How to access random item in list?

Question

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?

Answer 1

ArrayList ar = new ArrayList();
        ar.Add(1);
        ar.Add(5);
        ar.Add(25);
        ar.Add(37);
        ar.Add(6);
        ar.Add(11);
        ar.Add(35);
        Random r = new Random();
        int index = r.Next(0,ar.Count-1);
        MessageBox.Show(ar[index].ToString());

Answer 2

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]);
   

Answer 3

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 4

I have been using this ExtensionMethod for a while:

public static IEnumerable<T> GetRandom<T>(this IEnumerable<T> list, int count)
{
    if (count <= 0)
      yield break;
    var r = new Random();
    int limit = (count * 10);
    foreach (var item in list.OrderBy(x => r.Next(0, limit)).Take(count))
      yield return item;
}

Answer 5

Create a Random instance:

Random rnd = new Random();

Fetch a random string:

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

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 6

Printing randomly country name from JSON file.
Model:

public class Country
    {
        public string Name { get; set; }
        public string Code { get; set; }
    }

Implementaton:

string filePath = Path.GetFullPath(Path.Combine(Environment.CurrentDirectory, @"..\..\..\")) + @"Data\Country.json";
            string _countryJson = File.ReadAllText(filePath);
            var _country = JsonConvert.DeserializeObject<List<Country>>(_countryJson);


            int index = random.Next(_country.Count);
            Console.WriteLine(_country[index].Name);