I'm trying to use the shapely.geometry.Polygon module to find the area of polygons but it performs all calculations on the xy plane. This is fine for some of my polygons but others have a z dimension too so it's not quite doing what I'd like.
Is there a package which will either give me the area of a planar polygon from xyz coordinates, or alternatively a package or algorithm to rotate the polygon to the xy plane so that i can use shapely.geometry.Polygon().area?
The polygons are represented as a list of tuples in the form [(x1,y1,z1),(x2,y2,z3),...(xn,yn,zn)].
Here is the derivation of a formula for calculating the area of a 3D planar polygon
Here is Python code that implements it:
#determinant of matrix a
def det(a):
return a[0][0]*a[1][1]*a[2][2] + a[0][1]*a[1][2]*a[2][0] + a[0][2]*a[1][0]*a[2][1] - a[0][2]*a[1][1]*a[2][0] - a[0][1]*a[1][0]*a[2][2] - a[0][0]*a[1][2]*a[2][1]
#unit normal vector of plane defined by points a, b, and c
def unit_normal(a, b, c):
x = det([[1,a[1],a[2]],
[1,b[1],b[2]],
[1,c[1],c[2]]])
y = det([[a[0],1,a[2]],
[b[0],1,b[2]],
[c[0],1,c[2]]])
z = det([[a[0],a[1],1],
[b[0],b[1],1],
[c[0],c[1],1]])
magnitude = (x**2 + y**2 + z**2)**.5
return (x/magnitude, y/magnitude, z/magnitude)
#dot product of vectors a and b
def dot(a, b):
return a[0]*b[0] + a[1]*b[1] + a[2]*b[2]
#cross product of vectors a and b
def cross(a, b):
x = a[1] * b[2] - a[2] * b[1]
y = a[2] * b[0] - a[0] * b[2]
z = a[0] * b[1] - a[1] * b[0]
return (x, y, z)
#area of polygon poly
def area(poly):
if len(poly) < 3: # not a plane - no area
return 0
total = [0, 0, 0]
for i in range(len(poly)):
vi1 = poly[i]
if i is len(poly)-1:
vi2 = poly[0]
else:
vi2 = poly[i+1]
prod = cross(vi1, vi2)
total[0] += prod[0]
total[1] += prod[1]
total[2] += prod[2]
result = dot(total, unit_normal(poly[0], poly[1], poly[2]))
return abs(result/2)
And to test it, here's a 10x5 square that leans over:
>>> poly = [[0, 0, 0], [10, 0, 0], [10, 3, 4], [0, 3, 4]]
>>> poly_translated = [[0+5, 0+5, 0+5], [10+5, 0+5, 0+5], [10+5, 3+5, 4+5], [0+5, 3+5, 4+5]]
>>> area(poly)
50.0
>>> area(poly_translated)
50.0
>>> area([[0,0,0],[1,1,1]])
0
The problem originally was that I had oversimplified. It needs to calculate the unit vector normal to the plane. The area is half of the dot product of that and the total of all the cross products, not half of the sum of all the magnitudes of the cross products.
This can be cleaned up a bit (matrix and vector classes would make it nicer, if you have them, or standard implementations of determinant/cross product/dot product), but it should be conceptually sound.
This is the final code I've used. It doesn't use shapely, but implements Stoke's theorem to calculate the area directly. It builds on #Tom Smilack's answer which shows how to do it without numpy.
import numpy as np
#unit normal vector of plane defined by points a, b, and c
def unit_normal(a, b, c):
x = np.linalg.det([[1,a[1],a[2]],
[1,b[1],b[2]],
[1,c[1],c[2]]])
y = np.linalg.det([[a[0],1,a[2]],
[b[0],1,b[2]],
[c[0],1,c[2]]])
z = np.linalg.det([[a[0],a[1],1],
[b[0],b[1],1],
[c[0],c[1],1]])
magnitude = (x**2 + y**2 + z**2)**.5
return (x/magnitude, y/magnitude, z/magnitude)
#area of polygon poly
def poly_area(poly):
if len(poly) < 3: # not a plane - no area
return 0
total = [0, 0, 0]
N = len(poly)
for i in range(N):
vi1 = poly[i]
vi2 = poly[(i+1) % N]
prod = np.cross(vi1, vi2)
total[0] += prod[0]
total[1] += prod[1]
total[2] += prod[2]
result = np.dot(total, unit_normal(poly[0], poly[1], poly[2]))
return abs(result/2)
#pythonn code for polygon area in 3D (optimised version)
def polygon_area(poly):
#shape (N, 3)
if isinstance(poly, list):
poly = np.array(poly)
#all edges
edges = poly[1:] - poly[0:1]
# row wise cross product
cross_product = np.cross(edges[:-1],edges[1:], axis=1)
#area of all triangles
area = np.linalg.norm(cross_product, axis=1)/2
return sum(area)
if __name__ == "__main__":
poly = [[0+5, 0+5, 0+5], [10+5, 0+5, 0+5], [10+5, 3+5, 4+5], [0+5, 3+5, 4+5]]
print(polygon_area(poly))
The area of a 2D polygon can be calculated using Numpy as a one-liner...
poly_Area(vertices) = np.sum( [0.5, -0.5] * vertices * np.roll( np.roll(vertices, 1, axis=0), 1, axis=1) )
Fyi, here is the same algorithm in Mathematica, with a baby unit test
ClearAll[vertexPairs, testPoly, area3D, planeUnitNormal, pairwise];
pairwise[list_, fn_] := MapThread[fn, {Drop[list, -1], Drop[list, 1]}];
vertexPairs[Polygon[{points___}]] := Append[{points}, First[{points}]];
testPoly = Polygon[{{20, -30, 0}, {40, -30, 0}, {40, -30, 20}, {20, -30, 20}}];
planeUnitNormal[Polygon[{points___}]] :=
With[{ps = Take[{points}, 3]},
With[{p0 = First[ps]},
With[{qs = (# - p0) & /# Rest[ps]},
Normalize[Cross ## qs]]]];
area3D[p : Polygon[{polys___}]] :=
With[{n = planeUnitNormal[p], vs = vertexPairs[p]},
With[{areas = (Dot[n, #]) & /# pairwise[vs, Cross]},
Plus ## areas/2]];
area3D[testPoly]
Same as #Tom Smilack's answer, but in javascript
//determinant of matrix a
function det(a) {
return a[0][0] * a[1][1] * a[2][2] + a[0][1] * a[1][2] * a[2][0] + a[0][2] * a[1][0] * a[2][1] - a[0][2] * a[1][1] * a[2][0] - a[0][1] * a[1][0] * a[2][2] - a[0][0] * a[1][2] * a[2][1];
}
//unit normal vector of plane defined by points a, b, and c
function unit_normal(a, b, c) {
let x = math.det([
[1, a[1], a[2]],
[1, b[1], b[2]],
[1, c[1], c[2]]
]);
let y = math.det([
[a[0], 1, a[2]],
[b[0], 1, b[2]],
[c[0], 1, c[2]]
]);
let z = math.det([
[a[0], a[1], 1],
[b[0], b[1], 1],
[c[0], c[1], 1]
]);
let magnitude = Math.pow(Math.pow(x, 2) + Math.pow(y, 2) + Math.pow(z, 2), 0.5);
return [x / magnitude, y / magnitude, z / magnitude];
}
// dot product of vectors a and b
function dot(a, b) {
return a[0] * b[0] + a[1] * b[1] + a[2] * b[2];
}
// cross product of vectors a and b
function cross(a, b) {
let x = (a[1] * b[2]) - (a[2] * b[1]);
let y = (a[2] * b[0]) - (a[0] * b[2]);
let z = (a[0] * b[1]) - (a[1] * b[0]);
return [x, y, z];
}
// area of polygon poly
function area(poly) {
if (poly.length < 3) {
console.log("not a plane - no area");
return 0;
} else {
let total = [0, 0, 0]
for (let i = 0; i < poly.length; i++) {
var vi1 = poly[i];
if (i === poly.length - 1) {
var vi2 = poly[0];
} else {
var vi2 = poly[i + 1];
}
let prod = cross(vi1, vi2);
total[0] = total[0] + prod[0];
total[1] = total[1] + prod[1];
total[2] = total[2] + prod[2];
}
let result = dot(total, unit_normal(poly[0], poly[1], poly[2]));
return Math.abs(result/2);
}
}
Thanks for detailed answers, But I am little surprised there is no simple answer to get the area.
So, I am just posting a simplified approach for calculating area using 3d Coordinates of polygon or surface using pyny3d.
#Install pyny3d as:
pip install pyny3d
#Calculate area
import numpy as np
import pyny3d.geoms as pyny
coords_3d = np.array([[0, 0, 0],
[7, 0, 0],
[7, 10, 2],
[0, 10, 2]])
polygon = pyny.Polygon(coords_3d)
print(f'Area is : {polygon.get_area()}')
Related
I am implementing the Francis double step QR Iteration algorithm using the notes and psuedocode from lecture https://people.inf.ethz.ch/arbenz/ewp/Lnotes/chapter4.pdf - Algorithm 4.5
The psuedocode is provided in Matlab I believe.
Below is the implementation of my code.
# compute upper hessenberg form of matrix
def hessenberg(A):
m,n = A.shape
H = A.astype(np.float64)
for k in range(n-2):
x = H[k+1:, k]
v = np.concatenate([np.array([np.sign(x[0]) * np.linalg.norm(x)]), x[1:]])
v = v / np.linalg.norm(v)
H[k+1:, k:] -= 2 * np.outer(v, np.dot(v, H[k+1:, k:]))
H[:, k+1:] -= 2 * np.outer(np.dot(H[:, k+1:], v), v)
return(H)
# compute first three elements of M
def first_three_M(T,s,t):
x = T[0, 0]**2 + T[0, 1] * T[1, 0] - s * T[0, 0] + t
y = T[1, 0] * (T[0, 0] + T[1, 1] - s)
z = T[1, 0] * T[2, 1]
return(x,y,z)
# householder reflection
def householder_reflection_step(x_1):
v = x_1[0] + np.sign(x_1[0]) * np.linalg.norm(x_1)
v = v / np.linalg.norm(v)
P = np.eye(3) - 2 * np.outer(v, v)
return(P)
# update elements of M
def update_M(T,k,p):
x = T[k+1, k]
y = T[k+2, k]
if k < p - 3:
z = T[k+3, k]
else:
z = 0
return(x,y,z)
# givens rotation
def givens_step(T,x_2,x,y,p,q,n):
# calculate c and s
c = x / np.sqrt(x**2 + y**2)
s = -y / np.sqrt(x**2 + y**2)
P = np.array([[c, s], [-s, c]])
T[q-1:p, p-3:n] = P.T # T[q-1:p, p-3:n]
T[0:p, p-2:p] = T[0:p, p-2:p] # P
return(T)
# deflation step
def deflation_step(T,p,q,epsilon):
if abs(T[p-1, p-2]) < epsilon * (abs(T[p-2, p-2]) + abs(T[p-1, p-1])):
T[p-1, p-2] = 0
p = p - 1
q = p - 1
elif abs(T[p-2, p-3]) < epsilon * (abs(T[p-3, p-3]) + abs(T[p-2, p-2])):
T[p-2, p-3] = 0
p = p - 2
q = p - 1
return(T,p,q)
# francis qr step
def francis_step(H, epsilon=0.90):
n = H.shape[0]
T = H.copy().astype(np.float64)
p = n - 1
while p > 2:
q = p - 1
s = T[q, q] + T[p, p]
t = T[q, q] * T[p, p] - T[q, p] * T[p, q]
# Compute M
x,y,z = first_three_M(T,s,t)
x_1 = np.transpose([[x], [y], [z]])
# Bulge chasing
for k in range(p - 3):
# Compute Householder reflector
P = householder_reflection_step(x_1)
r = max(1, k-1)
T[k:k+3, r:] = P.T # T[k:k+3, r:]
r = min(k + 3, p)
T[0:r, k:k+3] = T[0:r, k:k+3] # P
# Update M
x,y,z = update_M(T,k,p)
x_2 = np.transpose([[x], [y]])
# Compute Givens rotation
T = givens_step(T,x_2,x,y,p,q,n)
# Check for convergence
T,p,q = deflation_step(T,p,q,epsilon)
return(T)
# francis qr iteration
def francis_qr_iteration(A):
m,n = A.shape
H = hessenberg(A)
eigvals = []
iters = 0
max_iters = 100
while iters<max_iters:
# Perform Francis step
T = francis_step(H)
eigvals.append(np.diag(T))
iters+=1
return(eigvals)
# for quick testing
A = np.array([[2, 2, 3, 4, 2],
[1, 2, 4, 2, 3],
[4, 1, 2, 1, 5],
[5, 2, 5, 2, 1],
[3, 6, 3, 1, 4]])
eigenvals = francis_qr_iteration(A)
#comparing our method to scipy - final eigvals obtained
print(len(eigenvals))
print(sorted(eigenvals[-1]))
print(sorted(scipy.linalg.eig(A)[0].real))
And this is the output I am getting.
100
[-4.421235127393854, -0.909209110641351, -0.8342390091346807, 3.7552499102751575, 8.215454029003958]
[-3.0411228516834217, -1.143605409373778, -1.143605409373778, 3.325396565009845, 14.002937105421134]
The matrix T is not changing and hence it does not converge to the Schur form through which I can obtain the eigenvalues by using np.diag(T). I believe the error is coming either from the Givens rotation step or the Householder reflection step. It could be an indexing issue since I tried to work in python using matlab psuedocode. Please let me know where I am going wrong so I can improve the code and make it converge.
I'm writing a program for university on python. I have to find the shortest distance between two lines in 3d given by two points (A B and C D) and find the points on both of these lines with the shortest distance between them. I'm bad at math so I can't understand how to find the points, only managed to find the formula of the minimal distance between two lines.
tried to write program, but it finds only point of intersection between lines and doesn't work correctly
#line 1
A = [1, 3, 1]
B = [0, -1, 2]
#line 2
C = [0, -2, 3]
D = [1, 0, 2]
def line_intersection(a, b, c, d):
v1 = [a_i - b_i for a_i, b_i in zip(B, A)]
v2 = [a_i - b_i for a_i, b_i in zip(D, C)]
if (-v1[0] * v2[1] + v1[1] * v2[0]) == 0:
t = ((c[1]-a[1]) * (-v2[2]) + (v2[1]) * (c[2]-a[2]))/(-v1[1] * v2[2] + v1[2] * v2[1])
else:
t = ((c[0]-a[0]) * (-v2[1]) + (v2[0]) * (c[1]-a[1]))/(-v1[0] * v2[1] + v1[1] * v2[0])
x = a[0] + t * v1[0]
y = a[1] + t * v1[1]
z = a[2] + t * v1[2]
return x, y, z
Currently I'm working on a project that implements cubic spline interpolation. So far I have managed to calculate coefficients for my equations.
Now I'm trying to return an interpolating function that for any x returns y.
Let's assume that we have
x = [1, 3, 5]
y = [6, -2, 4]
The coefficients that we get are as follow:
[ 6, -5.75, 0, 0.4375, -2, -0.5, 2.625, -0.4375]
It is equal to
[ a<sub>0</sub>, b<sub>0</sub>, c<sub>0</sub>, d<sub>0</sub>, a<sub>1</sub>, b<sub>1</sub>, c<sub>1</sub>, d<sub>1</sub>]
The interpolating polynomials are
S<sub>0</sub>(x) = a<sub>0</sub> + b<sub>0</sub>*x + c<sub>0</sub>*x<sup>2</sup> + d<sub>0</sub>*x<sup>3</sup> x ∈ [1, 3]
S<sub>1</sub>(x) = a<sub>1</sub> + b<sub>1</sub>*x + c<sub>1</sub>*x<sup>2</sup> + d<sub>1</sub>*x<sup>3</sup> x ∈ (3, 5]
And so on - it can be calculated for more than only 3 points
Right now I have implemented a method that works only if one x is given as an input.
def interpolate_spline(x, x_array, coefficients):
i = 1
while x_array[i] < x:
i += 1
i = i - 1
a = coefficients[4 * i]
b = coefficients[4 * i + 1]
c = coefficients[4 * i + 2]
d = coefficients[4 * i + 3]
return a + b * x + c * (x ** 2) + d * (x ** 3)
And coming back to my question: Is there any possibility that it can be vectorized or at least take whole array as an input?
I don't know if that matters but assume that x_array is sorted
I have a list of N unit-normalized 3D vectors p stored in a numpy ndarray with shape (N, 3). I have another such list, q. I want to calculate an ndarray U of shape (N, 3, 3) storing the rotation matrices that rotate each point in p to the corresponding point q.
The list of rotation matrices U should satisfy:
np.all(np.einsum('ijk,ik->ij', U, p) == q)
On a point-by-point basis, the problem reduces to being able to compute a rotation matrix for a rotation of some angle about some axis. Code solving the single-point case appears below:
def rotation_matrix(angle, direction):
direction = np.atleast_1d(direction).astype('f4')
sina = np.sin(angle)
cosa = np.cos(angle)
direction = direction/np.sqrt(np.sum(direction*direction))
R = np.diag([cosa, cosa, cosa])
R += np.outer(direction, direction) * (1.0 - cosa)
direction *= sina
R += np.array(((0.0, -direction[2], direction[1]),
(direction[2], 0.0, -direction[0]),
(-direction[1], direction[0], 0.0)))
return R
What I need is a function that behaves exactly as the above function, but instead of accepting a single angle and a single direction, it accepts an angles array of shape (npts, ) and a directions array of shape (npts, 3). The code below is only partially finished - the problem is that neither np.diag nor np.outer accept an axis argument
def rotation_matrices(angles, directions):
directions = np.atleast_2d(directions)
angles = np.atleast_1d(angles)
npts = directions.shape[0]
directions = directions/np.sqrt(np.sum(directions*directions, axis=1)).reshape((npts, 1))
sina = np.sin(angles)
cosa = np.cos(angles)
# Lines below require extension to 2d case - np.diag and np.outer do not support axis arguments
R = np.diag([cosa, cosa, cosa])
R += np.outer(directions, directions) * (1.0 - cosa)
directions *= sina
R += np.array(((0.0, -directions[2], directions[1]),
(directions[2], 0.0, -directions[0]),
(-directions[1], directions[0], 0.0)))
return R
Does either numpy or scipy have a compact vectorized function computing the appropriate rotation matrices in a way that avoids using for loops? The problem is that neither np.diag nor np.outer accept axis as an argument. My application will have N be very large, 1e7 or greater, so a vectorized function that keeps all the relevant axes aligned is necessary for performance reasons.
Dropping this here for now, will explain later. Using levi-cevita symbols from #jaime's answer here and the matrix form of the Rodrigues formula here and some algebra based on k = (a x b)/sin(theta)
def rotmatx(p, q):
eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1
d = (p * q).sum(-1)[:, None, None]
c = (p.dot(eijk) # q[..., None]).squeeze() # cross product (optimized)
cx = c.dot(eijk)
return np.eye(3) + cx + cx # cx / (1 + d)
EDIT: dang. question changed.
def rotation_matrices(angles, directions):
eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1
theta = angles[:, None, None]
K = directions.dot(eijk)
return np.eye(3) + K * np.sin(theta) + K # K * (1 - np.cos(theta))
Dropping another solution for bulk rotation of a Nx3x3 matrix. Where the 3x3 components represent vector components in
[[11, 12, 13],
[21, 22, 23],
[31, 32, 33]]
Now matrix rotation by np.einsum is:
data = np.random.uniform(size=(500, 3, 3))
rotmat = np.random.uniform(size=(3, 3))
data_rot = np.einsum('ij,...jk,lk->...il', rotmat, data, rotmat)
This is equivalent to
for data_mat in data:
np.dot(np.dot(rotmat, data_mat), rotmat.T)
Speedup over a np.dot-loop is around 250x.
Is there something like Matlab's procrustes function in NumPy/SciPy or related libraries?
For reference. Procrustes analysis aims to align 2 sets of points (in other words, 2 shapes) to minimize square distance between them by removing scale, translation and rotation warp components.
Example in Matlab:
X = [0 1; 2 3; 4 5; 6 7; 8 9]; % first shape
R = [1 2; 2 1]; % rotation matrix
t = [3 5]; % translation vector
Y = X * R + repmat(t, 5, 1); % warped shape, no scale and no distortion
[d Z] = procrustes(X, Y); % Z is Y aligned back to X
Z
Z =
0.0000 1.0000
2.0000 3.0000
4.0000 5.0000
6.0000 7.0000
8.0000 9.0000
Same task in NumPy:
X = arange(10).reshape((5, 2))
R = array([[1, 2], [2, 1]])
t = array([3, 5])
Y = dot(X, R) + t
Z = ???
Note: I'm only interested in aligned shape, since square error (variable d in Matlab code) is easily computed from 2 shapes.
I'm not aware of any pre-existing implementation in Python, but it's easy to take a look at the MATLAB code using edit procrustes.m and port it to Numpy:
def procrustes(X, Y, scaling=True, reflection='best'):
"""
A port of MATLAB's `procrustes` function to Numpy.
Procrustes analysis determines a linear transformation (translation,
reflection, orthogonal rotation and scaling) of the points in Y to best
conform them to the points in matrix X, using the sum of squared errors
as the goodness of fit criterion.
d, Z, [tform] = procrustes(X, Y)
Inputs:
------------
X, Y
matrices of target and input coordinates. they must have equal
numbers of points (rows), but Y may have fewer dimensions
(columns) than X.
scaling
if False, the scaling component of the transformation is forced
to 1
reflection
if 'best' (default), the transformation solution may or may not
include a reflection component, depending on which fits the data
best. setting reflection to True or False forces a solution with
reflection or no reflection respectively.
Outputs
------------
d
the residual sum of squared errors, normalized according to a
measure of the scale of X, ((X - X.mean(0))**2).sum()
Z
the matrix of transformed Y-values
tform
a dict specifying the rotation, translation and scaling that
maps X --> Y
"""
n,m = X.shape
ny,my = Y.shape
muX = X.mean(0)
muY = Y.mean(0)
X0 = X - muX
Y0 = Y - muY
ssX = (X0**2.).sum()
ssY = (Y0**2.).sum()
# centred Frobenius norm
normX = np.sqrt(ssX)
normY = np.sqrt(ssY)
# scale to equal (unit) norm
X0 /= normX
Y0 /= normY
if my < m:
Y0 = np.concatenate((Y0, np.zeros(n, m-my)),0)
# optimum rotation matrix of Y
A = np.dot(X0.T, Y0)
U,s,Vt = np.linalg.svd(A,full_matrices=False)
V = Vt.T
T = np.dot(V, U.T)
if reflection != 'best':
# does the current solution use a reflection?
have_reflection = np.linalg.det(T) < 0
# if that's not what was specified, force another reflection
if reflection != have_reflection:
V[:,-1] *= -1
s[-1] *= -1
T = np.dot(V, U.T)
traceTA = s.sum()
if scaling:
# optimum scaling of Y
b = traceTA * normX / normY
# standarised distance between X and b*Y*T + c
d = 1 - traceTA**2
# transformed coords
Z = normX*traceTA*np.dot(Y0, T) + muX
else:
b = 1
d = 1 + ssY/ssX - 2 * traceTA * normY / normX
Z = normY*np.dot(Y0, T) + muX
# transformation matrix
if my < m:
T = T[:my,:]
c = muX - b*np.dot(muY, T)
#transformation values
tform = {'rotation':T, 'scale':b, 'translation':c}
return d, Z, tform
There is a Scipy function for it: scipy.spatial.procrustes
I'm just posting its example here:
>>> import numpy as np
>>> from scipy.spatial import procrustes
>>> a = np.array([[1, 3], [1, 2], [1, 1], [2, 1]], 'd')
>>> b = np.array([[4, -2], [4, -4], [4, -6], [2, -6]], 'd')
>>> mtx1, mtx2, disparity = procrustes(a, b)
>>> round(disparity)
0.0
You can have both Ordinary Procrustes Analysis and Generalized Procrustes Analysis in python with something like this:
import numpy as np
def opa(a, b):
aT = a.mean(0)
bT = b.mean(0)
A = a - aT
B = b - bT
aS = np.sum(A * A)**.5
bS = np.sum(B * B)**.5
A /= aS
B /= bS
U, _, V = np.linalg.svd(np.dot(B.T, A))
aR = np.dot(U, V)
if np.linalg.det(aR) < 0:
V[1] *= -1
aR = np.dot(U, V)
aS = aS / bS
aT-= (bT.dot(aR) * aS)
aD = (np.sum((A - B.dot(aR))**2) / len(a))**.5
return aR, aS, aT, aD
def gpa(v, n=-1):
if n < 0:
p = avg(v)
else:
p = v[n]
l = len(v)
r, s, t, d = np.ndarray((4, l), object)
for i in range(l):
r[i], s[i], t[i], d[i] = opa(p, v[i])
return r, s, t, d
def avg(v):
v_= np.copy(v)
l = len(v_)
R, S, T = [list(np.zeros(l)) for _ in range(3)]
for i, j in np.ndindex(l, l):
r, s, t, _ = opa(v_[i], v_[j])
R[j] += np.arccos(min(1, max(-1, np.trace(r[:1])))) * np.sign(r[1][0])
S[j] += s
T[j] += t
for i in range(l):
a = R[i] / l
r = [np.cos(a), -np.sin(a)], [np.sin(a), np.cos(a)]
v_[i] = v_[i].dot(r) * (S[i] / l) + (T[i] / l)
return v_.mean(0)
For testing purposes, the output of each algorithm can be visualized as follows:
import matplotlib.pyplot as p; p.rcParams['toolbar'] = 'None';
def plt(o, e, b):
p.figure(figsize=(10, 10), dpi=72, facecolor='w').add_axes([0.05, 0.05, 0.9, 0.9], aspect='equal')
p.plot(0, 0, marker='x', mew=1, ms=10, c='g', zorder=2, clip_on=False)
p.gcf().canvas.set_window_title('%f' % e)
x = np.ravel(o[0].T[0])
y = np.ravel(o[0].T[1])
p.xlim(min(x), max(x))
p.ylim(min(y), max(y))
a = []
for i, j in np.ndindex(len(o), 2):
a.append(o[i].T[j])
O = p.plot(*a, marker='x', mew=1, ms=10, lw=.25, c='b', zorder=0, clip_on=False)
O[0].set(c='r', zorder=1)
if not b:
O[2].set_color('b')
O[2].set_alpha(0.4)
p.axis('off')
p.show()
# Fly wings example (Klingenberg, 2015 | https://en.wikipedia.org/wiki/Procrustes_analysis)
arr1 = np.array([[588.0, 443.0], [178.0, 443.0], [56.0, 436.0], [50.0, 376.0], [129.0, 360.0], [15.0, 342.0], [92.0, 293.0], [79.0, 269.0], [276.0, 295.0], [281.0, 331.0], [785.0, 260.0], [754.0, 174.0], [405.0, 233.0], [386.0, 167.0], [466.0, 59.0]])
arr2 = np.array([[477.0, 557.0], [130.129, 374.307], [52.0, 334.0], [67.662, 306.953], [111.916, 323.0], [55.119, 275.854], [107.935, 277.723], [101.899, 259.73], [175.0, 329.0], [171.0, 345.0], [589.0, 527.0], [591.0, 468.0], [299.0, 363.0], [306.0, 317.0], [406.0, 288.0]])
def opa_out(a):
r, s, t, d = opa(a[0], a[1])
a[1] = a[1].dot(r) * s + t
return a, d, False
plt(*opa_out([arr1, arr2, np.matrix.copy(arr2)]))
def gpa_out(a):
g = gpa(a, -1)
D = [avg(a)]
for i in range(len(a)):
D.append(a[i].dot(g[0][i]) * g[1][i] + g[2][i])
return D, sum(g[3])/len(a), True
plt(*gpa_out([arr1, arr2]))
Probably you want to try this package with various flavors of different Procrustes methods, https://github.com/theochem/procrustes.