Python: fit 3D ellipsoid (oblate/prolate) to 3D points

Question

Dear fellow stackoverflow users,

I face a problem as follows: I would like to fit a 3D ellipsoid to 3D data points within my python script.

The starting data are a set of x, y and z coordinates (cartesian coordinates). What I would like to get are a and c in the defining equation of the best-fit ellipsoid of the convex hull of the 3D data points.

The equation is, in the properly rotated and translated coordinate system:

ellipsoid equation

So the tasks I would ideally like to do are:

Find convex hull of 3D data points
Fit best-fit ellipsoid to the convex hull and get a and c

Do you know whether there is some library allowing to do this in Python with minimal lines of code? Or do I have to explicitly code every of these steps with my limited math knowledge (which essentially amounts to zero when it comes to find best fit ellipsoid)?

Asked By: Glxblt76

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Answer 1

All right, I found my solution by combining the convex hull algorithm of scipy with some python function found on this website.

Let us assume that you get a numpy vector of x coordinates, a numpy vector of y coordinates, and a numpy vector of z coordinates, named x, y and z. This worked for me:

from   scipy.spatial            
import ConvexHull, convex_hull_plot_2d
import numpy as np
from   numpy.linalg import eig, inv

def ls_ellipsoid(xx,yy,zz):                                  
    #finds best fit ellipsoid. Found at http://www.juddzone.com/ALGORITHMS/least_squares_3D_ellipsoid.html
    #least squares fit to a 3D-ellipsoid
    #  Ax^2 + By^2 + Cz^2 +  Dxy +  Exz +  Fyz +  Gx +  Hy +  Iz  = 1
    #
    # Note that sometimes it is expressed as a solution to
    #  Ax^2 + By^2 + Cz^2 + 2Dxy + 2Exz + 2Fyz + 2Gx + 2Hy + 2Iz  = 1
    # where the last six terms have a factor of 2 in them
    # This is in anticipation of forming a matrix with the polynomial coefficients.
    # Those terms with factors of 2 are all off diagonal elements.  These contribute
    # two terms when multiplied out (symmetric) so would need to be divided by two
    
    # change xx from vector of length N to Nx1 matrix so we can use hstack
    x = xx[:,np.newaxis]
    y = yy[:,np.newaxis]
    z = zz[:,np.newaxis]
    
    #  Ax^2 + By^2 + Cz^2 +  Dxy +  Exz +  Fyz +  Gx +  Hy +  Iz = 1
    J = np.hstack((x*x,y*y,z*z,x*y,x*z,y*z, x, y, z))
    K = np.ones_like(x) #column of ones
    
    #np.hstack performs a loop over all samples and creates
    #a row in J for each x,y,z sample:
    # J[ix,0] = x[ix]*x[ix]
    # J[ix,1] = y[ix]*y[ix]
    # etc.
    
    JT=J.transpose()
    JTJ = np.dot(JT,J)
    InvJTJ=np.linalg.inv(JTJ);
    ABC= np.dot(InvJTJ, np.dot(JT,K))

    # Rearrange, move the 1 to the other side
    #  Ax^2 + By^2 + Cz^2 +  Dxy +  Exz +  Fyz +  Gx +  Hy +  Iz - 1 = 0
    #    or
    #  Ax^2 + By^2 + Cz^2 +  Dxy +  Exz +  Fyz +  Gx +  Hy +  Iz + J = 0
    #  where J = -1
    eansa=np.append(ABC,-1)

    return (eansa)

def polyToParams3D(vec,printMe):                             
    #gets 3D parameters of an ellipsoid. Found at http://www.juddzone.com/ALGORITHMS/least_squares_3D_ellipsoid.html
    # convert the polynomial form of the 3D-ellipsoid to parameters
    # center, axes, and transformation matrix
    # vec is the vector whose elements are the polynomial
    # coefficients A..J
    # returns (center, axes, rotation matrix)
    
    #Algebraic form: X.T * Amat * X --> polynomial form
    
    if printMe: print('npolynomialn',vec)
    
    Amat=np.array(
    [
    [ vec[0],     vec[3]/2.0, vec[4]/2.0, vec[6]/2.0 ],
    [ vec[3]/2.0, vec[1],     vec[5]/2.0, vec[7]/2.0 ],
    [ vec[4]/2.0, vec[5]/2.0, vec[2],     vec[8]/2.0 ],
    [ vec[6]/2.0, vec[7]/2.0, vec[8]/2.0, vec[9]     ]
    ])
    
    if printMe: print('nAlgebraic form of polynomialn',Amat)
    
    #See B.Bartoni, Preprint SMU-HEP-10-14 Multi-dimensional Ellipsoidal Fitting
    # equation 20 for the following method for finding the center
    A3=Amat[0:3,0:3]
    A3inv=inv(A3)
    ofs=vec[6:9]/2.0
    center=-np.dot(A3inv,ofs)
    if printMe: print('nCenter at:',center)
    
    # Center the ellipsoid at the origin
    Tofs=np.eye(4)
    Tofs[3,0:3]=center
    R = np.dot(Tofs,np.dot(Amat,Tofs.T))
    if printMe: print('nAlgebraic form translated to centern',R,'n')
    
    R3=R[0:3,0:3]
    R3test=R3/R3[0,0]
    # print('normed n',R3test)
    s1=-R[3, 3]
    R3S=R3/s1
    (el,ec)=eig(R3S)
    
    recip=1.0/np.abs(el)
    axes=np.sqrt(recip)
    if printMe: print('nAxes aren',axes  ,'n')
    
    inve=inv(ec) #inverse is actually the transpose here
    if printMe: print('nRotation matrixn',inve)
    return (center,axes,inve)


#let us assume some definition of x, y and z

#get convex hull
surface  = np.stack((conf.x,conf.y,conf.z), axis=-1)
hullV    = ConvexHull(surface)
lH       = len(hullV.vertices)
hull     = np.zeros((lH,3))
for i in range(len(hullV.vertices)):
    hull[i] = surface[hullV.vertices[i]]
hull     = np.transpose(hull)         
            
#fit ellipsoid on convex hull
eansa            = ls_ellipsoid(hull[0],hull[1],hull[2]) #get ellipsoid polynomial coefficients
print("coefficients:"  , eansa)
center,axes,inve = polyToParams3D(eansa,False)   #get ellipsoid 3D parameters
print("center:"        , center)
print("axes:"          , axes)
print("rotationMatrix:", inve)

Answered By: Glxblt76

Python: fit 3D ellipsoid (oblate/prolate) to 3D points

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