Rebecca - 8 months ago 84

Python Question

I fit a plane to a bunch of points in 3d and initially gave it an arbitrary size using np.meshgrid, but now I'm trying to plot a cylinder centered on that plane and oriented the same way (such that the plane fit would cut the height of the cylinder in half), but with a specified radius and height. The only examples of cylinders plotted in matplotlib I can find are hollow and usually open at the top and bottom. I want the one I plot to be solid so I can clearly see what points it's enclosing.

Here's a minimum working example with a randomly generated plane. Since the plane I'm using is always given by a point and a normal vector, the cylinder should be based off of those things as well (plus a provided radius, and height to extend above and below the plane).

`from __future__ import division #Enables new-style division`

import matplotlib.pyplot as plt

from mpl_toolkits.mplot3d import Axes3D

import seaborn as sns

import numpy as np

cen_x = 0

cen_y = 0

cen_z = 0

origin = np.array([cen_x,cen_y,cen_z])

normal = np.array([np.random.uniform(-1,1),np.random.uniform(-1,1),np.random.uniform(0,1)])

a = normal[0]

b = normal[1]

c = normal[2]

#equation for a plane is a*x+b*y+c*z+d=0 where [a,b,c] is the normal

#so calculate d from the normal

d = -origin.dot(normal)

# create x,y meshgrid

xx, yy = np.meshgrid(np.arange(cen_x-1,cen_x+1,0.01),np.arange(cen_y-1,cen_y+1,0.01))

# calculate corresponding z

zz = (-a * xx - b * yy - d) * 1./c

halo_x = [-0.3, -0.9, 0.8, 1.3, -0.1, 0.5]

halo_y = [0.8, 1.1, -0.5, -0.7, -1.2, 0.1]

halo_z = [1.0, -0.4, 0.3, -1.2, 0.9, 1.2]

fig = plt.figure(figsize=(9,9))

plt3d = fig.gca(projection='3d')

plt3d.plot_surface(xx, yy, zz, color='r', alpha=0.4)

plt3d.set_xlim3d(cen_x-3,cen_x+3)

plt3d.set_ylim3d(cen_y-3,cen_y+3)

plt3d.set_zlim3d(cen_z-3,cen_z+3)

plt3d.set_xlabel('X')

plt3d.set_ylabel('Y')

plt3d.set_zlabel('Z')

plt.show()

Answer

I've modified a solution to a question How to add colors to each individual face of a cylinder using matplotlib, removing the fancy shading and adding end caps. If you want to show the enclosed points, you can use alpha=0.5 or somesuch to make the cylinder semi-transparent.

The orientation of the cylinder is defined by a unit vector v with length mag, which could be the normal to your surface.

```
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Sun Oct 2 18:33:10 2016
Modified from http://stackoverflow.com/questions/38076682/how-to-add-colors-to-each-individual-face-of-a-cylinder-using-matplotlib
to add "end caps" and to undo fancy coloring.
@author: astrokeat
"""
import numpy as np
from matplotlib import pyplot as plt
from scipy.linalg import norm
#axis and radius
p0 = np.array([1, 3, 2]) #point at one end
p1 = np.array([8, 5, 9]) #point at other end
R = 5
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 0, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 2)
theta = np.linspace(0, 2 * np.pi, 100)
rsample = np.linspace(0, R, 2)
#use meshgrid to make 2d arrays
t, theta2 = np.meshgrid(t, theta)
rsample,theta = np.meshgrid(rsample, theta)
#generate coordinates for surface
# "Tube"
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta2) * n1[i] + R * np.cos(theta2) * n2[i] for i in [0, 1, 2]]
# "Bottom"
X2, Y2, Z2 = [p0[i] + rsample[i] * np.sin(theta) * n1[i] + rsample[i] * np.cos(theta) * n2[i] for i in [0, 1, 2]]
# "Top"
X3, Y3, Z3 = [p0[i] + v[i]*mag + rsample[i] * np.sin(theta) * n1[i] + rsample[i] * np.cos(theta) * n2[i] for i in [0, 1, 2]]
ax=plt.subplot(111, projection='3d')
ax.plot_surface(X, Y, Z, color='blue')
ax.plot_surface(X2, Y2, Z2, color='blue')
ax.plot_surface(X3, Y3, Z3, color='blue')
plt.show()
```

The result:

Source (Stackoverflow)