C ++函数用于从列表中挑选，其中每个元素具有不同的概率 [英] C++ function for picking from a list where each element has a distinct probability

问题描述

我有一个结构体数组，结构体中的一个字段是一个浮点数。我想选择一个结构体，其中选择它的概率是相对于float的值。即

  struct s {
 float probability; 
 ... 
} 
 
 s sArray [50]; 
  

最快的方法是决定选择哪一个？有这样的功能吗？如果我知道所有概率字段的总和（注意它不会是1），那么我可以迭代通过每个s，并与随机数比较 probability / total_probability 每个s的随机数？

  if（（float）（rand（）/ RAND_MAX）< probability）... 
  
 
 
解决方案
 
 static_cast< float>（RAND_MAX））* total_probability; 
 s * current =& sArray [0]; 
 while（（p  -  = current-> probability）> 0）
 ++ current; 
 //`current`现在指向你选择的目标
  
 
I have an array of structs and one of the fields in the struct is a float.  I want to pick one of the structs where the probability of picking it is relative to the value of the float.  ie
struct s{
  float probability;
  ...
}

s sArray[50];
What is the fastest way to decide which s to pick?  Is there a function for this?  If I knew the sum of all the probability fields (Note it will not be 1), then could I iterate through each s and compare probability/total_probability with a random number, changing the random number for each s? ie
if( (float) (rand() / RAND_MAX) < probability)...

解决方案
float p = (rand() / static_cast<float>(RAND_MAX)) * total_probability;
s* current = &sArray[0];
while ( (p -= current->probability) > 0)
    ++current;
// `current` now points to your chosen target


                        
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