How to display color as a function of distance to the origin in matplotlib's 3D surface plot?

user_13241623 0
Viewed 338

I am trying to plot a certain function Q(theta, phi) in spherical coordinate using python's matplotlib package. I want the color to show up as a function of distance to the origin instead of the z coordinate. Is there a way to do this? Here is the code I use to produce the 3D surface plot:

%matplotlib notebook

fig = plt.figure(figsize=(12,12))

ax = fig.add_subplot(111, projection='3d')
X, Y = Q*np.sin(Theta)*np.cos(Phi), Q*np.sin(Theta)*np.sin(Phi)
Z = Q*np.cos(Theta)
ax.plot_surface(X, Y, Z, cmap=plt.cm.YlGnBu_r)
ax.set_zlim(-1, 1)
ax.set_ylim(-1, 1)
ax.set_xlim(-1, 1)
ax.set_xlabel('J_x')
ax.set_ylabel('J_y')
ax.set_zlabel('J_z')

plt.show()

I have not earned enough reputations to post the plot I created. It is similar to the plot in this link: https://matplotlib.org/3.1.1/gallery/mplot3d/surface3d.html. As you can see, the color is a function of the z coordinate, not of the distance from the origin.

matplotlib python
1 Answers

You should specify facecolor in plot_surface, not cmap.

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


fig = plt.figure()
ax = fig.gca(projection='3d')

# Make data.
X = np.arange(-5, 5, 0.25)
xlen = len(X)
Y = np.arange(-5, 5, 0.25)
ylen = len(Y)
X, Y = np.meshgrid(X, Y)
R = np.sqrt(X**2 + Y**2)
Z = np.sin(R)

d = np.sqrt(X**2+Y**2+Z**2)
d = d/d.max()

# Plot the surface with face colors taken from the array we made.
surf = ax.plot_surface(X, Y, Z, facecolors=plt.cm.viridis(d), linewidth=0)

# Customize the z axis.
ax.set_zlim(-1, 1)
ax.w_zaxis.set_major_locator(LinearLocator(6))

enter image description here

user_58551310
Related