Question

如何使用python和scipy计算凸包的质心？我发现的只是计算面积和体积的方法。

的问候，坦率。

Answer 1

假设您使用scipy.spatial.ConvexHull构造了凸包，则返回的对象应该具有点的位置，因此质心可能很简单，

import numpy as np
from scipy.spatial import ConvexHull

points = np.random.rand(30, 2)   # 30 random points in 2-D
hull = ConvexHull(points)

#Get centoid
cx = np.mean(hull.points[hull.vertices,0])
cy = np.mean(hull.points[hull.vertices,1])

您可以如下绘制，

import matplotlib.pyplot as plt
#Plot convex hull
for simplex in hull.simplices:
    plt.plot(points[simplex, 0], points[simplex, 1], 'k-')

#Plot centroid
plt.plot(cx, cy,'x',ms=20)
plt.show()

scipy凸包基于Qhull，它应该有方法中心，来自Qhull docs，

中心是小平面超平面上的一个点。中心是小平面顶点的平均值。如果每个中心位于相邻小平面的超平面之下，则相邻小平面是凸的。

其中centrum与简单构面的质心相同，

对于具有d个顶点的单纯面，中心相当于质心或重心。

由于scipy似乎没有提供这个，你可以在子类中定义自己的hull，

class CHull(ConvexHull):

    def __init__(self, points):
        ConvexHull.__init__(self, points)

    def centrum(self):

        c = []
        for i in range(self.points.shape[1]):
            c.append(np.mean(self.points[self.vertices,i]))

        return c

 hull = CHull(points)
 c = hull.centrum()

Answer 2

要找到船体顶点的几何中心，只需使用

# Calculate geometric centroid of convex hull 
hull = ConvexHull(points)   
centroid = np.mean(points[hull.vertices, :], axis=0)

绘制船体试试，

import numpy as np
import pylab as pl
import scipy as sp
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d as a3

#  Plot polyhedra    
ax = a3.Axes3D(pl.figure())
facetCol = sp.rand(3)
for simplex in hull.simplices:
    vtx = [points[simplex[0],:], points[simplex[1],:], points[simplex[2],:]]
    tri = a3.art3d.Poly3DCollection([vtx], linewidths = 2, alpha = 0.8)
    tri.set_color(facetCol)
    tri.set_edgecolor('k')
    ax.add_collection3d(tri)

# Plot centroid      
ax.scatter(centroid0], centroid[1], centroid[2]) 
plt.axis('equal')
plt.axis('off')
plt.show()

Answer 3

如果船体的点在周长上不规则间隔，或者至少会给出倾斜的答案，那么简单的“均值”方法是不正确的。我使用的最好方法是计算船体点的Delaunay三角形的质心。这将对计算进行加权，以将形心计算为形状的COM，而不仅仅是顶点的平均值：


def _centroid_poly(poly):

    T = spatial.Delaunay(poly).simplices
    n = T.shape[0]
    W = np.zeros(n)
    C = 0

    for m in range(n):
        sp = poly[T[m,:],:]
        W[m] = spatial.ConvexHull(sp).volume
        C += W[m] +np.mean(sp, axis = 0)

    return C / np.sum(W)

类似的东西应该起作用

Answer 4

如果船体周围有不规则间隔的点，此解决方案也是正确的。

import numpy as np
from scipy.spatial import ConvexHull

def areaPoly(points):
    area = 0
    nPoints = len(points)
    j = nPoints - 1
    i = 0
    for point in points:
        p1 = points[i]
        p2 = points[j]
        area += (p1[0]*p2[1])
        area -= (p1[1]*p2[0])
        j = i
        i += 1

    area /= 2
    return area

def centroidPoly(points):
    nPoints = len(points)
    x = 0
    y = 0
    j = nPoints - 1
    i = 0

    for point in points:
        p1 = points[i]
        p2 = points[j]
        f = p1[0]*p2[1] - p2[0]*p1[1]
        x += (p1[0] + p2[0])*f
        y += (p1[1] + p2[1])*f
        j = i
        i += 1

    area = areaPoly(hull_points)
    f = area*6
    return [x/f, y/f]

# 'points' is an array of tuples (x, y)
points = np.array(points)
hull = ConvexHull(points)
hull_points = points[hull.vertices]
centroid = centroidPoly(hull_points)

Scipy：凸壳的质心

4 个答案: