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

Answer 1

How about something simple and readable:

ints[randgen.Next(ints.Length)];

Seriously, why obfuscate your code with lambdas .OrderBy and .First and .Skip and so forth!?

Answer 2

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).

Answer 3

No, that's basically the easiest way. Of course, that's only semi-random, but I think it fits most needs.

EDIT: Huge Point Here...

If you only want ONE value randomly chosen from the list... then just do this:

var myRandomValue = ints[(new Random()).Next(0, ints.Length)];

That's a O(1) operation.

Answer 4

Sorting will be far less efficient. Just use Skip(n) and First():

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

ints.Skip(x=> randgen.Next(0, ints.Count())).First();

ints.ElementAt(x=> randgen.Next(0, ints.Count()));