Cython代码比Python / Numpy代码慢3-4倍？
[英]Cython code 3-4 times slower than Python / Numpy code?
问题描述
我试图将我的Python / Numpy代码转换为Cython代码以加速。 但是，Cython比Python / Numpy代码慢很多（3-4倍）。 我正确使用Cython吗？ 我在Cython代码中正确地将参数传递给myc_rb_etc（）吗？ 当我调用集成函数时呢？ 预先感谢您的帮助。 这是我的Python / Numpy代码：
from pylab import *
import pylab as pl
from numpy import *
import numpy as np
from scipy import integrate
def myc_rb_e2f(y,t,k,d):
M = y[0]
E = y[1]
CD = y[2]
CE = y[3]
R = y[4]
RP = y[5]
RE = y[6]
S = 0.01
if t > 300:
S = 5.0
#if t > 400
#S = 0.01
t1 = k[0]*S/(k[7]+S);
t2 = k[1]*(M/(k[14]+M))*(E/(k[15]+E));
t3 = k[5]*M/(k[14]+M);
t4 = k[11]*CD*RE/(k[16]+RE);
t5 = k[12]*CE*RE/(k[17]+RE);
t6 = k[2]*M/(k[14]+M);
t7 = k[3]*S/(k[7]+S);
t8 = k[6]*E/(k[15]+E);
t9 = k[13]*RP/(k[18]+RP);
t10 = k[9]*CD*R/(k[16]+R);
t11 = k[10]*CE*R/(k[17]+R);
dM = t1-d[0]*M
dE = t2+t3+t4+t5-k[8]*R*E-d[1]*E
dCD = t6+t7-d[2]*CD
dCE = t8-d[3]*CE
dR = k[4]+t9-k[8]*R*E-t10-t11-d[4]*R
dRP = t10+t11+t4+t5-t9-d[5]*RP
dRE = k[8]*R*E-t4-t5-d[6]*RE
dy = [dM,dE,dCD,dCE,dR,dRP,dRE]
return dy
t = np.zeros(10000)
t = np.linspace(0.,3000.,10000.)
# Initial concentrations of [M,E,CD,CE,R,RP,RE]
y0 = np.array([0.,0.,0.,0.,0.4,0.,0.25])
E_simulated = np.zeros([10000,5000])
E_avg = np.zeros([10000])
k = np.zeros([19])
d = np.zeros([7])
for i in range (0,5000):
k[0] = 1.+0.1*randn(1)
k[1] = 0.15+0.05*randn(1)
k[2] = 0.2+0.05*randn(1)
k[3] = 0.2+0.05*randn(1)
k[4] = 0.35+0.05*randn(1)
k[5] = 0.001+0.0001*randn(1)
k[6] = 0.5+0.05*randn(1)
k[7] = 0.3+0.05*randn(1)
k[8] = 30.+5.*randn(1)
k[9] = 18.+3.*randn(1)
k[10] = 18.+3.*randn(1)
k[11] = 18.+3.*randn(1)
k[12] = 18.+3.*randn(1)
k[13] = 3.6+0.5*randn(1)
k[14] = 0.15+0.05*randn(1)
k[15] = 0.15+0.05*randn(1)
k[16] = 0.92+0.1*randn(1)
k[17] = 0.92+0.1*randn(1)
k[18] = 0.01+0.001*randn(1)
d[0] = 0.7+0.05*randn(1)
d[1] = 0.25+0.025*randn(1)
d[2] = 1.5+0.05*randn(1)
d[3] = 1.5+0.05*randn(1)
d[4] = 0.06+0.01*randn(1)
d[5] = 0.06+0.01*randn(1)
d[6] = 0.03+0.005*randn(1)
r = integrate.odeint(myc_rb_e2f,y0,t,args=(k,d))
E_simulated[:,i] = r[:,1]
for i in range(0,10000):
E_avg[i] = sum(E_simulated[i,:])/5000.
pl.plot(t,E_avg,'-ro')
pl.show()
这是转换成Cython的代码：
cimport numpy as np
import numpy as np
from numpy import *
import pylab as pl
from pylab import *
from scipy import integrate
def myc_rb_e2f(y,t,k,d):
cdef double M = y[0]
cdef double E = y[1]
cdef double CD = y[2]
cdef double CE = y[3]
cdef double R = y[4]
cdef double RP = y[5]
cdef double RE = y[6]
cdef double S = 0.01
if t > 300.0:
S = 5.0
#if t > 400
#S = 0.01
cdef double t1 = k[0]*S/(k[7]+S)
cdef double t2 = k[1]*(M/(k[14]+M))*(E/(k[15]+E))
cdef double t3 = k[5]*M/(k[14]+M)
cdef double t4 = k[11]*CD*RE/(k[16]+RE)
cdef double t5 = k[12]*CE*RE/(k[17]+RE)
cdef double t6 = k[2]*M/(k[14]+M)
cdef double t7 = k[3]*S/(k[7]+S)
cdef double t8 = k[6]*E/(k[15]+E)
cdef double t9 = k[13]*RP/(k[18]+RP)
cdef double t10 = k[9]*CD*R/(k[16]+R)
cdef double t11 = k[10]*CE*R/(k[17]+R)
cdef double dM = t1-d[0]*M
cdef double dE = t2+t3+t4+t5-k[8]*R*E-d[1]*E
cdef double dCD = t6+t7-d[2]*CD
cdef double dCE = t8-d[3]*CE
cdef double dR = k[4]+t9-k[8]*R*E-t10-t11-d[4]*R
cdef double dRP = t10+t11+t4+t5-t9-d[5]*RP
cdef double dRE = k[8]*R*E-t4-t5-d[6]*RE
dy = [dM,dE,dCD,dCE,dR,dRP,dRE]
return dy
def main():
cdef np.ndarray[double,ndim=1] t = np.zeros(10000)
t = np.linspace(0.,3000.,10000.)
# Initial concentrations of [M,E,CD,CE,R,RP,RE]
cdef np.ndarray[double,ndim=1] y0 = np.array([0.,0.,0.,0.,0.4,0.,0.25])
cdef np.ndarray[double,ndim=2] E_simulated = np.zeros([10000,5000])
cdef np.ndarray[double,ndim=2] r = np.zeros([10000,7])
cdef np.ndarray[double,ndim=1] E_avg = np.zeros([10000])
cdef np.ndarray[double,ndim=1] k = np.zeros([19])
cdef np.ndarray[double,ndim=1] d = np.zeros([7])
cdef int i
for i in range (0,5000):
k[0] = 1.+0.1*randn(1)
k[1] = 0.15+0.05*randn(1)
k[2] = 0.2+0.05*randn(1)
k[3] = 0.2+0.05*randn(1)
k[4] = 0.35+0.05*randn(1)
k[5] = 0.001+0.0001*randn(1)
k[6] = 0.5+0.05*randn(1)
k[7] = 0.3+0.05*randn(1)
k[8] = 30.+5.*randn(1)
k[9] = 18.+3.*randn(1)
k[10] = 18.+3.*randn(1)
k[11] = 18.+3.*randn(1)
k[12] = 18.+3.*randn(1)
k[13] = 3.6+0.5*randn(1)
k[14] = 0.15+0.05*randn(1)
k[15] = 0.15+0.05*randn(1)
k[16] = 0.92+0.1*randn(1)
k[17] = 0.92+0.1*randn(1)
k[18] = 0.01+0.001*randn(1)
d[0] = 0.7+0.05*randn(1)
d[1] = 0.25+0.025*randn(1)
d[2] = 1.5+0.05*randn(1)
d[3] = 1.5+0.05*randn(1)
d[4] = 0.06+0.01*randn(1)
d[5] = 0.06+0.01*randn(1)
d[6] = 0.03+0.005*randn(1)
r = integrate.odeint(myc_rb_e2f,y0,t,args=(k,d))
E_simulated[:,i] = r[:,1]
for i in range(0,10000):
E_avg[i] = sum(E_simulated[i,:])/5000.
pl.plot(t,E_avg,'-ro')
pl.show()
以下是我的Python / Numpy代码中cProfile的一些pstats：
ncalls tottime percall cumtime percall
5000 82.505 0.017 236.760 0.047 {scipy.integrate._odepack.odeint}
1 1.504 1.504 238.949 238.949 myc_rb_e2f.py:1(<module>)
5000 0.025 0.000 236.855 0.047 C:\\Python27\\lib\\site-packages\\scipy\\integrate\\odepack.py:18(odeint)
12291237 154.255 0.000 154.255 0.000 myc_rb_e2f.py:7(myc_rb_e2f)
2 个解决方案
解决方案1
10 已采纳 2012-10-24 08:18:53
更改函数定义以包括参数的类型：
def myc_rb_e2f(np.ndarray[double,ndim=1]y, double t, np.ndarray[double, ndim=1] k, np.ndarray[double, ndim=1] d):
这将使numpy实现的运行时间大约增加3倍，而初始cython实现的运行时间则增加6到7倍。 仅供参考，我降低了迭代次数，这样我就不必在测试时永远等待它完成。 它应该按您所需的迭代次数进行扩展。
[pkerp@plastilin so]$ time python run_numpy.py
real 0m47.572s
user 0m45.702s
sys 0m0.049s
[pkerp@plastilin so]$ time python run_cython1.py
real 1m14.851s
user 1m12.308s
sys 0m0.135s
[pkerp@plastilin so]$ time python run_cython2.py
real 0m15.774s
user 0m14.115s
sys 0m0.105s
编辑：
此外，每次要从myc_rb_e2f返回结果时，都不必创建新数组。 您可以在main声明一个结果数组，在每次调用时传入它，然后将其填入。它将为您节省大量不必要的分配。 这比以前的最佳运行时间减少了一半：
[pkerp@plastilin so]$ time python run_cython3.py
real 0m6.165s
user 0m4.818s
sys 0m0.152s
和代码：
cimport numpy as np
import numpy as np
from numpy import *
import pylab as pl
from pylab import *
from scipy import integrate
def myc_rb_e2f(np.ndarray[double,ndim=1]y, double t, np.ndarray[double, ndim=1] k, np.ndarray[double, ndim=1] d, np.ndarray[double, ndim=1] res):
cdef double S = 0.01
if t > 300.0:
S = 5.0
#if t > 400
#S = 0.01
cdef double t1 = k[0]*S/(k[7]+S)
cdef double t2 = k[1]*(y[0]/(k[14]+y[0]))*(y[1]/(k[15]+y[1]))
cdef double t3 = k[5]*y[0]/(k[14]+y[0])
cdef double t4 = k[11]*y[2]*y[6]/(k[16]+y[6])
cdef double t5 = k[12]*y[3]*y[6]/(k[17]+y[6])
cdef double t6 = k[2]*y[0]/(k[14]+y[0])
cdef double t7 = k[3]*S/(k[7]+S)
cdef double t8 = k[6]*y[1]/(k[15]+y[1])
cdef double t9 = k[13]*y[5]/(k[18]+y[5])
cdef double t10 = k[9]*y[2]*y[4]/(k[16]+y[4])
cdef double t11 = k[10]*y[3]*y[4]/(k[17]+y[4])
cdef double dM = t1-d[0]*y[0]
cdef double dE = t2+t3+t4+t5-k[8]*y[4]*y[1]-d[1]*y[1]
cdef double dCD = t6+t7-d[2]*y[2]
cdef double dCE = t8-d[3]*y[3]
cdef double dR = k[4]+t9-k[8]*y[4]*y[1]-t10-t11-d[4]*y[4]
cdef double dRP = t10+t11+t4+t5-t9-d[5]*y[5]
cdef double dRE = k[8]*y[4]*y[1]-t4-t5-d[6]*y[6]
res[0] = dM
res[1] = dE
res[2] = dCD
res[3] = dCE
res[4] = dR
res[5] = dRP
res[6] = dRE
return res
def main():
cdef np.ndarray[double,ndim=1] t = np.zeros(467)
cdef np.ndarray[double,ndim=1] results = np.zeros(7)
t = np.linspace(0.,3000.,467.)
# Initial concentrations of [M,E,CD,CE,R,RP,RE]
cdef np.ndarray[double,ndim=1] y0 = np.array([0.,0.,0.,0.,0.4,0.,0.25])
cdef np.ndarray[double,ndim=2] E_simulated = np.zeros([467,554])
cdef np.ndarray[double,ndim=2] r = np.zeros([467,7])
cdef np.ndarray[double,ndim=1] E_avg = np.zeros([467])
cdef np.ndarray[double,ndim=1] k = np.zeros([19])
cdef np.ndarray[double,ndim=1] d = np.zeros([7])
cdef int i
for i in range (0,554):
k[0] = 1.+0.1*randn(1)
k[1] = 0.15+0.05*randn(1)
k[2] = 0.2+0.05*randn(1)
k[3] = 0.2+0.05*randn(1)
k[4] = 0.35+0.05*randn(1)
k[5] = 0.001+0.0001*randn(1)
k[6] = 0.5+0.05*randn(1)
k[7] = 0.3+0.05*randn(1)
k[8] = 30.+5.*randn(1)
k[9] = 18.+3.*randn(1)
k[10] = 18.+3.*randn(1)
k[11] = 18.+3.*randn(1)
k[12] = 18.+3.*randn(1)
k[13] = 3.6+0.5*randn(1)
k[14] = 0.15+0.05*randn(1)
k[15] = 0.15+0.05*randn(1)
k[16] = 0.92+0.1*randn(1)
k[17] = 0.92+0.1*randn(1)
k[18] = 0.01+0.001*randn(1)
d[0] = 0.7+0.05*randn(1)
d[1] = 0.25+0.025*randn(1)
d[2] = 1.5+0.05*randn(1)
d[3] = 1.5+0.05*randn(1)
d[4] = 0.06+0.01*randn(1)
d[5] = 0.06+0.01*randn(1)
d[6] = 0.03+0.005*randn(1)
r = integrate.odeint(myc_rb_e2f,y0,t,args=(k,d,results))
E_simulated[:,i] = r[:,1]
for i in range(0,467):
E_avg[i] = sum(E_simulated[i,:])/554.
#pl.plot(t,E_avg,'-ro')
#pl.show()
if __name__ == "__main__":
main()
解决方案2
1 2017-01-04 08:31:57
免责声明：我是上述提到的工具的核心开发者之一
作为替代方案，将myc_rb_e2f函数解myc_rb_e2f到单个文件并使用pythran编译它会产生x5加速。
添加了提取的python代码（只有一个类型签名）：
#pythran export myc_rb_e2f(float[], float, float[], float[])
def myc_rb_e2f(y,t,k,d):
M = y[0]
E = y[1]
CD = y[2]
CE = y[3]
R = y[4]
RP = y[5]
RE = y[6]
S = 0.01
if t > 300:
S = 5.0
#if t > 400
#S = 0.01
t1 = k[0]*S/(k[7]+S);
t2 = k[1]*(M/(k[14]+M))*(E/(k[15]+E));
t3 = k[5]*M/(k[14]+M);
t4 = k[11]*CD*RE/(k[16]+RE);
t5 = k[12]*CE*RE/(k[17]+RE);
t6 = k[2]*M/(k[14]+M);
t7 = k[3]*S/(k[7]+S);
t8 = k[6]*E/(k[15]+E);
t9 = k[13]*RP/(k[18]+RP);
t10 = k[9]*CD*R/(k[16]+R);
t11 = k[10]*CE*R/(k[17]+R);
dM = t1-d[0]*M
dE = t2+t3+t4+t5-k[8]*R*E-d[1]*E
dCD = t6+t7-d[2]*CD
dCE = t8-d[3]*CE
dR = k[4]+t9-k[8]*R*E-t10-t11-d[4]*R
dRP = t10+t11+t4+t5-t9-d[5]*RP
dRE = k[8]*R*E-t4-t5-d[6]*RE
dy = [dM,dE,dCD,dCE,dR,dRP,dRE]
return dy
编译：
> pythran myc.py
并from myc import myc_rb_e2f代替函数定义。
原始代码在我的笔记本电脑上运行103.46s ，调用pythran编译的例程运行在22.23s 。
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