我需要檢查一個點是否位于邊界長方體內(nèi)部。長方體的數(shù)量非常多（~4M）。我想出的代碼是：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)運行需要8秒， 在我的電腦上np.all運行需要3秒np.nonzero。有什么想法可以加快代碼速度嗎？
1 回答
白衣非少年
TA貢獻1155條經(jīng)驗 獲得超0個贊
我們可以減少內(nèi)存擁塞all-reduction沿著slicing的最小軸長度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])
時間安排 -
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)
添加回答
舉報
0/150
提交
取消