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How to use mgrid to interpolate between a rectangle and a circle

I am trying to create a 3D surface that has a 1/4 rectangle for the exterior and 1/4 circle for the interior. I had help before to create the 3D surface with an ellipse as an exterior but I cannot do this for a rectangle for some reason. I have done the math by hand which makes sense, but my code does not. I would greatly appreciate any help with this.
import numpy as np
import pyvista as pv
# parameters for the waveguide
# diameter of the inner circle
waveguide_throat = 30
# axes of the outer ellipse
ellipse_x = 250
ellipse_y = 170
# shape parameters for the z profile
depth_factor = 4
angle_factor = 40
# number of grid points in radial and angular direction
array_length = 100
phase_plug = 0
phase_plug_dia = 20
plug_offset = 5
dome_dia = 28
# theta is angle where x and y intersect
theta = np.arctan(ellipse_x / ellipse_y)
# chi is for x direction and lhi is for y direction
chi = np.linspace(0, theta, 100)
lhi = np.linspace(theta, np.pi/2, 100)
# mgrid to create structured grid
r, phi = np.mgrid[0:1:array_length*1j, 0:np.pi/2:array_length*1j]
# Rectangle exterior, circle interior
x = (ellipse_y * np.tan(chi)) * r + ((waveguide_throat / 2 * (1 - r)) * np.cos(phi))
y = (ellipse_x / np.tan(lhi)) * r + ((waveguide_throat / 2 * (1 - r)) * np.sin(phi))
# compute z profile
angle_factor = angle_factor / 10000
z = (ellipse_x / 2 * r / angle_factor) ** (1 / depth_factor)
plotter = pv.Plotter()
waveguide_mesh = pv.StructuredGrid(x, y, z)
plotter.add_mesh(waveguide_mesh)
plotter.show()

The linear interpolation you're trying to use is a general tool that should work (with one small caveat). So the issue is first with your rectangular edge.
Here's a sanity check which plots your interior and exterior lines:
# debugging: plot interior and exterior
exterior_points = np.array([
ellipse_y * np.tan(chi),
ellipse_x / np.tan(lhi),
np.zeros_like(chi)
]).T
phi_aux = np.linspace(0, np.pi/2, array_length)
interior_points = np.array([
waveguide_throat / 2 * np.cos(phi_aux),
waveguide_throat / 2 * np.sin(phi_aux),
np.zeros_like(phi_aux)
]).T
plotter = pv.Plotter()
plotter.add_mesh(pv.wrap(exterior_points))
plotter.add_mesh(pv.wrap(interior_points))
plotter.show()
The bottom left is your interior circle, looks good. The top right is what's supposed to be a rectangle, but isn't.
To see why your original surface looks the way it does, we have to note one more thing (this is the small caveat I mentioned): the orientation of your curves is also the opposite. This implies that you interpolate the "top" (in the screenshot) point of your interior curve with the "bottom" point of the exterior curve. This explains the weird fan shape.
So you need to fix the exterior curve, and make sure the orientation of the two edges is the same. Note that you can just create the two 1d arrays for the two edges, and then interpolate them. You don't have to come up with a symbolic formula that you plug into the interpolation step. If you have 1d arrays of the same shape x_interior, y_interior, x_exterior, y_exterior then you can then do x_exterior * r + x_interior * (1 - r) and the same for y. This means removing the mgrid call, only using an array r of shape (n, 1), and making use of array broadcasting to do the interpolation. This means doing r = np.linspace(0, 1, array_length)[:, None].
So the question is how to define your rectangle. You need to have the same number of points on the rectangular curve than what you have on the circle (I would strongly recommend using the array_length parameter everywhere to ensure this!). Since you want to span the whole rectangle, I believe you have to choose an array index (i.e. a certain angle in the circular arc) which will map to the corner of the rectangle. Then it's a simple matter of varying only y for the points until that index, and x for the rest (or vice versa).
Here's what I mean: you know that the rectangle's corner is at angle theta in your code (although I think you have x and y mixed up if we assume the conventional relationship between "x", "y" and the tangent of the angle). Since theta goes from 0 to pi/2, and your phi values also go from 0 to pi/2, you should choose index (array_length * (2*theta/np.pi)).round().astype(int) - 1 (or something similar) that will map to the rectangle's corner. If you have a square, this gives you theta = pi/4, and consequently (array_length / 2).round().astype(int) - 1. For array_length = 3 this is index (2 - 1) == 1, which is the middle index for 3-length arrays. (The more points you have along the edge, the less it will matter if you commit an off-by-one error here.)
The only remaining complication then is that we have to explicitly broadcast the 1d z array to the common shape. And we can use the same math you used to get a rectangular edge that is equidistant in angles.
Your code fixed with this suggestion (note that I've added 1 to the corner index because I'm using it as a right-exclusive range index):
import numpy as np
import pyvista as pv
# parameters for the waveguide
# diameter of the inner circle
waveguide_throat = 30
# axes of the outer ellipse
ellipse_x = 250
ellipse_y = 170
# shape parameters for the z profile
depth_factor = 4
angle_factor = 40
# number of grid points in radial and angular direction
array_length = 100
# quarter circle interior line
phi = np.linspace(0, np.pi/2, array_length)
x_interior = waveguide_throat / 2 * np.cos(phi)
y_interior = waveguide_throat / 2 * np.sin(phi)
# theta is angle where x and y intersect
theta = np.arctan2(ellipse_y, ellipse_x)
# find array index which maps to the corner of the rectangle
corner_index = (array_length * (2*theta/np.pi)).round().astype(int)
# construct rectangular coordinates manually
x_exterior = np.zeros_like(x_interior)
y_exterior = x_exterior.copy()
phi_aux = np.linspace(0, theta, corner_index)
x_exterior[:corner_index] = ellipse_x
y_exterior[:corner_index] = ellipse_x * np.tan(phi_aux)
phi_aux = np.linspace(np.pi/2, theta, array_length - corner_index, endpoint=False)[::-1] # mind the reverse!
x_exterior[corner_index:] = ellipse_y / np.tan(phi_aux)
y_exterior[corner_index:] = ellipse_y
# interpolate between two curves
r = np.linspace(0, 1, array_length)[:, None] # shape (array_length, 1) for broadcasting
x = x_exterior * r + x_interior * (1 - r)
y = y_exterior * r + y_interior * (1 - r)
# debugging: plot interior and exterior
exterior_points = np.array([
x_exterior,
y_exterior,
np.zeros_like(x_exterior),
]).T
interior_points = np.array([
x_interior,
y_interior,
np.zeros_like(x_interior),
]).T
plotter = pv.Plotter()
plotter.add_mesh(pv.wrap(exterior_points))
plotter.add_mesh(pv.wrap(interior_points))
plotter.show()
# compute z profile
angle_factor = angle_factor / 10000
z = (ellipse_x / 2 * r / angle_factor) ** (1 / depth_factor)
# explicitly broadcast to the shape of x and y
z = np.broadcast_to(z, x.shape)
plotter = pv.Plotter()
waveguide_mesh = pv.StructuredGrid(x, y, z)
plotter.add_mesh(waveguide_mesh, style='wireframe')
plotter.show()
The curves look reasonable:
As does the interpolated surface:

How to rotate a set of 2D points to minimize vertical span in numpy?

I have a collection of 2D points and I want to rotate them so as to display them spanning the least vertical height. Is there a way to find the angle without using optimization?
As a brief illustration:
import numpy as np
# some initial points spread out vertically
xy = np.random.randn(2, 100)
xy[1, :] *= 2
def rotate(xy, theta):
"""Return a rotated set of points.
"""
s = np.sin(theta)
c = np.cos(theta)
xyr = np.empty_like(xy)
xyr[0, :] = c * xy[0, :] - s * xy[1, :]
xyr[1, :] = s * xy[0, :] + c * xy[1, :]
return xyr
def span(xy):
"""Return the vertical span of the points.
"""
return xy[1, :].max() - xy[1, :].min()
def plot(xy):
"""2D plot with fixed aspect ratio.
"""
import matplotlib.pyplot as plt
fig, ax = plt.subplots(figsize=(3, 3))
ax.scatter(*xy, alpha=0.3)
ax.set_aspect(1)
plt.show()
plt.show()
Have a look at the original points:
>>> span(xy)
11.503342270923472
>>> plot(xy)
Have a look at some rotated points:
>>> xyr = rotate(xy, np.pi / 2)
>>> span(xyr)
4.594620173868735
>>> plot(xyr)
The optimal answer (which should here be around pi / 2) is easy to find by invoking scipy:
>>> from scipy.optimize import minimize_scalar
>>> minimize_scalar(lambda theta: span(rotate(xy, theta)))
fun: 4.523188831276214
nfev: 38
nit: 34
success: True
x: 1.590391370976612
But surely there is a simpler way that doesn't require scipy - and 38 function evaluations!

This is probably not necessarily optimal, but a good enough simple solution may be to take the main components of the data and rotate the data to align largest main component with the horizontal axis.
import numpy as np
# Posted code...
# Find eigenvector with largest eigenvalue
c = (xy # xy.T) / (xy.shape[1] - 1)
evals, evecs = np.linalg.eig(c)
imax = np.abs(evals).argmax()
v = evecs[:, imax]
# Correction angle is opposite of vector angle
theta = -np.arctan2(v[1], v[0])
# Rotate
xy2 = rotate(xy, theta)
# Plot
plot(xy2)
Output:

Apply rotation defined by Euler angles to 3D image, in python

I'm working with 3D images and have to rotate them according to Euler angles (phi,psi,theta) in 'zxz' convention (these Euler angles are part of a dataset, so I have to use that convention). I found the function scipy.ndimage.rotate that seems useful in that regard.
arrayR = scipy.ndimage.rotate(array , phi, axes=(0,1), reshape=False)
arrayR = scipy.ndimage.rotate(arrayR, psi, axes=(1,2), reshape=False)
arrayR = scipy.ndimage.rotate(arrayR, the, axes=(0,1), reshape=False)
Sadly, this does not do what intended. This is why:
Definition:
In the z-x-z convention, the x-y-z frame is rotated three times: first
about the z-axis by an angle phi; then about the new x-axis by an
angle psi; then about the newest z-axis by an angle theta.
However with above code, the rotations are always with respect to the original axes. Which is why obtained rotations are not correct. Anyone has a suggestion to obtain correct rotations, as explained in the definition?
In other words, in the present 'zxz' convention the rotations are intrinsic (rotations about the axes of the rotating coordinate system XYZ, solidary with the moving body, which changes its orientation after each elemental rotation). If I use the above code, the rotations are extrinsic (rotations about the axes xyz of the original coordinate system, which is assumed to remain motionless). I need a way for doing extrinsic rotations, in python.

I found a satisfying solution following this link: https://nbviewer.jupyter.org/gist/lhk/f05ee20b5a826e4c8b9bb3e528348688
This method uses np.meshgrid, scipy.ndimage.map_coordinates. The above link uses some third party library for generating the rotation matrix, however I use scipy.spatial.transform.Rotation. This function allows to define both intrinsic and extrinsic rotations: see description of scipy.spatial.transform.Rotation.from_euler.
Here is my function:
import numpy as np
from scipy.spatial.transform import Rotation as R
from scipy.ndimage import map_coordinates
# Rotates 3D image around image center
# INPUTS
# array: 3D numpy array
# orient: list of Euler angles (phi,psi,the)
# OUTPUT
# arrayR: rotated 3D numpy array
# by E. Moebel, 2020
def rotate_array(array, orient):
phi = orient[0]
psi = orient[1]
the = orient[2]
# create meshgrid
dim = array.shape
ax = np.arange(dim[0])
ay = np.arange(dim[1])
az = np.arange(dim[2])
coords = np.meshgrid(ax, ay, az)
# stack the meshgrid to position vectors, center them around 0 by substracting dim/2
xyz = np.vstack([coords[0].reshape(-1) - float(dim[0]) / 2, # x coordinate, centered
coords[1].reshape(-1) - float(dim[1]) / 2, # y coordinate, centered
coords[2].reshape(-1) - float(dim[2]) / 2]) # z coordinate, centered
# create transformation matrix
r = R.from_euler('zxz', [phi, psi, the], degrees=True)
mat = r.as_matrix()
# apply transformation
transformed_xyz = np.dot(mat, xyz)
# extract coordinates
x = transformed_xyz[0, :] + float(dim[0]) / 2
y = transformed_xyz[1, :] + float(dim[1]) / 2
z = transformed_xyz[2, :] + float(dim[2]) / 2
x = x.reshape((dim[1],dim[0],dim[2]))
y = y.reshape((dim[1],dim[0],dim[2]))
z = z.reshape((dim[1],dim[0],dim[2])) # reason for strange ordering: see next line
# the coordinate system seems to be strange, it has to be ordered like this
new_xyz = [y, x, z]
# sample
arrayR = map_coordinates(array, new_xyz, order=1)
Note:
You can also use this function for intrinsic rotations, simply adapt the first argument of 'from_euler' to your Euler convention. In this case, you obtain equivalent result than in my 1st post (using scipy.ndimage.rotate). However I noticed that the present code is 3x faster (0.01s for 40^3 volume) than when using scipy.ndimage.rotate (0.03s for 40^3 volume).
Hope this will help someone!

There seem to be a bit confusion about the "axes" parameter in your first post. To do a rotation about the x axis, the plane of rotation would be the yz plane which means your "axes" parameter should be set to (1,2). Also the first and the third rotations are, presumably about the x and z axes. But, both your rotations are in the xy plane. Could these be possibly the reasons behind the discrepancies in your answers? I am not convinced by your explanations about the new and original axes. The independent calls to the "rotate" function does not have access to the old data in any form or shape. It only sees the new axes and rotated array.

I check the code https://nbviewer.jupyter.org/gist/lhk/f05ee20b5a826e4c8b9bb3e528348688
There is a minor bug. The tested image is square, but if doing rectangular image, it will encounter some problems. below are correct ones for 2D and 3D rotations (noted that the Euler angle sequence used in my example is 'ZYZ', you should define this before using it):
def rotate_array_2D(array, orient):
# create a transformation matrix
angle=orient/180.*np.pi
c=np.cos(angle)
s=np.sin(angle)
mat=np.array([[c,s],[-s,c]])
# create meshgrid
dim = array.shape
ax = np.arange(dim[0])
ay = np.arange(dim[1])
coords = np.meshgrid(ax, ay)
# stack the meshgrid to position vectors, center them around 0 by substracting dim/2
xy = np.vstack([coords[0].reshape(-1) - float(dim[0]) / 2, # x coordinate, centered
coords[1].reshape(-1) - float(dim[1]) / 2]) # y coordinate, centered
# apply transformation
transformed_xy = np.dot(mat, xy)
# extract coordinates
x = transformed_xy[0, :] + float(dim[0]) / 2
y = transformed_xy[1, :] + float(dim[1]) / 2
x = x.reshape((dim[1],dim[0]))
y = y.reshape((dim[1],dim[0]))
new_xy = [x,y]
# sample
arrayR = map_coordinates(array, new_xy, order=1).T
return arrayR
def rotate_array_3D(array, orient):
rot = orient[0]
tilt = orient[1]
phi = orient[2]
# create meshgrid
dim = array.shape
ax = np.arange(dim[0])
ay = np.arange(dim[1])
az = np.arange(dim[2])
coords = np.meshgrid(ax, ay, az)
# stack the meshgrid to position vectors, center them around 0 by substracting dim/2
xyz = np.vstack([coords[0].reshape(-1) - float(dim[0]) / 2, # x coordinate, centered
coords[1].reshape(-1) - float(dim[1]) / 2, # y coordinate, centered
coords[2].reshape(-1) - float(dim[2]) / 2]) # z coordinate, centered
# create transformation matrix
r = R.from_euler('ZYZ', [rot, tilt, phi], degrees=True)
mat = r.as_matrix()
# apply transformation
transformed_xyz = np.dot(mat, xyz)
# extract coordinates
x = transformed_xyz[0, :] + float(dim[0]) / 2
y = transformed_xyz[1, :] + float(dim[1]) / 2
z = transformed_xyz[2, :] + float(dim[2]) / 2
x = x.reshape((dim[1],dim[0],dim[2]))
y = y.reshape((dim[1],dim[0],dim[2]))
z = z.reshape((dim[1],dim[0],dim[2])) # I test the rotation in 2D and this strange thing can be explained
new_xyz = [x,y,z]
arrayR = map_coordinates(array, new_xyz, order=1).T
return arrayR

how to generate new points as offset with gaussian distribution for some points in spherical coordinates in python

I am working with some points in spherical coordinates. I need to generate new points as the error points for them and a kind of offset for the old points.
The new point should be in a specific distance from the old one which distributing by gaussian distribution. The angle of new point compared to old one is not important.I am trying to generate new points for r direction. no matter what are phi and theta (Spherical coordinates)
To generate the new point distributing by gaussian function, I tried the numpy.rand.normal(mean,std,..). But It is generating 1D random points over mean value and this mean value is a real number. In my case I need an approach to specify the position of the old point and I have one given standard deviation for this distance from the original points.
Honesty, I dont have a copy of my code. It is on the university's server. But let's assume I have an array of size 100*3 including the spherical (or cartesian) coordinates of some points on a surface of a cylinder. In spherical case, the first column presents the radius value, the second column is theta and third one shows the phi for the points. now I want to generate random points from them using gaussian distribution. there is a given standard deviation for the gaussian distribution. The only important thing is that the new points generated by gaussian distribution are limited in r value. No matter the position of points in term of theta and phi.
When I tried numpy.rand.normal(mean,std,..), this generate some random points over the mean value. It does not help me. I want new points over my old ones with the given STD.
any idea would be appreciated.
This is a code, similar to mine written By Ophion How to generate regular points on cylindrical surface
def make_cylinder(radius, length, nlength, alpha, nalpha, center, orientation):
#Create the length array
I = np.linspace(0, length, nlength)
#Create alpha array avoid duplication of endpoints
#Conditional should be changed to meet your requirements
if int(alpha) == 360:
A = np.linspace(0, alpha, num=nalpha, endpoint=False)/180*np.pi
else:
A = np.linspace(0, alpha, num=nalpha)/180*np.pi
#Calculate X and Y
X = radius * np.cos(A)
Y = radius * np.sin(A)
#Tile/repeat indices so all unique pairs are present
pz = np.tile(I, nalpha)
px = np.repeat(X, nlength)
py = np.repeat(Y, nlength)
points = np.vstack(( pz, px, py )).T
#Shift to center
shift = np.array(center) - np.mean(points, axis=0)
points += shift
#Orient tube to new vector
#Grabbed from an old unutbu answer
def rotation_matrix(axis,theta):
a = np.cos(theta/2)
b,c,d = -axis*np.sin(theta/2)
return np.array([[a*a+b*b-c*c-d*d, 2*(b*c-a*d), 2*(b*d+a*c)],
[2*(b*c+a*d), a*a+c*c-b*b-d*d, 2*(c*d-a*b)],
[2*(b*d-a*c), 2*(c*d+a*b), a*a+d*d-b*b-c*c]])
ovec = orientation / np.linalg.norm(orientation)
cylvec = np.array([1,0,0])
if np.allclose(cylvec, ovec):
return points
#Get orthogonal axis and rotation
oaxis = np.cross(ovec, cylvec)
rot = np.arccos(np.dot(ovec, cylvec))
R = rotation_matrix(oaxis, rot)
return points.dot(R)
now calling the function:
points = make_cylinder(3, 5, 5, 360, 10, [0,2,0], [1,0,0])
sigma = 0.5 # given STD
ossfet_points = numpy.random.normal(np.mean(point[:,0]), sigma, size=(n,3))

If I'm not mistaken, you want random points on a spherical manifold with a gaussian distribution of distances from the center. If so, then you have the latter problem solved by sampling gaussian values of the radius using numpy.rand.normal
To get random spherical points is a little bit more tricky, but here's some code to do it (and a description of the math behind it at Wolfram MathWorld):
import numpy as np
num_points = 500
U = np.random.random(num_points)
V = np.random.random(num_points)
import math as m
def spherical_to_cartesian(vec):
'''
Convert spherical polar coordinates to cartesian coordinates:
See the definition of spherical_cartesian_to_polar.
#param vec: A vector of the 3 polar coordinates (r, u, v)
#return: (x, y, z)
'''
(r, u, v) = vec
x = r * m.sin(u) * m.cos(v)
y = r * m.sin(u) * m.sin(v)
z = r * m.cos(u)
return [x, y, z]
radius = 1.
points = np.array([spherical_to_cartesian([radius, 2 * np.pi * u, np.arccos(2*v - 1)]) for u,v in zip(U,V)])
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
ax = Axes3D(fig)
ax.plot(points[:,0], points[:,1], points[:,2], 'o')
Which will give you points like this:
Now if you want them to have normally distributed radii, you just need to substitute your randomly generated values in the list comprehension which uses the variable radius like this:
radii = np.random.normal(10, 3, 100)
points = np.array([spherical_to_cartesian([r, 2 * np.pi * u, np.arccos(2*v - 1)]) for r,u,v in zip(radii, U,V)])
Is this more or less what you're looking for?

Mask a circular sector in a numpy array

I have a code that slices a numpy array into a circle. I wish to recover only the values included in a certain range of angles from the circle and mask the array. For example: mask the original array with the (x,y) positions comprised between 0 and 45 degrees of the circle.
Is there a pythonic way for doing so?
Here's my (simplified) original code:
import numpy as np
matrix = np.zeros((500,500))
x = 240
y = 280
radius = 10
mask=np.ogrid[x-radius:x+radius+1,y-radius:y+radius+1]
matrix[mask]
Thanks in advance
Edit: I omitted that radius can vary.

I would do this by converting from cartesian to polar coordinates and constructing boolean masks for the circle and for the range of angles you want:
import numpy as np
def sector_mask(shape,centre,radius,angle_range):
"""
Return a boolean mask for a circular sector. The start/stop angles in
`angle_range` should be given in clockwise order.
"""
x,y = np.ogrid[:shape[0],:shape[1]]
cx,cy = centre
tmin,tmax = np.deg2rad(angle_range)
# ensure stop angle > start angle
if tmax < tmin:
tmax += 2*np.pi
# convert cartesian --> polar coordinates
r2 = (x-cx)*(x-cx) + (y-cy)*(y-cy)
theta = np.arctan2(x-cx,y-cy) - tmin
# wrap angles between 0 and 2*pi
theta %= (2*np.pi)
# circular mask
circmask = r2 <= radius*radius
# angular mask
anglemask = theta <= (tmax-tmin)
return circmask*anglemask
For example:
from matplotlib import pyplot as pp
from scipy.misc import lena
matrix = lena()
mask = sector_mask(matrix.shape,(200,100),300,(0,50))
matrix[~mask] = 0
pp.imshow(matrix)
pp.show()

Same approach for centered circles in square matrices:
def circleMask(mat, r=0):
if mat.shape[0] != mat.shape[1]:
raise TypeError('Matrix has to be square')
if not isinstance(r, int):
raise TypeError('Radius has to be of type int')
s = mat.shape[0]
d = num.abs(num.arange(-s/2 + s%2, s/2 + s%2))
dm = num.sqrt(d[:, num.newaxis]**2 + d[num.newaxis, :]**2)
return num.logical_and(dm >= r-.5, dm < r+.5)
looping over this implicit function is costly!