将平面投影到新坐标系上

Question

说我有一个网格

xGrid = np.linspace(0.1, 1, 10)
yGrid = np.linspace(5, 10, 5)

以及该网格上的一些数据：

X, Y = np.meshgrid(xGrid, yGrid, indexing='ij')
Z = X*Y + 1

我现在可以轻松地绘制Z(x, y)。现在，有一个转换t(x, y)：

T = X+1+Y/2

，我想绘制Z(t(x, y), y)。为此，我需要将Z数据投影到t(x,y)-y平面上。最好的方法是什么？

由于我最终想要绘制数据并且不再做任何进一步的工作，因此也接受在matplotlib中执行此操作的直接方法（但实际上是绘制正确的新坐标，而不仅仅是重新标记刻度线）

Answer 1

You can use interpolation to compute the values in the projection, for example with scipy.interpolate.RectBivariateSpline:

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

xGrid = np.linspace(0.1, 1, 10)
yGrid = np.linspace(5, 10, 5)
X, Y = np.meshgrid(xGrid, yGrid, indexing='ij')
Z = X * Y + 1
T = X + 1 + Y / 2
# Interpolate values
interp = scipy.interpolate.RectBivariateSpline(xGrid, yGrid, Z)
Zt = interp.ev(T.ravel(), Y.ravel()).reshape(Z.shape)
# Plot
fig = plt.figure(figsize=(8, 10))
ax1 = fig.add_subplot(211, projection='3d')
ax1.set_title('Original')
ax1.plot_surface(X, Y, Z)
ax2 = fig.add_subplot(212, projection='3d')
ax2.set_title('Projected')
ax2.plot_surface(T, Y, Zt)
fig.tight_layout()

Output:

Answer 2

如果我了解您的问题，则可以使用pcolormesh来用于非常规网格

In [8]: import numpy as np 
   ...: import matplotlib.pyplot as plt 
   ...: from matplotlib.collections import PatchCollection, QuadMesh 
   ...: from matplotlib.patches import Rectangle 
   ...:  
   ...: np.random.seed(2018) 
   ...: xGrid = np.linspace(0.1, 1, 10) 
   ...: yGrid = np.linspace(5, 10, 6) 
   ...: X, Y = np.meshgrid(xGrid, yGrid, indexing='ij') 
   ...: Z = X*Y + 1 
   ...: T = X+1+Y/2 
   ...: Zt = T*Y + 1 
   ...: plt.pcolormesh(T, Y, Zt) 
   ...: plt.colorbar()                                                                           
Out[8]: <matplotlib.colorbar.Colorbar at 0x7fda83cd4ef0>

产生

如果乐队太丑陋，请使用plt.pcolormesh(T, Y, Zt, shading='gouraud')