快速(或更多)numpy花式索引编制和减少? [英] Fast(er) numpy fancy indexing and reduction?

问题描述

我正在尝试使用和加速花式索引来连接"两个数组并在结果轴之一上求和.

I'm trying to use and accelerate fancy indexing to "join" two arrays and sum over one of results' axis.

类似这样的东西:

$ ipython
In [1]: import numpy as np
In [2]: ne, ds = 12, 6
In [3]: i = np.random.randn(ne, ds).astype('float32')
In [4]: t = np.random.randint(0, ds, size=(1e5, ne)).astype('uint8')

In [5]: %timeit i[np.arange(ne), t].sum(-1)
10 loops, best of 3: 44 ms per loop

是否有一种简单的方法来加速In [5]中的语句?我应该使用OpenMP还是scipy.weave或Cython的prange之类的东西吗?

Is there a simple way to accelerate the statement in In [5] ? Should I go with OpenMP and something like scipy.weave or Cython's prange ?

推荐答案

numpy.take由于某种原因比花哨的索引要快得多.唯一的技巧是将数组视为平面.

numpy.take is much faster than fancy indexing for some reason. The only trick is that it treats the array as flat.

In [1]: a = np.random.randn(12,6).astype(np.float32)

In [2]: c = np.random.randint(0,6,size=(1e5,12)).astype(np.uint8)

In [3]: r = np.arange(12)

In [4]: %timeit a[r,c].sum(-1)
10 loops, best of 3: 46.7 ms per loop

In [5]: rr, cc = np.broadcast_arrays(r,c)

In [6]: flat_index = rr*a.shape[1] + cc

In [7]: %timeit a.take(flat_index).sum(-1)
100 loops, best of 3: 5.5 ms per loop

In [8]: (a.take(flat_index).sum(-1) == a[r,c].sum(-1)).all()
Out[8]: True

我认为，除此以外，您还将看到速度改进的唯一其他方法是使用

I think the only other way you're going to see much of a speed improvement beyond this would be to write a custom kernel for a GPU using something like PyCUDA.

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