如何加速 numpy.all 和 numpy.nonzero()？

Question

我需要检查一个点是否位于边界长方体内部。长方体的数量非常多（~4M）。我想出的代码是：

import numpy as np

# set the numbers of points and cuboids

n_points = 64

n_cuboid = 4000000

# generate the test data

points = np.random.rand(1, 3, n_points)*512

cuboid_min = np.random.rand(n_cuboid, 3, 1)*512

cuboid_max = cuboid_min + np.random.rand(n_cuboid, 3, 1)*8

# main body: check if the points are inside the cuboids

inside_cuboid = np.all((points > cuboid_min) & (points < cuboid_max), axis=1)

indices = np.nonzero(inside_cuboid)

运行需要8秒， 在我的电脑上np.all运行需要3秒np.nonzero。有什么想法可以加快代码速度吗？

Answer 1

我们可以减少内存拥塞all-reduction沿着slicing的最小轴长度3得到inside_cuboid-out = (points[0,0,:] > cuboid_min[:,0]) & (points[0,0,:] < cuboid_max[:,0]) & \&nbsp; &nbsp; &nbsp; (points[0,1,:] > cuboid_min[:,1]) & (points[0,1,:] < cuboid_max[:,1]) & \&nbsp; &nbsp; &nbsp; (points[0,2,:] > cuboid_min[:,2]) & (points[0,2,:] < cuboid_max[:,2])时间安排 -In [43]: %timeit np.all((points > cuboid_min) & (points < cuboid_max), axis=1)2.49 s ± 20 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)In [51]: %%timeit&nbsp; &nbsp; ...: out = (points[0,0,:] > cuboid_min[:,0]) & (points[0,0,:] < cuboid_max[:,0]) & \&nbsp; &nbsp; ...:&nbsp; &nbsp; &nbsp; &nbsp;(points[0,1,:] > cuboid_min[:,1]) & (points[0,1,:] < cuboid_max[:,1]) & \&nbsp; &nbsp; ...:&nbsp; &nbsp; &nbsp; &nbsp;(points[0,2,:] > cuboid_min[:,2]) & (points[0,2,:] < cuboid_max[:,2])1.95 s ± 10.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)