我正在尝试使用SciPy wrapper for QHull获取一组点的凸包的量。
根据documentation of QHull，我应该传递"FA"选项以获取总表面积和体积。
这就是我得到的。我在做什么错？> pts
[(494.0, 95.0, 0.0), (494.0, 95.0, 1.0) ... (494.0, 100.0, 4.0), (494.0, 100.0, 5.0)]
> hull = spatial.ConvexHull(pts, qhull_options="FA")
> dir(hull)
['__class__', '__del__', '__delattr__', '__dict__', '__doc__', '__format__', '__getattribute__', '__hash__', '__init__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__', '_qhull', '_update', 'add_points', 'close', 'coplanar', 'equations', 'max_bound', 'min_bound', 'ndim', 'neighbors', 'npoints', 'nsimplex', 'points', 'simplices']
> dir(hull._qhull)
['__class__', '__delattr__', '__doc__', '__format__', '__getattribute__', '__hash__', '__init__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__']
最佳答案
不管传入什么参数，似乎都没有任何明显的方法可以直接获得所要获得的结果。如果您使用 ConvexHull (而不是Delaunay，它也提供了大多数功能，凸包的相关信息)。
def tetrahedron_volume(a, b, c, d):
return np.abs(np.einsum('ij,ij->i', a-d, np.cross(b-d, c-d))) / 6
from scipy.spatial import Delaunay
pts = np.random.rand(10, 3)
dt = Delaunay(pts)
tets = dt.points[dt.simplices]
vol = np.sum(tetrahedron_volume(tets[:, 0], tets[:, 1],
tets[:, 2], tets[:, 3]))
编辑根据评论，以下是获取凸包体积的更快方法:
def convex_hull_volume(pts):
ch = ConvexHull(pts)
dt = Delaunay(pts[ch.vertices])
tets = dt.points[dt.simplices]
return np.sum(tetrahedron_volume(tets[:, 0], tets[:, 1],
tets[:, 2], tets[:, 3]))
def convex_hull_volume_bis(pts):
ch = ConvexHull(pts)
simplices = np.column_stack((np.repeat(ch.vertices[0], ch.nsimplex),
ch.simplices))
tets = ch.points[simplices]
return np.sum(tetrahedron_volume(tets[:, 0], tets[:, 1],
tets[:, 2], tets[:, 3]))
使用一些组合数据，第二种方法似乎要快大约2倍，并且数值精度似乎非常好(小数点后15位!!!)，尽管病理情况要多得多:
pts = np.random.rand(1000, 3)
In [26]: convex_hull_volume(pts)
Out[26]: 0.93522518081853867
In [27]: convex_hull_volume_bis(pts)
Out[27]: 0.93522518081853845
In [28]: %timeit convex_hull_volume(pts)
1000 loops, best of 3: 2.08 ms per loop
In [29]: %timeit convex_hull_volume_bis(pts)
1000 loops, best of 3: 1.08 ms per loop