绘制立方体上的曲面

Question

我想将曲面绘制成一个立方体matplotlib.我正在尝试使用ax.plot_surface(X, Y, Z),但我有点困惑.它应该是什么X,Y并Z表示为2D数组？

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

points = np.array([[-1, -1, -1],
                      [1, -1, -1 ],
                      [1, 1, -1],
                      [-1, 1, -1],
                      [-1, -1, 1],
                      [1, -1, 1 ],
                      [1, 1, 1],
                      [-1, 1, 1]])

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
# ax.plot_surface(X, Y, Z)  # how?
ax.scatter3D(points[:, 0], points[:, 1], points[:, 2])
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
plt.show()

Answer 1

修复新的 matplotlib

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

points = np.array([[-1, -1, -1],
                      [1, -1, -1 ],
                      [1, 1, -1],
                      [-1, 1, -1],
                      [-1, -1, 1],
                      [1, -1, 1 ],
                      [1, 1, 1],
                      [-1, 1, 1]])

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
r = [-1,1]
X, Y = np.meshgrid(r, r)
one = np.ones(4).reshape(2, 2)
ax.plot_wireframe(X,Y,one, alpha=0.5)
ax.plot_wireframe(X,Y,-one, alpha=0.5)
ax.plot_wireframe(X,-one,Y, alpha=0.5)
ax.plot_wireframe(X,one,Y, alpha=0.5)
ax.plot_wireframe(one,X,Y, alpha=0.5)
ax.plot_wireframe(-one,X,Y, alpha=0.5)
ax.scatter3D(points[:, 0], points[:, 1], points[:, 2])
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
plt.show()

Answer 2

plot_surface要在绘制长方体时使用单个调用，可以在两个变量上对表面进行参数化，例如使用球坐标。

下面绘制一个边长为 a、b、c 的长方体：

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

def get_cube():   
    phi = np.arange(1,10,2)*np.pi/4
    Phi, Theta = np.meshgrid(phi, phi)

    x = np.cos(Phi)*np.sin(Theta)
    y = np.sin(Phi)*np.sin(Theta)
    z = np.cos(Theta)/np.sqrt(2)
    return x,y,z


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

a = 1
b = 2
c = 3
x,y,z = get_cube()


ax.plot_surface(x*a, y*b, z*c)

ax.set_xlim(-2,2)
ax.set_ylim(-2,2)
ax.set_zlim(-2,2)
plt.show()

另请参阅使用 matplotlib 表示体素

Answer 3

立方体的每个面都是一个表面,您可以自己定义每个角,或使用meshgrid:

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

points = np.array([[-1, -1, -1],
                      [1, -1, -1 ],
                      [1, 1, -1],
                      [-1, 1, -1],
                      [-1, -1, 1],
                      [1, -1, 1 ],
                      [1, 1, 1],
                      [-1, 1, 1]])

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
r = [-1,1]
X, Y = np.meshgrid(r, r)
ax.plot_surface(X,Y,1, alpha=0.5)
ax.plot_surface(X,Y,-1, alpha=0.5)
ax.plot_surface(X,-1,Y, alpha=0.5)
ax.plot_surface(X,1,Y, alpha=0.5)
ax.plot_surface(1,X,Y, alpha=0.5)
ax.plot_surface(-1,X,Y, alpha=0.5)
ax.scatter3D(points[:, 0], points[:, 1], points[:, 2])
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
plt.show()

X,Y和Z是(相同的)2D点列表:

>>> numpy.meshgrid([-1,1], [-1,1])
[array([[-1,  1],
       [-1,  1]]), array([[-1, -1],
       [ 1,  1]])]