Add cylinder to plot

Question

I would like to add a transparent cylinder to my 3D scatter plot. How can I do it?

This is the code I am using to make the plot:

fig = plt.figure(2, figsize=(8, 6))
ax = fig.add_subplot(111, projection='3d')

ax.scatter(X, Y, Z, c=Z,cmap=plt.cm.Paired)
ax.set_xlabel("X")
ax.set_ylabel("Y")
ax.set_zlabel("Z")
plt.xticks()

score 25 · Accepted Answer · answered Mar 16 '18 at 00:29

Today I have to do the same thing in my project about adding a transparent cylinder in the result. This is the code I get finally. So I share it with you guys just for learning

import numpy as np

def data_for_cylinder_along_z(center_x,center_y,radius,height_z):
    z = np.linspace(0, height_z, 50)
    theta = np.linspace(0, 2*np.pi, 50)
    theta_grid, z_grid=np.meshgrid(theta, z)
    x_grid = radius*np.cos(theta_grid) + center_x
    y_grid = radius*np.sin(theta_grid) + center_y
    return x_grid,y_grid,z_grid

import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

Xc,Yc,Zc = data_for_cylinder_along_z(0.2,0.2,0.05,0.1)
ax.plot_surface(Xc, Yc, Zc, alpha=0.5)

plt.show()

And you will get this beautiful figure.

+1, other solutions use Cartesian coordinates to get Y as a function of X, which leads to non-uniform distribution of points in the surface and, more importantly, can cause issues when determining all edges of the cylinder (you might end up with nan values due to numerical approximations). This solution works well in all cases — FernAndr, Oct 02 '18 at 10:49

score 16 · Answer 2 · answered Nov 18 '14 at 08:22

One possible method is to use the plot_surface. Adapting the solution given in this blog post then have

import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# Scatter graph
N = 100
X = np.random.uniform(-1, 1, N)
Y = np.random.uniform(-1, 1, N)
Z = np.random.uniform(-2, 2, N)
ax.scatter(X, Y, Z)

# Cylinder
x=np.linspace(-1, 1, 100)
z=np.linspace(-2, 2, 100)
Xc, Zc=np.meshgrid(x, z)
Yc = np.sqrt(1-Xc**2)

# Draw parameters
rstride = 20
cstride = 10
ax.plot_surface(Xc, Yc, Zc, alpha=0.2, rstride=rstride, cstride=cstride)
ax.plot_surface(Xc, -Yc, Zc, alpha=0.2, rstride=rstride, cstride=cstride)

ax.set_xlabel("X")
ax.set_ylabel("Y")
ax.set_zlabel("Z")
plt.show()

enter image description here

I've added some minimal configuration of the surface, better can be achieved by consulting the docs.

score 7 · Answer 3 · answered Nov 13 '15 at 08:41

I improved on @Greg's answer and made a solid 3D cylinder with a top and bottom surface and rewrote the equation so that you can translate in the x, y,and z

from mpl_toolkits.mplot3d import Axes3D
import mpl_toolkits.mplot3d.art3d as art3d
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import Circle

def plot_3D_cylinder(radius, height, elevation=0, resolution=100, color='r', x_center = 0, y_center = 0):
    fig=plt.figure()
    ax = Axes3D(fig, azim=30, elev=30)

    x = np.linspace(x_center-radius, x_center+radius, resolution)
    z = np.linspace(elevation, elevation+height, resolution)
    X, Z = np.meshgrid(x, z)

    Y = np.sqrt(radius**2 - (X - x_center)**2) + y_center # Pythagorean theorem

    ax.plot_surface(X, Y, Z, linewidth=0, color=color)
    ax.plot_surface(X, (2*y_center-Y), Z, linewidth=0, color=color)

    floor = Circle((x_center, y_center), radius, color=color)
    ax.add_patch(floor)
    art3d.pathpatch_2d_to_3d(floor, z=elevation, zdir="z")

    ceiling = Circle((x_center, y_center), radius, color=color)
    ax.add_patch(ceiling)
    art3d.pathpatch_2d_to_3d(ceiling, z=elevation+height, zdir="z")

    ax.set_xlabel('x-axis')
    ax.set_ylabel('y-axis')
    ax.set_zlabel('z-axis')

    plt.show()

# params
radius = 3
height = 10
elevation = -5
resolution = 100
color = 'r'
x_center = 3
y_center = -2

plot_3D_cylinder(radius, height, elevation=elevation, resolution=resolution, color=color, x_center=x_center, y_center=y_center)

Add cylinder to plot

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