complexity of .Count() and a random .ElementAt() call; however, both special-case for arrays and lists, so in those cases it will be O(1).