我检查了 QHull 版本(从上面)和线性规划解决方案(例如，请参阅 这个问题).到目前为止，使用 QHull 似乎是***的选择，尽管我可能会遗漏一些对 scipy.spatial LP 的优化.

I checked out the QHull version (from above) and the linear programming solution (e.g. see this question). So far, using QHull seems to be the best bet, although I might be missing some optimizations with the scipy.spatial LP.

import numpy
import numpy.random
from numpy import zeros, ones, arange, asarray, concatenate
from scipy.optimize import linprog

from scipy.spatial import ConvexHull

def pnt_in_cvex_hull_1(hull, pnt):
    '''
    Checks if `pnt` is inside the convex hull.
    `hull` -- a QHull ConvexHull object
    `pnt` -- point array of shape (3,)
    '''
    new_hull = ConvexHull(concatenate((hull.points, [pnt])))
    if numpy.array_equal(new_hull.vertices, hull.vertices): 
        return True
    return False


def pnt_in_cvex_hull_2(hull_points, pnt):
    '''
    Given a set of points that defines a convex hull, uses simplex LP to determine
    whether point lies within hull.
    `hull_points` -- (N, 3) array of points defining the hull
    `pnt` -- point array of shape (3,)
    '''
    N = hull_points.shape[0]
    c = ones(N)
    A_eq = concatenate((hull_points, ones((N,1))), 1).T   # rows are x, y, z, 1
    b_eq = concatenate((pnt, (1,)))
    result = linprog(c, A_eq=A_eq, b_eq=b_eq)
    if result.success and c.dot(result.x) == 1.:
        return True
    return False


points = numpy.random.rand(8, 3)
hull = ConvexHull(points, incremental=True)
hull_points = hull.points[hull.vertices, :]
new_points = 1. * numpy.random.rand(1000, 3)

哪里

%%time
in_hull_1 = asarray([pnt_in_cvex_hull_1(hull, pnt) for pnt in new_points], dtype=bool)

产生:

CPU times: user 268 ms, sys: 4 ms, total: 272 ms
Wall time: 268 ms

和

%%time
in_hull_2 = asarray([pnt_in_cvex_hull_2(hull_points, pnt) for pnt in new_points], dtype=bool)

生产

CPU times: user 3.83 s, sys: 16 ms, total: 3.85 s
Wall time: 3.85 s