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根据法线向量和Matlab或matplotlib中的点绘制平面

[英]Plot a plane based on a normal vector and a point in Matlab or matplotlib

如何从法向矢量和一个点在matlab或matplotlib中绘制平面?

For all the copy/pasters out there, here is similar code for Python using matplotlib: 对于那里的所有复制/粘贴,以下是使用matplotlib的Python类似代码:

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

point  = np.array([1, 2, 3])
normal = np.array([1, 1, 2])

# a plane is a*x+b*y+c*z+d=0
# [a,b,c] is the normal. Thus, we have to calculate
# d and we're set
d = -point.dot(normal)

# create x,y
xx, yy = np.meshgrid(range(10), range(10))

# calculate corresponding z
z = (-normal[0] * xx - normal[1] * yy - d) * 1. /normal[2]

# plot the surface
plt3d = plt.figure().gca(projection='3d')
plt3d.plot_surface(xx, yy, z)
plt.show()

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For Matlab: 对于Matlab:

point = [1,2,3];
normal = [1,1,2];

%# a plane is a*x+b*y+c*z+d=0
%# [a,b,c] is the normal. Thus, we have to calculate
%# d and we're set
d = -point*normal'; %'# dot product for less typing

%# create x,y
[xx,yy]=ndgrid(1:10,1:10);

%# calculate corresponding z
z = (-normal(1)*xx - normal(2)*yy - d)/normal(3);

%# plot the surface
figure
surf(xx,yy,z)

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Note: this solution only works as long as normal(3) is not 0. If the plane is parallel to the z-axis, you can rotate the dimensions to keep the same approach: 注意:仅当normal(3)不为0时,此解决方案才有效。如果平面平行于z轴,则可以旋转尺寸以保持相同的方法:

z = (-normal(3)*xx - normal(1)*yy - d)/normal(2); %% assuming normal(3)==0 and normal(2)~=0

%% plot the surface
figure
surf(xx,yy,z)

%% label the axis to avoid confusion
xlabel('z')
ylabel('x')
zlabel('y')

For copy-pasters wanting a gradient on the surface: 对于需要在表面上渐变的复制粘贴器:

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

point = np.array([1, 2, 3])
normal = np.array([1, 1, 2])

# a plane is a*x+b*y+c*z+d=0
# [a,b,c] is the normal. Thus, we have to calculate
# d and we're set
d = -point.dot(normal)

# create x,y
xx, yy = np.meshgrid(range(10), range(10))

# calculate corresponding z
z = (-normal[0] * xx - normal[1] * yy - d) * 1. / normal[2]

# plot the surface
plt3d = plt.figure().gca(projection='3d')

Gx, Gy = np.gradient(xx * yy)  # gradients with respect to x and y
G = (Gx ** 2 + Gy ** 2) ** .5  # gradient magnitude
N = G / G.max()  # normalize 0..1

plt3d.plot_surface(xx, yy, z, rstride=1, cstride=1,
                   facecolors=cm.jet(N),
                   linewidth=0, antialiased=False, shade=False
)
plt.show()

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The above answers are good enough. 以上答案已经足够了。 One thing to mention is, they are using the same method that calculate the z value for given (x,y). 值得一提的是,他们使用相同的方法来计算给定(x,y)的z值。 The draw back comes that they meshgrid the plane and the plane in space may vary (only keeping its projection the same). 缺点是他们将网格划分为网格,并且空间中的平面可能会发生变化(仅保持其投影不变)。 For example, you cannot get a square in 3D space (but a distorted one). 例如,您无法在3D空间中得到一个正方形(而是一个扭曲的正方形)。

To avoid this, there is a different way by using the rotation. 为避免这种情况,可以使用旋转方式。 If you first generate data in xy plane (can be any shape), then rotate it by equal amount ([0 0 1] to your vector) , then you will get what you want. 如果首先在xy平面上生成数据(可以是任何形状),然后将其旋转相等的量(对矢量[0 0 1]),那么您将获得所需的内容。 Simply run below code for your reference. 只需运行下面的代码以供参考。

point = [1,2,3];
normal = [1,2,2];
t=(0:10:360)';
circle0=[cosd(t) sind(t) zeros(length(t),1)];
r=vrrotvec2mat(vrrotvec([0 0 1],normal));
circle=circle0*r'+repmat(point,length(circle0),1);
patch(circle(:,1),circle(:,2),circle(:,3),.5);
axis square; grid on;
%add line
line=[point;point+normr(normal)]
hold on;plot3(line(:,1),line(:,2),line(:,3),'LineWidth',5)

It get a circle in 3D: 它会以3D形式显示一个圆圈: 产生的图片

A cleaner Python example that also works for tricky $z,y,z$ situations, 一个更干净的Python示例,也适用于棘手的$ z,y,z $情况,

from mpl_toolkits.mplot3d import axes3d
from matplotlib.patches import Circle, PathPatch
import matplotlib.pyplot as plt
from matplotlib.transforms import Affine2D
from mpl_toolkits.mplot3d import art3d
import numpy as np

def plot_vector(fig, orig, v, color='blue'):
   ax = fig.gca(projection='3d')
   orig = np.array(orig); v=np.array(v)
   ax.quiver(orig[0], orig[1], orig[2], v[0], v[1], v[2],color=color)
   ax.set_xlim(0,10);ax.set_ylim(0,10);ax.set_zlim(0,10)
   ax = fig.gca(projection='3d')  
   return fig

def rotation_matrix(d):
    sin_angle = np.linalg.norm(d)
    if sin_angle == 0:return np.identity(3)
    d /= sin_angle
    eye = np.eye(3)
    ddt = np.outer(d, d)
    skew = np.array([[    0,  d[2],  -d[1]],
                  [-d[2],     0,  d[0]],
                  [d[1], -d[0],    0]], dtype=np.float64)

    M = ddt + np.sqrt(1 - sin_angle**2) * (eye - ddt) + sin_angle * skew
    return M

def pathpatch_2d_to_3d(pathpatch, z, normal):
    if type(normal) is str: #Translate strings to normal vectors
        index = "xyz".index(normal)
        normal = np.roll((1.0,0,0), index)

    normal /= np.linalg.norm(normal) #Make sure the vector is normalised
    path = pathpatch.get_path() #Get the path and the associated transform
    trans = pathpatch.get_patch_transform()

    path = trans.transform_path(path) #Apply the transform

    pathpatch.__class__ = art3d.PathPatch3D #Change the class
    pathpatch._code3d = path.codes #Copy the codes
    pathpatch._facecolor3d = pathpatch.get_facecolor #Get the face color    

    verts = path.vertices #Get the vertices in 2D

    d = np.cross(normal, (0, 0, 1)) #Obtain the rotation vector    
    M = rotation_matrix(d) #Get the rotation matrix

    pathpatch._segment3d = np.array([np.dot(M, (x, y, 0)) + (0, 0, z) for x, y in verts])

def pathpatch_translate(pathpatch, delta):
    pathpatch._segment3d += delta

def plot_plane(ax, point, normal, size=10, color='y'):    
    p = Circle((0, 0), size, facecolor = color, alpha = .2)
    ax.add_patch(p)
    pathpatch_2d_to_3d(p, z=0, normal=normal)
    pathpatch_translate(p, (point[0], point[1], point[2]))


o = np.array([5,5,5])
v = np.array([3,3,3])
n = [0.5, 0.5, 0.5]

from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.gca(projection='3d')  
plot_plane(ax, o, n, size=3)    
ax.set_xlim(0,10);ax.set_ylim(0,10);ax.set_zlim(0,10)
plt.show()

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