C ++-根据范围内的正态分布生成随机数

本文介绍了C ++-根据范围内的正态分布生成随机数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我需要生成一个服从正态分布的随机数，该随机数应介于1000和11000之间，平均值为7000.我想使用 c ++ 11库函数，但我不了解如何在间隔内生成数字.有人可以帮忙吗?

I need to generate random numbers that follow a normal distribution which should lie within the interval of 1000 and 11000 with a mean of 7000. I want to use the c++11 library function but I am not understanding how to generate the numbers within the interval. Can someone help?

推荐答案

您未指定标准差.假设给定间隔的标准偏差为2000，您可以尝试以下操作:

You don't specify the standard deviation. Assuming a standard deviation of 2000 for the given interval you can try this:

#include <iostream>
#include <random>

class Generator {
    std::default_random_engine generator;
    std::normal_distribution<double> distribution;
    double min;
    double max;
public:
    Generator(double mean, double stddev, double min, double max):
        distribution(mean, stddev), min(min), max(max)
    {}

    double operator ()() {
        while (true) {
            double number = this->distribution(generator);
            if (number >= this->min && number <= this->max)
                return number;
        }
    }
};

int main() {
    Generator g(7000.0, 2000.0, 1000.0, 11000.0);
    for (int i = 0; i < 10; i++)
        std::cout << g() << std::endl;
}

可能的输出:

如果只想指定min和max值，那么我们可以假设平均值为(min + max) / 2.我们还可以假设min和max与平均值相差3个标准差.使用这些设置，我们将只丢弃生成值的0.3％.因此，您可以添加以下构造函数:

If you want to specify only the min and max values, then we can assume that the mean value is (min + max) / 2. Also we can assume that min and max are 3 standard deviations away from the mean value. With these settings we are going to throw away only 0.3% of the generated values. So you can add the following constructor:

Generator(double min, double max):
        distribution((min + max) / 2, (max - min) / 6), min(min), max(max)
    {}

并将生成器初始化为:

Generator g(1000.0, 11000.0);

这篇关于C ++-根据范围内的正态分布生成随机数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

We