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I am trying to convert my Python / Numpy code to Cython code for speedup purposes. However, Cython is MUCH slower (3-4 times) than the Python / Numpy code. Am I using Cython correctly? Am I passing arguments correctly to myc_rb_etc() in my Cython code? What about when I call the integrate function? Thank you in advance for your help. Here is my Python / Numpy code: 

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()

Here is the code converted into Cython:

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()

Here are some pstats from cProfile on my Python / Numpy code:

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)

4

2 回答 2

11

Change the function definition to include the types of the parameters:

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

This will improve the running time about 3 times over the numpy implementation and 6 - 7 times over your initial cython implementation. Just FYI, I lowered the number of iterations so that I didn't have to wait forever for it to finish while testing. It should scale up with your desired number of iterations.

[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

Edit:

Also, you don't have to create a new array each time you want to return the results from myc_rb_e2f. You can just declare a results array in main, pass it in on every call, and then fill it in. It will save you a lot of unnecessary allocation. This more than halves the previous best running time:

[pkerp@plastilin so]$ time python run_cython3.py

real    0m6.165s
user    0m4.818s
sys     0m0.152s

And the code:

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()
于 2012-10-24T08:18:53.357 回答
1

disclaimer: I'm one of the core dev of the sub-mentioned tool

As an alternative, extracting the myc_rb_e2f function to a single file and compiling it with pythran yields a x5 speedup on my side.

The extracted python code (only a type signature) added:

#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

compiled with:

> pythran myc.py

And imported with from myc import myc_rb_e2f in place of the function definition.

The original code runs in 103.46s on my laptop, the one calling the pythran-compiled routine runs in 22.23s.

于 2017-01-04T08:31:57.717 回答