How to access random item in list?

问题:

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?

回答1:

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]);
   

回答2:

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>();

回答3:

You can do:

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

回答4:

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.

回答5:

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.

回答6:

Why not:

public static T GetRandom<T>(this IEnumerable<T> list)
{
   return list.ElementAt(new Random(DateTime.Now.Millisecond).Next(list.Count()));
}

回答7:

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());

回答8:

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;
}

回答9:

I needed to more item instead of just one. So, I wrote this:

public static TList GetSelectedRandom<TList>(this TList list, int count)
       where TList : IList, new()
{
    var r = new Random();
    var rList = new TList();
    while (count > 0 && list.Count > 0)
    {
        var n = r.Next(0, list.Count);
        var e = list[n];
        rList.Add(e);
        list.RemoveAt(n);
        count--;
    }

    return rList;
}

With this, you can get elements how many you want as randomly like this:

var _allItems = new List<TModel>()
{
    // ...
    // ...
    // ...
}

var randomItemList = _allItems.GetSelectedRandom(10); 

回答10:

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);

回答11:

Why not[2]:

public static T GetRandom<T>(this List<T> list)
{
     return list[(int)(DateTime.Now.Ticks%list.Count)];
}