高效的python外产品 [英] Efficient outer product in python

问题描述

当我们必须处理维数为10k的向量时，python的外部乘积似乎很慢.有人可以给我一些想法，如何在python中加快该操作的速度吗?

Outer product in python seems quite slow when we have to deal with vectors of dimension of order 10k. Could someone please give me some idea how could I speed up this operation in python?

代码如下:

 In [8]: a.shape
 Out[8]: (128,)

 In [9]: b.shape
 Out[9]: (32000,)

 In [10]: %timeit np.outer(b,a)
 100 loops, best of 3: 15.4 ms per loop

由于我必须多次执行此操作，所以我的代码越来越慢.

Since I have to do this operation several times, my code is getting slower.

推荐答案

并没有比这更快的速度，这些是您的选择:

It doesn't really get any faster than that, these are your options:

numpy.outer

>>> %timeit np.outer(a,b)
100 loops, best of 3: 9.79 ms per loop

numpy.einsum

>>> %timeit np.einsum('i,j->ij', a, b)
100 loops, best of 3: 16.6 ms per loop

数字

from numba.decorators import autojit

@autojit
def outer_numba(a, b):
    m = a.shape[0]
    n = b.shape[0]
    result = np.empty((m, n), dtype=np.float)
    for i in range(m):
        for j in range(n):
            result[i, j] = a[i]*b[j]
    return result

>>> %timeit outer_numba(a,b)
100 loops, best of 3: 9.77 ms per loop

鹦鹉

from parakeet import jit

@jit
def outer_parakeet(a, b):
   ... same as numba

>>> %timeit outer_parakeet(a, b)
100 loops, best of 3: 11.6 ms per loop

cython

cimport numpy as np
import numpy as np
cimport cython
ctypedef np.float64_t DTYPE_t

@cython.boundscheck(False)
@cython.wraparound(False)
def outer_cython(np.ndarray[DTYPE_t, ndim=1] a, np.ndarray[DTYPE_t, ndim=1] b):
    cdef int m = a.shape[0]
    cdef int n = b.shape[0]
    cdef np.ndarray[DTYPE_t, ndim=2] result = np.empty((m, n), dtype=np.float64)
    for i in range(m):
        for j in range(n):
            result[i, j] = a[i]*b[j]
    return result

>>> %timeit outer_cython(a, b)
100 loops, best of 3: 10.1 ms per loop

theano

from theano import tensor as T
from theano import function

x = T.vector()
y = T.vector()

outer_theano = function([x, y], T.outer(x, y))

>>> %timeit outer_theano(a, b)
100 loops, best of 3: 17.4 ms per loop

pypy

# Same code as the `outer_numba` function
>>> timeit.timeit("outer_pypy(a,b)", number=100, setup="import numpy as np;a = np.random.rand(128,);b = np.random.rand(32000,);from test import outer_pypy;outer_pypy(a,b)")*1000 / 100.0
16.36 # ms

结论:

╔═══════════╦═══════════╦═════════╗
║  method   ║ time(ms)* ║ version ║
╠═══════════╬═══════════╬═════════╣
║ numba     ║ 9.77      ║ 0.16.0  ║
║ np.outer  ║ 9.79      ║ 1.9.1   ║
║ cython    ║ 10.1      ║ 0.21.2  ║
║ parakeet  ║ 11.6      ║ 0.23.2  ║
║ pypy      ║ 16.36     ║ 2.4.0   ║
║ np.einsum ║ 16.6      ║ 1.9.1   ║
║ theano    ║ 17.4      ║ 0.6.0   ║
╚═══════════╩═══════════╩═════════╝
* less time = faster

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