Question

我正在尝试将我的实验数据与一个非常著名的药代动力学模型进行拟合。方程组非常复杂：

dC1/dt = k1*Cp - (k2+k3)*C1 + k4*C2
dC2/dt = k3*C1 - k4*C2
Ctissue = (1-vB)*(C1 + C2) + vB*Cp

vB是一个常数，Cp是一个数组（通过测量已经知道的因变量），k1，k2，k3，k4是不同部分之间的动力学常数，也是我希望从拟合中获得的参数。 Ctissue是我要适合真实数据的东西。 C1和C2是执行拟合后我应该能够计算的两个数组。有一个商业软件（PKIN）可以计算此方程组，因此我确定这是可能的，但我不知道如何用python实现。

这是我的代码

tini = np.array([  15.,   45.,   75.,  120.,  180.,  240.,  300.,  360.,  450.,
        570.,  690.,  810.,  930., 1080., 1260., 1440., 1650., 1890.,
       2130., 2400., 2700., 3000., 3300., 3525.])

Ctissue = np.array([  1.00229754,  25.06505484,  60.0265695 ,  82.87576127,
        68.07901198,  67.10795788,  81.42071546,  81.05644343,
       100.6740041 ,  90.43091176, 111.7861611 , 110.3851624 ,
       116.4682562 , 126.7390119 , 133.8460856 , 132.8657165 ,
       145.3951029 , 141.4012821 , 156.7317122 , 159.8293774 ,
       163.609847  , 175.7823822 , 168.5340708 , 171.5013387 ])

Cp = np.array([ 13.99461153, 559.5563251 , 914.7457277 , 782.4498718 ,
       574.7527458 , 521.4668956 , 412.9772775 , 421.5475443 ,
       403.2700613 , 368.6237412 , 355.8405377 , 340.0395723 ,
       306.9848032 , 295.0192494 , 295.0294368 , 240.9861338 ,
       245.9420067 , 217.3042524 , 229.6231028 , 196.4563327 ,
       190.8358096 , 190.161142  , 182.2021123 , 169.1384708 ])

vB = 0.05

# initial conditions
x10 = 0.1
x20 = 0.1
y0 = [x10, x20]
guess = [0.1,0.1,0.1,0.1]

import scipy as sp
import numpy as np
import pandas as pd
import matplotlib as mpl
from matplotlib import pyplot as plt
import math as m
from scipy.integrate import odeint
from lmfit import minimize, Parameters, Parameter, report_fit
from scipy.interpolate import interp1d

def myCp( t ):
    cp = interp1d( tini, Cp )
    if np.all(t < tini[0]):  
        out = Cp[0] 
    elif np.all(t > tini[-1]):
        out = 0 
    else:
        out = cp( t )
    return out

def f(y, t, paras):
#define differential equations
    x1 = y[0]
    x2 = y[1]

    try:
        k1 = paras['k1'].value
        k2 = paras['k2'].value
        k3 = paras['k3'].value 
        k4 = paras['k4'].value

    except KeyError:
        k1, k2, k3, k4 = paras
    f1 = k1*myCp( t ) - (k2+k3)*x1 + k4*x2
    f2 = k3*x1 - k4*x2
    return [f1, f2]

def g(t, x0, paras):
    x = odeint(f, x0, t, args=(paras,))
    return x

def tis2comp3(t, paras):
    x0 = params['x10'].value, params['x20'].value
    model = g(t, x0, paras)
    x1_model = model[:, 0]
    x2_model = model[:, 1]
    Ct = (1-vB)*(x1_model + x2_model) + vB*myCp( t )
    return Ct

def residual(paras, t, data):
    Ct = tis2comp3(t, params)
    return (Ct - data).ravel()

# set parameters
params = Parameters()
params.add('x10', value=x10, vary=False)
params.add('x20', value=x20, vary=False)
params.add('k1', value=guess[0], min=0.0001, max=2.)
params.add('k2', value=guess[1], min=0.0001, max=2.)
params.add('k3', value=guess[2], min=0.0001, max=2.)
params.add('k4', value=guess[3], min=0.0001, max=2.)

# fit model
result = minimize(residual, params, args=(tini, Ctissue), method='leastsq')  # leastsq nelder
# check results of the fit
xfit = np.linspace(15., 3525., 100)
yfit = tis2comp3(xfit, result.params)

#plot the final optimization results
figopt = plt.figure(figsize=(10,6))
lineini, = plt.plot(tini,Ctissue, 'b', linestyle='none', marker='o', markersize=7, label='data')
lineopt, = plt.plot(xfit,yfit, 'r-', label='optimized curve')
plt.legend(handles=[lineini,lineopt])

拟合运行平稳，但拟合曲线不令人满意。你们还有其他意见和建议吗？

Answer 1

您的问题似乎是，如果您查看{{1}的第一个输出，则使用Cp和准连续图像混合使用odeint和f的离散图像}在合适的迭代中，您会看到第一个（即f）是一个数组，而第二个是一个数字。因此存在概念错误。

将您的f1更改为以下内容：

应该工作。

当import numpy as np from matplotlib import pyplot as plt from scipy.integrate import odeint from scipy.interpolate import interp1d tini = np.array([ 15., 45., 75., 120., 180., 240., 300., 360., 450., 570., 690., 810., 930., 1080., 1260., 1440., 1650., 1890., 2130., 2400., 2700., 3000., 3300., 3525.]) Ctissue = np.array([ 1.00229754, 25.06505484, 60.0265695 , 82.87576127, 68.07901198, 67.10795788, 81.42071546, 81.05644343, 100.6740041, 90.43091176, 111.7861611 , 110.3851624 , 116.4682562, 126.7390119, 133.8460856 , 132.8657165 , 145.3951029, 141.4012821, 156.7317122 , 159.8293774 , 163.609847, 175.7823822, 168.5340708 , 171.5013387 ]) Cp = np.array([ 13.99461153, 559.5563251 , 914.7457277 , 782.4498718 , 574.7527458 , 521.4668956 , 412.9772775 , 421.5475443 , 403.2700613 , 368.6237412 , 355.8405377 , 340.0395723 , 306.9848032 , 295.0192494 , 295.0294368 , 240.9861338 , 245.9420067 , 217.3042524 , 229.6231028 , 196.4563327 , 190.8358096 , 190.161142 , 182.2021123 , 169.1384708 ]) vB = 0.05 def myCp( t ): cp = interp1d( tini, Cp ) if t < tini[0]: # does this makes sense out = Cp[0] # may require to be refined elif t > tini[-1]: out = Cp[-1] # same here. else: out = cp( t ) return out def f( y, t, paras ): #define differential equations x1 = y[0] x2 = y[1] k1, k2, k3, k4 = paras f1 = k1 * myCp( t ) - ( k2 + k3 ) * x1 + k4 * x2 f2 = k3 * x1 - k4 * x2 return [ f1, f2 ] paras=[.1, .12, .14, .15 ] sol = odeint( f, [ .1, .2], tini, args=( paras, ) ) print sol[ :, 0 ] print sol[ :, 1 ]检查超出odeint限制的值时，您必须弄清楚合理的推断是什么样子。

用lmfit的2组织3室模型拟合数据

1 个答案: