使用python，numpy和matplotlib绘制蒙面曲面图

Question

我正在使用matplotlib 1.1.0绘制曲面。

绘图Z轴被掩盖如下：

Zm = ma.masked_where((abs(z_grid) < 1.09) & (abs(z_grid) > 0.91), (z_surface))
surf = ax.plot_surface(X, Y,Zm, rstride=2, cstride=2, cmap=colors,linewidth=0, antialiased=False)

但是我没有看到在剧情上应用的面具。我将面具本身绘制为子图

surf = ax.plot_surface(X, Y,ma.getmask(Zm), rstride=2, cstride=2, cmap=colors,linewidth=0, antialiased=False)

哪个有效，所以我知道我的面具实际上包含真值。

完整代码：

from pylab import *
import matplotlib.pyplot as plt
from matplotlib.widgets import Button
import numpy
from mpl_toolkits.mplot3d.axes3d import Axes3D
from  matplotlib import patches
from matplotlib.figure import Figure
from matplotlib import rcParams


fig = plt.figure(figsize=plt.figaspect(0.5))
ax = fig.add_subplot(1, 2, 1,projection='3d')

pole_positions_orig = [-0.6+0.73j];
zero_positions_orig = [0.29-0.41j];

surface_limit = 1.7;
min_val = -surface_limit;
max_val = surface_limit;

surface_resolution = 0.0333;

X = numpy.arange(min_val,max_val,surface_resolution)
Y = numpy.arange(min_val,max_val,surface_resolution)
X, Y = numpy.meshgrid(X, Y)

z_grid = X + Y*1j;
z_surface = z_grid*0;

pole_positions = numpy.round(pole_positions_orig,1) + surface_resolution/2+(surface_resolution/2)*1j;
zero_positions = numpy.round(zero_positions_orig,1) + surface_resolution/2 +(surface_resolution/2)*1j;

for k in range(0, len(zero_positions)):
    z_surface = z_surface + 20*log10((z_grid - zero_positions[k].real - zero_positions[k].imag*1j));
    z_surface = z_surface + 20*log10((z_grid - zero_positions[k].real + zero_positions[k].imag*1j));

for k in range(0, len(pole_positions)):
    z_surface = z_surface - 20*log10((z_grid - pole_positions[k].real - pole_positions[k].imag*1j));
    z_surface = z_surface - 20*log10((z_grid - pole_positions[k].real + pole_positions[k].imag*1j));    


colors = cm.jet;
colors.set_bad('k');


Zm = ma.masked_where((abs(z_grid) < 1.09) & (abs(z_grid) > 0.91), (z_surface))

z_surface = Zm;

surf = ax.plot_surface(X, Y,z_surface, rstride=2, cstride=2, cmap=colors,linewidth=0, antialiased=False)


ticks = [-1, 1]; 
z_ticks = [-30,-20,-10,0,10,20,30]; 
ax.set_xticks(ticks);
ax.set_yticks(ticks);   
ax.set_zticks(z_ticks);

ax.set_xlabel('Re')
ax.set_ylabel('Im')
ax.set_zlabel('Mag(db)',ha='left')
plt.setp(ax.get_zticklabels(), fontsize=7)
plt.setp(ax.get_xticklabels(), fontsize=7)  
plt.setp(ax.get_yticklabels(), fontsize=7)

ax = fig.add_subplot(1, 2, 2,projection='3d')
surf = ax.plot_surface(X, Y,ma.getmask(z_surface), rstride=2, cstride=2, cmap=colors,linewidth=0, antialiased=False)

ax.grid(b=None);
show();

这就是我所拥有的：

这就是我想要的（来自matlab）：

我错过了什么？

Answer 1

Fraxel提到surface_plot不支持屏蔽。为了解决这个问题，我就这样做了：

我基本上通过将每个屏蔽值设置为numpy.nan来手动屏蔽z轴数据，如下所示：

Zm = ma.masked_where((abs(z_grid) < 1.02) & (abs(z_grid) > 0.98), (z_surface))
z_surface[where(ma.getmask(Zm)==True)] = numpy.nan

然而，它搞砸了我的色彩图缩放。为了解决这个问题，我做了这个：

cmap = cm.jet
lev = numpy.arange(-30,30,1);
norml = colors.BoundaryNorm(lev, 256)

surf = ax.plot_surface(X, Y, z_surface,...,norm = norml)

不是我想要的100％，但仍然是一个很好的妥协。

Answer 2

你可以这样做，但你需要自己手工着色表面;

cmap函数占用0到1之间的nubmer，所以我们只需要在调用cmap函数之前对这些值进行规范化。

z_surface = numpy.real(z_surface)
min_z, max_z = z_surface.min(), z_surface.max()
colours = numpy.zeros_like(z_surface, dtype=object)

for i in range(len(z_surface)):
  for j in range(len(z_surface[0])):
    if 0.91 < numpy.sqrt(X[i,j]**2 + Y[i,j]**2) < 1.09:
      colours[i,j] = "red"  
    else:
      colours[i,j] = plt.get_cmap("jet")((z_surface[i,j]-min_z) / (max_z - min_z))


surf = ax.plot_surface(X, Y, z_surface, rstride=2, cstride=2, facecolors=colours, linewidth=0, antialiased=False)

我还应该指出matplotlib正在将你的z数组投射到真实状态 - 无论你是否有意利用这个，尽管我不知道。