问题描述
限时送ChatGPT账号..我需要在一张图片中绘制许多小球和大球.以下代码有效,但运行时间非常长.
将 matplotlib.pyplot 导入为 plt从 mpl_toolkits.mplot3d 导入 Axes3D导入 numpyfig = plt.figure()ax = fig.gca(projection='3d')ax.set_aspect('相等')u = numpy.linspace(0, 2*numpy.pi, 100)v = numpy.linspace(0, numpy.pi, 100)x = numpy.outer(numpy.cos(u), numpy.sin(v))y = numpy.outer(numpy.sin(u), numpy.sin(v))z = numpy.outer(numpy.ones(numpy.size(u)), numpy.cos(v))对于范围内的 k(200):c = numpy.random.rand(3)r = numpy.random.rand(1)ax.plot_surface(r*x + c[0]、r*y + c[1]、r*z + c[2]、颜色='#1f77b4',阿尔法=0.5,线宽=0)plt.show()
我正在寻找更有效的解决方案.也许在 matplotlib 中有一个我没有找到的本地领域艺术家?
解决方案不,没有领域艺术家"这样的东西.即使有,绘制它也不会花费更少的时间.
您在问题中提出的解决方案是绘制许多球体的明智方法.但是,您可能需要考虑在球体上使用更少的点,
u = numpy.linspace(0, 2*numpy.pi, 12)v = numpy.linspace(0, numpy.pi, 7)
一个应该始终考虑的选项是不使用 matplotlib 进行 3D 绘图,因为它是
I need to draw many spheres, small and large, in one picture. The following code works, but takes awfully long to run.
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.set_aspect('equal')
u = numpy.linspace(0, 2*numpy.pi, 100)
v = numpy.linspace(0, numpy.pi, 100)
x = numpy.outer(numpy.cos(u), numpy.sin(v))
y = numpy.outer(numpy.sin(u), numpy.sin(v))
z = numpy.outer(numpy.ones(numpy.size(u)), numpy.cos(v))
for k in range(200):
c = numpy.random.rand(3)
r = numpy.random.rand(1)
ax.plot_surface(
r*x + c[0], r*y + c[1], r*z + c[2],
color='#1f77b4',
alpha=0.5,
linewidth=0
)
plt.show()
I'm looking for a more efficient solution. Perhaps there is a native sphere artist in matplotlib that I didn't find?
解决方案No, there is no such thing as a "sphere artist". And even if there was, it would not take less time to draw it.
The solution you present in the question is a sensible way to draw many spheres. However, you might want to consider using a lot less points on the sphere,
u = numpy.linspace(0, 2*numpy.pi, 12)
v = numpy.linspace(0, numpy.pi, 7)
An option one should always consider is not to use matplotlib for 3D plotting, as it is not actually been designed for it; and use Mayavi instead. The above in mayavi would look like
from mayavi import mlab
import numpy as np
[phi,theta] = np.mgrid[0:2*np.pi:12j,0:np.pi:12j]
x = np.cos(phi)*np.sin(theta)
y = np.sin(phi)*np.sin(theta)
z = np.cos(theta)
def plot_sphere(p):
r,a,b,c = p
return mlab.mesh(r*x+a, r*y+b, r*z+c)
for k in range(200):
c = np.random.rand(4)
c[0] /= 10.
plot_sphere(c)
mlab.show()
While the calculation takes a similar time, interactively zooming or panning is much faster in Mayavi.
Furthermore Mayavi actually provides something like a "sphere artist", which is called points3d
from mayavi import mlab
import numpy as np
c = np.random.rand(200,3)
r = np.random.rand(200)/10.
mlab.points3d(c[:,0],c[:,1],c[:,2],r)
mlab.show()
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