【发布时间】：2016-05-07 13:09:55
【问题描述】：
我将 Python 加速器（Numba、Cython、f2py）与针对特定问题的简单 For 循环和 Numpy 的 einsum 进行比较（见下文）。到目前为止，Numpy 在这个问题上是最快的（速度快了 6 倍），但是如果我应该尝试其他优化，或者我做错了什么，我想要一些反馈。这个简单的代码基于一个更大的代码，它有许多这样的 einsum 调用，但没有显式的 for 循环。我正在检查这些加速器是否可以做得更好。
在 Mac OS X Yosemite 上使用 Python 2.7.9 完成计时，并从 Homebrew 安装 gcc-5.3.0 (--with-fortran --without-multilib)。也做了 %timeit 电话；这些单次调用时间相当准确。
In [1]: %run -i test_numba.py
test_numpy: 0.0805640220642
Matches Numpy output: True
test_dumb: 1.43043899536
Matches Numpy output: True
test_numba: 0.464295864105
Matches Numpy output: True
test_cython: 0.627640008926
Matches Numpy output: True
test_f2py: 5.01890516281
Matches Numpy output: True
test_f2py_order: 2.31424307823
Matches Numpy output: True
test_f2py_reorder: 0.507861852646
Matches Numpy output: True
主要代码：
import numpy as np
import numba
import time
import test_f2py as tf2py
import pyximport
pyximport.install(setup_args={'include_dirs':np.get_include()})
import test_cython as tcyth
def test_dumb(f,b):
fnew = np.empty((f.shape[1],f.shape[2]))
for i in range(f.shape[0]):
for l in range(f.shape[3]):
fnew += f[i,:,:,l] * b[i,l]
return fnew
def test_dumber(f,b):
fnew = np.empty((f.shape[1],f.shape[2]))
for i in range(f.shape[0]):
for j in range(f.shape[1]):
for k in range(f.shape[2]):
for l in range(f.shape[3]):
fnew[j,k] += f[i,j,k,l] * b[i,l]
return fnew
@numba.jit(nopython=True)
def test_numba(f,b):
fnew = np.zeros((f.shape[1],f.shape[2])) #NOTE: can't be empty, gives errors
for i in range(f.shape[0]):
for j in range(f.shape[1]):
for k in range(f.shape[2]):
for l in range(f.shape[3]):
fnew[j,k] += f[i,j,k,l] * b[i,l]
return fnew
def test_numpy(f,b):
return np.einsum('i...k,ik->...',f,b)
def test_f2py(f,b):
return tf2py.test_f2py(f,b)
def test_f2py_order(f,b):
return tf2py.test_f2py(f,b)
def test_f2py_reorder(f,b):
return tf2py.test_f2py_reorder(f,b)
def test_cython(f,b):
return tcyth.test_cython(f,b)
if __name__ == '__main__':
#goal is to create: fnew = sum f*b over dim 0 and 3.
f = np.random.rand(32,33,2000,64)
b = np.random.rand(32,64)
f1 = np.asfortranarray(f)
b1 = np.asfortranarray(b)
f2 = np.asfortranarray(np.transpose(f,[1,2,0,3]))
funcs = [test_dumb,test_numba, test_cython, \
test_f2py,test_f2py_order,test_f2py_reorder]
tstart = time.time()
fnew_numpy= test_numpy(f,b)
tstop = time.time()
print test_numpy.__name__+': '+str(tstop-tstart)
print 'Matches Numpy output: '+str(np.allclose(fnew_numpy,fnew_numpy))
print ''
for func in funcs:
tstart = time.time()
if func.__name__ == 'test_f2py_order':
fnew = func(f1,b1)
elif func.__name__ == 'test_f2py_reorder':
fnew = func(f2,b1)
else:
fnew = func(f,b)
tstop = time.time()
print func.__name__+': '+str(tstop-tstart)
print 'Matches Numpy output: '+str(np.allclose(fnew,fnew_numpy))
print ''
f2py 文件（使用 f2py -c -m test_f2py test_f2py.F90 编译）：
!file: test_f2py
subroutine test_f2py(f,b,fnew,n1,n2,n3,n4)
integer :: n1,n2,n3,n4
real(8), dimension(n1,n2,n3,n4) :: f
real(8), dimension(n1,n4) :: b
real(8), dimension(n2,n3) :: fnew
!f2py intent(in) f
!f2py intent(in) b
!f2py intent(out) fnew
!f2py intent(in) n1
!f2py intent(in) n2
!f2py intent(in) n3
!f2py intent(in) n4
integer :: i1,i2,i3,i4
do i1=1,n1
do i2=1,n2
do i3=1,n3
do i4=1,n4
fnew(i2,i3) = fnew(i2,i3) + f(i1,i2,i3,i4)*b(i1,i4)
enddo
enddo
enddo
enddo
end subroutine test_f2py
subroutine test_f2py_reorder(f,b,fnew,n1,n2,n3,n4)
integer :: n1,n2,n3,n4
real(8), dimension(n1,n2,n3,n4) :: f
real(8), dimension(n3,n4) :: b
real(8), dimension(n1,n2) :: fnew
!f2py intent(in) f
!f2py intent(in) b
!f2py intent(out) fnew
!f2py intent(in) n1
!f2py intent(in) n2
!f2py intent(in) n3
!f2py intent(in) n4
integer :: i1,i2,i3,i4
do i3=1,n3
do i4=1,n4
do i1=1,n1
do i2=1,n2
fnew(i1,i2) = fnew(i1,i2) + f(i1,i2,i3,i4)*b(i3,i4)
enddo
enddo
enddo
enddo
end subroutine test_f2py_reorder
还有 Cython .pyx 文件（在主程序中使用 pyximport 编译）：
#/usr/bin python
import numpy as np
cimport numpy as np
def test_cython(np.ndarray[np.float64_t,ndim=4] f, np.ndarray[np.float64_t,ndim=2] b):
# cdef np.ndarray[np.float64_t,ndim=4] f
# cdef np.ndarray[np.float64_t,ndim=2] b
cdef np.ndarray[np.float64_t,ndim=2] fnew = np.empty((f.shape[1],f.shape[2]),dtype=np.float64)
cdef int i,j,k,l
cdef int Ni = f.shape[0]
cdef int Nj = f.shape[1]
cdef int Nk = f.shape[2]
cdef int Nl = f.shape[3]
for i in range(Ni):
for j in range(Nj):
for k in range(Nk):
for l in range(Nl):
fnew[j,k] += f[i,j,k,l] * b[i,l]
return fnew
【问题讨论】：
-
既然你已经有了工作代码，你的问题可能更适合CodeReview.SE
-
在我的笔记本电脑 (OSX 10.9.5) 上运行 Numba 0.23.1
test_numpy()使用%timeit每次循环需要 75.5 毫秒，test_numba()每次循环需要 123 毫秒，因此差异不会看起来和你的测试一样极端。在对您调用一次的 numba 代码进行基准测试时要特别小心，以便在基准测试之外实际 jit 代码，否则您将在您的数字中包含该成本，而随后的每次调用都会快得多。
标签： python numpy cython numba f2py