Access random item in list.

Question

Hi All,

I have an ArrayList, and I need to be able to click a button and then randomly pick out a string from that list and display it in a messagebox.

How would I go about doing this? Any help at all is greatly appreciated.

Thank you

Bael

Answer 1

Create a Random instance:

Random rnd = new Random();

Fetch a random string:

string s = arraylist[rnd.Next(arraylist.Length)];

Remember though, that if you do this frequently you should re-use the Random object. Put it as a static field in the class so it's initialized only once and then access it.

Answer 2

You want to adapt that and store the random instance somewhere. The concept is:

var itemToShow = yourList[new Random().Next(yourList.Count-1)];

Answer 3

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:

   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]);
   

Answer 4

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>();

Answer 5

You can do:

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