计算百分比分布中的下一个项目

Question

我正在开展一个项目,涉及将电话转移到多个目的地.

例如,我想:

• 10%的电话前往目的地A.
• 20%的电话前往目的地B.
• 30%的电话转到目的地C.
• 40%的电话前往目的地D.

目的地数量及其百分比必须是可配置的.

---

我一直在考虑如何做到这一点,玩电子表格和一些代码,我想出了这个:

对于每个目的地,取一个随机数,乘以百分比,然后选择编号最大的目的地.像这样:

Item: RANDOM * PERCENTAGE = RESULT
   A:   48   *     10     =   480
   B:   33   *     20     =   660
   C:   81   *     30     =  2430  <--- Highest number, select C
   D:    5   *     40     =   200

我以为我已经解决了,因为D显然会被选中最多,其次是C,然后是B,最不是A.

但它不起作用.如果我这样做5000次,并计算每个目的地被选中的实际百分比,我得到这个:

• 1%的电话转到目的地A.
• 12%的电话前往目的地B.
• 31%的电话转到目的地C.
• 56%的电话前往目的地D.

---

这是我用来测试这个的代码:

// Initialise item weighting percentages
Dictionary<string, int> weighting = new Dictionary<string, int>();
weighting["A"] = 10; //10%
weighting["B"] = 20; //20%
weighting["C"] = 30; //30%
weighting["D"] = 40; //40% (total = 100%)

// Initialise data set used for each iteration
Dictionary<string, int> data = new Dictionary<string, int>();

// Initialise counts of the selected items
Dictionary<string, int> count = new Dictionary<string, int>();
count["A"] = 0;
count["B"] = 0;
count["C"] = 0;
count["D"] = 0;

Random rand = new Random();

// Loop 5000 times
for (int i = 0; i < 5000; i++) {

    // For each item, get a random number between 0 and 99
    // and multiply it by the percentage to get a
    // weighted random number.
    data["A"] = rand.Next(100) * weighting["A"];
    data["B"] = rand.Next(100) * weighting["B"];
    data["C"] = rand.Next(100) * weighting["C"];
    data["D"] = rand.Next(100) * weighting["D"];

    // Find which item came out on top and increment the count
    string sel = data.First(x => x.Value == data.Max(y => y.Value)).Key;
    count[sel]++;

    // Log, so you can see whats going on...
    if (i < 15)
        Console.WriteLine("A:{0:00000}  B:{1:00000}  C:{2:00000}  D:{3:00000}  SELECTED:{4}",
            data["A"], data["B"], data["C"], data["D"], sel);
    else if (i == 15) Console.WriteLine("...");

}

// Output the results, showing the percentage of the number
// occurrances of each item.
Console.WriteLine();
Console.WriteLine("Results: ");
Console.WriteLine("    A = {0}%", 100 * ((double)count["A"] / (double)count.Sum(z => z.Value)));
Console.WriteLine("    B = {0}%", 100 * ((double)count["B"] / (double)count.Sum(z => z.Value)));
Console.WriteLine("    C = {0}%", 100 * ((double)count["C"] / (double)count.Sum(z => z.Value)));
Console.WriteLine("    D = {0}%", 100 * ((double)count["D"] / (double)count.Sum(z => z.Value)));

该结果是:

A:00780  B:00300  C:01740  D:03680  SELECTED:D
A:00600  B:00660  C:00060  D:03400  SELECTED:D
A:00900  B:01880  C:00510  D:00720  SELECTED:B
A:00260  B:01380  C:00540  D:01520  SELECTED:D
A:00220  B:01960  C:00210  D:02080  SELECTED:D
A:00020  B:01400  C:01530  D:00120  SELECTED:C
A:00980  B:00400  C:01560  D:03280  SELECTED:D
A:00330  B:00300  C:01500  D:03680  SELECTED:D
A:00590  B:00460  C:02730  D:02400  SELECTED:C
A:00580  B:01900  C:02040  D:01320  SELECTED:C
A:00620  B:01320  C:00750  D:01760  SELECTED:D
A:00320  B:01040  C:01350  D:03640  SELECTED:D
A:00340  B:01520  C:02010  D:03880  SELECTED:D
A:00850  B:01420  C:00480  D:03400  SELECTED:D
A:00560  B:00680  C:00030  D:00000  SELECTED:B
...

Results: 
    A = 1.44%
    B = 11.54%
    C = 30.6%
    D = 56.42%

任何人都可以建议一种方法来解决这个问题,以便真正的百分比按照配置出来吗？

---

对于奖励积分,任何人都可以建议一种类似但不使用随机数的方法,以便明确定义所选目的地的顺序.使用上面的示例每次都会输出此序列:

ABCDBCDCDD ABCDBCDCDD ABCDBCDCDD ABCDBCDCDD ...

(注意序列均匀分布)

谢谢.本

Answer 1

好吧,我之前已经在模拟中做了很多次,所以这里是我使用的基本方法(没有正确的错误检查):

你需要想象一下整个页面的线条画面从0到100.现在我们正在做的是将这条线按比例划分到你的目的地之间.然后我们使用随机数在这一行上选择一个点.具有该行区域的目的地是所选择的目的地.

编辑:尝试线图

|-----------------------------------------------------|   Line 1 to 100
|-----|----------|---------------|--------------------|   Line split proportionally
0  A  10    B    30     C        60      D           100

我们可以这样做.

假设您的目标百分比在数组中,而不是在单独的变量中.

int totalPercentages = 0; 
int destinationsIndex = -1;
int randomNumberBetween0and100 = GetRandomNumber();
for(int i = 0; i < destinationPercentageArrays.Length; i++)
{
    totalPercentages += destinationPercentageArrays[i];
    if (totalPercentages > randomNumberBetween0and100)
    {
        destinationIndex = i;
        break;
    }
}

if (destinationIndex == -1)
{
   throw new Exception("Something went badly wrong.");
}

现在变量destinationIndex指向所选目标.