procrustes | R Documentation |
procrustes
solves for two matrices A
and B
the
‘Procrustes Problem’ of finding an orthogonal matrix Q
such that
A-B*Q
has the minimal Frobenius norm.
kabsch
determines a best rotation of a given vector set into a
second vector set by minimizing the weighted sum of squared deviations.
The order of vectors is assumed fixed.
procrustes(A, B)
kabsch(A, B, w = NULL)
A, B |
two numeric matrices of the same size. |
w |
weights , influence the distance of points |
The function procrustes(A,B)
uses the svd
decomposition
to find an orthogonal matrix Q
such that A-B*Q
has a
minimal Frobenius norm, where this norm for a matrix C
is defined
as sqrt(Trace(t(C)*C))
, or norm(C,'F')
in R.
Solving it with B=I
means finding a nearest orthogonal matrix.
kabsch
solves a similar problem and uses the Procrustes procedure
for its purpose. Given two sets of points, represented as columns of the
matrices A
and B
, it determines an orthogonal matrix
U
and a translation vector R
such that U*A+R-B
is minimal.
procrustes
returns a list with components P
, which is
B*Q
, then Q
, the orthogonal matrix, and d
, the
Frobenius norm of A-B*Q
.
kabsch
returns a list with U
the orthogonal matrix applied,
R
the translation vector, and d
the least root mean square
between U*A+R
and B
.
The kabsch
function does not take into account scaling of the sets,
but this could easily be integrated.
Golub, G. H., and Ch. F. van Loan (1996). Matrix Computations. 3rd Edition, The John Hopkins University Press, Baltimore London. [Sect. 12.4, p. 601]
Kabsch, W. (1976). A solution for the best rotation to relate two sets of vectors. Acta Cryst A, Vol. 32, p. 9223.
svd
## Procrustes
U <- randortho(5) # random orthogonal matrix
P <- procrustes(U, eye(5))
## Kabsch
P <- matrix(c(0, 1, 0, 0, 1, 1, 0, 1,
0, 0, 1, 0, 1, 0, 1, 1,
0, 0, 0, 1, 0, 1, 1, 1), nrow = 3, ncol = 8, byrow = TRUE)
R <- c(1, 1, 1)
phi <- pi/4
U <- matrix(c(1, 0, 0,
0, cos(phi), -sin(phi),
0, sin(phi), cos(phi)), nrow = 3, ncol = 3, byrow = TRUE)
Q <- U %*% P + R
K <- kabsch(P, Q)
# K$R == R and K$U %*% P + c(K$R) == Q
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