How can a colormap of a Surface be mapped to a scalar function?

Question

How can a colormap of a Surface be mapped to a scalar function?

Translation of the question I asked in the SO :

I have a scalar function that represents the electrical potential on a spherical surface. I want to plot, for a given radius, the surface and map its color points based on the potential function.

How can I map such a function to the surface? I suspect it has to do with the arguments passed to the function ax.plot_surface . I tried using the argument: facecolors=potencial(x,y,z), ma I got ValueError: Invalid RGBA argument. looking at the source code from the third example , there are:

# Create an empty array of strings with the same shape as the meshgrid, and
# populate it with two colors in a checkerboard pattern.
colortuple = ('y', 'b')
colors = np.empty(X.shape, dtype=str)
for y in range(ylen):
    for x in range(xlen):
        colors[x, y] = colortuple[(x + y) % len(colortuple)]

Which I didn't understand, nor have any idea how link with a scalar function.

My code

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

def potencial(x,y,z, a=1., v=1.):
    r = np.sqrt( np.square(x) + np.square(y) + np.square(z) )    
    p = r/z #cos(theta)
    asr = a/r
    s=0
    s += np.polyval(special.legendre(1), x) * 3/2*np.power(asr, 2)
    s += np.polyval(special.legendre(3), x) * -7/8*np.power(asr, 4)
    s += np.polyval(special.legendre(5), x) * 11/16*np.power(asr, 6)    
    return v*s

# criar dados
def sphere_surface(r):
    u = np.linspace(0, 2 * np.pi, 100)
    v = np.linspace(0, np.pi, 100)
    x = r * np.outer(np.cos(u), np.sin(v))
    y = r * np.outer(np.sin(u), np.sin(v))
    z = r * np.outer(np.ones(np.size(u)), np.cos(v))
    return x,y,z

x,y,z = sphere_surface(1.5)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# Plotar a superficie
surf = ax.plot_surface(x,y,z, cmap=cm.coolwarm,
                       linewidth=0, antialiased=False)
fig.colorbar(surf, shrink=0.5, aspect=5)
# Está mapeado aos valores do eixo-z

ax.set_xlabel("x")
ax.set_ylabel("y")
ax.set_zlabel("z")
plt.show()

2

python matplotlib 3d plot

Author: Comunidade, 2017-05-01

Source

1 answers

score 1 · Answer 1

Translated answer (original):

In principle there are two ways to color a surface plot in matplotlib.

use the argument cmap to specify a colormap . In this case the color will be chosen according to the array z. In this case this is not desired,
use the argument facecolors. It expects a array of colors in the same shape ( shape ) as z.

So in this case we need to choose Option 2 and build a array of colors. For this purpose, one can choose a colormap. A colormap maps values between 0 and 1 to a color. Since the potential has values far above and below this range, it must be normalized to the range [0,1]. Matplotlib already provides a Help function to do this normalization and since the potential has a 1/x dependency, a logarithmic color scale may be appropriate.

At the end as facecolors can be given by an array

colors = cmap(norm(potential(...)))

The missing part now is the Color Bar (colorbar ). For it to be linked colours from plot surface, you need to manually mount a ScalarMappable in the colormap, instance normalization, which we will then provide the colorbar.

sm = plt.cm.ScalarMappable(cmap=plt.cm.coolwarm, norm=norm)
sm.set_array(pot)
fig.colorbar(sm, shrink=0.5, aspect=5)

Here is a complete example

from __future__ import division
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import matplotlib.colors
import numpy as np
from scipy import special    

def potencial(x,y,z, a=1., v=1.):
    r = np.sqrt( np.square(x) + np.square(y) + np.square(z) )    
    p = z/r #cos(theta)
    asr = a/r
    s=0
    s += np.polyval(special.legendre(1), p) * 3/2*np.power(asr, 2)
    s += np.polyval(special.legendre(3), p) * -7/8*np.power(asr, 4)
    s += np.polyval(special.legendre(5), p) * 11/16*np.power(asr, 6)    
    return v*s

# Make data
def sphere_surface(r):
    u = np.linspace(0, 2 * np.pi, 100)
    v = np.linspace(0, np.pi, 100)
    x = r * np.outer(np.cos(u), np.sin(v))
    y = r * np.outer(np.sin(u), np.sin(v))
    z = r * np.outer(np.ones(np.size(u)), np.cos(v))
    return x,y,z

x,y,z = sphere_surface(1.5)
pot = potencial(x,y,z)


norm=matplotlib.colors.SymLogNorm(1,vmin=pot.min(),vmax=pot.max())
colors=plt.cm.coolwarm(norm(pot))

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# Plot the surface
surf = ax.plot_surface(x,y,z, facecolors=colors,
                       linewidth=0, antialiased=False)
# Set up colorbar
sm = plt.cm.ScalarMappable(cmap=plt.cm.coolwarm, norm=norm)
sm.set_array(pot)
fig.colorbar(sm, shrink=0.5, aspect=5)


ax.set_xlabel("x")
ax.set_ylabel("y")
ax.set_zlabel("z")
plt.show()