CODEWITHSUNDEEP
pythonmatplotlibplot
randomal
randomal

Reputation: 6612

Add cylinder to plot

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()

Upvotes: 14

Views: 36985

Answers (3)

Yong Yang
Yong Yang

Reputation: 1322

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. enter image description here

Upvotes: 29

crypdick
crypdick

Reputation: 19864

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)

Solid cylinder

Upvotes: 8

Greg
Greg

Reputation: 12234

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.

Upvotes: 16

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