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Plot density function on sphere surface using plotly (python)

I'm interested in plotting a real-valued function f(x,y,z)=a, where (x,y,z) is a 3D point on the sphere and a is a real number. I calculate the Cartesian coordinates of the points of the sphere as follows, but I have no clue on how to visualize the value of f on each of those points.
import plotly.graph_objects as go
import numpy as np
fig = go.Figure(layout=go.Layout(title=go.layout.Title(text=title), hovermode=False))
# Create mesh grid for spherical coordinates
phi, theta = np.mgrid[0.0:np.pi:100j, 0.0:2.0 * np.pi:100j]
# Get Cartesian mesh grid
x = np.sin(phi) * np.cos(theta)
y = np.sin(phi) * np.sin(theta)
z = np.cos(phi)
# Plot sphere surface
self.fig.add_surface(x=x, y=y, z=z, opacity=0.35)
fig.show()
I would imagine/expect/like a visualization like this
Additionally, I also have the gradient of f calculated in closed-form (i.e., for each (x,y,z) I calculate the 3D-dimensional gradient of f). Is there a way of plotting this vector field, similarly to what is shown in the figure above?

Here's an answer that's far from perfect, but hopefully that's enough for you to build on.
For the sphere itself, I don't know of any "shortcut" to do something like that in plotly, so my approach is simply to manually create a sphere mesh. Generating the vertices is simple, for example like you did - the slightly more tricky part is figuring out the vertex indices for the triangles (which depends on the vertex generation scheme). There are various algorithms to do that smoothly (i.e. generating a sphere with no "tip"), I hacked something crude just for the demonstration. Then we can use the Mesh3d object to display the sphere along with the intensities and your choice of colormap:
N = 100 # Sphere resolution (both rings and segments, can be separated to different constants)
theta, z = np.meshgrid(np.linspace(-np.pi, np.pi, N), np.linspace(-1, 1, N))
r = np.sqrt(1 - z ** 2)
x = r * np.cos(theta)
y = r * np.sin(theta)
x = x.ravel()
y = y.ravel()
z = z.ravel()
# Triangle indices
indices = np.arange(N * (N - 1) - 1)
i1 = np.concatenate([indices, (indices // N + 1) * N + (indices + 1) % N])
i2 = np.concatenate([indices + N, indices // N * N + (indices + 1) % N])
i3 = np.concatenate([(indices // N + 1) * N + (indices + 1) % N, indices])
# Point intensity function
def f(x, y, z):
return (np.cos(x * 2) + np.sin(y ** 2) + np.sin(z) + 3) / 6
fig = go.Figure(data=[
go.Mesh3d(
x=x,
y=y,
z=z,
colorbar_title='f(x, y, z)',
colorscale=[[0, 'gold'],
[0.5, 'mediumturquoise'],
[1, 'magenta']],
intensity = f(x, y, z),
i = i1,
j = i2,
k = i3,
name='y',
showscale=True
)
])
fig.show()
This yields the following interactive plot:
To add the vector field you can use the Cone plot; this requires some tinkering because when I simply draw the cones at the same x, y, z position as the sphere, some of the cones are partially or fully occluded by the sphere. So I generate another sphere, with a slightly larger radius, and place the cones there. I also played with some lighting parameters to make it black like in your example. The full code looks like this:
N = 100 # Sphere resolution (both rings and segments, can be separated to different constants)
theta, z = np.meshgrid(np.linspace(-np.pi, np.pi, N), np.linspace(-1, 1, N))
r = np.sqrt(1 - z ** 2)
x = r * np.cos(theta)
y = r * np.sin(theta)
x = x.ravel()
y = y.ravel()
z = z.ravel()
# Triangle indices
indices = np.arange(N * (N - 1) - 1)
i1 = np.concatenate([indices, (indices // N + 1) * N + (indices + 1) % N])
i2 = np.concatenate([indices + N, indices // N * N + (indices + 1) % N])
i3 = np.concatenate([(indices // N + 1) * N + (indices + 1) % N, indices])
# Point intensity function
def f(x, y, z):
return (np.cos(x * 2) + np.sin(y ** 2) + np.sin(z) + 3) / 6
# Vector field function
def grad_f(x, y, z):
return np.stack([np.cos(3 * y + 5 * x),
np.sin(z * y),
np.cos(4 * x - 3 * y + z * 7)], axis=1)
# Second sphere for placing cones
N2 = 50 # Smaller resolution (again rings and segments combined)
R2 = 1.05 # Slightly larger radius
theta2, z2 = np.meshgrid(np.linspace(-np.pi, np.pi, N2), np.linspace(-R2, R2, N2))
r2 = np.sqrt(R2 ** 2 - z2 ** 2)
x2 = r2 * np.cos(theta2)
y2 = r2 * np.sin(theta2)
x2 = x2.ravel()
y2 = y2.ravel()
z2 = z2.ravel()
uvw = grad_f(x2, y2, z2)
fig = go.Figure(data=[
go.Mesh3d(
x=x,
y=y,
z=z,
colorbar_title='f(x, y, z)',
colorscale=[[0, 'gold'],
[0.5, 'mediumturquoise'],
[1, 'magenta']],
intensity = f(x, y, z),
i = i1,
j = i2,
k = i3,
name='y',
showscale=True
),
go.Cone(
x=x2, y=y2, z=z2, u=uvw[:, 0], v=uvw[:, 1], w=uvw[:, 2], sizemode='absolute', sizeref=2, anchor='tail',
lighting_ambient=0, lighting_diffuse=0, opacity=.2
)
])
fig.show()
And yields this plot:
Hope this helps. There are a lot of tweaks to the display, and certainly better ways to construct a sphere mesh (e.g. see this article), so there should be a lot of freedom there (albeit at the cost of some work).
Good luck!

How can I make a 3D plot in matplotlib of an ellipsoid defined by a quadratic equation?

I have the general formula of an ellipsoid:
A*x**2 + C*y**2 + D*x + E*y + B*x*y + F + G*z**2 = 0
where A,B,C,D,E,F,G are constant factors.
How can I plot this equation as a 3D plot in matplotlib? (A wireframe would be best.)
I saw this example but it is in parametric form and I am not sure how to put the z-coordinates in this code. Is there a way to keep the general form to plot this without the parametric form?
I started to put this in some kind of code like this:
from mpl_toolkits import mplot3d
%matplotlib notebook
import numpy as np
import matplotlib.pyplot as plt
def f(x, y):
return ((A*x**2 + C*y**2 + D*x + E*y + B*x*y + F))
def f(z):
return G*z**2
x = np.linspace(-2200, 1850, 30)
y = np.linspace(-100, 60, 30)
z = np.linspace(-100, 60, 30)
X, Y, Z = np.meshgrid(x, y, z)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_wireframe(X, Y, Z, rstride=10, cstride=10)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z');
I got this error:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-1-95b1296ae6a4> in <module>()
18 fig = plt.figure()
19 ax = fig.add_subplot(111, projection='3d')
---> 20 ax.plot_wireframe(X, Y, Z, rstride=10, cstride=10)
21 ax.set_xlabel('x')
22 ax.set_ylabel('y')
C:\Program Files (x86)\Microsoft Visual Studio\Shared\Anaconda3_64\lib\site-packages\mpl_toolkits\mplot3d\axes3d.py in plot_wireframe(self, X, Y, Z, *args, **kwargs)
1847 had_data = self.has_data()
1848 if Z.ndim != 2:
-> 1849 raise ValueError("Argument Z must be 2-dimensional.")
1850 # FIXME: Support masked arrays
1851 X, Y, Z = np.broadcast_arrays(X, Y, Z)
ValueError: Argument Z must be 2-dimensional.

Side note, but what you have is not the most general equation for a 3d ellipsoid. Your equation can be rewritten as
A*x**2 + C*y**2 + D*x + E*y + B*x*y = - G*z**2 - F,
which means that in effect for each value of z you get a different level of a 2d ellipse, and the slices are symmetric with respect to the z = 0 plane. This shows how your ellipsoid is not general, and it helps check the results to make sure that what we get makes sense.
Assuming we take a general point r0 = [x0, y0, z0], you have
r0 # M # r0 + b0 # r0 + c0 == 0
where
M = [ A B/2 0
B/2 C 0
0 0 G],
b0 = [D, E, 0],
c0 = F
where # stands for matrix-vector or vector-vector product.
You could take your function and plot its isosurface, but that would be suboptimal: you would need a gridded approximation for your function which is very expensive to do to sufficient resolution, and you'd have to choose the domain for this sampling wisely.
Instead you can perform a principal axis transformation on your data to generalize the parametric plot of a canonical ellipsoid that you yourself linked.
The first step is to diagonalize M as M = V # D # V.T, where D is diagonal. Since it's a real symmetric matrix this is always possible and V is orthogonal. Then we have
r0 # V # D # V.T # r0 + b0 # r0 + c0 == 0
which we can regroup as
(V.T # r0) # D # (V.T # r0) + b0 # V # (V.T # r0) + c0 == 0
which motivates the definition of the auxiliary coordinates r1 = V.T # r0 and vector b1 = b0 # V, for which we get
r1 # D # r1 + b1 # r1 + c0 == 0.
Since D is a symmetric matrix with the eigenvalues d1, d2, d3 in its diagonal, the above is the equation
d1 * x1**2 + d2 * x2**2 + d3 * x3**3 + b11 * x1 + b12 * x2 + b13 * x3 + c0 == 0
where r1 = [x1, x2, x3] and b1 = [b11, b12, b13].
What's left is to switch from r1 to r2 such that we remove the linear terms:
d1 * (x1 + b11/(2*d1))**2 + d2 * (x2 + b12/(2*d2))**2 + d3 * (x3 + b13/(2*d3))**2 - b11**2/(4*d1) - b12**2/(4*d2) - b13**2/(4*d3) + c0 == 0
So we define
r2 = [x2, y2, z2]
x2 = x1 + b11/(2*d1)
y2 = y1 + b12/(2*d2)
z2 = z1 + b13/(2*d3)
c2 = b11**2/(4*d1) b12**2/(4*d2) b13**2/(4*d3) - c0.
For these we finally have
d1 * x2**2 + d2 * y2**2 + d3 * z2**2 == c2,
d1/c2 * x2**2 + d2/c2 * y2**2 + d3/c2 * z2**2 == 1
which is the canonical form of a second-order surface. In order for this to meaningfully correspond to an ellipsoid we must ensure that d1, d2, d3 and c2 are all strictly positive. If this is guaranteed then the semi-major axes of the canonical form are sqrt(c2/d1), sqrt(c2/d2) and sqrt(c2/d3).
So here's what we do:
ensure that the parameters correspond to an ellipsoid
generate a theta and phi mesh for polar and azimuthal angles
compute the transformed coordinates [x2, y2, z2]
shift them back (by r2 - r1) to get [x1, y1, z1]
transform the coordinates back by V to get r0, the actual [x, y, z] coordinates we're interested in.
Here's how I'd implement this:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
def get_transforms(A, B, C, D, E, F, G):
""" Get transformation matrix and shift for a 3d ellipsoid
Assume A*x**2 + C*y**2 + D*x + E*y + B*x*y + F + G*z**2 = 0,
use principal axis transformation and verify that the inputs
correspond to an ellipsoid.
Returns: (d, V, s) tuple of arrays
d: shape (3,) of semi-major axes in the canonical form
(X/d1)**2 + (Y/d2)**2 + (Z/d3)**2 = 1
V: shape (3,3) of the eigensystem
s: shape (3,) shift from the linear terms
"""
# construct original matrix
M = np.array([[A, B/2, 0],
[B/2, C, 0],
[0, 0, G]])
# construct original linear coefficient vector
b0 = np.array([D, E, 0])
# constant term
c0 = F
# compute eigensystem
D, V = np.linalg.eig(M)
if (D <= 0).any():
raise ValueError("Parameter matrix is not positive definite!")
# transform the shift
b1 = b0 # V
# compute the final shift vector
s = b1 / (2 * D)
# compute the final constant term, also has to be positive
c2 = (b1**2 / (4 * D)).sum() - c0
if c2 <= 0:
print(b1, D, c0, c2)
raise ValueError("Constant in the canonical form is not positive!")
# compute the semi-major axes
d = np.sqrt(c2 / D)
return d, V, s
def get_ellipsoid_coordinates(A, B, C, D, E, F, G, n_theta=20, n_phi=40):
"""Compute coordinates of an ellipsoid on an ellipsoidal grid
Returns: x, y, z arrays of shape (n_theta, n_phi)
"""
# get canonical grid
theta,phi = np.mgrid[0:np.pi:n_theta*1j, 0:2*np.pi:n_phi*1j]
r2 = np.array([np.sin(theta) * np.cos(phi),
np.sin(theta) * np.sin(phi),
np.cos(theta)]) # shape (3, n_theta, n_phi)
# get transformation data
d, V, s = get_transforms(A, B, C, D, E, F, G) # could be *args I guess
# shift and transform back the coordinates
r1 = d[:,None,None]*r2 - s[:,None,None] # broadcast along first of three axes
r0 = (V # r1.reshape(3, -1)).reshape(r1.shape) # shape (3, n_theta, n_phi)
return r0 # unpackable to x, y, z of shape (n_theta, n_phi)
Here's an example with an ellipsoid and proof that it works:
A,B,C,D,E,F,G = args = 2, -1, 2, 3, -4, -3, 4
x,y,z = get_ellipsoid_coordinates(*args)
print(np.allclose(A*x**2 + C*y**2 + D*x + E*y + B*x*y + F + G*z**2, 0)) # True
The actual plotting from here is trivial. Using the 3d scaling hack from this answer to preserve equal axes:
# create 3d axes
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# plot the data
ax.plot_wireframe(x, y, z)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
# scaling hack
bbox_min = np.min([x, y, z])
bbox_max = np.max([x, y, z])
ax.auto_scale_xyz([bbox_min, bbox_max], [bbox_min, bbox_max], [bbox_min, bbox_max])
plt.show()
Here's how the result looks:
Rotating it around it's nicely visible that the surface is indeed reflection symmetric with respect to the z = 0 plane, which was evident from the equation.
You can change the n_theta and n_phi keyword arguments to the function to generate a grid with a different mesh. The fun thing is that you can take any scattered points lying on the unit sphere and plug it into the definition of r2 in the function get_ellipsoid_coordinates (as long as this array has a first dimension of size 3), and the output coordinates will have the same shape, but they will be transformed onto the actual ellipsoid.
You can also use other libraries to visualize the surface, for instance mayavi where you can either plot the surface we just computed, or compare it with an isosurface which is built-in there.

Points on sphere

I am new in Python and I have a sphere of radius (R) and centred at (x0,y0,z0). Now, I need to find those points which are either on the surface of the sphere or inside the sphere e.g. points (x1,y1,z1) which satisfy ((x1-x0)**2+(y1-y0)**2+(z1-x0)*82)**1/2 <= R. I would like to print only those point's coordinates in a form of numpy array. Output would be something like this-[[x11,y11,z11],[x12,y12,z12],...]. I have the following code so far-
import numpy as np
import math
def create_points_around_atom(number,atom_coordinates):
n= number
x0 = atom_coordinates[0]
y0 = atom_coordinates[1]
z0 = atom_coordinates[2]
R = 1.2
for i in range(n):
phi = np.random.uniform(0,2*np.pi,size=(n,))
costheta = np.random.uniform(-1,1,size=(n,))
u = np.random.uniform(0,1,size=(n,))
theta = np.arccos(costheta)
r = R * np.cbrt(u)
x1 = r*np.sin(theta)*np.cos(phi)
y1 = r*np.sin(theta)*np.sin(phi)
z1 = r*np.cos(theta)
dist = np.sqrt((x1-x0)**2+(y1-y0)**2+(z1-z0)**2)
distance = list(dist)
point_on_inside_sphere = []
for j in distance:
if j <= R:
point_on_inside_sphere.append(j)
print('j:',j,'\tR:',R)
print('The list is:', point_on_inside_sphere)
print(len(point_on_inside_sphere))
kk =0
for kk in range(len(point_on_inside_sphere)):
for jj in point_on_inside_sphere:
xx = np.sqrt(jj**2-y1**2-z1**2)
yy = np.sqrt(jj**2-x1**2-z1**2)
zz = np.sqrt(jj**2-y1**2-x1**2)
print("x:", xx, "y:", yy,"z:", zz)
kk +=1
And I am running it-
create_points_around_atom(n=2,structure[1].coords)
where, structure[1].coords is a numpy array of three coordinates.

To sum up what has been discussed in the comments, and some other points:
There is no need to filter the points because u <= 1, which means np.cbrt(u) <= 1 and hence r = R * np.cbrt(u) <= R, i.e. all points will already be inside or on the surface of the sphere.
Calling np.random.uniform with size=(n,) creates an array of n elements, so there's no need to do this n times in a loop.
You are filtering distances from the atom_coordinate, but the points you are generating are centered on [0, 0, 0], because you are not adding this offset.
Passing R as an argument seems more sensible than hard-coding it.
There's no need to "pre-load" arguments in Python like one would sometimes do in C.
Since sin(theta) is non-negative over the sphere, you can directly calculate it from the costheta array using the identity cos²(x) + sin²(x) = 1.
Sample implementation:
# pass radius as an argument
def create_points_around_atom(number, center, radius):
# generate the random quantities
phi = np.random.uniform( 0, 2*np.pi, size=(number,))
theta_cos = np.random.uniform(-1, 1, size=(number,))
u = np.random.uniform( 0, 1, size=(number,))
# calculate sin(theta) from cos(theta)
theta_sin = np.sqrt(1 - theta_cos**2)
r = radius * np.cbrt(u)
# use list comprehension to generate the coordinate array without a loop
# don't forget to offset by the atom's position (center)
return np.array([
np.array([
center[0] + r[i] * theta_sin[i] * np.cos(phi[i]),
center[1] + r[i] * theta_sin[i] * np.sin(phi[i]),
center[2] + r[i] * theta_cos[i]
]) for i in range(number)
])

How to get the x,y coordinates of a offset spline from a x,y list of points and offset distance

I need to make an offset parallel enclosure of an airfoil profile curve, but I cant figure out how to make all the points be equidistant to the points on the primary profile curve at desired distance.
this is my example airfoil profile
this is my best and not good approach
EDIT #Patrick Solution for distance 0.2

You'll have to special-case slopes of infinity/zero, but the basic approach is to use interpolation to calculate the slope at a point, and then find the perpendicular slope, and then calculate the point at that distance.
I have modified the example from here to add a second graph. It works with the data file you provided, but you might need to change the sign calculation for a different envelope.
EDIT As per your comments about wanting the envelope to be continuous, I have added a cheesy semicircle at the end that gets really close to doing this for you. Essentially, when creating the envelope, the rounder and more convex you can make it, the better it will work. Also, you need to overlap the beginning and the end, or you'll have a gap.
Also, it could almost certainly be made more efficient -- I am not a numpy expert by any means, so this is just pure Python.
def offset(coordinates, distance):
coordinates = iter(coordinates)
x1, y1 = coordinates.next()
z = distance
points = []
for x2, y2 in coordinates:
# tangential slope approximation
try:
slope = (y2 - y1) / (x2 - x1)
# perpendicular slope
pslope = -1/slope # (might be 1/slope depending on direction of travel)
except ZeroDivisionError:
continue
mid_x = (x1 + x2) / 2
mid_y = (y1 + y2) / 2
sign = ((pslope > 0) == (x1 > x2)) * 2 - 1
# if z is the distance to your parallel curve,
# then your delta-x and delta-y calculations are:
# z**2 = x**2 + y**2
# y = pslope * x
# z**2 = x**2 + (pslope * x)**2
# z**2 = x**2 + pslope**2 * x**2
# z**2 = (1 + pslope**2) * x**2
# z**2 / (1 + pslope**2) = x**2
# z / (1 + pslope**2)**0.5 = x
delta_x = sign * z / ((1 + pslope**2)**0.5)
delta_y = pslope * delta_x
points.append((mid_x + delta_x, mid_y + delta_y))
x1, y1 = x2, y2
return points
def add_semicircle(x_origin, y_origin, radius, num_x = 50):
points = []
for index in range(num_x):
x = radius * index / num_x
y = (radius ** 2 - x ** 2) ** 0.5
points.append((x, -y))
points += [(x, -y) for x, y in reversed(points)]
return [(x + x_origin, y + y_origin) for x, y in points]
def round_data(data):
# Add infinitesimal rounding of the envelope
assert data[-1] == data[0]
x0, y0 = data[0]
x1, y1 = data[1]
xe, ye = data[-2]
x = x0 - (x0 - x1) * .01
y = y0 - (y0 - y1) * .01
yn = (x - xe) / (x0 - xe) * (y0 - ye) + ye
data[0] = x, y
data[-1] = x, yn
data.extend(add_semicircle(x, (y + yn) / 2, abs((y - yn) / 2)))
del data[-18:]
from pylab import *
with open('ah79100c.dat', 'rb') as f:
f.next()
data = [[float(x) for x in line.split()] for line in f if line.strip()]
t = [x[0] for x in data]
s = [x[1] for x in data]
round_data(data)
parallel = offset(data, 0.1)
t2 = [x[0] for x in parallel]
s2 = [x[1] for x in parallel]
plot(t, s, 'g', t2, s2, 'b', lw=1)
title('Wing with envelope')
grid(True)
axes().set_aspect('equal', 'datalim')
savefig("test.png")
show()

If you are willing (and able) to install a third-party tool, I'd highly recommend the Shapely module. Here's a small sample that offsets both inward and outward:
from StringIO import StringIO
import matplotlib.pyplot as plt
import numpy as np
import requests
import shapely.geometry as shp
# Read the points
AFURL = 'http://m-selig.ae.illinois.edu/ads/coord_seligFmt/ah79100c.dat'
afpts = np.loadtxt(StringIO(requests.get(AFURL).content), skiprows=1)
# Create a Polygon from the nx2 array in `afpts`
afpoly = shp.Polygon(afpts)
# Create offset airfoils, both inward and outward
poffafpoly = afpoly.buffer(0.03) # Outward offset
noffafpoly = afpoly.buffer(-0.03) # Inward offset
# Turn polygon points into numpy arrays for plotting
afpolypts = np.array(afpoly.exterior)
poffafpolypts = np.array(poffafpoly.exterior)
noffafpolypts = np.array(noffafpoly.exterior)
# Plot points
plt.plot(*afpolypts.T, color='black')
plt.plot(*poffafpolypts.T, color='red')
plt.plot(*noffafpolypts.T, color='green')
plt.axis('equal')
plt.show()
And here's the output; notice how the 'bowties' (self-intersections) on the inward offset are automatically removed:

orthogonal projection with numpy

I have a list of 3D-points for which I calculate a plane by numpy.linalg.lstsq - method. But Now I want to do a orthogonal projection for each point into this plane, but I can't find my mistake:
from numpy.linalg import lstsq
def VecProduct(vek1, vek2):
return (vek1[0]*vek2[0] + vek1[1]*vek2[1] + vek1[2]*vek2[2])
def CalcPlane(x, y, z):
# x, y and z are given in lists
n = len(x)
sum_x = sum_y = sum_z = sum_xx = sum_yy = sum_xy = sum_xz = sum_yz = 0
for i in range(n):
sum_x += x[i]
sum_y += y[i]
sum_z += z[i]
sum_xx += x[i]*x[i]
sum_yy += y[i]*y[i]
sum_xy += x[i]*y[i]
sum_xz += x[i]*z[i]
sum_yz += y[i]*z[i]
M = ([sum_xx, sum_xy, sum_x], [sum_xy, sum_yy, sum_y], [sum_x, sum_y, n])
b = (sum_xz, sum_yz, sum_z)
a,b,c = lstsq(M, b)[0]
'''
z = a*x + b*y + c
a*x = z - b*y - c
x = -(b/a)*y + (1/a)*z - c/a
'''
r0 = [-c/a,
0,
0]
u = [-b/a,
1,
0]
v = [1/a,
0,
1]
xn = []
yn = []
zn = []
# orthogonalize u and v with Gram-Schmidt to get u and w
uu = VecProduct(u, u)
vu = VecProduct(v, u)
fak0 = vu/uu
erg0 = [val*fak0 for val in u]
w = [v[0]-erg0[0],
v[1]-erg0[1],
v[2]-erg0[2]]
ww = VecProduct(w, w)
# P_new = ((x*u)/(u*u))*u + ((x*w)/(w*w))*w
for i in range(len(x)):
xu = VecProduct([x[i], y[i], z[i]], u)
xw = VecProduct([x[i], y[i], z[i]], w)
fak1 = xu/uu
fak2 = xw/ww
erg1 = [val*fak1 for val in u]
erg2 = [val*fak2 for val in w]
erg = [erg1[0]+erg2[0], erg1[1]+erg2[1], erg1[2]+erg2[2]]
erg[0] += r0[0]
xn.append(erg[0])
yn.append(erg[1])
zn.append(erg[2])
return (xn,yn,zn)
This returns me a list of points which are all in a plane, but when I display them, they are not at the positions they should be.
I believe there is already a certain built-in method to solve this problem, but I couldn't find any =(

You are doing a very poor use of np.lstsq, since you are feeding it a precomputed 3x3 matrix, instead of letting it do the job. I would do it like this:
import numpy as np
def calc_plane(x, y, z):
a = np.column_stack((x, y, np.ones_like(x)))
return np.linalg.lstsq(a, z)[0]
>>> x = np.random.rand(1000)
>>> y = np.random.rand(1000)
>>> z = 4*x + 5*y + 7 + np.random.rand(1000)*.1
>>> calc_plane(x, y, z)
array([ 3.99795126, 5.00233364, 7.05007326])
It is actually more convenient to use a formula for your plane that doesn't depend on the coefficient of z not being zero, i.e. use a*x + b*y + c*z = 1. You can similarly compute a, b and c doing:
def calc_plane_bis(x, y, z):
a = np.column_stack((x, y, z))
return np.linalg.lstsq(a, np.ones_like(x))[0]
>>> calc_plane_bis(x, y, z)
array([-0.56732299, -0.70949543, 0.14185393])
To project points onto a plane, using my alternative equation, the vector (a, b, c) is perpendicular to the plane. It is easy to check that the point (a, b, c) / (a**2+b**2+c**2) is on the plane, so projection can be done by referencing all points to that point on the plane, projecting the points onto the normal vector, subtract that projection from the points, then referencing them back to the origin. You could do that as follows:
def project_points(x, y, z, a, b, c):
"""
Projects the points with coordinates x, y, z onto the plane
defined by a*x + b*y + c*z = 1
"""
vector_norm = a*a + b*b + c*c
normal_vector = np.array([a, b, c]) / np.sqrt(vector_norm)
point_in_plane = np.array([a, b, c]) / vector_norm
points = np.column_stack((x, y, z))
points_from_point_in_plane = points - point_in_plane
proj_onto_normal_vector = np.dot(points_from_point_in_plane,
normal_vector)
proj_onto_plane = (points_from_point_in_plane -
proj_onto_normal_vector[:, None]*normal_vector)
return point_in_plane + proj_onto_plane
So now you can do something like:
>>> project_points(x, y, z, *calc_plane_bis(x, y, z))
array([[ 0.13138012, 0.76009389, 11.37555123],
[ 0.71096929, 0.68711773, 13.32843506],
[ 0.14889398, 0.74404116, 11.36534936],
...,
[ 0.85975642, 0.4827624 , 12.90197969],
[ 0.48364383, 0.2963717 , 10.46636903],
[ 0.81596472, 0.45273681, 12.57679188]])

You can simply do everything in matrices is one option.
If you add your points as row vectors to a matrix X, and y is a vector, then the parameters vector beta for the least squares solution are:
import numpy as np
beta = np.linalg.inv(X.T.dot(X)).dot(X.T.dot(y))
but there's an easier way, if we want to do projections: QR decomposition gives us an orthonormal projection matrix, as Q.T, and Q is itself the matrix of orthonormal basis vectors. So, we can first form QR, then get beta, then use Q.T to project the points.
QR:
Q, R = np.linalg.qr(X)
beta:
# use R to solve for beta
# R is upper triangular, so can use triangular solver:
beta = scipy.solve_triangular(R, Q.T.dot(y))
So now we have beta, and we can project the points using Q.T very simply:
X_proj = Q.T.dot(X)
Thats it!
If you want more information and graphical piccies and stuff, I made a whole bunch of notes, whilst doing something similar, at: https://github.com/hughperkins/selfstudy-IBP/blob/9dedfbb93f4320ac1bfef60db089ae0dba5e79f6/test_bases.ipynb
(Edit: note that if you want to add a bias term, so the best-fit doesnt have to pass through the origin, you can simply add an additional column, with all-1s, to X, which acts as the bias term/feature)

This web page has a pretty great code base. It implements the theory expounded by Maple in numpy quite well, as follows:
# import numpy to perform operations on vector
import numpy as np
# vector u
u = np.array([2, 5, 8])
# vector n: n is orthogonal vector to Plane P
n = np.array([1, 1, 7])
# Task: Project vector u on Plane P
# finding norm of the vector n
n_norm = np.sqrt(sum(n**2))
# Apply the formula as mentioned above
# for projecting a vector onto the orthogonal vector n
# find dot product using np.dot()
proj_of_u_on_n = (np.dot(u, n)/n_norm**2)*n
# subtract proj_of_u_on_n from u:
# this is the projection of u on Plane P
print("Projection of Vector u on Plane P is: ", u - proj_of_u_on_n)