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Regular Distribution of Points in the Volume of a Sphere

I'm trying to generate a regular n number of points within the volume of a sphere. I found this similar answer (https://scicomp.stackexchange.com/questions/29959/uniform-dots-distribution-in-a-sphere) on generating a uniform regular n number of points on the surface of a sphere, with the following code:
import numpy as np
n = 5000
r = 1
z = []
y = []
x = []
alpha = 4.0*np.pi*r*r/n
d = np.sqrt(alpha)
m_nu = int(np.round(np.pi/d))
d_nu = np.pi/m_nu
d_phi = alpha/d_nu
count = 0
for m in range (0,m_nu):
nu = np.pi*(m+0.5)/m_nu
m_phi = int(np.round(2*np.pi*np.sin(nu)/d_phi))
for n in range (0,m_phi):
phi = 2*np.pi*n/m_phi
xp = r*np.sin(nu)*np.cos(phi)
yp = r*np.sin(nu)*np.sin(phi)
zp = r*np.cos(nu)
x.append(xp)
y.append(yp)
z.append(zp)
count = count +1
which works as intended:
How can I modify this to generate a regular set of n points in the volume of a sphere?

Another method to do this, yielding uniformity in volume:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
dim_len = 30
spacing = 2 / dim_len
point_cloud = np.mgrid[-1:1:spacing, -1:1:spacing, -1:1:spacing].reshape(3, -1).T
point_radius = np.linalg.norm(point_cloud, axis=1)
sphere_radius = 0.5
in_points = point_radius < sphere_radius
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(point_cloud[in_points, 0], point_cloud[in_points, 1], point_cloud[in_points, 2], )
plt.show()
Output (matplotlib mixes up the view but it is a uniformly sampled sphere (in volume))
Uniform sampling, then checking if points are in the sphere or not by their radius.
Uniform sampling reference [see this answer's edit history for naiive sampling].
This method has the drawback of generating redundant points which are then discarded.
It has the upside of vectorization, which probably makes up for the drawback. I didn't check.
With fancy indexing, one could generate the same points as this method without generating redundant points, but I doubt it can be easily (or at all) vectorized.

Sample uniformly along X. For every value of X, you draw two Y from X²+Y²=1. Sample uniformly between these two Y. Then for every (X, Y) pair, you draw two Z from X²+Y²+Z²=1. Sample uniformly between these two Z.

Test whether points are inside ellipses, without using Matplotlib?

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()

How to find the intervals that are intersection of a circle with a rectangle

I need to integrate over the arcs that are resulted from the intersection of a circle with a rectangle and fall inside the rectangle. I can find the intersection points using the shapely package. However, I don't know how to obtain integration intervals. For example, in the below figure my code returns [-2.1562, 2.1562] in radians (with respect to the center of the circle), while it should be able to automatically understand that the integration intervals that falls inside the rectangle are [[2.1562, 3.1415],[-3.1415, -2.1562]] (assuming pi = 3.1415).
Here is another example:
My code returns [-0.45036, -0.29576, 0.29576, 0.45036] and the expected intervals will be [[0.29576, 0.45036], [-0.45036, -0.29576]].
The code should also work for any other location that the circle is located (with any radius), whether its center is outside or inside the rectangle.
Here is my code, written using iPython:
import matplotlib.pyplot as plt
import math
import numpy as np
from shapely.geometry import LineString, MultiPoint
from shapely.geometry import Polygon
from shapely.geometry import Point
# Utilities
def cart2pol(xy, center):
x,y = xy
x_0,y_0 = center
rho = np.sqrt((x-x_0)**2 + (y-y_0)**2)
phi = np.arctan2(y-y_0, x-x_0)
return(rho, phi)
def pol2cart(rho, phi, center):
x_0,y_0 = center
x = rho * np.cos(phi)+x_0
y = rho * np.sin(phi)+y_0
return(x, y)
def distance(A,B):
return math.sqrt((A[0]-B[0])**2+(A[1]-B[1])**2)
#######################
rad = 6
center = (-1,5)
p = Point(center)
c = p.buffer(rad).boundary
A = (10,0)
B = (0,0)
C = (0,10)
D = (10,10)
coords = [Point(A), Point(B), Point(C), Point(D)]
poly = MultiPoint(coords).convex_hull
i=c.intersection(poly)
lines = [LineString([A, D]), LineString([D, C]),
LineString([C, B]), LineString([B, A])]
points = []
for l in lines:
i = c.intersection(l)
if not i.is_empty:
if i.geom_type == 'MultiPoint':
for j in range(len(i.geoms)):
points.append(i.geoms[j].coords[0])
else:
points.append(i.coords[0])
# Repeat the tangential points
for k, point in enumerate(points.copy()):
if abs(distance(center, point)**2 + distance(point, B)**2 - distance(B, center)**2) < 1e-4:
points.insert(k+1,point)
elif abs(distance(center, point)**2 + distance(point, D)**2 -distance(D, center)**2) < 1e-4:
points.insert(k+1,point)
# Sort points in polar coordinates
phis = [cart2pol(point,center)[1] for point in points]
phis.sort()
print(phis)
# Plot the shapes
x,y = c.xy
plt.plot(*c.xy)
for l in lines:
plt.plot(*l.xy, 'b')
plt.gca().set_aspect('equal', adjustable='box')
I tried to sort the intersection points according to their angle in a way that each two adjacent items in the list of intersection points corresponds to an arc. The problem is that there will be a jump in the angles from -pi to pi when rotating along the unit circle. Also I don't know how to find that whether an arc is inside the rectangle or not given its 2 end points.

Dealing with angle ranges is not straightforward.
1) select a non-ambiguous representation range, such as [-π, π) radians.
2) write a function that finds the intersections of the circle with a (h/v) half-plane and returns an angle interval. It the interval straddles the ±π border, split it in two.
3) write a function that finds the intersection between two lists of intervals (this is a modified merging problem).
4) process the four edges and intersect the resulting intervals.
5) possibly merge intervals that straddle the ±π border.

Find the area between two curves plotted in matplotlib (fill_between area)

I have a list of x and y values for two curves, both having weird shapes, and I don't have a function for any of them. I need to do two things:
Plot it and shade the area between the curves like the image below.
Find the total area of this shaded region between the curves.
I'm able to plot and shade the area between those curves with fill_between and fill_betweenx in matplotlib, but I have no idea on how to calculate the exact area between them, specially because I don't have a function for any of those curves.
Any ideas?
I looked everywhere and can't find a simple solution for this. I'm quite desperate, so any help is much appreciated.
Thank you very much!
EDIT: For future reference (in case anyone runs into the same problem), here is how I've solved this: connected the first and last node/point of each curve together, resulting in a big weird-shaped polygon, then used shapely to calculate the polygon's area automatically, which is the exact area between the curves, no matter which way they go or how nonlinear they are. Works like a charm! :)
Here is my code:
from shapely.geometry import Polygon
x_y_curve1 = [(0.121,0.232),(2.898,4.554),(7.865,9.987)] #these are your points for curve 1 (I just put some random numbers)
x_y_curve2 = [(1.221,1.232),(3.898,5.554),(8.865,7.987)] #these are your points for curve 2 (I just put some random numbers)
polygon_points = [] #creates a empty list where we will append the points to create the polygon
for xyvalue in x_y_curve1:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append all xy points for curve 1
for xyvalue in x_y_curve2[::-1]:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append all xy points for curve 2 in the reverse order (from last point to first point)
for xyvalue in x_y_curve1[0:1]:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append the first point in curve 1 again, to it "closes" the polygon
polygon = Polygon(polygon_points)
area = polygon.area
print(area)
EDIT 2: Thank you for the answers. Like Kyle explained, this only works for positive values. If your curves go below 0 (which is not my case, as showed in the example chart), then you would have to work with absolute numbers.

The area calculation is straightforward in blocks where the two curves don't intersect: thats the trapezium as has been pointed out above. If they intersect, then you create two triangles between x[i] and x[i+1], and you should add the area of the two. If you want to do it directly, you should handle the two cases separately. Here's a basic working example to solve your problem. First, I will start with some fake data:
#!/usr/bin/python
import numpy as np
# let us generate fake test data
x = np.arange(10)
y1 = np.random.rand(10) * 20
y2 = np.random.rand(10) * 20
Now, the main code. Based on your plot, looks like you have y1 and y2 defined at the same X points. Then we define,
z = y1-y2
dx = x[1:] - x[:-1]
cross_test = np.sign(z[:-1] * z[1:])
cross_test will be negative whenever the two graphs cross. At these points, we want to calculate the x coordinate of the crossover. For simplicity, I will calculate x coordinates of the intersection of all segments of y. For places where the two curves don't intersect, they will be useless values, and we won't use them anywhere. This just keeps the code easier to understand.
Suppose you have z1 and z2 at x1 and x2, then we are solving for x0 such that z = 0:
# (z2 - z1)/(x2 - x1) = (z0 - z1) / (x0 - x1) = -z1/(x0 - x1)
# x0 = x1 - (x2 - x1) / (z2 - z1) * z1
x_intersect = x[:-1] - dx / (z[1:] - z[:-1]) * z[:-1]
dx_intersect = - dx / (z[1:] - z[:-1]) * z[:-1]
Where the curves don't intersect, area is simply given by:
areas_pos = abs(z[:-1] + z[1:]) * 0.5 * dx # signs of both z are same
Where they intersect, we add areas of both triangles:
areas_neg = 0.5 * dx_intersect * abs(z[:-1]) + 0.5 * (dx - dx_intersect) * abs(z[1:])
Now, the area in each block x[i] to x[i+1] is to be selected, for which I use np.where:
areas = np.where(cross_test < 0, areas_neg, areas_pos)
total_area = np.sum(areas)
That is your desired answer. As has been pointed out above, this will get more complicated if the both the y graphs were defined at different x points. If you want to test this, you can simply plot it (in my test case, y range will be -20 to 20)
negatives = np.where(cross_test < 0)
positives = np.where(cross_test >= 0)
plot(x, y1)
plot(x, y2)
plot(x, z)
plt.vlines(x_intersect[negatives], -20, 20)

Define your two curves as functions f and g that are linear by segment, e.g. between x1 and x2, f(x) = f(x1) + ((x-x1)/(x2-x1))*(f(x2)-f(x1)).
Define h(x)=abs(g(x)-f(x)). Then use scipy.integrate.quad to integrate h.
That way you don't need to bother about the intersections. It will do the "trapeze summing" suggested by ch41rmn automatically.

Your set of data is quite "nice" in the sense that the two sets of data share the same set of x-coordinates. You can therefore calculate the area using a series of trapezoids.
e.g. define the two functions as f(x) and g(x), then, between any two consecutive points in x, you have four points of data:
(x1, f(x1))-->(x2, f(x2))
(x1, g(x1))-->(x2, g(x2))
Then, the area of the trapezoid is
A(x1-->x2) = ( f(x1)-g(x1) + f(x2)-g(x2) ) * (x2-x1)/2 (1)
A complication arises that equation (1) only works for simply-connected regions, i.e. there must not be a cross-over within this region:
|\ |\/|
|_| vs |/\|
The area of the two sides of the intersection must be evaluated separately. You will need to go through your data to find all points of intersections, then insert their coordinates into your list of coordinates. The correct order of x must be maintained. Then, you can loop through your list of simply connected regions and obtain a sum of the area of trapezoids.
EDIT:
For curiosity's sake, if the x-coordinates for the two lists are different, you can instead construct triangles. e.g.
.____.
| / \
| / \
| / \
|/ \
._________.
Overlap between triangles must be avoided, so you will again need to find points of intersections and insert them into your ordered list. The lengths of each side of the triangle can be calculated using Pythagoras' formula, and the area of the triangles can be calculated using Heron's formula.

The area_between_two_curves function in pypi library similaritymeasures (released in 2018) might give you what you need. I tried a trivial example on my side, comparing the area between a function and a constant value and got pretty close tie-back to Excel (within 2%). Not sure why it doesn't give me 100% tie-back, maybe I am doing something wrong. Worth considering though.

I had the same problem.The answer below is based on an attempt by the question author. However, shapely will not directly give the area of the polygon in purple. You need to edit the code to break it up into its component polygons and then get the area of each. After-which you simply add them up.
Area Between two lines
Consider the lines below:
Sample Two lines
If you run the code below you will get zero for area because it takes the clockwise and subtracts the anti clockwise area:
from shapely.geometry import Polygon
x_y_curve1 = [(1,1),(2,1),(3,3),(4,3)] #these are your points for curve 1
x_y_curve2 = [(1,3),(2,3),(3,1),(4,1)] #these are your points for curve 2
polygon_points = [] #creates a empty list where we will append the points to create the polygon
for xyvalue in x_y_curve1:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append all xy points for curve 1
for xyvalue in x_y_curve2[::-1]:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append all xy points for curve 2 in the reverse order (from last point to first point)
for xyvalue in x_y_curve1[0:1]:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append the first point in curve 1 again, to it "closes" the polygon
polygon = Polygon(polygon_points)
area = polygon.area
print(area)
The solution is therefore to split the polygon into smaller pieces based on where the lines intersect. Then use a for loop to add these up:
from shapely.geometry import Polygon
x_y_curve1 = [(1,1),(2,1),(3,3),(4,3)] #these are your points for curve 1
x_y_curve2 = [(1,3),(2,3),(3,1),(4,1)] #these are your points for curve 2
polygon_points = [] #creates a empty list where we will append the points to create the polygon
for xyvalue in x_y_curve1:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append all xy points for curve 1
for xyvalue in x_y_curve2[::-1]:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append all xy points for curve 2 in the reverse order (from last point to first point)
for xyvalue in x_y_curve1[0:1]:
polygon_points.append([xyvalue[0],xyvalue[1]]) #append the first point in curve 1 again, to it "closes" the polygon
polygon = Polygon(polygon_points)
area = polygon.area
x,y = polygon.exterior.xy
# original data
ls = LineString(np.c_[x, y])
# closed, non-simple
lr = LineString(ls.coords[:] + ls.coords[0:1])
lr.is_simple # False
mls = unary_union(lr)
mls.geom_type # MultiLineString'
Area_cal =[]
for polygon in polygonize(mls):
Area_cal.append(polygon.area)
Area_poly = (np.asarray(Area_cal).sum())
print(Area_poly)

A straightforward application of the area of a general polygon (see Shoelace formula) makes for a super-simple and fast, vectorized calculation:
def area(p):
# for p: 2D vertices of a polygon:
# area = 1/2 abs(sum(p0 ^ p1 + p1 ^ p2 + ... + pn-1 ^ p0))
# where ^ is the cross product
return np.abs(np.cross(p, np.roll(p, 1, axis=0)).sum()) / 2
Application to area between two curves. In this example, we don't even have matching x coordinates!
np.random.seed(0)
n0 = 10
n1 = 15
xy0 = np.c_[np.linspace(0, 10, n0), np.random.uniform(0, 10, n0)]
xy1 = np.c_[np.linspace(0, 10, n1), np.random.uniform(0, 10, n1)]
p = np.r_[xy0, xy1[::-1]]
>>> area(p)
4.9786...
Plot:
plt.plot(*xy0.T, 'b-')
plt.plot(*xy1.T, 'r-')
p = np.r_[xy0, xy1[::-1]]
plt.fill(*p.T, alpha=.2)
Speed
For both curves having 1 million points:
n = 1_000_000
xy0 = np.c_[np.linspace(0, 10, n), np.random.uniform(0, 10, n)]
xy1 = np.c_[np.linspace(0, 10, n), np.random.uniform(0, 10, n)]
%timeit area(np.r_[xy0, xy1[::-1]])
# 42.9 ms ± 140 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Simple viz of polygon area calculation
# say:
p = np.array([[0, 3], [1, 0], [3, 3], [1, 3], [1, 2]])
p_closed = np.r_[p, p[:1]]
fig, axes = plt.subplots(ncols=2, figsize=(10, 5), subplot_kw=dict(box_aspect=1), sharex=True)
ax = axes[0]
ax.set_aspect('equal')
ax.plot(*p_closed.T, '.-')
ax.fill(*p_closed.T, alpha=0.6)
center = p.mean(0)
txtkwargs = dict(ha='center', va='center')
ax.text(*center, f'{area(p):.2f}', **txtkwargs)
ax = axes[1]
ax.set_aspect('equal')
for a, b in zip(p_closed, p_closed[1:]):
ar = 1/2 * np.cross(a, b)
pos = ar >= 0
tri = np.c_[(0,0), a, b, (0,0)].T
# shrink a bit to make individual triangles easier to visually identify
center = tri.mean(0)
tri = (tri - center)*0.95 + center
c = 'b' if pos else 'r'
ax.plot(*tri.T, 'k')
ax.fill(*tri.T, c, alpha=0.2, zorder=2 - pos)
t = ax.text(*center, f'{ar:.1f}', color=c, fontsize=8, **txtkwargs)
t.set_bbox(dict(facecolor='white', alpha=0.8, edgecolor='none'))
plt.tight_layout()

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?