如何基于一定概率获得值 [英] How to obtain a value based on a certain probability

问题描述

我有一些函数可以生成double，float，short，long随机值.我有另一个函数，该函数将数据类型传递给该函数，并且应该返回一个随机值.现在，我需要在该函数中根据传递的数据类型选择返回值.例如，如果我通过float，则需要:

I have some functions which generate double, float, short, long random values. I have another function to which I pass the datatype and which should return a random value. Now I need to choose in that function the return value based on the passed datatype. For example, if I pass float, I need:

返回为浮点型的概率为70％，返回为double，short或long的概率分别为10％.我可以调用另一个函数来生成相应的随机值，但是如何使概率权重适合最终的收益呢?我的代码是C ++.

the probability that the return is a float is 70%, the probability that the return is a double, short or long is 10% each. I can make calls to the other function for generating the corresponding random values, but how do I fit in the probabilistic weights for the final return? My code is in C++.

一些指针是值得赞赏的.

Some pointers are appreciated.

谢谢.

推荐答案

C ++随机数具有均匀分布.如果您需要随机变量. org/wiki/Probability_distribution"rel =" nofollow noreferrer>分布，您需要基于统一分布来建立其数学公式.

C++ random numbers have uniform distribution. If you need random variables of another distribution you need to base its mathematical formula on uniform distribution.

如果您的随机变量没有数学公式，则可以执行以下操作:

If you don't have a mathematical formula for your random variable you can do something like this:

int x = rand() % 10;
if (x < 7)
{
 // return float
}
else (if x == 7)
{
 // return double
}
else (if x == 8)
{
 // return short
}
else (if x == 9)
{
 // return long
}

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