加速Python/Cython循环。
我试图让python中的循环尽可能快地运行。所以我就跳进了NumPy和Cython。 以下是原始Python代码:加速Python/Cython循环。,python,arrays,performance,numpy,cython,Python,Arrays,Performance,Numpy,Cython,我试图让python中的循环尽可能快地运行。所以我就跳进了NumPy和Cython。 以下是原始Python代码: def calculate_bsf_u_loop(uvel,dy,dz): """ Calculate barotropic stream function from zonal velocity uvel (t,z,y,x) dy (y,x) dz (t,z,y,x) bsf (t,y,x) """ nt = uve
def calculate_bsf_u_loop(uvel,dy,dz):
"""
Calculate barotropic stream function from zonal velocity
uvel (t,z,y,x)
dy (y,x)
dz (t,z,y,x)
bsf (t,y,x)
"""
nt = uvel.shape[0]
nz = uvel.shape[1]
ny = uvel.shape[2]
nx = uvel.shape[3]
bsf = np.zeros((nt,ny,nx))
for jn in range(0,nt):
for jk in range(0,nz):
for jj in range(0,ny):
for ji in range(0,nx):
bsf[jn,jj,ji] = bsf[jn,jj,ji] + uvel[jn,jk,jj,ji] * dz[jn,jk,jj,ji] * dy[jj,ji]
return bsf
这只是k个指数的和。数组大小为nt=12、nz=75、ny=559、nx=1442,因此约有7.25亿个元素。
这花了68秒。现在,我已经在cython做过了
import numpy as np
cimport numpy as np
cimport cython
@cython.boundscheck(False) # turn off bounds-checking for entire function
@cython.wraparound(False) # turn off negative index wrapping for entire function
## Use cpdef instead of def
## Define types for arrays
cpdef calculate_bsf_u_loop(np.ndarray[np.float64_t, ndim=4] uvel, np.ndarray[np.float64_t, ndim=2] dy, np.ndarray[np.float64_t, ndim=4] dz):
"""
Calculate barotropic stream function from zonal velocity
uvel (t,z,y,x)
dy (y,x)
dz (t,z,y,x)
bsf (t,y,x)
"""
## cdef the constants
cdef int nt = uvel.shape[0]
cdef int nz = uvel.shape[1]
cdef int ny = uvel.shape[2]
cdef int nx = uvel.shape[3]
## cdef loop indices
cdef ji,jj,jk,jn
## cdef. Note that the cdef is followed by cython type
## but the np.zeros function as python (numpy) type
cdef np.ndarray[np.float64_t, ndim=3] bsf = np.zeros([nt,ny,nx], dtype=np.float64)
for jn in xrange(0,nt):
for jk in xrange(0,nz):
for jj in xrange(0,ny):
for ji in xrange(0,nx):
bsf[jn,jj,ji] += uvel[jn,jk,jj,ji] * dz[jn,jk,jj,ji] * dy[jj,ji]
return bsf
这花了49秒。
但是,将循环替换为
for jn in range(0,nt):
for jk in range(0,nz):
bsf[jn,:,:] = bsf[jn,:,:] + uvel[jn,jk,:,:] * dz[jn,jk,:,:] * dy[:,:]
只需要0.29秒!不幸的是,我不能在我的完整代码中这样做
为什么NumPy切片比Cython循环快得多?
我以为NumPy跑得很快,因为它是引擎盖下的Cython。那么它们不应该有相似的速度吗
如您所见,我在cython中禁用了边界检查,并且还使用“快速数学”进行编译。然而,这只会带来很小的加速。
是否有任何方法可以使循环的速度与NumPy切片的速度相似,或者循环总是比切片的速度慢
非常感谢您的帮助!
/Joakim鉴于您正在执行
元素乘法
,然后在4D
乘积数组的第二个轴上执行求和
,代码迫切需要用户的干预,这对于
allynumpy.einsum
以高效的方式进行。要解决您的问题,您可以通过两种方式使用numpy.einsum
-
bsf = np.einsum('ijkl,ijkl,kl->ikl',uvel,dz,dy)
bsf = np.einsum('ijkl,ijkl->ikl',uvel,dz)*dy
运行时测试和验证输出-
In [100]: # Take a (1/5)th of original input shapes
...: original_shape = [12,75, 559,1442]
...: m,n,p,q = (np.array(original_shape)/5).astype(int)
...:
...: # Generate random arrays with given shapes
...: uvel = np.random.rand(m,n,p,q)
...: dy = np.random.rand(p,q)
...: dz = np.random.rand(m,n,p,q)
...:
In [101]: bsf = calculate_bsf_u_loop(uvel,dy,dz)
In [102]: print(np.allclose(bsf,np.einsum('ijkl,ijkl,kl->ikl',uvel,dz,dy)))
True
In [103]: print(np.allclose(bsf,np.einsum('ijkl,ijkl->ikl',uvel,dz)*dy))
True
In [104]: %timeit calculate_bsf_u_loop(uvel,dy,dz)
1 loops, best of 3: 2.16 s per loop
In [105]: %timeit np.einsum('ijkl,ijkl,kl->ikl',uvel,dz,dy)
100 loops, best of 3: 3.94 ms per loop
In [106]: %timeit np.einsum('ijkl,ijkl->ikl',uvel,dz)*dy
100 loops, best of 3: 3.96 ms per loo
您忘记声明循环变量的类型。在当前Cython版本中不需要声明循环变量。@jakevdp在这种情况下,不声明它们似乎可以让它正确地推断类型,但如果不声明类型而将它们声明为
cdef ji,jj,jk,jn
,则会将其丢弃。对–我的意思是cdef ji,jj,jk,jn
行可以完全删除,这可能会提高执行时间。非常感谢!将循环索引声明为cdefintji、jj、jk、jn
产生了所有的不同!这是一个令人印象深刻的加速。我将研究einsum方法。谢谢