Matplotlib 3D Quiver绘图单元向量指向随机方向的问题

Question

您好，我想生成一个指向 3D 空间中随机方向的单位向量，但我需要有关 quiver 3d 的帮助，因为这段代码给了我ValueError: need at least one array to concatenate。谢谢！

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
np.random.seed(420)
campione = 10 # I want generate 10 vectors to plot
phi = 2*np.pi * np.random.random(size=campione)
theta = np.arccos(1 - 2*np.random.random(size=campione))
vett3D = np.array([ np.sin(theta)*np.cos(phi), np.sin(theta)*np.sin(phi), np.cos(theta) ])    
origine = np.zeros((3,campione))
fig = plt.figure(num=None, figsize=(8, 6), dpi=800, facecolor='w', edgecolor='k')
ax = fig.gca(projection='3d')
ax.quiver(origine, vett3D)
plt.show()

Answer 1

以此为指导：https://matplotlib.org/stable/gallery/mplot3d/quiver3d.html

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(1)

campione = 10 # I want generate 10 vectors to plot

phi = np.pi * np.random.random(size=campione)
theta = 2 * np.pi * np.random.random(size=campione)

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

# Make the grid
points3d = np.random.uniform(low=-3, high=3, size=(campione, 3)) #Each row is a point x, y, z

# # Make the direction data for the arrows
u = np.cos(theta) * np.sin(phi) #Using spherical coordinates
v = np.sin(theta) * np.sin(phi)
w = np.cos(phi)

ax.quiver(points3d[:, 0], points3d[:, 1], points3d[:, 2], u, v, w, normalize=False)

ax.set_xlim(-4, 4)
ax.set_ylim(-4, 4)
ax.set_zlim(-5, 5)

plt.show()