如何绘制随机飞机

Question

我正在使用以下代码绘制通过原点的三维随机飞机。

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

#Number of hyperplanes 
n = 20 
#Dimension of space 
d = 3 

plt3d = plt.figure().gca(projection='3d') 
for i in xrange(n): 
    #Create random point on unit sphere 
    v = np.random.normal(size = d) 
    v = v/np.sqrt(np.sum(v**2)) 
    # create x,y 
    xx, yy = np.meshgrid(range(-5,5), range(-5,5)) 
    z = (-v[0] * xx - v[1] * yy)/v[2] 
    # plot the surface 
    plt3d.plot_surface(xx, yy, z, alpha = 0.5) 
plt.show() 

但看着图片我不相信他们一律选择。我究竟做错了什么？

Answer 1

您的代码生成与随机分布的法线的平面内。他们只是不这样看，因为z尺度比x尺度和y尺度大得多。

您可以通过生成平面上均匀分布的点来生成更好看的图像。为此，按照 新坐标（u，v）对平面进行参数化，然后在均匀间隔的网格 （u，v）点上对平面进行采样。然后将这些（u，v）点转换为（x，y，z）空间中的点。

from __future__ import division 
import numpy as np 
import matplotlib.pyplot as plt 
from mpl_toolkits.mplot3d import Axes3D 
import math 
import itertools as IT 

def points_on_sphere(dim, N, norm=np.random.normal): 
    """ 
    http://en.wikipedia.org/wiki/N-sphere#Generating_random_points 
    """ 
    normal_deviates = norm(size=(N, dim)) 
    radius = np.sqrt((normal_deviates ** 2).sum(axis=0)) 
    points = normal_deviates/radius 
    return points 

# Number of hyperplanes 
n = 10 
# Dimension of space 
d = 3 

fig, ax = plt.subplots(subplot_kw=dict(projection='3d')) 
points = points_on_sphere(n, d).T 
uu, vv = np.meshgrid([-5, 5], [-5, 5], sparse=True) 
colors = np.linspace(0, 1, len(points)) 
cmap = plt.get_cmap('jet') 
for nhat, c in IT.izip(points, colors): 
    u = (0, 1, 0) if np.allclose(nhat, (1, 0, 0)) else np.cross(nhat, (1, 0, 0)) 
    u /= math.sqrt((u ** 2).sum()) 
    v = np.cross(nhat, u) 
    u = u[:, np.newaxis, np.newaxis] 
    v = v[:, np.newaxis, np.newaxis] 
    xx, yy, zz = u * uu + v * vv 
    ax.plot_surface(xx, yy, zz, alpha=0.5, color=cmap(c)) 
ax.set_xlim3d([-5,5]) 
ax.set_ylim3d([-5,5]) 
ax.set_zlim3d([-5,5])   
plt.show() 

或者，您也可以通过使用Till Hoffmann's pathpatch_2d_to_3d utility function避免毛毛数学：

for nhat, c in IT.izip(points, colors): 
    p = patches.Rectangle((-2.5, -2.5), 5, 5, color=cmap(c), alpha=0.5) 
    ax.add_patch(p) 
    pathpatch_2d_to_3d(p, z=0, normal=nhat) 

ax.set_xlim3d([-5,5]) 
ax.set_ylim3d([-5,5]) 
ax.set_zlim3d([-5,5])   
plt.show() 

Answer 2

看起来并不是一切。你最好再测量一次： - ]。它似乎是非随机分布的，因为你没有固定轴。因此，你会看到一架主飞机，由于规模，看起来非常相似，而且没有随机分布。

这个怎么样代码：

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

#Number of hyperplanes 
n = 20 
#Dimension of space 
d = 3 

plt3d = plt.figure().gca(projection='3d') 
for i in xrange(n): 
    #Create random point on unit sphere 
    v = np.random.normal(size = d) 
    v = v/np.sqrt(np.sum(v**2)) 
    # create x,y 
    xx, yy = np.meshgrid(range(-1,1), range(-1,1)) 
    z = (-v[0] * xx - v[1] * yy)/v[2] 
    # plot the surface 
    plt3d.plot_surface(xx, yy, z, alpha = 0.5) 

plt3d.set_xlim3d([-1,1]) 
plt3d.set_ylim3d([-1,1]) 
plt3d.set_zlim3d([-1,1]) 
plt.show() 

它不是完美的，但它似乎更随意，现在...

Answer 3

我想这一次，也许这是一个更好的方式来创建统一飞机。我随机选取两个不同的角度作为球面坐标系，并将其转换为笛卡尔坐标以获得平面的法向量。另外，当你绘制时，你应该知道你的飞机的中点不在原点上。

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

fig = plt.figure() 
ax = Axes3D(fig) 

for i in range(20): 
    theta = 2*np.pi*np.random.uniform(-1,1)  
    psi = 2*np.pi*np.random.uniform(-1,1) 
    normal = np.array([np.sin(theta)*np.cos(psi),np.sin(theta)*np.sin(psi), 
         np.cos(theta)]) 
    xx, yy = np.meshgrid(np.arange(-1,1), np.arange(-1,1)) 
    z = (-normal[0] * xx - normal[1] * yy)/normal[2] 
    ax.plot_surface(xx, yy, z, alpha=0.5)  

Answer 4

我建议你检查一下你的坐标轴。你的计算使Z轴的方式太大，这意味着你有一个荒谬偏见的观点。

首先检查你的法线均匀地涂在圈分布：

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

#Number of hyperplanes 
n = 1000 
#Dimension of space 
d = 3 

plt3d = plt.figure().gca(projection='3d') 
for i in xrange(n): 
    #Create random point on unit sphere 
    v = np.random.normal(size = d) 
    v = v/np.sqrt(np.sum(v**2)) 
    v *= 10 

    plt3d.scatter(v[0], v[1], v[2]) 

plt3d.set_aspect(1) 
plt3d.set_xlim(-10, 10) 
plt3d.set_ylim(-10, 10) 
plt3d.set_zlim(-10, 10) 

plt.show() 

然后检查是否被正确创建你的飞机：

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

#Number of hyperplanes 
n = 1 
#Dimension of space 
d = 3 

plt3d = plt.figure().gca(projection='3d') 
for i in xrange(n): 
    #Create random point on unit sphere 
    v = np.random.normal(size = d) 
    v = v/np.sqrt(np.sum(v**2)) 
    v *= 10 

    # create x,y 
    xx, yy = np.meshgrid(np.arange(-5,5,0.3), np.arange(-5,5,0.3)) 
    xx = xx.flatten() 
    yy = yy.flatten() 
    z = (-v[0] * xx - v[1] * yy)/v[2] 

    # Hack to keep the plane small 
    filter = xx**2 + yy**2 + z**2 < 5**2 
    xx = xx[filter] 
    yy = yy[filter] 
    z = z[filter] 

    # plot the surface 
    plt3d.scatter(xx, yy, z, alpha = 0.5) 

    for i in np.arange(0.1, 1, 0.1): 
     plt3d.scatter(i*v[0], i*v[1], i*v[2]) 

plt3d.set_aspect(1) 
plt3d.set_xlim(-10, 10) 
plt3d.set_ylim(-10, 10) 
plt3d.set_zlim(-10, 10) 

plt.show() 

然后你可以看到你实际上已经有了很好的结果！

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

#Number of hyperplanes 
n = 100 
#Dimension of space 
d = 3 

plt3d = plt.figure().gca(projection='3d') 
for i in xrange(n): 
    #Create random point on unit sphere 
    v = np.random.normal(size = d) 
    v = v/np.sqrt(np.sum(v**2)) 
    v *= 10 

    # create x,y 
    xx, yy = np.meshgrid(np.arange(-5,5,0.3), np.arange(-5,5,0.3)) 
    xx = xx.flatten() 
    yy = yy.flatten() 
    z = (-v[0] * xx - v[1] * yy)/v[2] 

    # Hack to keep the plane small 
    filter = xx**2 + yy**2 + z**2 < 5**2 
    xx = xx[filter] 
    yy = yy[filter] 
    z = z[filter] 

    # plot the surface 
    plt3d.scatter(xx, yy, z, alpha = 0.5) 

plt3d.set_aspect(1) 
plt3d.set_xlim(-10, 10) 
plt3d.set_ylim(-10, 10) 
plt3d.set_zlim(-10, 10) 

plt.show()