[英]C++ TR1: how to use the normal_distribution?
I'm trying to use the C++ STD TechnicalReport1 extensions to generate numbers following a normal distribution, but this code (adapted from this article ): 我正在尝试使用C ++ STD TechnicalReport1扩展来生成正常分布后的数字,但是这段代码(改编自本文 ):
mt19937 eng;
eng.seed(SEED);
normal_distribution<double> dist;
// XXX if I use the one below it exits the for loop
// uniform_int<int> dist(1, 52);
for (unsigned int i = 0; i < 1000; ++i) {
cout << "Generating " << i << "-th value" << endl;
cout << dist(eng) << endl;
}
only prints 1 "Generating..." log message, then never exits the for loop ! 只打印1“Generating ...”日志消息,然后永远不会退出for循环 ! If I use the distribution I commented out instead, it terminates, so I'm wondering what I'm doing wrong. 如果我使用我注释掉的发行版,它会终止,所以我想知道我做错了什么。 Any idea? 任何的想法?
Thanks a lot! 非常感谢!
I have had the same issue with the code originally posted and investigated the GNU implementation of 我最初发布的代码遇到了同样的问题并调查了GNU的实现
first some observations: with g++-4.4 and using the code hangs, with g++-4.5 and using -std=c++0x (ie not TR1 but the real thing) above code works 首先是一些观察结果:使用g ++ - 4.4并使用代码挂起,使用g ++ - 4.5并使用-std = c ++ 0x(即不是TR1但是真实的东西)上面的代码可以工作
IMHO, there was a change between TR1 and c++0x with regard to adaptors between random number generation and consumption of random numbers -- mt19937 produces integers, normal_distribution consumes doubles 恕我直言,TR1和c ++ 0x之间有关于随机数生成和随机数消耗之间的适配器的变化 - mt19937产生整数,normal_distribution消耗双精度
the c++0x uses adaption automatically, the g++ TR1 code does not c ++ 0x自动使用自适应,g ++ TR1代码没有
in order to get your code working with g++-4.4 and TR1, do the following 为了让您的代码使用g ++ - 4.4和TR1,请执行以下操作
std::tr1::mt19937 prng(seed);
std::tr1::normal_distribution<double> normal;
std::tr1::variate_generator<std::tr1::mt19937, std::tr1::normal_distribution<double> > randn(prng,normal);
double r = randn();
This definitely would not hang the program. 这肯定不会挂起程序。 But, not sure if it really meets your needs. 但是,不确定它是否真的满足您的需求。
#include <random>
#include <iostream>
using namespace std;
typedef std::tr1::ranlux64_base_01 Myeng;
typedef std::tr1::normal_distribution<double> Mydist;
int main()
{
Myeng eng;
eng.seed(1000);
Mydist dist(1,10);
dist.reset(); // discard any cached values
for (int i = 0; i < 10; i++)
{
std::cout << "a random value == " << (int)dist(eng) << std::endl;
}
return (0);
}
If your TR1 random number generation implementation is buggy, you can avoid TR1 by writing your own normal generator as follows. 如果您的TR1随机数生成实现有问题,您可以通过编写自己的普通生成器来避免TR1,如下所示。
Generate two uniform (0, 1) random samples u and v using any random generator you trust. 使用您信任的任何随机生成器生成两个均匀(0,1)随机样本u和v。 Then let r = sqrt( -2 log(u) ) and return x = r sin(2 pi v). 然后让r = sqrt(-2 log(u))并返回x = r sin(2 pi v)。 (This is called the Box-Mueller method.) (这被称为Box-Mueller方法。)
If you need normal samples samples with mean mu and standard deviation sigma, return sigma*x + mu instead of just x. 如果您需要具有平均μ和标准偏差西格玛的正常样品样品,则返回sigma * x + mu而不是x。
While this appears to be a bug, a quick confirmation would be to pass the default 0.0, 1.0 parameters. 虽然这似乎是一个错误,但快速确认将传递默认的0.0,1.0参数。 normal_distribution<double>::normal_distribution()
should equal normal_distribution<double>::normal_distribution(0.0, 1.0)
normal_distribution<double>::normal_distribution()
应该等于normal_distribution<double>::normal_distribution(0.0, 1.0)
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