Develop Reference

How to return pairs of (x,y) coordinates defining an implicit curve?

I have implemented this Heart curve (Wolfram Mathworld) via an implicit function script I found here Plot implicit equations in Python 3
import matplotlib.pyplot as plt
import numpy as np
delta = 0.025
xrange = np.arange(-2, 2, delta)
yrange = np.arange(-2, 2, delta)
X, Y = np.meshgrid(xrange,yrange)
# F is one side of the equation, G is the other
F = X**2
G = 1- (Y - (np.abs(X))**(2/3))**2
plt.contour((F - G), [0])
plt.show()
How would I extract the resulting array of x,y pairs defining this curve?

One could say this is "cheating" a bit, but a rough, concise recipe can be getting points in the mesh that roughly satisfy the condition and getting their coordinates as follows:
# subset points that are very close to satifying the exact condition
# and get their indices
x_id, y_id = np.nonzero(abs(F-G) < 8e-3)
# get coordinates corresponding to the indices
plt.scatter(xrange[x_id], yrange[y_id])
xrange[x_id], yrange[y_id] are the desired arrays of coordinates.

I'm not really a matplotlib user and there may be much, much better ways of doing what you ask - so please wait and see if someone comes up with a proper answer...
However, for the moment, you appear to be plotting the array F-G, so I saved that as a TIF (because it can store floats) using PIL like this:
from PIL import Image
...
your code
...
contour = (F-G).astype(np.float32)
Image.fromarray(contour).save('contour.tif')
that gives this:
Inspection of contour shows that the range is -1 to 16:
print(contour.min(), contour.max())
-1.0 15.869446307662544
So, I deduce that your points are probably the ones near zero:
points = (contour>-0.1) & (contour<0.1)
So, I can get the list of X,Y with:
Y, X = np.where(points)

Converting gaussian to histogram

I'm running a model of particles, and I want to have initial conditions for the particle locations mimicking a gaussian distribution.
If I have N number of particles on 1D grid from -10 to 10, I want them to be distributed on the grid according to a gaussian with a known mean and standard deviation. It's basically creating a histogram where each bin width is 1 (the x-axis of locations resolution is 1), and the frequency of each bin should be how many particles are in it, which should all add up to N.
My strategy was to plot a gaussian function on the x-axis grid, and then just approximate the value of each point for the number of particles:
def gaussian(x, mu, sig):
return 1./(np.sqrt(2.*np.pi)*sig)*np.exp(-np.power((x - mu)/sig, 2.)/2)
mean = 0
sigma = 1
x_values = np.arange(-10, 10, 1)
y = gaussian(x_values, mean, sigma)
However, I have normalization issues (the sum doesn't add up to N), and the number of particles in each point should be an integer (I thought about converting the y array to integers but again, because of the normalization issue I get a flat line).
Usually, the problem is fitting a gaussian to histogram, but in my case, I need to do the reverse - and I couldn't find a solution for it yet. I will appreciate any help!
Thank you!!!

You can use numpy.random.normal to sample this distribution. You can get N points inside range (-10, 10) that follows Gaussian distribution with the following code.
import numpy as np
import matplotlib.pyplot as plt
N = 10000
mean = 5
sigma = 3
bin_edges = np.arange(-10, 11, 1)
x_values = (bin_edges[1:] + bin_edges[:-1]) / 2
points = np.random.normal(mean, sigma, N * 10)
mask = np.logical_and(points < 10, points > -10)
points = points[mask] # drop points outside range
points = points[:N] # only use the first N points
y, _ = np.histogram(points, bins=bin_edges)
plt.scatter(x_values, y)
plt.show()
The idea is to generate a lot of random numbers (10 N in the code), and ignores the points outside your desired range.

Mapping the color scale of 3D isosurface on a scalar field

Let's say we have some 3D complex valued function f(x,y,z). Using Plotly, I'm trying to plot isosurfaces of the magnitude |f(x,y,z)| of such function. So far, everything is OK and my code seems to do well, please find below a working example on atomic orbitals functions :
import chart_studio.plotly as py
import plotly.graph_objs as go
import scipy.special as scispe
import numpy as np
import math
a=5.29e-11 # Bohr radius (m)
def orbital(n,l,m,r,theta,phi): # Complex function I want to plot
L=scispe.genlaguerre(n-l-1,2*l+1) # Laguerre polynomial
radial= (2/(n*a))**(3/2) * np.sqrt(math.factorial(n-l-1)/(2*n*math.factorial(n+l))) * np.exp(-2*r/n) * (2*r/n)**l * L(2*r/n)
wavefunction = radial * scispe.sph_harm(m,l, phi, theta)
return wavefunction
#Quantum numbers
n=2
l=1
m=0
goodspan = (3 * n**2 - l * (l+1))/2 #Plot span adpated to the mean electron position
x, y, z = np.mgrid[-goodspan:goodspan:40j, -goodspan:goodspan:40j, -goodspan:goodspan:40j] #in units of a
r = np.sqrt(x**2 + y**2 + z**2) #Function has to be evaluated in spherical coordinates
theta = np.arccos(z/r)
phi = np.arctan(y/x)
AO=orbital(n,l,m,r,theta,phi)
magnitude = abs(AO) # Compute the magnitude of the function
phase = np.angle(AO) # Compute the phase of the function
isoprob = np.amax(magnitude)/2 # Set value the isosurface
fig = go.Figure(data=go.Isosurface(
x=x.flatten(),
y=y.flatten(),
z=z.flatten(),
value=magnitude.flatten(),
opacity=0.5,
isomin=isoprob,
isomax=isoprob,
surface_count=1,
caps=dict(x_show=True, y_show=True)
))
fig.show()
which gives me this :
At this point, the color scale of the graph is attributed depending on the value of the magnitude |f(x,y,z)|, so that a single isosurface is always uniform in color.
Now, instead to have a color scale mapped on the magnitude |f(x,y,z)|, I would like it to be mapped on the value of the phase Ф(x,y,z) = arg(f(x,y,z)), so that the color of each point of a ploted isosurface tells us about the value of the field Ф(x,y,z) (which would be distributed on [-π,π] ideally) instead of |f(x,y,z)| in thsi point.
Basically, I would like to do this with Plotly instead of Mayavi if it's possible.
It seems to me that all of that has something to do with a special way to set the cmin and cmax parameters of the function Isosurface, but I can't figure out how to do this.

As #gnodab mentioned in his comment, plotly isosurfaces do not really support colouring the surfaces by a fifth dimension (at least there is no obvious way to do it). I am also not sure if it might be possible to extract the data describing the isosurface somehow to be re-plotted as a regular surface.
In this post, however, they describe how to generate an isosurface with skimage.measure.marching_cubes_lewiner which is then plotted and coloured by a custom colorscale with plotly as 'mesh3d' trace. This might be what you want. If I find the time, I'll give that a try and edit my answer later.

Given #Jan Joswig's answer and the link they provided, the quick/compact way of doing it will be:
import plotly.graph_objects as go
from skimage import measure
import numpy as np
xyz_shape = vol.shape
verts, faces = measure.marching_cubes(vol, .5)[:2] # iso-surface at .5 level
x, y, z = verts.T
I, J, K = faces.T
fig = go.Figure(
data=[go.Mesh3d(
x=x,
y=y,
z=z,
color='lightpink',
opacity=0.50,
i=I,
j=J,
k=K, )])
fig.show()

How to randomly generate continuous functions

My objective is to randomly generate good looking continuous functions, good looking meaning that functions which can be recovered from their plots.
Essentially I want to generate a random time series data for 1 second with 1024 samples per second. If I randomly choose 1024 values, then the plot looks very noisy and nothing meaningful can be extracted out of it. In the end I have attached plots of two sinusoids, one with a frequency of 3Hz and another with a frequency of 100Hz. I consider 3Hz cosine as a good function because I can extract back the timeseries by looking at the plot. But the 100 Hz sinusoid is bad for me as I cant recover the timeseries from the plot. So in the above mentioned meaning of goodness of a timeseries, I want to randomly generate good looking continuos functions/timeseries.
The method I am thinking of using is as follows (python language):
(1) Choose 32 points in x-axis between 0 to 1 using x=linspace(0,1,32).
(2) For each of these 32 points choose a random value using y=np.random.rand(32).
(3) Then I need an interpolation or curve fitting method which takes as input (x,y) and outputs a continuos function which would look something like func=curve_fit(x,y)
(4) I can obtain the time seires by sampling from the func function
Following are the questions that I have:
1) What is the best curve-fitting or interpolation method that I can
use. They should also be available in python.
2) Is there a better method to generate good looking functions,
without using curve fitting or interpolation.
Edit
Here is the code I am using currently for generating random time-series of length 1024. In my case I need to scale the function between 0 and 1 in the y-axis. Hence for me l=0 and h=0. If that scaling is not needed you just need to uncomment a line in each function to randomize the scaling.
import numpy as np
from scipy import interpolate
from sklearn.preprocessing import MinMaxScaler
import matplotlib.pyplot as plt
## Curve fitting technique
def random_poly_fit():
l=0
h=1
degree = np.random.randint(2,11)
c_points = np.random.randint(2,32)
cx = np.linspace(0,1,c_points)
cy = np.random.rand(c_points)
z = np.polyfit(cx, cy, degree)
f = np.poly1d(z)
y = f(x)
# l,h=np.sort(np.random.rand(2))
y = MinMaxScaler(feature_range=(l,h)).fit_transform(y.reshape(-1, 1)).reshape(-1)
return y
## Cubic Spline Interpolation technique
def random_cubic_spline():
l=0
h=1
c_points = np.random.randint(4,32)
cx = np.linspace(0,1,c_points)
cy = np.random.rand(c_points)
z = interpolate.CubicSpline(cx, cy)
y = z(x)
# l,h=np.sort(np.random.rand(2))
y = MinMaxScaler(feature_range=(l,h)).fit_transform(y.reshape(-1, 1)).reshape(-1)
return y
func_families = [random_poly_fit, random_cubic_spline]
func = np.random.choice(func_families)
x = np.linspace(0,1,1024)
y = func()
plt.plot(x,y)
plt.show()

Add sin and cosine signals
from numpy.random import randint
x= np.linspace(0,1,1000)
for i in range(10):
y = randint(0,100)*np.sin(randint(0,100)*x)+randint(0,100)*np.cos(randint(0,100)*x)
y = MinMaxScaler(feature_range=(-1,1)).fit_transform(y.reshape(-1, 1)).reshape(-1)
plt.plot(x,y)
plt.show()
Output:
convolve sin and cosine signals
for i in range(10):
y = np.convolve(randint(0,100)*np.sin(randint(0,100)*x), randint(0,100)*np.cos(randint(0,100)*x), 'same')
y = MinMaxScaler(feature_range=(-1,1)).fit_transform(y.reshape(-1, 1)).reshape(-1)
plt.plot(x,y)
plt.show()
Output:

How to generate a random sample of points from a 3-D ellipsoid using Python?

I am trying to sample around 1000 points from a 3-D ellipsoid, uniformly. Is there some way to code it such that we can get points starting from the equation of the ellipsoid?
I want points on the surface of the ellipsoid.

Theory
Using this excellent answer to the MSE question How to generate points uniformly distributed on the surface of an ellipsoid? we can
generate a point uniformly on the sphere, apply the mapping f :
(x,y,z) -> (x'=ax,y'=by,z'=cz) and then correct the distortion
created by the map by discarding the point randomly with some
probability p(x,y,z).
Assuming that the 3 axes of the ellipsoid are named such that
0 < a < b < c
We discard a generated point with
p(x,y,z) = 1 - mu(x,y,y)/mu_max
probability, ie we keep it with mu(x,y,y)/mu_max probability where
mu(x,y,z) = ((acy)^2 + (abz)^2 + (bcx)^2)^0.5
and
mu_max = bc
Implementation
import numpy as np
np.random.seed(42)
# Function to generate a random point on a uniform sphere
# (relying on https://stackoverflow.com/a/33977530/8565438)
def randompoint(ndim=3):
vec = np.random.randn(ndim,1)
vec /= np.linalg.norm(vec, axis=0)
return vec
# Give the length of each axis (example values):
a, b, c = 1, 2, 4
# Function to scale up generated points using the function `f` mentioned above:
f = lambda x,y,z : np.multiply(np.array([a,b,c]),np.array([x,y,z]))
# Keep the point with probability `mu(x,y,z)/mu_max`, ie
def keep(x, y, z, a=a, b=b, c=c):
mu_xyz = ((a * c * y) ** 2 + (a * b * z) ** 2 + (b * c * x) ** 2) ** 0.5
return mu_xyz / (b * c) > np.random.uniform(low=0.0, high=1.0)
# Generate points until we have, let's say, 1000 points:
n = 1000
points = []
while len(points) < n:
[x], [y], [z] = randompoint()
if keep(x, y, z):
points.append(f(x, y, z))
Checks
Check if all points generated satisfy the ellipsoid condition (ie that x^2/a^2 + y^2/b^2 + z^2/c^2 = 1):
for p in points:
pscaled = np.multiply(p,np.array([1/a,1/b,1/c]))
assert np.allclose(np.sum(np.dot(pscaled,pscaled)),1)
Runs without raising any errors. Visualize the points:
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(projection="3d")
points = np.array(points)
ax.scatter(points[:, 0], points[:, 1], points[:, 2])
# set aspect ratio for the axes using https://stackoverflow.com/a/64453375/8565438
ax.set_box_aspect((np.ptp(points[:, 0]), np.ptp(points[:, 1]), np.ptp(points[:, 2])))
plt.show()
These points seem evenly distributed.
Problem with currently accepted answer
Generating a point on a sphere and then just reprojecting it without any further corrections to an ellipse will result in a distorted distribution. This is essentially the same as setting this posts's p(x,y,z) to 0. Imagine an ellipsoid where one axis is orders of magnitude bigger than another. This way, it is easy to see, that naive reprojection is not going to work.

Consider using Monte-Carlo simulation: generate a random 3D point; check if the point is inside the ellipsoid; if it is, keep it. Repeat until you get 1,000 points.
P.S. Since the OP changed their question, this answer is no longer valid.

J.F. Williamson, "Random selection of points distributed on curved surfaces", Physics in Medicine & Biology 32(10), 1987, describes a general method of choosing a uniformly random point on a parametric surface. It is an acceptance/rejection method that accepts or rejects each candidate point depending on its stretch factor (norm-of-gradient). To use this method for a parametric surface, several things have to be known about the surface, namely—
x(u, v), y(u, v) and z(u, v), which are functions that generate 3-dimensional coordinates from two dimensional coordinates u and v,
The ranges of u and v,
g(point), the norm of the gradient ("stretch factor") at each point on the surface, and
gmax, the maximum value of g for the entire surface.
The algorithm is then:
Generate a point on the surface, xyz.
If g(xyz) >= RNDU01()*gmax, where RNDU01() is a uniform random variate in [0, 1), accept the point. Otherwise, repeat this process.
Chen and Glotzer (2007) apply the method to the surface of a prolate spheroid (one form of ellipsoid) in "Simulation studies of a phenomenological model for elongated virus capsid formation", Physical Review E 75(5), 051504 (preprint).

Here is a generic function to pick a random point on a surface of a sphere, spheroid or any triaxial ellipsoid with a, b and c parameters. Note that generating angles directly will not provide uniform distribution and will cause excessive population of points along z direction. Instead, phi is obtained as an inverse of randomly generated cos(phi).
import numpy as np
def random_point_ellipsoid(a,b,c):
u = np.random.rand()
v = np.random.rand()
theta = u * 2.0 * np.pi
phi = np.arccos(2.0 * v - 1.0)
sinTheta = np.sin(theta);
cosTheta = np.cos(theta);
sinPhi = np.sin(phi);
cosPhi = np.cos(phi);
rx = a * sinPhi * cosTheta;
ry = b * sinPhi * sinTheta;
rz = c * cosPhi;
return rx, ry, rz
This function is adopted from this post: https://karthikkaranth.me/blog/generating-random-points-in-a-sphere/

One way of doing this whch generalises for any shape or surface is to convert the surface to a voxel representation at arbitrarily high resolution (the higher the resolution the better but also the slower). Then you can easily select the voxels randomly however you want, and then you can select a point on the surface within the voxel using the parametric equation. The voxel selection should be completely unbiased, and the selection of the point within the voxel will suffer the same biases that come from using the parametric equation but if there are enough voxels then the size of these biases will be very small.
You need a high quality cube intersection code but with something like an elipsoid that can optimised quite easily. I'd suggest stepping through the bounding box subdivided into voxels. A quick distance check will eliminate most cubes and you can do a proper intersection check for the ones where an intersection is possible. For the point within the cube I'd be tempted to do something simple like a random XYZ distance from the centre and then cast a ray from the centre of the elipsoid and the selected point is where the ray intersects the surface. As I said above, it will be biased but with small voxels, the bias will probably be small enough.
There are libraries that do convex shape intersection very efficiently and cube/elipsoid will be one of the options. They will be highly optimised but I think the distance culling would probably be worth doing by hand whatever. And you will need a library that differentiates between a surface intersection and one object being totally inside the other.
And if you know your elipsoid is aligned to an axis then you can do the voxel/edge intersection very easily as a stack of 2D square intersection elipse problems with the set of squares to be tested defined as those that are adjacent to those in the layer above. That might be quicker.
One of the things that makes this approach more managable is that you do not need to write all the code for edge cases (it is a lot of work to get around issues with floating point inaccuracies that can lead to missing or doubled voxels at the intersection). That's because these will be very rare so they won't affect your sampling.
It might even be quicker to simply find all the voxels inside the elipse and then throw away all the voxels with 6 neighbours... Lots of options. It all depends how important performance is. This will be much slower than the opther suggestions but if you want ~1000 points then ~100,000 voxels feels about the minimum for the surface, so you probably need ~1,000,000 voxels in your bounding box. However even testing 1,000,000 intersections is pretty fast on modern computers.

Depending on what "uniformly" refers to, different methods are applicable. In any case, we can use the parametric equations using spherical coordinates (from Wikipedia):
where s = 1 refers to the ellipsoid given by the semi-axes a > b > c. From these equations we can derive the relevant volume/area element and generate points such that their probability of being generated is proportional to that volume/area element. This will provide constant volume/area density across the surface of the ellipsoid.
1. Constant volume density
This method generates points on the surface of an ellipsoid such that their volume density across the surface of the ellipsoid is constant. A consequence of this is that the one-dimensional projections (i.e. the x, y, z coordinates) are uniformly distributed; for details see the plot below.
The volume element for a triaxial ellipsoid is given by (see here):
and is thus proportional to sin(theta) (for 0 <= theta <= pi). We can use this as the basis for a probability distribution that indicates "how many" points should be generated for a given value of theta: where the area density is low/high, the probability for generating a corresponding value of theta should be low/high, too.
Hence, we can use the function f(theta) = sin(theta)/2 as our probability distribution on the interval [0, pi]. The corresponding cumulative distribution function is F(theta) = (1 - cos(theta))/2. Now we can use Inverse transform sampling to generate values of theta according to f(theta) from a uniform random distribution. The values of phi can be obtained directly from a uniform distribution on [0, 2*pi].
Example code:
import matplotlib.pyplot as plt
import numpy as np
from numpy import sin, cos, pi
rng = np.random.default_rng(seed=0)
a, b, c = 10, 3, 1
N = 5000
phi = rng.uniform(0, 2*pi, size=N)
theta = np.arccos(1 - 2*rng.random(size=N))
x = a*sin(theta)*cos(phi)
y = b*sin(theta)*sin(phi)
z = c*cos(theta)
fig = plt.figure()
ax = fig.add_subplot(projection='3d')
ax.scatter(x, y, z, s=2)
plt.show()
which produces the following plot:
The following plot shows the one-dimensional projections (i.e. density plots of x, y, z):
import seaborn as sns
sns.kdeplot(data=dict(x=x, y=y, z=z))
plt.show()
2. Constant area density
This method generates points on the surface of an ellipsoid such that their area density is constant across the surface of the ellipsoid.
Again, we start by calculating the corresponding area element. For simplicity we can use SymPy:
from sympy import cos, sin, symbols, Matrix
a, b, c, t, p = symbols('a b c t p')
x = a*sin(t)*cos(p)
y = b*sin(t)*sin(p)
z = c*cos(t)
J = Matrix([
[x.diff(t), x.diff(p)],
[y.diff(t), y.diff(p)],
[z.diff(t), z.diff(p)],
])
print((J.T # J).det().simplify())
This yields
-a**2*b**2*sin(t)**4 + a**2*b**2*sin(t)**2 + a**2*c**2*sin(p)**2*sin(t)**4 - b**2*c**2*sin(p)**2*sin(t)**4 + b**2*c**2*sin(t)**4
and further simplifies to (dividing by (a*b)**2 and taking the sqrt):
sin(t)*np.sqrt(1 + ((c/b)**2*sin(p)**2 + (c/a)**2*cos(p)**2 - 1)*sin(t)**2)
Since for this case the area element is more complex, we can use rejection sampling:
import matplotlib.pyplot as plt
import numpy as np
from numpy import cos, sin
def f_redo(t, p):
return (
sin(t)*np.sqrt(1 + ((c/b)**2*sin(p)**2 + (c/a)**2*cos(p)**2 - 1)*sin(t)**2)
< rng.random(size=t.size)
)
rng = np.random.default_rng(seed=0)
N = 5000
a, b, c = 10, 3, 1
t = rng.uniform(0, np.pi, size=N)
p = rng.uniform(0, 2*np.pi, size=N)
redo = f_redo(t, p)
while redo.any():
t[redo] = rng.uniform(0, np.pi, size=redo.sum())
p[redo] = rng.uniform(0, 2*np.pi, size=redo.sum())
redo[redo] = f_redo(t[redo], p[redo])
x = a*np.sin(t)*np.cos(p)
y = b*np.sin(t)*np.sin(p)
z = c*np.cos(t)
fig = plt.figure()
ax = fig.add_subplot(projection='3d')
ax.scatter(x, y, z, s=2)
plt.show()
which yields the following distribution:
The following plot shows the corresponding one-dimensional projections (x, y, z):