从整数列表中获取随机数

Question

如果我有一个整数列表:

List<int> myValues = new List<int>(new int[] { 1, 2, 3, 4, 5, 6 } );

如何从该列表中获得3个随机整数？

Answer 1

一个简单的方法:

Random r = new Random();
IEnumerable<int> threeRandom = myValues.OrderBy(x => r.Next()).Take(3);

更好的方式:费雪耶茨洗牌:

public static class EnumerableExtensions
{
    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source)
    {
        return source.Shuffle(new Random());
    }

    public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng)
    {
        if (source == null) throw new ArgumentNullException("source");
        if (rng == null) throw new ArgumentNullException("rng");

        return source.ShuffleIterator(rng);
    }

    private static IEnumerable<T> ShuffleIterator<T>(
        this IEnumerable<T> source, Random rng)
    {
        List<T> buffer = source.ToList();
        for (int i = 0; i < buffer.Count; i++)
        {
            int j = rng.Next(i, buffer.Count);
            yield return buffer[j];

            buffer[j] = buffer[i];
        }
    }
}

你如何使用它:

IEnumerable<int> threeRandom = myValues.Shuffle().Take(3);

Answer 2

最简单的方法是这样的:

var r = new Random();
var myValues = new int[] { 1, 2, 3, 4, 5, 6 }; // Will work with array or list
var randomValues = Enumerable.Range(0, 3)
    .Select(e => myValues[r.Next(myValues.Length)]);

但是一个更好的方法,如果你想确保没有重复,就是使用一个改组算法,比如Fisher-Yates算法,然后取前3项:

public static T[] Shuffle<T>(IEnumerable<T> items)
{
    var result = items.ToArray();
    var r = new Random();
    for (int i = items.Length; i > 1; i--)
    {
        int j = r.Next(i);
        var t = result[j];
        result[j] = result[i - 1];
        result[i - 1] = t;
    }

    return result;
}

var myValues = new int[] { 1, 2, 3, 4, 5, 6 }; // Will work with any enumerable
var randomValues = myValues.Shuffle().Take(3);