Python绘制平行六面体 [英] python draw parallelepiped

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问题描述

我正在尝试绘制一个平行六面体。其实我是从python脚本开始绘制一个多维数据集的:

  import numpy as mpl_toolkits.mplot3d import nx 
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]])


图= 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()

为了获得平行六面体,我将点矩阵乘以以下矩阵:

  P = 

[[2.06498904e-01 -6.30755443e-07 1.07477548e-03]

[1.61535574e-06 1.18897198e-01 7.85307721e-06]

[7.08353661e- 02 4.48415767e-06 2.05395893e-01]]

为:

  Z = np.zeros((8,3))

对于范围(8)中的i:

Z [i ,:] = np.dot(points [i,:],P)

Z = 10.0 * Z

我的想法如下:

  ax.scatter3D (Z [:, 0],Z [:, 1],Z [:, 2])

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')

plt.show()

这就是我得到的:





然后如何在这些不同的点上放置曲面以形成平行六面体(以上面的多维数据集的方式)?

解决方案

使用3D PolyCollection绘制表面(


I am trying to draw a parallelepiped. Actually I started from the python script drawing a cube as:

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()

In order to obtain a parallelepiped, I have multiplied the points matrix by the following matrix:

P = 

[[2.06498904e-01  -6.30755443e-07   1.07477548e-03]

 [1.61535574e-06   1.18897198e-01   7.85307721e-06]

 [7.08353661e-02   4.48415767e-06   2.05395893e-01]]

as:

Z = np.zeros((8,3))

for i in range(8):

   Z[i,:] = np.dot(points[i,:],P)

Z = 10.0*Z

My idea is then to represent as follows:

ax.scatter3D(Z[:, 0], Z[:, 1], Z[:, 2])

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')

plt.show()

And this is what I get:

How can I then put surfaces on these different points to form the parallelepiped (in the way of the cube above)?

解决方案

Plot surfaces with 3D PolyCollection (example)

import numpy as np
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection, Line3DCollection
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]])

P = [[2.06498904e-01 , -6.30755443e-07 ,  1.07477548e-03],
 [1.61535574e-06 ,  1.18897198e-01 ,  7.85307721e-06],
 [7.08353661e-02 ,  4.48415767e-06 ,  2.05395893e-01]]

Z = np.zeros((8,3))
for i in range(8): Z[i,:] = np.dot(points[i,:],P)
Z = 10.0*Z

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

r = [-1,1]

X, Y = np.meshgrid(r, r)
# plot vertices
ax.scatter3D(Z[:, 0], Z[:, 1], Z[:, 2])

# list of sides' polygons of figure
verts = [[Z[0],Z[1],Z[2],Z[3]],
 [Z[4],Z[5],Z[6],Z[7]], 
 [Z[0],Z[1],Z[5],Z[4]], 
 [Z[2],Z[3],Z[7],Z[6]], 
 [Z[1],Z[2],Z[6],Z[5]],
 [Z[4],Z[7],Z[3],Z[0]]]

# plot sides
ax.add_collection3d(Poly3DCollection(verts, 
 facecolors='cyan', linewidths=1, edgecolors='r', alpha=.25))

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')

plt.show()

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