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