Volume of convex hull with QHull from SciPy

问题

I'm trying to get the volume of the convex hull of a set of points using the SciPy wrapper for QHull.

According to the documentation of QHull, I should be passing the "FA" option to get the total surface area and volume.

Here is what I get.. What am I doing wrong?

> 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__']

回答1:

There does not seem to be any obvious way of directly getting the results you are after, regardless of what parameters you pass in. It shouldn't be too hard to compute yourself if, instead of ConvexHull, you use Delaunay (which also provides most of the convex hull related info).

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]))

EDIT As per the comments, the following are faster ways of obtaining the convex hull volume:

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]))

With some made up data, the second method seems to be about 2x faster, and numerical accuracy seems very good (15 decimal places!!!) although there has to be some much more pathological cases:

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

回答2:

Although this question has celebrated its second birthday, I would like to point out that now, the scipy wrapper automatically reports the volume (and area) computed by Qhull.

来源：https://stackoverflow.com/questions/24733185/volume-of-convex-hull-with-qhull-from-scipy

标签

python

scipy

convex-hull

qhull