随机函数和计算百分比

Question

使用具有这些功能的随机库：

randomChance(p) Returns true with the probability indicated by p.

randomInteger(low, high) Returns a random integer in the range low to high, inclusive.

实现“随机选择器”的最简单方法是什么，它考虑了百分比，1/4或1/3等...我得到了一个带键/值配对的数组。例如，"a" migth的值为2，"b"的值为2 1/2。

最大值将是数组的大小，因为它只包含唯一的项目。 randomChance()函数的范围介于0.0 - 1.0之间，其中1 = 100％。如果我的数组大小是4，那么“让4为1”的最佳方法是什么。

Answer 1

让我们说：

a = 2, b = 2, c = 1, d = 3

现在成功：

a = 2, b = 4, c = 5, d = 8

创建一个从1到MaxVal的随机数（最后一个键的值，本例中为8）。选择第一个Key，其中Value＆gt; = RandomNum

修改

我做了一个小的VB.Net来显示算法及其工作原理。代码并不意味着：良好，优雅，高效或可读。

Module Module1

    Private Class Value
        Public vOrg, vRecalc, HitCount As Integer
        Public Key As String
        Public Sub New(s, v1, v2, c)
            Key = s : vOrg = v1 : vRecalc = v2 : HitCount = c
        End Sub
    End Class

    Sub Main()

        ' set initial values
        Dim KVP() As Value = {New Value("A", 2, 0, 0),
                              New Value("B", 2, 0, 0),
                              New Value("C", 1, 0, 0),
                              New Value("D", 3, 0, 0)}
        ' recalc values
        For i = 0 To KVP.Length - 1
            If i = 0 Then KVP(0).vRecalc = KVP(0).vOrg Else KVP(i).vRecalc = KVP(i).vOrg + KVP(i - 1).vRecalc
        Next
        ' do test
        Dim r As New Random
        Dim runs As Integer = 1000 * 1000, maxval As Integer = KVP(KVP.Length - 1).vRecalc
        For i = 1 To runs
            Dim RandVal = r.Next(1, maxval + 1)
            Dim chosen As Integer = (From j In Enumerable.Range(0, KVP.Length) Where KVP(j).vRecalc >= RandVal Take 1 Select j)(0)
            KVP(chosen).HitCount += 1
        Next
        ' ouput results
        For Each kv In KVP
            Console.WriteLine("{0} was chosen with {1:F3} propability, expected was {2:F3}", kv.Key, kv.HitCount / CDbl(runs), kv.vOrg / CDbl(maxval))
        Next

        Console.ReadLine()

    End Sub

End Module

输出样本：

A was chosen with 0.250 propability, expected was 0.250
B was chosen with 0.251 propability, expected was 0.250
C was chosen with 0.124 propability, expected was 0.125
D was chosen with 0.375 propability, expected was 0.375

Answer 2

将randomChance（）结果与数组长度相乘。它将为您提供可用于访问数组的范围[0，array_length-1]中的索引

array_index = (unsigned int)(randomChance(p) * (array_length - 1));

也许你的意思是“让3成为1”（而不是4）。长度为4的数组的最后一个索引是3。