Question

我在做什么：我修改了僵尸入侵系统中的代码，以演示应如何编写并尝试使用fmin函数优化最小二乘误差（定义为得分函数）。

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部分中，当我限制lb <= P，d，B，G，A <= ub等lb的“ rates”值时，有什么方法可以获得最佳的bestrates值。 =下限和ub =上限，并设法在该限制区域中获得最低分？它不一定是最优化的值。 fmin使用Nelder-Mead（简单）算法。

我对此很陌生，所以在正确方向上的任何帮助都会很棒。如有任何疑问，请随时提出疑问，我将尽我所能回答。。谢谢。

Answer 1

我不确定Adventures in Python : Fitting a Differential Equation System to Data的原始作者为什么会跳过箍以获取与给定数据点相对应的样本，通过将时间数组传递给ind <- order(df$var2, df$var1, decreasing = T) df$rank <- seq(nrow(df))[order(ind)] df # var1 var2 rank # 1 234 1456 1 # 2 24 456 3 # 3 34 456 2 # 4 68 343 4可以大大简化该过程而不是其施工参数

eq

此后可以称为

#=======================================================
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
#=======================================================

但也仅产生T = np.linspace(0, 5.0, 1000+1) S,Z,R=eq(rates,y0,T)函数所需的值

score

然后将分数函数简化为

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

现在，例如，如果您要强烈拒绝否定参数，则可以将返回值更改为

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:])

的确使所有参数为正（以前在我的笔记本中第一个参数为负）。

有没有一种方法可以优化输出以限制来自scipy.optimize的fmin中的参数

1 个答案: