我认为我已将此作为随机化列表的最简单且可单元测试的方法,但有兴趣听到任何改进.
public static IList<T> RandomiseList<T>(IList<T> list, int seed)
{
Random random = new Random(seed);
List<T> takeFrom = new List<T>(list);
List<T> ret = new List<T>(takeFrom.Count);
while (takeFrom.Count > 0)
{
int pos = random.Next(0, takeFrom.Count - 1);
T item = takeFrom[pos];
takeFrom.RemoveAt(pos);
ret.Add(item);
}
return ret;
}
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Joe*_*orn 17
你想要一个洗牌,最好的办法是Fisher-Yates shuffle:
public static IList<T> Randomise<T>(IList<T> list, int seed)
{
Random rng = new Random(seed);
List<T> ret = new List<T>(list);
int n = ret.Length;
while (n > 1)
{
n--;
int k = rng.Next(n + 1);
// Simple swap of variables
T tmp = list[k];
ret[k] = ret[n];
ret[n] = tmp;
}
return ret;
}
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Guf*_*ffa 15
我喜欢Dennis Palmers想要返回一个洗牌的IEnumerable,而不是将列表放在适当位置,但是使用RemoveAt方法会让它变慢.这是一个没有RemoveAt方法的替代方法:
public static IEnumerable<T> Shuffle<T>(IEnumerable<T> list, int seed) {
Random rnd = new Random(seed);
List<T> items = new List<T>(list);
for (int i = 0; i < items.Count; i++) {
int pos = rnd.Next(i, items.Count);
yield return items[pos];
items[pos] = items[i];
}
}
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我用10000整数来扼杀它,它的速度提高了大约30倍.