在matplotlib中显示最大表面? [英] Display the maximum surface in matplotlib?

问题描述

我正在使用matplotlib在同一图形上绘制多个曲面，并且我只想看到最顶部的曲面，如matlab所示.

I'm plotting multiple surfaces on the same figure using matplotlib, and I'd like to see only the topmost surface, as matlab shows.

Matlab 3D视图:

Matlab 3D view:

Matlab顶视图:

Matlab top view:

Matplotlib 3D视图:

Matplotlib 3D view:

Matplotlib顶视图:

Matplotlib top view:

如何让Matplotlib显示类似于Matlab的结果，在该结果中，最顶层的类显示在顶部，而不是一个类优先于另一个类?

How can I get Matplotlib to show a result similar to Matlab, in which whatever topmost class is shown at the top, as opposed to one single class taking precedence over the other?

推荐答案

我很想考虑一些肮脏的骇客，例如在答案中提到了mgab，但后来决定走一条简单得多的路线:

I was goign to think about some dirty hacks like mgab mentions in their answer, but then decided just to go a considerably simpler route:

您完全可以通过使用透明度来获得相似的效果，只需确保透明度足够低，否则仍然会发生明显的重叠现象:

You can get a similar effect purely by using transparency, you just have to make sure the transparency is low enough, otherwise you still get obvious overlapping things happening:

from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import matplotlib.pyplot as plt
import numpy as np
from scipy.special import erf

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

X = np.arange(0, 6, 0.25)
Y = np.arange(0, 6, 0.25)
X, Y = np.meshgrid(X, Y)

Z1 = np.zeros_like(X)
Z2 = np.ones_like(X)

for i in range(len(X)):
  for j in range(len(X[0])):
    Z1[i,j] = 0.5*(erf((X[i,j]+Y[i,j]-4.5)*0.5)+1)
    Z2[i,j] = 0.5*(erf((-X[i,j]-Y[i,j]+4.5)*0.5)+1)


alpha = 0.25

surf1 = ax.plot_surface(X, Y, Z1, cstride=2, rstride=1, cmap=cm.Oranges, linewidth=0, antialiased=False, alpha=alpha)

surf2 = ax.plot_surface(X, Y, Z2, cstride=2, rstride=1, cmap=cm.Blues, linewidth=0, antialiased=False, alpha=alpha)

ax.zaxis.set_major_locator(LinearLocator(10))
ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))

fig.colorbar(surf1, shrink=0.5, aspect=5)
fig.colorbar(surf2, shrink=0.5, aspect=5)

plt.show()

添加一个相交线将是一个不错的选择，不过目前我还没有一种简单的方法可以添加它.

Adding an intersection line would be a nice addition, i don't have a simple way to add that in at the moment though.

使用mgab的桥"解决方案从mgab的答案中大量偷窃，然后还使用颜色映射表的表面，并通过使用RGBA元组将桥面设置为透明，您几乎可以得到完全一样的结果想要

Stealing heavily from mgab's answer, using his "bridge" solution, but then also using colour maps for the surfaces, and setting the bridge faces to be transparent by using RGBA tuples, you can get almost exactly what you want:

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

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

X = np.arange(0, 6, 0.25)
Y = np.arange(0, 6, 0.25)
X, Y = np.meshgrid(X, Y)

Z1 = np.empty_like(X)
Z2 = np.empty_like(X)
C1 = np.empty_like(X, dtype=object)
C2 = np.empty_like(X, dtype=object)

for i in range(len(X)):
  for j in range(len(X[0])):
    z1 = 0.5*(erf((X[i,j]+Y[i,j]-4.5)*0.5)+1)
    z2 = 0.5*(erf((-X[i,j]-Y[i,j]+4.5)*0.5)+1)
    Z1[i,j] = z1
    Z2[i,j] = z2

    # If you want to grab a colour from a matplotlib cmap function, 
    # you need to give it a number between 0 and 1. z1 and z2 are 
    # already in this range, so it just works.
    C1[i,j] = plt.get_cmap("Oranges")(z1)
    C2[i,j] = plt.get_cmap("Blues")(z2)


# Create a transparent bridge region
X_bridge = np.vstack([X[-1,:],X[-1,:]])
Y_bridge = np.vstack([Y[-1,:],Y[-1,:]])
Z_bridge = np.vstack([Z1[-1,:],Z2[-1,:]])
color_bridge = np.empty_like(Z_bridge, dtype=object)

color_bridge.fill((1,1,1,0)) # RGBA colour, onlt the last component matters.

# Join the two surfaces flipping one of them (using also the bridge)
X_full = np.vstack([X, X_bridge, np.flipud(X)])
Y_full = np.vstack([Y, Y_bridge, np.flipud(Y)])
Z_full = np.vstack([Z1, Z_bridge, np.flipud(Z2)])
color_full = np.vstack([C1, color_bridge, np.flipud(C2)])

surf_full = ax.plot_surface(X_full, Y_full, Z_full, rstride=1, cstride=1,
                            facecolors=color_full, linewidth=0,
                            antialiased=False)


plt.show()

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