概率分布的随机数发生器 [英] Distributed probability random number generator

问题描述

欲生成基于分布式概率的数字。例如，只说有每个数字下面OCCURENCES：

I want to generate a number based on a distributed probability. For example, just say there are the following occurences of each numbers:

Number| Count           
1    |  150                
2    |  40          
3    |  15          
4    |  3  

with a total of (150+40+15+3) = 208     
then the probability of a 1 is 150/208= 0.72    
and the probability of a 2 is 40/208 = 0.192    

我如何做一个随机数生成器返回基于此概率分布的数字？

How do I make a random number generator that returns be numbers based on this probability distribution?

我很高兴这是基于一个静态的，硬$ C $光盘一套，但现在我最终还是希望它得到数据库查询的概率分布。

I'm happy for this to be based on a static, hardcoded set for now but I eventually want it to derive the probability distribution from a database query.

我见过像类似的例子这一个，但他们都不是很普通的。有什么建议？

I've seen similar examples like this one but they are not very generic. Any suggestions?

推荐答案

的一般方法是从0..1区间饲料均匀分布的随机数到<一个href=\"http://en.wikipedia.org/wiki/Cumulative_distribution_function#Inverse_distribution_function_.28quantile_function.29\">the你所需的分布的累积分布函数的倒数。

The general approach is to feed uniformly distributed random numbers from 0..1 interval into the inverse of the cumulative distribution function of your desired distribution.

因此​​，在你的情况下，只需使用一个随机数x，从0..1（例如使用<一个href=\"http://msdn.microsoft.com/en-us/library/system.random.nextdouble.aspx\"><$c$c>Random.NextDouble())并根据其价值回归

Thus in your case, just draw a random number x from 0..1 (for example with Random.NextDouble()) and based on its value return

• 1，如果0℃= X＆LT;二百零八分之一百五十零，


• 2，如果二百零八分之一百五十零＆LT; = X＆LT;二百零八分之一百九十零，


• 3，如果二百零八分之一百九十〇＆LT; = X＆LT;二百零八分之二百零五和


• 4，否则。

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