在球形体积内采样均匀分布的随机点 [英] Sampling uniformly distributed random points inside a spherical volume

问题描述

我希望能够生成随机均匀的粒子位置样本，这些样本属于球形体积。

下面的图片（由给出。我似乎无法破译或理解他们做了什么。

 函数X = randsphere（m，n，r）
 
％这个函数返回一个m n数组X，其中
％中的每一行具有n个笛卡尔坐标
％均匀分布在具有$的n维超球的
％内部的随机点b $ b％半径r并且在原点处居中。函数
％'randn'最初用于生成m组具有独立多元
％正态分布的n 
％随机变量，均值为0，方差为1. 
％Then不完全伽马函数'gammainc'，
％被用于径向映射这些点，以适应具有统一的％空间分布的有限半径r的
％超球面。 
％Roger Stafford  -  12/23/05 
 
 X = randn（m，n）; 
 s2 = sum（X. ^ 2,2）; 
 X = X. * repmat（r *（gammainc（s2 / 2，n / 2）。^（1 / n））./ sqrt（s2），1，n）。 
  

我非常感谢在Python中从球形体生成真正统一样本的任何建议。 p>

似乎有很多示例显示如何从统一的球形壳中进行采样，但似乎更容易解决问题。这个问题与缩放有关 - 在0.1的半径处应该有比在1.0的半径处更少的粒子以从球体的体积产生均匀的样本。

< 编辑 ：修正并删除了我通常询问的事实，而我的意思是统一的。

解决方案

尽管我更喜欢球体的丢弃方法，但为了完整性

在球形坐标系中，利用 stackoverflow.com/questions/2106503/pseudorandom-number-generator-exponential-distribution/2106568#2106568\">抽样规则：

 <$随机（0,1）
 
 theta = arccos（costheta） 
r = R * cuberoot（u）
  

现在你有一个（r，theta，phi）组，它可以通常的方式转换为（x，y，z）

  x = r * sin（theta）* cos（phi）
y = r * sin（theta）* sin（phi） 
z = r * cos（theta）
  

I am looking to be able to generate a random uniform sample of particle locations that fall within a spherical volume.

The image below (courtesy of http://nojhan.free.fr/metah/) shows what I am looking for. This is a slice through the sphere, showing a uniform distribution of points:

This is what I am currently getting:

You can see that there is a cluster of points at the center due to the conversion between spherical and Cartesian coordinates.

The code I am using is:

def new_positions_spherical_coordinates(self):
   radius = numpy.random.uniform(0.0,1.0, (self.number_of_particles,1)) 
   theta = numpy.random.uniform(0.,1.,(self.number_of_particles,1))*pi
   phi = numpy.arccos(1-2*numpy.random.uniform(0.0,1.,(self.number_of_particles,1)))
   x = radius * numpy.sin( theta ) * numpy.cos( phi )
   y = radius * numpy.sin( theta ) * numpy.sin( phi )
   z = radius * numpy.cos( theta )
   return (x,y,z)

Below is some MATLAB code that supposedly creates a uniform spherical sample, which is similar to the equation given by http://nojhan.free.fr/metah. I just can't seem to decipher it or understand what they did.

function X = randsphere(m,n,r)

% This function returns an m by n array, X, in which 
% each of the m rows has the n Cartesian coordinates 
% of a random point uniformly-distributed over the 
% interior of an n-dimensional hypersphere with 
% radius r and center at the origin.  The function 
% 'randn' is initially used to generate m sets of n 
% random variables with independent multivariate 
% normal distribution, with mean 0 and variance 1.
% Then the incomplete gamma function, 'gammainc', 
% is used to map these points radially to fit in the 
% hypersphere of finite radius r with a uniform % spatial distribution.
% Roger Stafford - 12/23/05

X = randn(m,n);
s2 = sum(X.^2,2);
X = X.*repmat(r*(gammainc(s2/2,n/2).^(1/n))./sqrt(s2),1,n);

I would greatly appreciate any suggestions on generating a truly uniform sample from a spherical volume in Python.

There seem to be plenty of examples showing how to sample from a uniform spherical shell, but that seems to be easier an easier problem. The issue has to do with the scaling - there should be fewer particles at a radius of 0.1 than at a radius of 1.0 to generate a uniform sample from the volume of the sphere.

Edit: Fixed and removed the fact I asked for normally and I meant uniform.

解决方案

While I prefer the discarding method for spheres, for completeness I offer the exact solution.

In spherical coordinates, taking advantage of the sampling rule:

phi = random(0,2pi)
costheta = random(-1,1)
u = random(0,1)

theta = arccos( costheta )
r = R * cuberoot( u )

now you have a (r, theta, phi) group which can be transformed to (x, y, z) in the usual way

x = r * sin( theta) * cos( phi )
y = r * sin( theta) * sin( phi )
z = r * cos( theta )

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