在球体表面绘制点
问题描述 投票：0回答：3
我正在尝试生成球体图，其中一些点绘制在球体表面上。 （具体来说，这些点是 Lebedev 正交点）我希望我的图看起来与我在网上找到的这个类似：
我继续绘制球面，然后用散点图覆盖它。然而，这导致我的大部分点被下面的球体“吸收”，使得它们很难看到。看看：
如何防止我的点被球体遮挡？这是我用来生成该图的脚本：
import matplotlib.pyplot as plt
from matplotlib import cm, colors
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
# Create a sphere
r = 1
pi = np.pi
cos = np.cos
sin = np.sin
phi, theta = np.mgrid[0.0:pi:100j, 0.0:2.0*pi:100j]
x = r*sin(phi)*cos(theta)
y = r*sin(phi)*sin(theta)
z = r*cos(phi)
#Import data
data = np.genfromtxt('leb.txt')
xx, yy, zz = np.hsplit(data, 3)
#Set colours and render
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(
x, y, z, rstride=1, cstride=1, color='c', alpha=0.6, linewidth=0)
ax.scatter(xx,yy,zz,color="k",s=20)
ax.set_xlim([-1,1])
ax.set_ylim([-1,1])
ax.set_zlim([-1,1])
ax.set_aspect("equal")
plt.tight_layout()
#plt.show()
编辑
我找到了一种使用 Python 的 mayavi 来做到这一点的方法。这是我得到的：
这是我使用的代码：
from mayavi import mlab
import numpy as np
# Create a sphere
r = 1.0
pi = np.pi
cos = np.cos
sin = np.sin
phi, theta = np.mgrid[0:pi:101j, 0:2 * pi:101j]
x = r*sin(phi)*cos(theta)
y = r*sin(phi)*sin(theta)
z = r*cos(phi)
mlab.figure(1, bgcolor=(1, 1, 1), fgcolor=(0, 0, 0), size=(400, 300))
mlab.clf()
data = np.genfromtxt('leb.txt')
xx, yy, zz = np.hsplit(data, 3)
mlab.mesh(x , y , z, color=(0.0,0.5,0.5))
mlab.points3d(xx, yy, zz, scale_factor=0.05)
mlab.show()
3个回答
21
投票
投票
如果您认为这些点显示得不够好，您可以降低球体的 Alpha。但是，我认为您可能错误地将数据处理为 x、y、z 坐标。我从这里得到了一个点列表：http://people.sc.fsu.edu/~jburkardt/m_src/sphere_lebedev_rule_display/sphere_lebedev_rule_display.html，我的球体的点看起来有点像你的，直到我意识到文件包含 theta 和 phi 的值，我需要将度数转换为弧度。
import matplotlib.pyplot as plt
from matplotlib import cm, colors
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
# Create a sphere
r = 1
pi = np.pi
cos = np.cos
sin = np.sin
phi, theta = np.mgrid[0.0:pi:100j, 0.0:2.0*pi:100j]
x = r*sin(phi)*cos(theta)
y = r*sin(phi)*sin(theta)
z = r*cos(phi)
#Import data
data = np.genfromtxt('leb.txt')
theta, phi, r = np.hsplit(data, 3)
theta = theta * pi / 180.0
phi = phi * pi / 180.0
xx = sin(phi)*cos(theta)
yy = sin(phi)*sin(theta)
zz = cos(phi)
#Set colours and render
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(
x, y, z, rstride=1, cstride=1, color='c', alpha=0.3, linewidth=0)
ax.scatter(xx,yy,zz,color="k",s=20)
ax.set_xlim([-1,1])
ax.set_ylim([-1,1])
ax.set_zlim([-1,1])
ax.set_aspect("equal")
plt.tight_layout()
plt.show()
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1
投票
投票
尝试使用 zorder 参数。在下面给出的示例中，3D 线图将显示在 3D trisurf 图的顶部。 zorder 从 0 到 10 而不是 0 到 1 的原因在here给出。
plt_axes.plot_trisurf(x, y, z, shade=False, color='blue', cmap='Blues', zorder=0)
plt_axes.plot(x, y, z, marker='.', linestyle='None', label='Label', color='red', zorder=10)
0
投票
投票
基本上我使用了与您相同的代码，并将散点图的半径更改为1.05。那么它看起来很适合我。
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