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October 23, 2013 12:47
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Umeyama algorithm for absolute orientation problem in Python
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""" | |
RALIGN - Rigid alignment of two sets of points in k-dimensional | |
Euclidean space. Given two sets of points in | |
correspondence, this function computes the scaling, | |
rotation, and translation that define the transform TR | |
that minimizes the sum of squared errors between TR(X) | |
and its corresponding points in Y. This routine takes | |
O(n k^3)-time. | |
Inputs: | |
X - a k x n matrix whose columns are points | |
Y - a k x n matrix whose columns are points that correspond to | |
the points in X | |
Outputs: | |
c, R, t - the scaling, rotation matrix, and translation vector | |
defining the linear map TR as | |
TR(x) = c * R * x + t | |
such that the average norm of TR(X(:, i) - Y(:, i)) | |
is minimized. | |
""" | |
""" | |
Copyright: Carlo Nicolini, 2013 | |
Code adapted from the Mark Paskin Matlab version | |
from http://openslam.informatik.uni-freiburg.de/data/svn/tjtf/trunk/matlab/ralign.m | |
""" | |
import numpy as np | |
def ralign(X,Y): | |
m, n = X.shape | |
mx = X.mean(1) | |
my = Y.mean(1) | |
Xc = X - np.tile(mx, (n, 1)).T | |
Yc = Y - np.tile(my, (n, 1)).T | |
sx = np.mean(np.sum(Xc*Xc, 0)) | |
sy = np.mean(np.sum(Yc*Yc, 0)) | |
Sxy = np.dot(Yc, Xc.T) / n | |
U,D,V = np.linalg.svd(Sxy,full_matrices=True,compute_uv=True) | |
V=V.T.copy() | |
#print U,"\n\n",D,"\n\n",V | |
r = np.rank(Sxy) | |
d = np.linalg.det(Sxy) | |
S = np.eye(m) | |
if r > (m - 1): | |
if ( np.det(Sxy) < 0 ): | |
S[m, m] = -1; | |
elif (r == m - 1): | |
if (np.det(U) * np.det(V) < 0): | |
S[m, m] = -1 | |
else: | |
R = np.eye(2) | |
c = 1 | |
t = np.zeros(2) | |
return R,c,t | |
R = np.dot( np.dot(U, S ), V.T) | |
c = np.trace(np.dot(np.diag(D), S)) / sx | |
t = my - c * np.dot(R, mx) | |
return R,c,t | |
# Run an example test | |
# We have 3 points in 3D. Every point is a column vector of this matrix A | |
A=np.array([[0.57215 , 0.37512 , 0.37551] ,[0.23318 , 0.86846 , 0.98642],[ 0.79969 , 0.96778 , 0.27493]]) | |
# Deep copy A to get B | |
B=A.copy() | |
# and sum a translation on z axis (3rd row) of 10 units | |
B[2,:]=B[2,:]+10 | |
# Reconstruct the transformation with ralign.ralign | |
R, c, t = ralign(A,B) | |
print "Rotation matrix=\n",R,"\nScaling coefficient=",c,"\nTranslation vector=",t |
I added to @jvanvugt the option to estimate without any scaling
def estimate_similarity_transformation(source: np.ndarray, target: np.ndarray, with_scaling = False) -> np.ndarray:
"""
Estimate similarity transformation (rotation, scale, translation) from source to target (such as the Sim3 group).
"""
k, n = source.shape
mx = source.mean(axis=1)
my = target.mean(axis=1)
source_centered = source - np.tile(mx, (n, 1)).T
target_centered = target - np.tile(my, (n, 1)).T
sx = np.mean(np.sum(source_centered**2, axis=0))
sy = np.mean(np.sum(target_centered**2, axis=0))
Sxy = (target_centered @ source_centered.T) / n
U, D, Vt = np.linalg.svd(Sxy, full_matrices=True, compute_uv=True)
V = Vt.T
rank = np.linalg.matrix_rank(Sxy)
if rank < k:
raise ValueError("Failed to estimate similarity transformation")
S = np.eye(k)
if np.linalg.det(Sxy) < 0:
S[k - 1, k - 1] = -1
R = U @ S @ V.T
if with_scaling:
s = np.trace(np.diag(D) @ S) / sx
t = my - s * (R @ mx)
else:
t = my - (R @ mx)
s = 1.0
return R, s, t
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In case anyone finds this now, the above algorithm seems to contain a couple of bugs, so I've rewritten it and cleaned it up a bit. Not sure if it's fully correct but it passes my randomly generated test cases.