Dear fellow stackoverflow users,
I am trying to calculate the normal vectors over an arbitrary (but smooth) surface defined by a set of 3D points. For this, I am using a plane fitting algorithm that finds the local least square plane based on the 10 nearest neighbors of the point at which I'm calculating the normal vector.
However, it does not always find what seems to be the best plane. Thus, I'm wondering whether there is a flaw in my implementation or a flaw in my algorithm. I'm using Singular Value Decomposition as I found recommended in several links on the subject of plane fitting. Here's a code that reproduces the behavior on my machine:
#library imports
import numpy as np
import math
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
#values used for best plane fit
xyz = np.array([[-1.04194694, -1.17965867, 1.09517722],
[-0.39947906, -1.37104542, 1.36019265],
[-1.0634807 , -1.35020616, 0.46773962],
[-0.48640524, -1.64476106, 0.2726187 ],
[-0.05720509, -1.6791781 , 0.76964551],
[-1.27522669, -1.10240358, 0.33761405],
[-0.61274031, -1.52709874, -0.09945502],
[-1.402693 , -0.86807757, 0.88866091],
[-0.72520241, -0.86800727, 1.69729388]])
''' best plane fit'''
#1.calculate centroid of points and make points relative to it
centroid = xyz.mean(axis = 0)
xyzT = np.transpose(xyz)
xyzR = xyz - centroid #points relative to centroid
xyzRT = np.transpose(xyzR)
#2. calculate the singular value decomposition of the xyzT matrix and get the normal as the last column of u matrix
u, sigma, v = np.linalg.svd(xyzRT)
normal = u[2]
normal = normal / np.linalg.norm(normal) #we want normal vectors normalized to unity
'''matplotlib display'''
#prepare normal vector for display
forGraphs = list()
forGraphs.append(np.array([centroid[0],centroid[1],centroid[2],normal[0],normal[1], normal[2]]))
#get d coefficient to plane for display
d = normal[0] * centroid[0] + normal[1] * centroid[1] + normal[2] * centroid[2]
# create x,y for display
minPlane = int(math.floor(min(min(xyzT[0]), min(xyzT[1]), min(xyzT[2]))))
maxPlane = int(math.ceil(max(max(xyzT[0]), max(xyzT[1]), max(xyzT[2]))))
xx, yy = np.meshgrid(range(minPlane,maxPlane), range(minPlane,maxPlane))
# calculate corresponding z for display
z = (-normal[0] * xx - normal[1] * yy + d) * 1. /normal[2]
#matplotlib display code
forGraphs = np.asarray(forGraphs)
X, Y, Z, U, V, W = zip(*forGraphs)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(xx, yy, z, alpha=0.2)
ax.scatter(xyzT[0],xyzT[1],xyzT[2])
ax.quiver(X, Y, Z, U, V, W)
ax.set_xlim([min(xyzT[0])- 0.1, max(xyzT[0]) + 0.1])
ax.set_ylim([min(xyzT[1])- 0.1, max(xyzT[1]) + 0.1])
ax.set_zlim([min(xyzT[2])- 0.1, max(xyzT[2]) + 0.1])
plt.show()
The result is:
I would expect it to be more like:
(sorry for the sketchy drawings)
So, what's wrong here? Could it be a display error in my matplotlib code?
All the best!
In the wiki article you can read that it is the right singular vector that minimizes the "orthogonal". So I guess you do not want to transpose and use v[2] instead of u[2]; works for me. Note, that using the second, i.e. last element relies on the fact that numpy (LAPACK) returns the singular values in descending order.
来源：https://stackoverflow.com/questions/53591350/plane-fit-of-3d-points-with-singular-value-decomposition