# Sampling from a multivariate von Mises-Fisher distribution in Python

I am looking for a simple way to sample from a multivariate von Mises-Fisher distribution in Python. I have looked in the stats module in scipy and the numpy module but only found the univariate von Mises distribution. Is there any code available? I have not found yet.

-- edit. Apparently, Wood (1994) has designed an algorithm for sampling from the vMF distribution according to this link, but I can't find the paper.

• I'm voting to close this question as off-topic because this is a stats question, migrate to stats.stackexchange.com Jun 16 '15 at 16:44
• @Spacedman Thank you for your opinion. Initially, stats.stackexchange.com wanted to do the same for being off-topic because this was considered a stat-software question.
– mic
Jun 16 '15 at 18:43
• I think it's close enough for this site since it's more about doing this is in software, and DS is more of the overlap between stats and engineering. Jun 17 '15 at 7:33

It looks like you can sample the von Mises-Fisher distribution with that R package. Have you thought about calling R from within Python using the rpy2 package? I haven't tried this for myself, but could you try the following?

from numpy import *
import scipy as sp
from pandas import *
from rpy2.robjects.packages import importr
import rpy2.robjects as ro
import pandas.rpy.common as com
from rpy2.robjects.packages import importr

# import the movMF R package
movMF = importr('movMF')
# call the rmovMF sampling function from the R package
print(movMF.rmovMF(10, 3 * c(1, -1) / sqrt(2)))


Thanks to your help. I finally got my code working, plus some bibliography.

I put my hands on Directional Statistics (Mardia and Jupp, 1999) and on the Ulrich-Wood's algorithm for sampling. I post here what I understood from it, i.e. my code (in Python), with a 'movMF' flavour.

The rejection sampling scheme:

def rW(n,kappa,m):
dim = m-1
b = dim / (np.sqrt(4*kappa*kappa + dim*dim) + 2*kappa)
x = (1-b) / (1+b)
c = kappa*x + dim*np.log(1-x*x)

y = []
for i in range(0,n):
done = False
while not done:
z = sc.stats.beta.rvs(dim/2,dim/2)
w = (1 - (1+b)*z) / (1 - (1-b)*z)
u = sc.stats.uniform.rvs()
if kappa*w + dim*np.log(1-x*w) - c >= np.log(u):
done = True
y.append(w)
return y


Then, the desired sampling is $v \sqrt{1-w^2} + w \mu$, where $w$ is the result from the rejection sampling scheme, and $v$ is uniformly sampled over the hypersphere.

def rvMF(n,theta):
dim = len(theta)
kappa = np.linalg.norm(theta)
mu = theta / kappa

result = []
for sample in range(0,n):
w = rW(kappa,dim)
v = np.random.randn(dim)
v = v / np.linalg.norm(v)

result.append(np.sqrt(1-w**2)*v + w*mu)

return result


And, for effectively sampling with this code, here is an example:

import numpy as np
import scipy as sc
import scipy.stats

n = 10
kappa = 100000
direction = np.array([1,-1,1])
direction = direction / np.linalg.norm(direction)

res_sampling = rvMF(n, kappa * direction)