我有一个相对简单的问题(我认为).我正在研究一段Cython代码,当给定应变和特定方向时(即,对于一定量的应变,平行于给定方向的半径),计算应变椭圆的半径.在每个程序运行期间,此函数被称为几百万次,并且分析显示该功能是性能方面的限制因素.这是代码:
# importing math functions from a C-library (faster than numpy)
from libc.math cimport sin, cos, acos, exp, sqrt, fabs, M_PI
cdef class funcs:
cdef inline double get_r(self, double g, double omega):
# amount of strain: g, angle: omega
cdef double l1, l2, A, r, g2, gs # defining some variables
if g == 0: return 1 # no strain means the strain ellipse is a circle
omega = omega*M_PI/180 # converting angle omega to radians
g2 = g*g
gs = g*sqrt(4 + g2)
l1 = 0.5*(2 + g2 + gs) # l1 and l2: eigenvalues of the Cauchy strain tensor
l2 = 0.5*(2 + g2 - gs)
A = acos(g/sqrt(g2 + (1 - l2)**2)) # orientation of the long axis of the ellipse
r = 1./sqrt(sqrt(l2)*(cos(omega - A)**2) + sqrt(l1)*(sin(omega - A)**2)) # the radius parallel to omega
return r # return of the jedi
运行此代码每次调用大约需要0.18微秒,我认为这对于这样一个简单的函数来说有点长.另外,math.h有一个square(x)函数,但是我不能从libc.math库中导入它,谁知道怎么做?有什么其他建议可以进一步改善这段小代码的性能吗?
更新2013/09/04:
似乎有更多的在发挥,而不是满足眼睛.当我分析一个调用get_r 10万次的函数时,我获得的性能与调用另一个函数不同.我添加了部分代码的更新版本.当我使用get_r_profile进行性能分析时,每次调用get_r得到0.073微秒,而MC_criterion_profile给我约0.164微秒/ get_r调用,50%的差异似乎与返回r的开销成本有关.
from libc.math cimport sin, cos, acos, exp, sqrt, fabs, M_PI
cdef class thesis_funcs:
cdef inline double get_r(self, double g, double omega):
cdef double l1, l2, A, r, g2, gs, cos_oa2, sin_oa2
if g == 0: return 1
omega = omega*SCALEDPI
g2 = g*g
gs = g*sqrt(4 + g2)
l1 = 0.5*(2 + g2 + gs)
l2 = l1 - gs
A = acos(g/sqrt(g2 + square(1 - l2)))
cos_oa2 = square(cos(omega - A))
sin_oa2 = 1 - cos_oa2
r = 1.0/sqrt(sqrt(l2)*cos_oa2 + sqrt(l1)*sin_oa2)
return r
@cython.profile(False)
cdef inline double get_mu(self, double r, double mu0, double mu1):
return mu0*exp(-mu1*(r - 1))
def get_r_profile(self): # Profiling through this guy gives me 0.073 microsec/call
cdef unsigned int i
for i from 0 <= i < 10000000:
self.get_r(3.0, 165)
def MC_criterion(self, double g, double omega, double mu0, double mu1, double C = 0.0):
cdef double r, mu, theta, res
r = self.get_r(g, omega)
mu = self.get_mu(r, mu0, mu1)
theta = 45 - omega
theta = theta*SCALEDPI
res = fabs(g*sin(2.0*theta)) - mu*(1 + g*cos(2.0*theta)) - C
return res
def MC_criterion_profile(self): # Profiling through this one gives 0.164 microsec/call
cdef double g, omega, mu0, mu1
cdef unsigned int i
omega = 165
mu0 = 0.6
mu1 = 2.0
g = 3.0
for i from 1 <= i < 10000000:
self.MC_criterion(g, omega, mu0, mu1)
我认为get_r_profile和MC_criterion之间可能存在根本区别,这会导致额外的开销成本.你能发现它吗?
最佳答案 根据你的评论,线路计算r是最昂贵的.如果是这种情况,那么我怀疑是触发性能的trig函数调用.
通过Pythagoras,cos(x)** 2 sin(x)** 2 == 1所以你可以通过计算跳过其中一个调用
cos_oa2 = cos(omega - A)**2
sin_oa2 = 1 - cos_oa2
r = 1. / sqrt(sqrt(l2) * cos_oa2 + sqrt(l1) * sin_oa2)
(或者可能翻转它们:在我的机器上,罪似乎比cos快.可能是一个NumPy小故障.)