在Python2.7中对循环有效

Question

我正在进行一项关于蚊帐对蚊子数量的影响以及蚊帐如何影响疟疾传播的研究。EIR告诉我们一个人一年中被传染的次数。为了计算这个，我必须在python中使用递归函数S（t）求和。级数在大约50次迭代后收敛，但是如果我将求和限制增加到20次以上，则需要很长的时间（大约一个小时）。我可能做错了什么，我会非常感谢你的帮助。下面显示了用python编写的代码。你知道吗

##IMPORTING PACKAGES
import sympy as sy
import matplotlib.pyplot as plt
import numpy as np
import profile

##Declaring the constants.
f = 1/3       ## frequency of feeding under zero net coverage
tau_1 = 0.69  ## time spent seeking blood under zero net coverage
tau_2 = 2.31  ## time spent resting during feeding cycle
Q_0 = 0.38    ## proportion of blood meals on humans under zero net coverage
mu_M0 = 0.096 ## daily mosquito mortality under zero net coverage
p_1 =0.91     ##probability of a mosquito survivng a blood meal at zero net coverage
p_2 = 0.74    ##probability of surviving resting phase at zero net coverage
phi_LLIN = 0.89   ##proportion of bites taken on humans when in bed
d_LLIN = 0.41     ##probability that a mosquito is killed by LLIN
s_LLIN = 0.03     ##probability that a mosquito feeds successfully with LLIN
r_LLIN = 0.56     ##probability that a mosquito is repelled by LLIN
phi_IRS = 0.97    ##proportion of bites taken on humans while indoors
psi = 0.86

#chi_LLIN = sy.symbols('chi_LLIN')
chi_LLIN=np.linspace(0.0,1.0,101)
w_LLIN =1. - Q_0*chi_LLIN*phi_LLIN*(1.-s_LLIN);w_LLIN

z_LLIN = Q_0 *chi_LLIN*phi_LLIN*r_LLIN; z_LLIN
tau_1chiLLIN = tau_1 /(1.-z_LLIN);tau_1chiLLIN
f_chiLLIN =1./ (tau_1chiLLIN + tau_2);f_chiLLIN
p1_chiLLIN = (p_1*w_LLIN)/(1.-(z_LLIN*p_1));p1_chiLLIN
p_chiLLIN = (p1_chiLLIN*p_2)**f_chiLLIN;p_chiLLIN
mu_MchiLLIN = -np.log(p_chiLLIN);mu_MchiLLIN

##Variable human blood index
Q_LLIN = Q_0 * ((1.+phi_LLIN*chi_LLIN*(s_LLIN-1.))/(1.+Q_0*phi_LLIN*chi_LLIN*(s_LLIN-1.) ))


##DEFINING E
totalx =39 ##limit of the summation
def E1(x):
    E =0.
    for t in xrange(1,x+1):
        E = E+((p_chiLLIN)**t)/f_chiLLIN
    return E


##DEFINING THE SPOROZOITE INFECTION PREVALENCE
n = 11 #the threshold
kappa=0.0297 ##human infectivity to mosquitoes

def S(x):
    if x<=n:    
        return 0
    elif x==n+1:    
        return 0.05
    else:
        return S(x-1) + ((kappa*Q_0*(1-S(x-1)))/f_chiLLIN)

B=0
for x in xrange(1,totalx+1):
    B = B + ((p_chiLLIN)**x)/f_chiLLIN*S(x)


beta_h = (Q_0/f_chiLLIN)*B

EIR = beta_h * E1(totalx)

EIR

Answer 1

def S(x):
    if x<=n:    
        return 0
    elif x==n+1:    
        return 0.05
    else:
        return S(x-1) + ((kappa*Q_0*(1-S(x-1)))/f_chiLLIN)
#              ^^^^^^                  ^^^^^^

在S(x)中，进行2次递归调用来计算S(x-1)，这4次递归调用来计算S(x-2)，这8次递归调用来计算S(x-3)，依此类推，直到达到递归的基本情况。您不需要重新计算太多内容；只需计算S(x-1)一次，然后使用值两次：

def S(x):
    if x<=n:    
        return 0
    elif x==n+1:    
        return 0.05
    else:
        s_lower = S(x-1)
        return s_lower + ((kappa*Q_0*(1-s_lower))/f_chiLLIN)

不过，这仍然会执行比所需更多的重新计算，因为每次调用S：

for x in xrange(1,totalx+1):
    B = B + ((p_chiLLIN)**x)/f_chiLLIN*S(x)

它必须再次将S的所有值向下计算到基本情况。考虑进一步的更改，例如memoizingS。你知道吗