Plotting function of 3 dimensions over given domain with matplotlib

Question

I am trying to visualize a function of 3 parameters over a cube in R^3 to get an idea of the smoothness of the function. An example of this problem is shown in the sample code below

%pylab
from mpl_toolkits.mplot3d import Axes3D
import itertools

x = np.linspace(0,10,50)
y = np.linspace(0,15,50)
z = np.linspace(0,8,50)

points = []
for element in itertools.product(x, y, z):
    points.append(element)

def f(vals):
    return np.cos(vals[0]) + np.sin(vals[1]) + vals[2]**0.5

fxyz = map(f, points)
xi, yi, zi = zip(*points)

fig = plt.figure(figsize=(12, 8))
ax = fig.add_subplot(111, projection='3d')
ax.scatter(xi, yi, zi, c=fxyz, alpha=0.5)
plt.show()

cube

The problem with this approach is that the inside of the cube cannot be visualized. Is there a better way to graph a function over some dense subset of R^3?

As mentioned, is a better option to use another software for this, like Mayavi. Two techniques that can be used are: isosurfacing and volume rendering. — nicoguaro
– nicoguaro, Commented Dec 21, 2014 at 13:11

mgilbert · Accepted Answer · 2014-12-21 19:54:03Z

7

As @HYRY and @nicoguaro suggested in the comments above, Mayavi is much better suited for this type of work. There is a good set of examples here that I used for reference. Here is what I came up with

import numpy as np
from mayavi import mlab

x = np.linspace(0,10,50)
y = np.linspace(0,15,50)
z = np.linspace(0,8,50)

X, Y, Z = np.meshgrid(x, y, z)

s = np.cos(X) + np.sin(Y) + Z**0.5
b1 = np.percentile(s, 20)
b2 = np.percentile(s, 80)
mlab.pipeline.volume(mlab.pipeline.scalar_field(s), vmin=b1, vmax=b2)
mlab.axes()
mlab.show()

After which I rotated the figure to desired angles with the GUI and saved desired views

angle1 angle2

answered Dec 21, 2014 at 19:54

mgilbert

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