计算该项目旁边的百分比分配百分比、该项目、分配
我的工作，涉及转接电话到多个目标的项目。
I'm working on a project that involves diverting phone calls to a number of destinations.
例如，我想：
呼叫 10％，去到目的地A 呼叫 20％，去到目的地乙 呼叫 30％，去到目的地ç 呼叫 40％，去到目的地ð 10% of calls to go to destination A 20% of calls to go to destination B 30% of calls to go to destination C 40% of calls to go to destination D目的地及比例的数目必须是可配置的。
The number of destinations and their percentages must be configurable.
我一直在思考如何做到这一点，玩弄以s preadsheets有的code，我想出了这一点：
I've been thinking about how to do this, playing around with spreadsheets and some code, and I came up with this:
对于每个目标，采取随机数，按比例乘以它，并选择具有最高数量的目的。像这样的：
For each destination, take a random number, multiply it by the percentage, and select the destination with the highest number. Like this:
Item: RANDOM * PERCENTAGE = RESULT
A: 48 * 10 = 480
B: 33 * 20 = 660
C: 81 * 30 = 2430 <--- Highest number, select C
D: 5 * 40 = 200
我想我的工作了为D显然是选择了最，其次是C，那么B，并至少所有A。
I thought I'd worked it out as D would clearly be selected the most, followed by C, then B, and least of all A.
不过，这是行不通的。如果我这样做5000次，并计算时代每个目的地被选中的实际比例，我得到这样的：
But it doesn't work. If I do this 5000 times, and calculate the actual percentage of times each destination was selected, I get this:
呼叫 1％去到目的地A 呼叫 12％，去到目的地乙 呼叫 31％，去到目的地ç 呼叫 56％，去到目的地ð 1% of calls to go to destination A 12% of calls to go to destination B 31% of calls to go to destination C 56% of calls to go to destination D下面是code我用来测试这一点：
Here is the code I used to test this:
// 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)));
借助结果是：
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%
任何人都可以提出一个方法来解决这个问题，这样的配置真正的百分比出来？
Can anyone suggest a way to fix this so that the real percentages come out as configured?
和奖励积分，任何人都可以提出一个方法做同样的事情，但没有使用的随机数，使所选目的地的顺序明确的规定。用上面的例子会输出这个序列中的每个时间：
And for bonus points, can anyone suggest a way to do something similar but not using random numbers, so that the sequence of selected destinations is clearly defined. Using the example above would output this sequence every time:
ABCDBCDCDD ABCDBCDCDD ABCDBCDCDD ABCDBCDCDD ...
（注意，该序列被均匀地分布）
(note that the sequence is evenly distributed)
感谢。 本
推荐答案
好吧，我以前在模拟做过无数次，这里的基本方法是我使用（不正确的错误检查）：
Ok, I have done this numerous times before in simulations so here is the basic method I use (without proper error checking):
您需要想象一下在页面上会从0到100。现在我们正在做的是分裂这一阵容当中比例的目的地线绘制。然后，我们使用随机数来选择这条线上的一个点。其具有线的那个区域的目的地是选择之一。
You need to imagine a line draw across the page going from 0 to 100. Now what we're doing is dividing this line up proportionally amongst your destinations. We then use random numbers to choose a point on this line. The destination which has that area of the line is the one selected.
编辑：尝试在线路图
|-----------------------------------------------------| Line 1 to 100
|-----|----------|---------------|--------------------| Line split proportionally
0 A 10 B 30 C 60 D 100
我们可以按如下做到这一点。
We can do this as follows.
假设你的目的地的百分数都是在一个阵列，而不是在单独的变量
Assume your destination percentages are in an array, instead of in separate variables.
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 指向选定的目标。
Now the variable destinationIndex points to the selected destination.
