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如何加速 numpy.all 和 numpy.nonzero()?

如何加速 numpy.all 和 numpy.nonzero()?

肥皂起泡泡 2023-08-22 16:29:47
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 cuboidsn_points = 64n_cuboid = 4000000# generate the test datapoints = np.random.rand(1, 3, n_points)*512cuboid_min = np.random.rand(n_cuboid, 3, 1)*512cuboid_max = cuboid_min + np.random.rand(n_cuboid, 3, 1)*8# main body: check if the points are inside the cuboidsinside_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?
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白衣非少年

TA贡献1155条经验 获得超0个赞

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]) & \

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

      (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

    ...: 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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反对 回复 2023-08-22
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