I'm working on a Python-based data analysis. I have some x-y data points, and some ellipses, and I want to determine whether points are inside any of the ellipses. The way that I've been doing this works, but it's kludgy. As I think about distributing my software to other people, I find myself wanting a cleaner way.
Right now, I'm using matplotlib.patches.Ellipse objects. Matplotlib Ellipses have a useful method called contains_point(). You can work in data coordinates on a Matplotlib Axes object by calling Axes.transData.transform().
The catch is that I have to create a Figure and an Axes object to hold the Ellipses. And when my program runs, an annoying Matplotlib Figure object will get rendered, showing the Ellipses, which I don't actually need to see. I have tried several methods to suppress this output. I have succeeded in deleting the Ellipses from the Axes, using Axes.clear(), resulting in an empty graph. But I can't get Matplolib's pyplot.close(fig_number) to delete the Figure itself before calling pyplot.show().
Any advice is appreciated, thanks!
Inspired by how a carpenter draws an ellipse using two nails and a piece of string, here is a numpy-friendly implementation to test whether points lie inside given ellipses.
One of the definitions of an ellipse, is that the sum of the distances to the two foci is constant, equal to the width (or height if it would be larger) of the ellipse. The distance between the center and the foci is sqrt(a*a - b*b), where a and b are half of the width and height. Using that distance and rotation by the desired angle finds the locations of the foci. numpy.linalg.norm can be used to calculate the distances using numpy's efficient array operations.
After the calculations, a plot is generated to visually check whether everything went correct.
import numpy as np
from numpy.linalg import norm # calculate the length of a vector
x = np.random.uniform(0, 40, 20000)
y = np.random.uniform(0, 20, 20000)
xy = np.dstack((x, y))
el_cent = np.array([20, 10])
el_width = 28
el_height = 17
el_angle = 20
# distance between the center and the foci
foc_dist = np.sqrt(np.abs(el_height * el_height - el_width * el_width) / 4)
# vector from center to one of the foci
foc_vect = np.array([foc_dist * np.cos(el_angle * np.pi / 180), foc_dist * np.sin(el_angle * np.pi / 180)])
# the two foci
el_foc1 = el_cent + foc_vect
el_foc2 = el_cent - foc_vect
# for each x,y: calculate z as the sum of the distances to the foci;
# np.ravel is needed to change the array of arrays (of 1 element) into a single array
z = np.ravel(norm(xy - el_foc1, axis=-1) + norm(xy - el_foc2, axis=-1) )
# points are exactly on the ellipse when the sum of distances is equal to the width
# z = np.where(z <= max(el_width, el_height), 1, 0)
# now create a plot to check whether everything makes sense
from matplotlib import pyplot as plt
from matplotlib import patches as mpatches
fig, ax = plt.subplots()
# show the foci as red dots
plt.plot(*el_foc1, 'ro')
plt.plot(*el_foc2, 'ro')
# create a filter to separate the points inside the ellipse
filter = z <= max(el_width, el_height)
# draw all the points inside the ellipse with the plasma colormap
ax.scatter(x[filter], y[filter], s=5, c=z[filter], cmap='plasma')
# draw all the points outside with the cool colormap
ax.scatter(x[~filter], y[~filter], s=5, c=z[~filter], cmap='cool')
# add the original ellipse to verify that the boundaries match
ellipse = mpatches.Ellipse(xy=el_cent, width=el_width, height=el_height, angle=el_angle,
facecolor='None', edgecolor='black', linewidth=2,
transform=ax.transData)
ax.add_patch(ellipse)
ax.set_aspect('equal', 'box')
ax.autoscale(enable=True, axis='both', tight=True)
plt.show()
The simplest solution here is to use shapely.
If you have an array of shape Nx2 containing a set of vertices (xy) then it is trivial to construct the appropriate shapely.geometry.polygon object and check if an arbitrary point or set of points (points) is contained within -
import shapely.geometry as geom
ellipse = geom.Polygon(xy)
for p in points:
if ellipse.contains(geom.Point(p)):
# ...
Alternatively, if the ellipses are defined by their parameters (i.e. rotation angle, semimajor and semiminor axis) then the array containing the vertices must be constructed and then the same process applied. I would recommend using the polar form relative to center as this is the most compatible with how shapely constructs the polygons.
import shapely.geometry as geom
from shapely import affinity
n = 360
a = 2
b = 1
angle = 45
theta = np.linspace(0, np.pi*2, n)
r = a * b / np.sqrt((b * np.cos(theta))**2 + (a * np.sin(theta))**2)
xy = np.stack([r * np.cos(theta), r * np.sin(theta)], 1)
ellipse = affinity.rotate(geom.Polygon(xy), angle, 'center')
for p in points:
if ellipse.contains(geom.Point(p)):
# ...
This method is advantageous because it supports any properly defined polygons - not just ellipses, it doesn't rely on matplotlib methods to perform the containment checking, and it produces a very readable code (which is often important when "distributing [one's] software to other people").
Here is a complete example (with added plotting to show it working)
import shapely.geometry as geom
from shapely import affinity
import matplotlib.pyplot as plt
import numpy as np
n = 360
theta = np.linspace(0, np.pi*2, n)
a = 2
b = 1
angle = 45.0
r = a * b / np.sqrt((b * np.cos(theta))**2 + (a * np.sin(theta))**2)
xy = np.stack([r * np.cos(theta), r * np.sin(theta)], 1)
ellipse = affinity.rotate(geom.Polygon(xy), angle, 'center')
x, y = ellipse.exterior.xy
# Create a Nx2 array of points at grid coordinates throughout
# the ellipse extent
rnd = np.array([[i,j] for i in np.linspace(min(x),max(x),50)
for j in np.linspace(min(y),max(y),50)])
# Filter for points which are contained in the ellipse
res = np.array([p for p in rnd if ellipse.contains(geom.Point(p))])
plt.plot(x, y, lw = 1, color='k')
plt.scatter(rnd[:,0], rnd[:,1], s = 50, color=(0.68, 0.78, 0.91)
plt.scatter(res[:,0], res[:,1], s = 15, color=(0.12, 0.67, 0.71))
plt.show()
Related
say we have a 2D grid that is projected on a 3D surface, resulting in a 3D numpy array, like the below image. What is the most efficient way to calculate a surface normal for each point of this grid?
I can give you an example with simulated data:
I showed your way, with three points. With three points you can always calculate the cross product to get the perpendicular vector based on the two vectors created from three points. Order does not matter.
I took the liberty to also add the PCA approach using predefined sklearn functions. You can create your own PCA, good exercise to understand what happens under the hood but this works fine. The benefit of the approach is that it is easy to increase the number of neighbors and you are still able to calculate the normal vector. It is also possible to select the neighbors within a range instead of N nearest neighbors.
If you need more explanation about the working of the code please let me know.
from functools import partial
import numpy as np
from sklearn.neighbors import KDTree
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
from sklearn.decomposition import PCA
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# Grab some test data.
X, Y, Z = axes3d.get_test_data(0.25)
X, Y, Z = map(lambda x: x.flatten(), [X, Y, Z])
plt.plot(X, Y, Z, '.')
plt.show(block=False)
data = np.array([X, Y, Z]).T
tree = KDTree(data, metric='minkowski') # minkowki is p2 (euclidean)
# Get indices and distances:
dist, ind = tree.query(data, k=3) #k=3 points including itself
def calc_cross(p1, p2, p3):
v1 = p2 - p1
v2 = p3 - p1
v3 = np.cross(v1, v2)
return v3 / np.linalg.norm(v3)
def PCA_unit_vector(array, pca=PCA(n_components=3)):
pca.fit(array)
eigenvalues = pca.explained_variance_
return pca.components_[ np.argmin(eigenvalues) ]
combinations = data[ind]
normals = list(map(lambda x: calc_cross(*x), combinations))
# lazy with map
normals2 = list(map(PCA_unit_vector, combinations))
## NEW ##
def calc_angle_with_xy(vectors):
'''
Assuming unit vectors!
'''
l = np.sum(vectors[:,:2]**2, axis=1) ** 0.5
return np.arctan2(vectors[:, 2], l)
dist, ind = tree.query(data, k=5) #k=3 points including itself
combinations = data[ind]
# map with functools
pca = PCA(n_components=3)
normals3 = list(map(partial(PCA_unit_vector, pca=pca), combinations))
print( combinations[10] )
print(normals3[10])
n = np.array(normals3)
n[calc_angle_with_xy(n) < 0] *= -1
def set_axes_equal(ax):
'''Make axes of 3D plot have equal scale so that spheres appear as spheres,
cubes as cubes, etc.. This is one possible solution to Matplotlib's
ax.set_aspect('equal') and ax.axis('equal') not working for 3D.
Input
ax: a matplotlib axis, e.g., as output from plt.gca().
FROM: https://stackoverflow.com/questions/13685386/matplotlib-equal-unit-length-with-equal-aspect-ratio-z-axis-is-not-equal-to
'''
x_limits = ax.get_xlim3d()
y_limits = ax.get_ylim3d()
z_limits = ax.get_zlim3d()
x_range = abs(x_limits[1] - x_limits[0])
x_middle = np.mean(x_limits)
y_range = abs(y_limits[1] - y_limits[0])
y_middle = np.mean(y_limits)
z_range = abs(z_limits[1] - z_limits[0])
z_middle = np.mean(z_limits)
# The plot bounding box is a sphere in the sense of the infinity
# norm, hence I call half the max range the plot radius.
plot_radius = 0.5*max([x_range, y_range, z_range])
ax.set_xlim3d([x_middle - plot_radius, x_middle + plot_radius])
ax.set_ylim3d([y_middle - plot_radius, y_middle + plot_radius])
ax.set_zlim3d([z_middle - plot_radius, z_middle + plot_radius])
u, v, w = n.T
fig = plt.figure()
ax = fig.add_subplot(projection='3d')
# ax.set_aspect('equal')
# Make the grid
ax.quiver(X, Y, Z, u, v, w, length=10, normalize=True)
set_axes_equal(ax)
plt.show()
The surface normal for a point cloud is not well defined. One way to define them is from the surface normal of a reconstructed mesh using triangulation (which can introduce artefacts regarding you specific input). A relatively simple and fast solution is to use VTK to do that, and more specifically, vtkSurfaceReconstructionFilter and vtkPolyDataNormals . Regarding your needs, it might be useful to apply other filters.
Problem
I am trying to implement Edelsbrunner's Algorithm for the alpha shape of a 3D point cloud in python as presented in this SO post. However, I'm having trouble plotting results. Half my sphere looks good, and the other half garbled.
I suspect it may have something to do with the fact that I have negative coordinates, but I'm not sure.
Definitions
I'm adding these so programmers can contribute without being bogged down by math. These are simplified definitions and not meant to be precise (Feel free to skip this part; for more, see see Introduction to Alpha Shapes and Discrete Differential Geometry):
Delaunay triangulation: for a 2D point set, a tesselation of the points into triangles (i.e. "triangulation") where the circumscribed circle (i.e. "circumcircle") about every triangle contains no other points in the set. For 3D points, replace "triangle" with "tetrahedron" and "circumcircle" with "circumsphere".
Affinely independent: a collection of points p0, ..., pk such that all vectors vi := pi-p0 are linearly independent (i.e. in 2D not collinear, in 3D not coplanar); also called "points in general position"
k-simplex: the convex hull of k+1 affinely-independent points; we call the points its vertices.
0-simplex = point (consists of 0+1 = 1 points)
1-simplex = line (consists of 1+1 = 2 points)
2-simplex = triangle (consists of 2+1 = 3 points)
3-simplex = tetrahedron (consists of 3+1 = 4 points)
face: any simplex whose vertices are a subset of the vertices of another simplex; i.e. "a part of a simplex"
(geometric) simplicial complex: a collection of simplices where (1) the intersection of two simplices is a simplex, and (2) every face of a simplex is in the complex; i.e. "a bunch of simplices"
alpha-exposed: a simplex within a point set where the circle (2D) or ball (3D) of radius alpha through its vertices doesn't contain any other point in the point set
alpha shape: the boundary of all alpha-exposed simplices of a point set
Algorithm
Edelsbrunner's Algorithm is as follows:
Given a point cloud pts:
Compute the Delaunay triangulation DT of the point cloud
Find the alpha-complex: search all simplices in the Delaunay triangulation and (a) if any ball around a simplex is empty and has
a radius less than alpha (called the "alpha test"), then add it to the
alpha complex
The boundary of the alpha complex is the alpha shape
Code
from scipy.spatial import Delaunay
import numpy as np
from collections import defaultdict
from matplotlib import pyplot as plt
import pyvista as pv
fig = plt.figure()
ax = plt.axes(projection="3d")
plotter = pv.Plotter()
def alpha_shape_3D(pos, alpha):
"""
Compute the alpha shape (concave hull) of a set of 3D points.
Parameters:
pos - np.array of shape (n,3) points.
alpha - alpha value.
return
outer surface vertex indices, edge indices, and triangle indices
"""
tetra = Delaunay(pos)
# Find radius of the circumsphere.
# By definition, radius of the sphere fitting inside the tetrahedral needs
# to be smaller than alpha value
# http://mathworld.wolfram.com/Circumsphere.html
tetrapos = np.take(pos,tetra.vertices,axis=0)
normsq = np.sum(tetrapos**2,axis=2)[:,:,None]
ones = np.ones((tetrapos.shape[0],tetrapos.shape[1],1))
a = np.linalg.det(np.concatenate((tetrapos,ones),axis=2))
Dx = np.linalg.det(np.concatenate((normsq,tetrapos[:,:,[1,2]],ones),axis=2))
Dy = -np.linalg.det(np.concatenate((normsq,tetrapos[:,:,[0,2]],ones),axis=2))
Dz = np.linalg.det(np.concatenate((normsq,tetrapos[:,:,[0,1]],ones),axis=2))
c = np.linalg.det(np.concatenate((normsq,tetrapos),axis=2))
r = np.sqrt(Dx**2+Dy**2+Dz**2-4*a*c)/(2*np.abs(a))
# Find tetrahedrals
tetras = tetra.vertices[r<alpha,:]
# triangles
TriComb = np.array([(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)])
Triangles = tetras[:,TriComb].reshape(-1,3)
Triangles = np.sort(Triangles,axis=1)
# Remove triangles that occurs twice, because they are within shapes
TrianglesDict = defaultdict(int)
for tri in Triangles:TrianglesDict[tuple(tri)] += 1
Triangles=np.array([tri for tri in TrianglesDict if TrianglesDict[tri] ==1])
#edges
EdgeComb=np.array([(0, 1), (0, 2), (1, 2)])
Edges=Triangles[:,EdgeComb].reshape(-1,2)
Edges=np.sort(Edges,axis=1)
Edges=np.unique(Edges,axis=0)
Vertices = np.unique(Edges)
return Vertices,Edges,Triangles
def ptcloud_sphere():
r = 3
phi = np.linspace(0, np.pi, 18)
theta = np.linspace(0, 2 * np.pi, 36)
PHI, THETA = np.meshgrid(phi, theta)
x = r * np.sin(PHI) * np.cos(THETA)
y = r * np.sin(PHI) * np.sin(THETA)
z = r * np.cos(PHI)
ax.scatter(x, y, z)
plt.show()
pts = np.stack((x.ravel(), y.ravel(), z.ravel()), axis=1)
return np.unique(pts, axis=0)
if __name__ == "__main__":
pts = ptcloud_sphere()
verts, edges, faces = alpha_shape_3D(pts, alpha=10)
faces_conn_list = np.insert(faces, 0, 3, axis=1)
num_faces = faces.shape[0]
mesh = pv.PolyData(pts[verts], faces_conn_list, n_faces=num_faces)
plotter.add_mesh(mesh, reset_camera=True)
plotter.show()
Output
Point Cloud:
Alpha Shape:
Update 09-SEP-2021
Per #akaszynski, the problem does indeed appear to be a combination of unique and negative points. He fixed this issue with the following:
pts = np.stack((x.ravel(), y.ravel(), z.ravel()), axis=1) + 10
return np.unique(pts, axis=0) - 10
However, if someone could perform a deeper investigation to determine why this is the problem, that would help.
Update 09-SEP-2021 #2
Per #AndrasDeak, both pyvista and VTK support the creation of 2d and 3d alpha shapes. The pyvista function delaunay_3d uses vtkDelaunay3D under the hood, which accepts an alpha parameter.
(See vtkDelaunay3D docs)
I’m trying to plot Riemann surface (imaginary part, real part is color-coded) for a complex valued function
using matplolib.
As it follows from nature of complex multivalued function, when some path passes a branch cut
its image jumps on f(z) plane.
As you can see in the code below, my mesh consists of circles
in polar coordinate system, which are the paths, passing the branch cut. It results in a jump that you can see in the figure below.
Question:
Is there any way to stitch the parts of the plot in a proper way.
To me a possible solution looks like either:
generate the values of mesh, bounded by the branch cut, and then concatenate the parts of them somehow (maybe even matplotlib itself has a function for it), or
sort out z-axis values being obtained on a simple mesh, which I use in the code
I've tried both but not succeeded.
If anybody has already faced such issues, or has any ideas, I would appreciate your comments!
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
k0 = 3 - 0.5j
# polar coordinates r and phi
Uphi = np.arange(0,2*np.pi,0.002) # phi from 0 to 2pi
Vr = np.arange(0,5,0.02) # r from 0 to 5
# polar grid
U, V = np.meshgrid(Uphi, Vr)
# coordinates on conplex plane: Z = X + iY; X,Y - cartezian coords
X = V*np.cos(U)
Y = V*np.sin(U)
Z1 = np.imag(-np.sqrt(-k0**2 + (X + 1j*Y)**2)) # first branch of the multivalued function
Z2 = np.imag(+np.sqrt(-k0**2 + (X + 1j*Y)**2)) # second branch
Z = np.zeros((Z1.shape[0],2*Z1.shape[1])) # resultng array to plot
Z[:,:Z1.shape[1]] = Z1
Z[:,Z1.shape[1]:] = Z2
# colormap -- color-coding the imaginary part of the 4D function -- another dimension
Z_cv = np.zeros((Z1.shape[0],2*Z1.shape[1]))
Z_cv[:,:Z1.shape[1]] = np.real(-np.sqrt(-k0**2 + (X + 1j*Y)**2))
Z_cv[:,Z1.shape[1]:] = np.real(+np.sqrt(-k0**2 + (X + 1j*Y)**2))
# expanding grid for two branches Z1 and Z2
X1 = np.zeros((Z1.shape[0],2*Z1.shape[1]))
Y1 = np.zeros((Z1.shape[0],2*Z1.shape[1]))
X1[:,:Z1.shape[1]] = X
X1[:,Z1.shape[1]:] = X
Y1[:,:Z1.shape[1]] = Y
Y1[:,Z1.shape[1]:] = Y
# plotting the surface
F = plt.figure(figsize=(12,8),dpi=100)
A = F.gca(projection='3d',autoscale_on=True)
# normalizing the values of color map
CV = (Z_cv + np.abs(np.min(Z_cv)))/(np.max(Z_cv) + np.abs(np.min(Z_cv)))
A.plot_surface(X1, Y1, Z, rcount=100, ccount=200, facecolors=cm.jet(CV,alpha=.3))
A.view_init(40,70)
m = cm.ScalarMappable(cmap=cm.jet)
m.set_array(Z_cv)
plt.colorbar(m)
plt.show()
The following figures represent two regular branches of the function on the Riemann surface using using Domain coloring approach. It more clearly shows those jumps.
import cplot
import numpy as np
cplot.show(lambda z: np.sqrt(z**2 - k0**2), -5, +5, -5, +5, 100, 100,alpha=0.7)
cplot.show(lambda z: -np.sqrt(z**2 - k0**2), -5, +5, -5, +5, 100, 100,alpha=0.7)
I would like to represent the elliptical orbit of a binary system of two stars. What I aim to, is something like this:
Where I have a grid of the sizes along the axes, an in-scale star at the focus, and the orbit of the secondary star. The decimal numbers along the orbit are the orbital phases. The arrow at the bottom is the Earth direction, and the thick part at the orbit is related to the observation for that specific case - I don't need it. What I want to change from this plot is:
Orbital phase: instead of numbers along the orbit, I would like "dashed rays" from the focus to the orbit, and the orbital phase above them:
I don't want the cross along (0, 0);
I would like to re-orient the orbit, in order that the 0.0 phase is around the top left part of the plot, and the Earth direction is an upward pointing straight arrow (the parameters of my system are different from the one plotted here).
I tried to look for python examples, but the only thing I came out with (from here), is a polar plot:
which is not really representative of what I want, but still is a beginning:
import numpy as np
import matplotlib.pyplot as plt
cos = np.cos
pi = np.pi
a = 10
e = 0.1
theta = np.linspace(0,2*pi, 360)
r = (a*(1-e**2))/(1+e*cos(theta))
fig = plt.figure()
ax = fig.add_subplot(111, polar=True)
ax.set_yticklabels([])
ax.plot(theta,r)
print(np.c_[r,theta])
plt.show()
Here's something that gets you very close. You do not need polar coordinates to plot a decent ellipse. There is a so-called artist you can readily utilize.
You probably have to customize the axis labels and maybe insert an arrow or two if you want:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse
# initializing the figure:
fig = plt.figure()
# the (carthesian) axis:
ax = fig.add_subplot(111,aspect='equal')
ax.grid(True)
# parameters of the ellipse:
a = 5.0
e = 4.0
b = np.sqrt(a**2.0 - e**2.0)
# the center of the ellipse:
x = 6.0
y = 6.0
# the angle by which the ellipse is rotated:
angle = -45.0
#angle = 0.0
# plotting the ellipse, using an artist:
ax.add_artist(Ellipse(xy=[x,y], width=2.0*a, height=2.0*b, \
angle=angle, facecolor='none'))
ax.set_xlim(0,2.0*x)
ax.set_ylim(0,2.0*y)
# marking the focus (actually, both)
# and accounting for the rotation of the ellipse by angle
xf = [x - e*np.cos(angle * np.pi/180.0),
x + e*np.cos(angle * np.pi/180.0)]
yf = [y - e*np.sin(angle * np.pi/180.0),
y + e*np.sin(angle * np.pi/180.0)]
ax.plot(xf,yf,'xr')
# plotting lines from the focus to the ellipse:
# these should be your "rays"
t = np.arange(np.pi,3.0*np.pi,np.pi/5.0)
p = b**2.0 / a
E = e / a
r = [p/(1-E*np.cos(ti)) for ti in t]
# converting the radius based on the focus
# into x,y coordinates on the ellipse:
xr = [ri*np.cos(ti) for ri,ti in zip(r,t)]
yr = [ri*np.sin(ti) for ri,ti in zip(r,t)]
# accounting for the rotation by anlge:
xrp = [xi*np.cos(angle * np.pi/180.0) - \
yi*np.sin(angle * np.pi/180.0) for xi,yi in zip(xr,yr)]
yrp = [xi*np.sin(angle * np.pi/180.0) + \
yi*np.cos(angle * np.pi/180.0) for xi,yi in zip(xr,yr)]
for q in range(0,len(t)):
ax.plot([xf[0], xf[0]+xrp[q]],[yf[0], yf[0]+yrp[q]],'--b')
# put labels outside the "rays"
offset = 0.75
rLabel = [ri+offset for ri in r]
xrl = [ri*np.cos(ti) for ri,ti in zip(rLabel,t)]
yrl = [ri*np.sin(ti) for ri,ti in zip(rLabel,t)]
xrpl = [xi*np.cos(angle * np.pi/180.0) - \
yi*np.sin(angle * np.pi/180.0) for xi,yi in zip(xrl,yrl)]
yrpl = [xi*np.sin(angle * np.pi/180.0) + \
yi*np.cos(angle * np.pi/180.0) for xi,yi in zip(xrl,yrl)]
# for fancy label rotation reduce the range of the angle t:
tlabel = [(ti -np.pi)*180.0/np.pi for ti in t]
for q in range(0,len(tlabel)):
if tlabel[q] >= 180.0:
tlabel[q] -= 180.0
# convert the angle t from radians into degrees:
tl = [(ti-np.pi)*180.0/np.pi for ti in t]
for q in range(0,len(t)):
rotate_label = angle + tlabel[q]
label_text = '%.1f' % tl[q]
ax.text(xf[0]+xrpl[q],yf[0]+yrpl[q],label_text,\
va='center', ha='center',rotation=rotate_label)
plt.show()
The example above will result in this figure:
Explanations:
You can use an artist to plot the ellipse, instead of using polar coordinates
The nomenclature is based on the definitions available on Wikipedia
The angle in the artist setup rotates the ellipse. This angle is later used to rotate coordinates for the rays and labels (this is just math)
The rays are derived from the polar form of the ellipse relative to a focus.
The angle t runs from pi to 3.0*pi because I assumed that this would correspond to your idea of where the rays should start. You get the same rays for 0 to 2.0*pi. I used np.arange instead of linspace because I wanted a defined increment (pi/5.0, or 36 degrees) in this example.
The labels at the end of the rays are placed as text, the variable offset controls the distance between the ellipse and the labels. Adjust this as needed.
For the alignment of the label text orientation with the rays, I reduced the angle, t, to the range 0 to 180 degrees. This makes for better readability compared to the full range of 0 to 360 degrees.
For the label text I just used the angle, t, for simplicity. Replace this with whatever information better suits your purpose.
The angle, t, was converted from radians to degrees before the loop that places the label. Inside the loop, each element of tl is converted to a string. This allows for more formatting control (e.g. %.3f if you needed 3 decimals)
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?