有没有一种方法可以优化scipy.optimize中fmin中限制参数的输出

Question

我正在做的事情：我修改了僵尸入侵系统中的代码，以演示如何编写它，并尝试使用fmin函数优化最小二乘误差(定义为分数函数(。

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
from scipy import integrate
from scipy.optimize import fmin
#=====================================================
#Notice we must import the Model Definition
from zombiewithdata import eq
#=====================================================
#1.Get Data
#====================================================
Td=np.array([0.5,1,1.5,2,2.2,3,3.5,4,4.5,5])#time
Zd=np.array([0,2,2,5,2,10,15,50,250,400])#zombie pop
#====================================================
#2.Set up Info for Model System
#===================================================
# model parameters
#----------------------------------------------------
P = 0       # birth rate
d = 0.0001  # natural death percent (per day)
B = 0.0095  # transmission percent  (per day)
G = 0.0001  # resurect percent (per day)
A = 0.0001  # destroy perecent (per day)
rates=(P,d,B,G,A)
# model initial conditions
#---------------------------------------------------
S0 = 500.               # initial population
Z0 = 0                  # initial zombie population
R0 = 0                  # initial death population
y0 = [S0, Z0, R0]      # initial condition vector
# model steps
#---------------------------------------------------
start_time=0.0
end_time=5.0
intervals=1000
mt=np.linspace(start_time,end_time,intervals)
# model index to compare to data
#----------------------------------------------------
findindex=lambda x:np.where(mt>=x)[0][0]
mindex=map(findindex,Td)
#=======================================================

#3.Score Fit of System
#=========================================================
def score(parms):
#a.Get Solution to system
F0,F1,F2,T=eq(parms,y0,start_time,end_time,intervals)
#b.Pick of Model Points to Compare
Zm=F1[mindex]
#c.Score Difference between model and data points
ss=lambda data,model:((data-model)**2).sum()
return ss(Zd,Zm)
#========================================================

#4.Optimize Fit
#=======================================================
fit_score=score(rates)
answ=fmin(score,(rates),full_output=1,maxiter=1000000)
bestrates=answ[0]
bestscore=answ[1]
P,d,B,G,A=answ[0]
newrates=(P,d,B,G,A)
#=======================================================
#5.Generate Solution to System
#=======================================================
F0,F1,F2,T=eq(newrates,y0,start_time,end_time,intervals)
Zm=F1[mindex]
Tm=T[mindex]
#======================================================

现在在#optimize fit部分，当我限制"rate"的值(如lb<=P、 d，B，G，A<=ub，其中lb=下限，ub=上限，并设法在该限制区域中获得最小分数？它不一定是最优化的值。fmin使用Nelder-Mead(单纯形(算法。

我对此还很陌生，所以在正确的方向上提供任何帮助都会很棒。请随时询问有关代码的任何疑问，我将尽我所知回答。非常感谢。

Answer 1

我不知道《Python历险记：将微分方程系统拟合到数据》的原作者为什么会跳过重重关卡来获取与给定数据点对应的样本，通过将时间数组而不是其构造参数传递给eq，可以大大简化该过程

#=======================================================
def eq(par,initial_cond,t):
#differential-eq-system----------------------
def funct(y,t):
Si, Zi, Ri=y
P,d,B,G,A=par
# the model equations (see Munz et al. 2009)
f0 = P - B*Si*Zi - d*Si
f1 = B*Si*Zi + G*Ri - A*Si*Zi
f2 = d*Si + A*Si*Zi - G*Ri
return [f0, f1, f2]
#integrate------------------------------------
ds = odeint(funct,initial_cond,t)
return ds.T
#=======================================================

这可以称为

T = np.linspace(0, 5.0, 1000+1)
S,Z,R=eq(rates,y0,T)

而且以仅产生CCD_ 2函数中所需的值的方式

Tm=np.append([0],Td)
Sm,Zm,Rm=eq(rates,y0,Tm)

这将分数函数简化为

def score(parms):
#a.Get Solution to system
Sm,Zm,Rm=eq(parms,y0,Tm)
#c.Score Difference between model and data points
ss=lambda data,model:((data-model)**2).sum()
return ss(Zd,Zm[1:])

现在，例如，如果你想强烈拒绝负参数，那么你可以将返回值更改为

return ss(Zd,Zm[1:]) + 1e6*sum(max(0,-x)**2 for x in parms)

这确实使所有参数都为正(之前在我的笔记本中有一个负的第一个参数(。