Sample uniformly in a multidimensional ring withou

问题:

The algorithm in this question tells us how to efficiently sample from a multidimensional ball. Is there a way to similarly efficiently sample from a multidimensional ring , i.e. have r1<r<r2

I hope that a not too complex modification of that scaling function r*(gammainc(s2/2,n/2).^(1/n))./sqrt(s2) is possible. (Mediocrity disclaimer: haven't even figured the algebra/geometry for the original scaling function yet).

Original matlab code copypasted:

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);

Equivalent python code with demo from Daniel's answer:

import numpy as np
from scipy.special import gammainc
from matplotlib import pyplot as plt
def sample(center,radius,n_per_sphere):
    r = radius
    ndim = center.size
    x = np.random.normal(size=(n_per_sphere, ndim))
    ssq = np.sum(x**2,axis=1)
    fr = r*gammainc(ndim/2,ssq/2)**(1/ndim)/np.sqrt(ssq)
    frtiled = np.tile(fr.reshape(n_per_sphere,1),(1,ndim))
    p = center + np.multiply(x,frtiled)
    return p

fig1 = plt.figure(1)
ax1 = fig1.gca()
center = np.array([0,0])
radius = 1
p = sample(center,radius,10000)
ax1.scatter(p[:,0],p[:,1],s=0.5)
ax1.add_artist(plt.Circle(center,radius,fill=False,color='0.5'))
ax1.set_xlim(-1.5,1.5)
ax1.set_ylim(-1.5,1.5)
ax1.set_aspect('equal')

回答1:

The last method here⁽¹⁾ is suitable for any dimensional sphere:

To pick a random point on a sphere:
- generate N Gaussian random variables x1,x2..xN
- get norm of x[i]

 L = Sqrt(x1*x1 + x2*x2 + .. + xn*xn)
 ux1 = x1 / L
 ux2 = x2 / L
 ...

Then the distribution of the vectors ux[i] is uniform over the surface S^N-1

To provide uniform distribution in the ring:
- generate uniform random in range

R_NPow = RandomUniform(R_Inner^N, R_Outer^N)

and get radius (like this 2D case)

R = R_NPow^1/N

then calculate resulting point coordinates:

res_x1 = R * ux1
res_x2 = R * ux2
...
res_xn = R * uxn

(1) Muller, M. E. "A Note on a Method for Generating Points Uniformly on -Dimensional Spheres." Comm. Assoc. Comput. Mach. 2, 19-20, Apr. 1959.

回答2:

I actually ended up using the inverse cdf method applied to points uniformly distributed on a sphere

like so

def random_uniform_ring(center=np.array([0,0]),R=1,r=0,nsamples=1):
    """
    generate point uniformly distributed in a ring
    """
    nd = len(center)
    x = np.random.normal(size = (nsamples,nd))
    x /=np.linalg.norm(x,axis=1)[:,np.newaxis] #generate on unit sphere
    # using the inverse cdf method
    u = np.random.uniform(size=(nsamples))
    sc = (u*(R**nd-r**nd)+r**nd)**(1/nd) #this is inverse the cdf of ring volume as a function of radius
    return x*sc[:,None]+center

To test

import numpy as np
from scipy.special import gammainc
from matplotlib import pyplot as plt
def test1():
    fig1 = plt.figure(1)
    ax1 = fig1.gca()
    # center = np.zeros((600))
    # center = np.array([0,0])
    center = np.array([2,1])
    r = 0.5
    R = 1.
    n = 1000
    p = random_uniform_ring(center,R,r,n)
    assert p.shape[0]==n
    ax1.scatter(p[:,0],p[:,1],s=0.5)
    ax1.add_artist(plt.Circle(center,R,fill=False,color='0.5'))
    ax1.add_artist(plt.Circle(center,r,fill=False,color='0.5'))
    ax1.set_xlim(-R-0.5+center[0],R+0.5+center[0])
    ax1.set_ylim(-R-0.5+center[1],R+0.5+center[1])
    ax1.set_aspect('equal')
    plt.show()


test1()

This might be equivalent to @Mbo's answer, but unfortunately I don't really have time to test. If somebody could test his answer, I would gladly accept.

回答3:

After some trial and error, I was able to get it done with the gammainc approach. The math behind it is beyond my depth, but I basically tweeked the coefficient 2 in gammainc to a power z to improve uniformity.

Also tested it in 3D and it seems to work fine.

(this had been in my todo list for a while, thanks for the ideas!)

import numpy as np
from scipy.special import gammainc
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

def sample_ring(center,r1,r2,n_points):
    nd = center.size
    x = np.random.normal(size=(n_points, nd))
    sq = np.sum(x**2,axis=1)
    z = (r2-r1)/r2
    fr = (r2-r1)*gammainc(nd/2**z,sq/2**z)**(1/nd)/np.sqrt(sq) + r1/np.sqrt(sq)
    frtiled = np.tile(fr.reshape(n_points,1),(1,nd))
    p = center + np.multiply(x,frtiled)
    return p

fig1 = plt.figure(1)
ax1 = fig1.gca()
center = np.array([0,0])
r1 = 1.5
R2 = 3
p = sample_ring(center,r1,R2,5000)
ax1.scatter(p[:,0],p[:,1],s=0.5)
ax1.add_artist(plt.Circle(center,r1,fill=False,color='0.5'))
ax1.add_artist(plt.Circle(center,R2,fill=False,color='0.5'))
ax1.set_xlim(-4,4)
ax1.set_ylim(-4,4)
ax1.set_aspect('equal')

fig3 = plt.figure(3)
ax3 = plt.gca(projection='3d')
ax3.set_aspect("equal")
theta, phi = np.mgrid[0:2*np.pi:10j, 0:np.pi:10j]
c_3d = np.array([0,0,0])
r1_3d = 0.5
x1 = c_3d[0] + r1_3d*np.cos(theta)*np.sin(phi)
y1 = c_3d[1] + r1_3d*np.sin(theta)*np.sin(phi)
z1 = c_3d[2] + r1_3d*np.cos(phi)
r2_3d = 1.4
x2 = c_3d[0] + r2_3d*np.cos(theta)*np.sin(phi)
y2 = c_3d[1] + r2_3d*np.sin(theta)*np.sin(phi)
z2 = c_3d[2] + r2_3d*np.cos(phi)
ax3.plot_wireframe(x1, y1, z1, color="r")
ax3.plot_wireframe(x2, y2, z2, color="r")
p = sample_ring(c_3d,r1_3d,r2_3d,1000)
ax3.scatter(p[:,0],p[:,1],p[:,2], c='b', marker='o')
ax3.set_xlim(-1.5, 1.5)
ax3.set_ylim(-1.5, 1.5)
ax3.set_zlim(-1.5, 1.5)