C ++ 11的随机数分布是跨平台并不一致 - 有什么替代品呢? [英] C++11 random number distributions are not consistent across platforms -- what alternatives are there?
问题描述
我找了一套标准的C ++ 11引擎,如`的std :: mt19937便携式分布(见的 HTTP://en.cp$p$pference.com/w/cpp/numeric/random )
I'm looking for a set of portable distributions for the standard C++11 engines like `std::mt19937' (see http://en.cppreference.com/w/cpp/numeric/random).
该发动机实现始终如一地执行(即相同的序列在不同平台上产生的 - 与锵和MSVC测试),但发行似乎在不同的平台上的实现方式不同。
The engine implementations perform consistently (i.e. same sequence generated on different platforms – tested with Clang and MSVC), but the distributions seem to be implemented differently on the different platforms.
因此,即使发动机产生相同的序列,似乎分配(例如,的std :: normal_distribution<双>
)不使用相同数量的样品在不同的平台,这是不能接受的在我的情况下(即产生不同的结果)。
So, even though the engines produce the same sequence, it seems that a distribution (for example, std::normal_distribution<double>
) does not use the same number of samples (i.e. produces different results) on the different platforms, which is not acceptable in my case.
有可能是第三方的lib我可以使用下面的C ++ 11的随机模板,但将在流行的平台上提供一致的值(纵观整个海湾合作委员会的支持,MSVC和锵/ LLVM)。
Is there maybe a 3rd party lib I can use that follows the C++11 random templates, but that will deliver consistent values across popular platforms (Looking at support across GCC, MSVC and Clang/llvm).
我看过迄今为止选项包括:
Options I have looked at so far are:
- Boost.random(有点重,但值得,因为它的C ++相匹配11同行相当不错)
- 从libstd克隆++(也犯不着和可能会移植,但是拉出具体功能可能不是直接的)
- 创建我自己的C ++ 11类随机分布
我需要统一的,正常的,毒物和瑞利。
I need uniform, normal, poison and Rayleigh.
推荐答案
我已经创建了自己的C ++ 11的分布:
I have created my own C++11 distributions:
template <typename T>
class UniformRealDistribution
{
public:
typedef T result_type;
public:
UniformRealDistribution(T _a = 0.0, T _b = 1.0)
:m_a(_a),
m_b(_b)
{}
void reset() {}
template <class Generator>
T operator()(Generator &_g)
{
double dScale = (m_b - m_a) / ((T)(_g.max() - _g.min()) + (T)1);
return (_g() - _g.min()) * dScale + m_a;
}
T a() const {return m_a;}
T b() const {return m_b;}
protected:
T m_a;
T m_b;
};
template <typename T>
class NormalDistribution
{
public:
typedef T result_type;
public:
NormalDistribution(T _mean = 0.0, T _stddev = 1.0)
:m_mean(_mean),
m_stddev(_stddev)
{}
void reset()
{
m_distU1.reset();
}
template <class Generator>
T operator()(Generator &_g)
{
// Use Box-Muller algorithm
const double pi = 3.14159265358979323846264338327950288419716939937511;
double u1 = m_distU1(_g);
double u2 = m_distU1(_g);
double r = sqrt(-2.0 * log(u1));
return m_mean + m_stddev * r * sin(2.0 * pi * u2);
}
T mean() const {return m_mean;}
T stddev() const {return m_stddev;}
protected:
T m_mean;
T m_stddev;
UniformRealDistribution<T> m_distU1;
};
均匀分布,似乎提供了良好的效果和正常分布提供了很好的效果:
The uniform distribution seems to deliver good results and the normal distribution delivers very good results:
100000价值观 - > 68.159%,1西格玛之内; 2西格玛内的95.437%; 3西格玛内99.747%
100000 values -> 68.159% within 1 sigma; 95.437% within 2 sigma; 99.747% within 3 sigma
正态分布采用箱穆勒的方法,它根据什么我迄今阅读,不是最快的方法,但它运行更多的速度不够快,我的应用程序。
The normal distribution uses the Box-Muller method, which according to what I have read so far, is not the fastest method, but it runs more that fast enough for my application.
无论是均匀和正态分布应该与任何C ++ 11的发动机(带的std :: mt19937测试)和工作的所有平台上提供相同的序列,这正是我想要的。
Both the uniform and normal distributions should work with any C++11 engine (tested with std::mt19937) and provides the same sequence on all platforms, which is exactly what I wanted.
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