Pick random element from a List<T>

Question

Pick random element from a List<T>

Asked 10 years, 4 months ago

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6

I have a List<int> numeros

It is possible to return a random element from this list?

4 answers

4

For a have that in .NET do:

IList<int> listaNumeros = new List<int>() { 95, 4, 9, 52, 40, 800, 90, 11, 2, 9, 4, 92, 8, 91, 120, 111 };
Random rand = new Random(DateTime.Now.Millisecond);
int resultado = listaNumeros[rand.Next(listaNumeros.Count)]

Example: Demo

References:

2

It is preferable to call the builder of Random no arguments. The argument-free constructor uses Environment.TickCount (2 31 possible values) like Seed, which has a much larger entropy than DateTime.Now.Millisecond (1000 possible values).

– dcastro

2014/07/10 at 13:01
2

It is better to use default behavior. Besides, the quantity of number is irrelevant - there is no performance degradation. Quite the contrary, there is a (slight) degradation of performance in using DateTime.Now.Millisecond because it implies parsing the date of the system, interpreting the time zone, and interpreting the milliseconds of the current time. Environment.TickCount and' a direct call to the kernel and e' therefore more efficient and ensures a more random result.

– dcastro

2014/07/10 at 13:33

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by Filipe Oliveira • **2,655** points · Answer 1 · 2014-07-09T14:25:26+00:00

You can do as follows in .net:

var lista = new List<int>{3,5,1,8,4,9};
var rnd = new Random();
var valorAleatorio = lista[rnd.Next(lista.Count)];

by Omni • **6,077** points · Answer 2 · 2014-07-09T16:10:40+00:00

An addition to the other answers, in the form of a extension method:

private static Random _randGen = new Random();
public static T GetRandomElement<T>(this IList<T> source)
{
    return source[_randGen.Next(0, source.Count)];
}

by electus • **971** points · Answer 3 · 2014-07-09T14:22:13+00:00

Yes. See the following example:

    List<Integer> numeros = new ArrayList<Integer>();
    numeros.add(1);
    numeros.add(2);
    numeros.add(3);
    numeros.add(4);
    numeros.add(5);
    numeros.add(6);
    numeros.add(7);

    Random gerador = new Random();
    int index = gerador.nextInt(numeros.size());

    System.out.println(numeros.get(index));