我已经编写了代码来在 matplotlib 中绘制抛物线的 3D 表面。
我将如何旋转图形以使图形保持原位(即没有垂直或水平移动),但是它会围绕 y = 0 和 z = 0 线旋转一个 theta 角(我已经突出显示了图形应该围绕的线以绿色旋转)。这是一个插图,可帮助形象化我所描述的内容:
例如,如果图形围绕线旋转 180 度角,那么这将导致图形“颠倒”翻转,因此原点处的点现在将成为最大点。
我还想旋转轴以保持颜色图。下面是绘制图形的代码:
#parabaloid
import numpy as np
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
#creating grid
y = np.linspace(-1,1,1000)
x = np.linspace(-1,1,1000)
x,y = np.meshgrid(x,y)
#set z values
z = x**2+y**2
#label axes
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
#plot figure
ax.plot_surface(x,y,z,linewidth=0, antialiased=False, shade = True, alpha = 0.5)
plt.show()
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Agu*_*guy 16
像这样的东西?
ax.view_init(-140, 30)
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在您的plt.show()
命令之前插入它。
根据我的评论:
import mayavi.mlab as mlab
import numpy as np
x,y = np.mgrid[-1:1:0.001, -1:1:0.001]
z = x**2+y**2
s = mlab.mesh(x, y, z)
alpha = 30 # degrees
mlab.view(azimuth=0, elevation=90, roll=-90+alpha)
mlab.show()
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或遵循@Tamas 的回答:
#parabaloid
import numpy as np
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from math import sin, cos, pi
import matplotlib.cm as cm
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
#creating grid
y = np.linspace(-1,1,200)
x = np.linspace(-1,1,200)
x,y = np.meshgrid(x,y)
#set z values
z0 = x**2+y**2
# rotate the samples by pi / 4 radians around y
a = pi / 4
t = np.transpose(np.array([x,y,z0]), (1,2,0))
m = [[cos(a), 0, sin(a)],[0,1,0],[-sin(a), 0, cos(a)]]
x,y,z = np.transpose(np.dot(t, m), (2,0,1))
# or `np.dot(t, m)` instead `t @ m`
#label axes
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
#plot figure
ax.plot_surface(x,y,z,linewidth=0, antialiased=False, shade = True, alpha = 0.5, facecolors=cm.viridis(z0))
plt.show()
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