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?

Navigation

#1 by (1 votes)

2

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 this function to the surface? I suspect there are arguments to the ax.plot_surface . I tried to use the argument: facecolors=potencial(x,y,z) , ma received ValueError: Invalid RGBA argument. Looking at the source code of the third example , there is:

# 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)]

That I did not understand, I have no idea how to 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()

python matplotlib plot 3d

asked by anonymous 01.05.2017 / 16:45

1 answer

Android Application Incompatible with Tablet (PlayStore Information) Problem with the Requests package

score 1 · Answer 1

Translated answer ( original ):

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

Use the cmap argument 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 facecolors argument. It expects an array of colors in the same way as z.

So, in this case we need to choose option 2 and construct an 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 well above and below this range, we need to normalize it to the interval [0,1]. Matplotlib already provides a help function to do this normalization and as the potential has a 1 / x dependency, a logarithmic color scale may be appropriate.

In the end, facecolors can be given by an array

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

The missing part is the colorbar ( colorbar ). In order for it to be linked to the surface colors of the plot , we need to manually mount a ScalarMappable with colormap and the normalization instance, which we can then provide 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()