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python - 如何加速 numpy.all 和 numpy.nonzero()？

问题描述

我需要检查一个点是否位于边界长方体内。长方体的数量非常大（~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)

在我的电脑上运行需要 8 秒，运行需要 np.all3 秒。np.nonzero有什么想法可以加快代码速度吗？

标签： pythonnumpy

解决方案

我们可以沿着最小的轴长度减少内存拥塞all-reduction以获得-slicing3inside_cuboid

out = (points[0,0,:] > cuboid_min[:,0]) & (points[0,0,:] < cuboid_max[:,0]) & \
      (points[0,1,:] > cuboid_min[:,1]) & (points[0,1,:] < cuboid_max[:,1]) & \
      (points[0,2,:] > cuboid_min[:,2]) & (points[0,2,:] < cuboid_max[:,2])

计时 -

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
    ...: out = (points[0,0,:] > cuboid_min[:,0]) & (points[0,0,:] < cuboid_max[:,0]) & \
    ...:       (points[0,1,:] > cuboid_min[:,1]) & (points[0,1,:] < cuboid_max[:,1]) & \
    ...:       (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)

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