散点图3D与标签和球体

Question

我试图用一些数据点(x,y,z,半径)制作一个散点图,这是我到目前为止的结果:

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

x = np.random.rand(20)
y = np.random.rand(20)
z = np.random.rand(20)
r = np.random.rand(20)


plt.rc('text', usetex=True)
plt.rcParams['text.latex.preamble']=[r"\usepackage{amsmath}"]

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

ax.scatter(x, y, z, s=np.pi*r**2*100, c='blue', alpha=0.75)

ax.set_xlabel(r'$x$ $\left[\frac{\text{Mpc}}{h}\right]$')
ax.set_ylabel(r'$y$ $\left[\frac{\text{Mpc}}{h}\right]$')
ax.set_zlabel(r'$z$ $\left[\frac{\text{Mpc}}{h}\right]$')

#plt.savefig('spheres.png')

plt.show()

如何改善此图以使x和y标签与抽搐没有重叠？

并且有可能在这个3D情节中制作球体而不是区域吗？

Answer 1

您可以创建由所描述的每个位置绘制一个使用领域,而不是圆标记的曲线在这里.这是一个例子:

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

def drawSphere(xCenter, yCenter, zCenter, r):
    #draw sphere
    u, v = np.mgrid[0:2*np.pi:20j, 0:np.pi:10j]
    x=np.cos(u)*np.sin(v)
    y=np.sin(u)*np.sin(v)
    z=np.cos(v)
    # shift and scale sphere
    x = r*x + xCenter
    y = r*y + yCenter
    z = r*z + zCenter
    return (x,y,z)


x = 10*np.random.rand(20)
y = 10*np.random.rand(20)
z = 10*np.random.rand(20)
r = np.random.rand(20)


fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# draw a sphere for each data point
for (xi,yi,zi,ri) in zip(x,y,z,r):
    (xs,ys,zs) = drawSphere(xi,yi,zi,ri)
    ax.plot_wireframe(xs, ys, zs, color="r")


plt.show()

要解决该标签定位问题,尝试添加第二行所描述的位置.