Question

我正在尝试将我的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()

以下是转换为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)

Answer 1

更改函数定义以包括参数类型：

def myc_rb_e2f(np.ndarray[double,ndim=1]y, double t, np.ndarray[double, ndim=1] k, np.ndarray[double, ndim=1] d):

这将比numpy实现大约3倍的运行时间和初始cython实现的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。它应该按照你想要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

修改

此外，每次要从myc_rb_e2f返回结果时，您都不必创建新数组。您可以在main中声明结果数组，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

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()

Answer 2

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函数解压缩到单个文件并使用pythran进行编译会产生x5加速。

添加了提取的python代码（只是一个类型签名）：

#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

使用from myc import myc_rb_e2f导入代替函数定义。

原始代码在我的笔记本电脑上的103.46s中运行，调用pythran编译例程的代码在22.23s中运行。

Cython代码比Python / Numpy代码慢3-4倍？

2 个答案: