用Matplotlib绘制椭球 [英] Plotting Ellipsoid with Matplotlib

问题描述

有人有绘制椭圆体的示例代码吗? matplotlib站点上的球体只有一个，而椭球体则没有.我正在尝试

Does anyone have sample code for plotting ellipsoids? There is one for sphere on matplotlib site, but nothing for ellipsoids. I am trying to plot

x**2 + 2*y**2 + 2*z**2 = c

其中，c是定义椭圆形的常数(如10).我尝试了meshgrid(x,y)路线，对方程进行了重新设计，使z在一侧，但是sqrt是一个问题. matplotlib球面示例可与角度u,v一起使用，但是我不确定如何对椭球进行处理.

where c is a constant (like 10) that defines an ellipsoid. I tried the meshgrid(x,y) route, reworked the equation so z is on one side, but the sqrt is a problem. The matplotlib sphere example works with angles, u,v, but I am not sure how to work that for ellipsoid.

推荐答案

以下是如何通过球坐标进行的操作:

Here is how you can do it via spherical coordinates:

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

fig = plt.figure(figsize=plt.figaspect(1))  # Square figure
ax = fig.add_subplot(111, projection='3d')

coefs = (1, 2, 2)  # Coefficients in a0/c x**2 + a1/c y**2 + a2/c z**2 = 1 
# Radii corresponding to the coefficients:
rx, ry, rz = 1/np.sqrt(coefs)

# Set of all spherical angles:
u = np.linspace(0, 2 * np.pi, 100)
v = np.linspace(0, np.pi, 100)

# Cartesian coordinates that correspond to the spherical angles:
# (this is the equation of an ellipsoid):
x = rx * np.outer(np.cos(u), np.sin(v))
y = ry * np.outer(np.sin(u), np.sin(v))
z = rz * np.outer(np.ones_like(u), np.cos(v))

# Plot:
ax.plot_surface(x, y, z,  rstride=4, cstride=4, color='b')

# Adjustment of the axes, so that they all have the same span:
max_radius = max(rx, ry, rz)
for axis in 'xyz':
    getattr(ax, 'set_{}lim'.format(axis))((-max_radius, max_radius))

plt.show()

生成的图类似于

上面的程序实际上产生了看起来更漂亮的正方形"图形.

The program above actually produces a nicer looking "square" graphics.

此解决方案的灵感来自示例 http://matplotlib.sourceforge.net/gallery.html"rel =" noreferrer> Matplotlib的画廊.

This solution is strongly inspired from the example in Matplotlib's gallery.

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