Plotting a solid cylinder centered on a plane in Matplotlib - python

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()

I have 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 some such 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 https://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:

Related

How to rotate a function by a desired angle?

How to rotate a function by the desired angle (for instance, 30 degrees)?
import matplotlib.pyplot as plt
import numpy as np
from numpy import exp, sin
def g(y):
return exp(-y)*sin(4*y)
y = np.linspace(0, 1.8, 501)
values = g(y)
fig, ax = plt.subplots(figsize=(5,5))
plt.plot(y, values)
plt.show()
Using the cosine and sine of the angle, you can create a rotation matrix. Multiplying each point (y, g(y)) with that matrix create a rotation around 0,0.
Here is some Python/numpy code to illustrate how everything could work together:
import matplotlib.pyplot as plt
import numpy as np
def g(y):
return np.exp(-y) * np.sin(4 * y)
y = np.linspace(0, 1.8, 501)
values = g(y)
theta = np.radians(30)
c, s = np.cos(theta), np.sin(theta)
rot_matrix = np.array(((c, s), (-s, c)))
xy = np.array([y, values]).T # rot_matrix
fig, ax = plt.subplots(figsize=(5, 5))
plt.plot(y, values)
plt.plot(xy[:, 0], xy[:, 1])
plt.axis('equal') # so angles on the screen look like the real angles
plt.show()
PS: To rotate around another point, first subtract the rotation center, do the rotation and then add it again:
center = np.array([0.9, 0])
xy = (np.array([y, values]).T - center) # rot_matrix + center

ploting the normal vector to a plane

The normal vector is calculated with the cross product of two vectors on the plane, so it shoud be perpendicular to the plane. But as you can seein the plot the normal vector produced with quiver isn't perpendicular.
Is the calculation of the plane wrong, my normal vector or the way i plot the normal vector?
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
points = [[3.2342, 1.8487, -1.8186],
[2.9829, 1.6434, -1.8019],
[3.4247, 1.5550, -1.8093]]
p0, p1, p2 = points
x0, y0, z0 = p0
x1, y1, z1 = p1
x2, y2, z2 = p2
ux, uy, uz = u = [x1-x0, y1-y0, z1-z0] #first vector
vx, vy, vz = v = [x2-x0, y2-y0, z2-z0] #sec vector
u_cross_v = [uy*vz-uz*vy, uz*vx-ux*vz, ux*vy-uy*vx] #cross product
point = np.array(p1)
normal = np.array(u_cross_v)
d = -point.dot(normal)
print('plane equation:\n{:1.4f}x + {:1.4f}y + {:1.4f}z + {:1.4f} = 0'.format(normal[0], normal[1], normal[2], d))
xx, yy = np.meshgrid(range(10), range(10))
z = (-normal[0] * xx - normal[1] * yy - d) * 1. / normal[2]
# plot the surface
plt3d = plt.figure().gca(projection='3d')
plt3d.quiver(x0, y0, z0, normal[0], normal[1], normal[2], color="m")
plt3d.plot_surface(xx, yy, z)
plt3d.set_xlabel("X", color='red', size=18)
plt3d.set_ylabel("Y", color='green', size=18)
plt3d.set_zlabel("Z", color='b', size=18)
plt.show()
Actually, your plot is 100% correct. The scale of your Z axis does not correspond to the same scale on X & Y axis. If you use a function to set the scale correct, you can see that:
...
plt3d.set_zlabel("Z", color='b', size=18)
# insert these lines
ax = plt.gca()
set_axis_equal(ax)
plt.show()
and the corresponding function from this post:
def set_axes_radius(ax, origin, radius):
'''
From StackOverflow question:
https://stackoverflow.com/questions/13685386/
'''
ax.set_xlim3d([origin[0] - radius, origin[0] + radius])
ax.set_ylim3d([origin[1] - radius, origin[1] + radius])
ax.set_zlim3d([origin[2] - radius, origin[2] + radius])
def set_axes_equal(ax, zoom=1.):
'''
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().
'''
limits = np.array([
ax.get_xlim3d(),
ax.get_ylim3d(),
ax.get_zlim3d(),
])
origin = np.mean(limits, axis=1)
radius = 0.5 * np.max(np.abs(limits[:, 1] - limits[:, 0])) / zoom
set_axes_radius(ax, origin, radius)

3D truncated cone in python

I want to plot a truncated cone by using exactly the same method used in
Plotting a solid cylinder centered on a plane in Matplotlib; which plots a cylinder when two points on the center of each base and the radius are known. On the other hand, I want to plot a truncated cone when the coordinates of the two points on the center of its bases and the radius of each base are known.
It seems that I just should change the second last line of the function in the following program which plots a cylinder, but I could not do this in all of my efforts.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
import pylab as pllt
fig = pllt.figure()
ax = fig.add_subplot(1,1,1, projection='3d')
#ax = pllt.subplot2grid((2,2), (0,0), rowspan=2, projection='3d')
#axis and radius
def cylinder(p0,p1,R,ccc):
#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, 1, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#print n1,'\t',norm(n1)
#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, 80)
theta = np.linspace(0, 2 * np.pi, 80)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z,color=ccc,linewidth=0, antialiased=False)
A0 = np.array([1, 3, 2])
A1 = np.array([8, 5, 9])
ax.set_xlim(0,10)
ax.set_ylim(0,10)
ax.set_zlim(0,10)
cylinder(A0,A1,1,'blue')
pllt.show()
I think I should change the radius as a function of v=p1-p0 as mentioned in:
http://mathworld.wolfram.com/ConicalFrustum.html to be able to do this.
Please let me know if there is any way to do this.
Instead of a constant radius, R, make it change from R0 to R1:
R = np.linspace(R0, R1, n)
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
import pylab as plt
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1, projection='3d')
def truncated_cone(p0, p1, R0, R1, color):
"""
Based on https://stackoverflow.com/a/39823124/190597 (astrokeat)
"""
# 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, 1, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
# make vector perpendicular to v
n1 = np.cross(v, not_v)
# print n1,'\t',norm(n1)
# 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
n = 80
t = np.linspace(0, mag, n)
theta = np.linspace(0, 2 * np.pi, n)
# use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
R = np.linspace(R0, R1, n)
# generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R *
np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z, color=color, linewidth=0, antialiased=False)
A0 = np.array([1, 3, 2])
A1 = np.array([8, 5, 9])
ax.set_xlim(0, 10)
ax.set_ylim(0, 10)
ax.set_zlim(0, 10)
truncated_cone(A0, A1, 1, 5, 'blue')
plt.show()

How to close the ends of a cylinder in matplotlib

I am trying to make a 'closed' cylinder in matplotlib but I am not sure how to go about doing this. So far I have a cylinder with the ends open, the code for this is as follows:
#make a cylinder without the ends closed
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import numpy as np
import math
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = [0,0,0]
#radius = R
p0 = np.array(origin)
p1 = np.array([8, 8, 8])
origin = np.array(origin)
R = 4
#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, 600)
theta = np.linspace(0, 2 * np.pi, 100)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
#make the color for the faces
col1 = plt.cm.autumn(np.ones(600)) # linear gradient along the t-axis
col1 = np.repeat(col1[np.newaxis,:, :], 100, axis=0) # expand over the theta-axis
ax.plot_surface(X, Y,Z, facecolors = col1, shade = True,edgecolors = "None", alpha = 0.4, linewidth = 0)
plt.show()
Running this code produces the following image
How would I close the ends of the cylinder with a solid circle (i.e. disk)?
A quick and easy way that's similar to your other code is to generate a surface using strips from r=0 to r=R. Right before plt.show() add the following lines:
R = np.array([0,R])
# cap at t=0
X, Y, Z = [p0[i] + np.outer(R, np.sin(theta)) * n1[i] + np.outer(R, np.cos(theta))*n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z, edgecolors = "r", alpha=.4, linewidth = .1)
# cap at t=mag
X, Y, Z = [p0[i] + v[i]*mag + np.outer(R, np.sin(theta)) * n1[i] + np.outer(R, np.cos(theta))*n2[i] for i in [0, 1, 2]]
ax.plot_surface(X, Y, Z, edgecolors = "r", alpha=.4, linewidth = .1)
Here the colors are more for illustrative purposes, mostly so you can see the strips. The result looks like:

Facecolor changing edgecolor in matplotlib

I am trying to remove the edge color in the plot of a cylinder where I have set an alpha and facecolors. However, if I also set the facecolors, I can still see the edge colors. If I remove the alpha = 0.5 statement then the problem is resolved, however I need the alpha to be <1 . Here is an example:
You can still see the blue edgecolors even tough I have set the edgecolor to None.
This is the code where I use plot_surface()
ax.plot_surface(X, Y,Z, edgecolor = "None", facecolors = col1, alpha = 0.5)
Yet the edge colors are still there? However, if I remove the facecolors statement inside plot_surface() then the edge colors are no longer there. Here is the complete code:
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import random
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = np.array([0, 0, 0])
#axis and radius
p0 = np.array([0, 0, 0])
p1 = np.array([8, 8, 8])
R = 4
#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, 200)
theta = np.linspace(0, 2 * np.pi, 100)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
col1 = plt.cm.Blues(np.linspace(0,1,200)) # linear gradient along the t-axis
col1 = np.repeat(col1[np.newaxis,:, :], 100, axis=0) # expand over the theta- axis
ax.plot_surface(X, Y,Z, edgecolor = None, facecolors = col1, alpha = 0.5)
#plot axis
ax.plot(*zip(p0, p1), color = 'red')
ax.set_xlim(0, 10)
ax.set_ylim(0, 10)
ax.set_zlim(0, 10)
plt.axis('off')
ax.axes.get_xaxis().set_visible(False)
ax.axes.get_yaxis().set_visible(False)
plt.show()
Setting linewidth=0 in plot_surface() solves this problem:
ax.plot_surface(X, Y, Z, edgecolor=None, facecolors=col1, alpha=0.5, linewidth=0)
p.s.: I didn't find this worth an answer, but per: Question with no answers, but issue solved in the comments (or extended in chat), I added it as a quick answer so the question can be marked as solved

Categories

Resources