如何在 3d 轴上绘制
imshow()
图像?我正在尝试这个post。在那篇文章中,表面图看起来与imshow()
图相同,但实际上它们不是。为了演示,这里我使用了不同的数据:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
# create a 21 x 21 vertex mesh
xx, yy = np.meshgrid(np.linspace(0,1,21), np.linspace(0,1,21))
# create vertices for a rotated mesh (3D rotation matrix)
X = xx
Y = yy
Z = 10*np.ones(X.shape)
# create some dummy data (20 x 20) for the image
data = np.cos(xx) * np.cos(xx) + np.sin(yy) * np.sin(yy)
# create the figure
fig = plt.figure()
# show the reference image
ax1 = fig.add_subplot(121)
ax1.imshow(data, cmap=plt.cm.BrBG, interpolation='nearest', origin='lower', extent=[0,1,0,1])
# show the 3D rotated projection
ax2 = fig.add_subplot(122, projection='3d')
ax2.plot_surface(X, Y, Z, rstride=1, cstride=1, facecolors=plt.cm.BrBG(data), shade=False)
这是我的情节:
我认为您在 3D 与 2D 表面颜色方面的错误是由于表面颜色的数据归一化造成的。如果您将传递给
plot_surface
facecolor 的数据标准化,facecolors=plt.cm.BrBG(data/data.max())
结果更接近您的预期。
如果你只是想要一个垂直于坐标轴的切片,而不是使用
imshow
,你可以使用 contourf
,从 matplotlib 1.1.0 开始支持 3D,
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
from matplotlib import cm
# create a 21 x 21 vertex mesh
xx, yy = np.meshgrid(np.linspace(0,1,21), np.linspace(0,1,21))
# create vertices for a rotated mesh (3D rotation matrix)
X = xx
Y = yy
Z = 10*np.ones(X.shape)
# create some dummy data (20 x 20) for the image
data = np.cos(xx) * np.cos(xx) + np.sin(yy) * np.sin(yy)
# create the figure
fig = plt.figure()
# show the reference image
ax1 = fig.add_subplot(121)
ax1.imshow(data, cmap=plt.cm.BrBG, interpolation='nearest', origin='lower', extent=[0,1,0,1])
# show the 3D rotated projection
ax2 = fig.add_subplot(122, projection='3d')
cset = ax2.contourf(X, Y, data, 100, zdir='z', offset=0.5, cmap=cm.BrBG)
ax2.set_zlim((0.,1.))
plt.colorbar(cset)
plt.show()
此代码生成此图像:
虽然这不适用于 3D 中任意位置的切片,但 imshow 解决方案 更好。