How to access random item in list?

Question

1. Create an instance of Random class somewhere. Note that it's pretty important not to create a new instance each time you need a random number. You should reuse the old instance to achieve uniformity in the generated numbers. You can have a static field somewhere (be careful about thread safety issues):

   static Random rnd = new Random();
   

2. Ask the Random instance to give you a random number with the maximum of the number of items in the ArrayList:

   int r = rnd.Next(list.Count);
   

3. Display the string:

   MessageBox.Show((string)list[r]);
   

I usually use this little collection of extension methods:

public static class EnumerableExtension
{
    public static T PickRandom<T>(this IEnumerable<T> source)
    {
        return source.PickRandom(1).Single();
    }

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

For a strongly typed list, this would allow you to write:

var strings = new List<string>();
var randomString = strings.PickRandom();

If all you have is an ArrayList, you can cast it:

var strings = myArrayList.Cast<string>();

You can do:

list.OrderBy(x => Guid.NewGuid()).FirstOrDefault()

Or simple extension class like this:

public static class CollectionExtension
{
    private static Random rng = new Random();

    public static T RandomElement<T>(this IList<T> list)
    {
        return list[rng.Next(list.Count)];
    }

    public static T RandomElement<T>(this T[] array)
    {
        return array[rng.Next(array.Length)];
    }
}

Then just call:

myList.RandomElement();

Works for arrays as well.

I would avoid calling OrderBy() as it can be expensive for larger collections. Use indexed collections like List<T> or arrays for this purpose.