以概率从列表中选择随机元素

Question

我有一个包含四个项目（A、B、C、D）的列表。每个项目都有被选择的概率。举例来说，A 有 74% 的机会被选中，B 15%，C 7%，D 4%。

我想创建一个函数，根据其概率随机选择一个项目。

请问有什么帮助吗？

Answer 1

为您的项目定义一个类，如下所示：

class Items<T>
{
    public double Probability { get; set; }
    public T Item { get; set; }
}

然后初始化它

var initial = new List<Items<string>>
{
    new Items<string> {Probability = 74 / 100.0, Item = "A"},
    new Items<string> {Probability = 15 / 100.0, Item = "B"},
    new Items<string> {Probability = 7 / 100.0, Item = "C"},
    new Items<string> {Probability = 4 / 100.0, Item = "D"},
};

然后你需要将其转换为聚合从 0 到 1 的概率之和

var converted = new List<Items<string>>(initial.Count);
var sum = 0.0;
foreach (var item in initial.Take(initial.Count - 1))
{
    sum += item.Probability;
    converted.Add(new Items<string> {Probability = sum, Item = item.Item});
}
converted.Add(new Items<string> {Probability = 1.0, Item = initial.Last().Item});

现在您可以从

converted

集合中根据概率选择一个项目：

var rnd = new Random();
while (true)
{
    var probability = rnd.NextDouble();
    var selected = converted.SkipWhile(i => i.Probability < probability).First();
    Console.WriteLine($"Selected item = {selected.Item}");
}

注意：我的实现非常复杂。您可以使用二分搜索对其进行优化（因为

O(n)

集合中的值已排序）

Answer 2

无论如何，这是我的尝试：

converted

用途：

public class WeightedItem<T>
{
    private T value;
    private int weight;
    private int cumulativeSum;
    private static Random rndInst = new Random();

    public WeightedItem(T value, int weight)
    {
        this.value = value;
        this.weight = weight;
    }

    public static T Choose(List<WeightedItem<T>> items)
    {
        int cumulSum = 0;
        int cnt = items.Count();

        for (int slot = 0; slot < cnt; slot++)
        {
            cumulSum += items[slot].weight;
            items[slot].cumulativeSum = cumulSum;
        }

        double divSpot = rndInst.NextDouble() * cumulSum;
        WeightedItem<T> chosen =  items.FirstOrDefault(i => i.cumulativeSum >= divSpot);
        if (chosen == null) throw new Exception("No item chosen - there seems to be a problem with the probability distribution.");
        return chosen.value;
    }
}

Answer 3