C# 普通随机数
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C# Normal Random Number
提问by J.Hendrix
I would like to create a function that accepts Double mean, Double deviationand returns a random number with a normal distribution.
我想创建一个函数，它接受Double mean，Double deviation并返回一个具有正态分布的随机数。
Example: if I pass in 5.00 as the mean and 2.00 as the deviation, 68% of the time I will get a number between 3.00 and 7.00
示例：如果我传入 5.00 作为平均值和 2.00 作为偏差，68% 的时间我会得到一个介于 3.00 和 7.00 之间的数字
My statistics is a little weak…. Anyone have an idea how I should approach this? My implementation will be C# 2.0 but feel free to answer in your language of choice as long as the math functions are standard.
我的统计有点弱...... 任何人都知道我应该如何处理这个问题？我的实现将是 C# 2.0，但只要数学函数是标准的，就可以用您选择的语言回答。
I think thismight actually be what I am looking for. Any help converting this to code?
我认为这实际上可能是我正在寻找的。任何帮助将其转换为代码？
Thanks in advance for your help.
在此先感谢您的帮助。
采纳答案by John D. Cook
See this CodeProject article: Simple Random Number Generation. The code is very short, and it generates samples from uniform, normal, and exponential distributions.
请参阅此 CodeProject 文章：简单随机数生成。代码很短，它从均匀分布、正态分布和指数分布中生成样本。
回答by Jason Punyon
回答by redcalx
回答by Martin Sherburn
Sorry I don't have any code for you but I can point you to some algorithms on Wikipedia. The algorithm you choose I guess depends on how accurate you want it and how fast it needs to be.
抱歉，我没有任何代码供您参考，但我可以为您指出Wikipedia上的一些算法。我猜你选择的算法取决于你想要它的准确度和它需要的速度。
回答by Justin R.
The MetaNumericslibrary, also .NET, will calculate a normal distribution (and just about anything else from statistics), super quickly. Have a look at the Feature's page for more details. The Codeplex page is here: http://metanumerics.codeplex.com/.
该MetaNumerics库，也.NET，将计算正态分布（和几乎从统计别的），超级快。查看功能页面以获取更多详细信息。Codeplex 页面在这里：http: //metanumerics.codeplex.com/。
回答by Ian W
Here is some C that returns two values (rand1 and rand2), just because the algorithm efficiently does so. It is the polar form of the Box-Muller transform.
这里有一些 C 返回两个值（rand1 和 rand2），只是因为算法有效地这样做了。它是Box-Muller 变换的极坐标形式。
void RandVal (double mean1, double sigma1, double *rand1, double mean2, double sigma2, double *rand2)
{
double u1, u2, v1, v2, s, z1, z2;
do {
u1 = Random (0., 1.); // a uniform random number from 0 to 1
u2 = Random (0., 1.);
v1 = 2.*u1 - 1.;
v2 = 2.*u2 - 1.;
s = v1*v1 + v2*v2;
} while (s > 1. || s==0.);
z1 = sqrt (-2.*log(s)/s)*v1;
z2 = sqrt (-2.*log(s)/s)*v2;
*rand1 = (z1*sigma1 + mean1);
*rand2 = (z2*sigma2 + mean2);
return;
}
}
回答by Ace
For those referencing this question, an easy solution might be:
对于那些引用这个问题的人，一个简单的解决方案可能是：
Random rand = new Random();
double normRand = alglib.invnormaldistribution(rand.NextDouble())
scale by mu and sigma as needed.
The alglib library is available at www.alglib.net
根据需要按 mu 和 sigma 进行缩放。
alglib 库可在www.alglib.net 上获得
回答by pascx64
I know this post is a little old but I would like to share a small project I created yesterday. I think the easier way is to use C++ 11 and create a .dll in Managed C++. There's a linkto the source and a zip containing the dll already compiled.
我知道这篇文章有点旧，但我想分享我昨天创建的一个小项目。我认为更简单的方法是使用 C++ 11 并在 Managed C++ 中创建一个 .dll。有一个指向源代码的链接和一个包含已编译的 dll 的 zip。
And the code I made :
我制作的代码：
// NormalDistributionRandom.h
#include <random>
#pragma once
using namespace System;
namespace NormalDistribution
{
class _NormalDistributionRandom
{
std::default_random_engine engine;
std::normal_distribution<double> distribution;
public:
_NormalDistributionRandom(double mean, double deviation) : distribution(mean, deviation)
{
}
double Next()
{
return distribution(engine);
}
};
public ref class NormalDistributionRandom
{
private:
void* Distribution;
public:
NormalDistributionRandom( double mean, double deviation)
{
Distribution = new _NormalDistributionRandom(mean, deviation);
}
double Next()
{
return ((_NormalDistributionRandom*)Distribution)->Next();
}
~NormalDistributionRandom()
{
this->!NormalDistributionRandom();
}
protected:
!NormalDistributionRandom()
{
if (Distribution != nullptr)
{
delete (_NormalDistributionRandom*)Distribution;
Distribution = nullptr;
}
}
};
}
回答by Declan Taylor
MathNet
数学网
from second top answer
从第二个最佳答案
public static double GenerateRandomVariant(double mean,double deviation,System.Random rand=null, int factor=1)
{
rand = rand ?? new Random();
double randNormal=(MathNet.Numerics.Distributions.Normal.Sample(rand, mean , deviation));
return factor * randNormal;
}
Box-Mueller Transform
Box-Mueller 变换
from top answer via link (twice as fast?)
从最佳答案通过链接（快两倍？）
by u/yoyoyoyosef Random Gaussian Variables
由 u/yoyoyoyosef随机高斯变量
public static double GenerateRandomVariant(double mean, double deviation, System.Random rand=null, int factor = 1)
{
rand = rand ?? new Random();
double u1 = 1.0 - rand.NextDouble(); //uniform(0,1] random doubles
double u2 = 1.0 - rand.NextDouble();
double randStdNormal = Math.Sqrt(-2.0 * Math.Log(u1)) *
Math.Sin(2.0 * Math.PI * u2); //random normal(0,1)
double randNormal=(
mean + deviation * randStdNormal); //random normal(mean,stdDev^2)
return randNormal * factor;
}
