Question

我想优化SIR模型的拟合。如果我只用60个数据点来拟合SIR模型，那么我会得到一个好的＆＃34;结果。＆＃34;良好的＆＃34;意味着，拟合的模型曲线接近数据点，直到t = 40。我的问题是，我怎样才能更好地适应，可能基于所有数据点？

ydata = ['1e-06', '1.49920166169172e-06', '2.24595472686361e-06', '3.36377954575331e-06', '5.03793663882291e-06', '7.54533628058909e-06', '1.13006564683911e-05', '1.69249500601052e-05', '2.53483161761933e-05', '3.79636391699325e-05', '5.68567547875179e-05', '8.51509649182741e-05', '0.000127522555808945', '0.000189928392105942', '0.000283447055673738', '0.000423064043409294', '0.000631295993246634', '0.000941024110897193', '0.00140281896645859', '0.00209085569326554', '0.00311449589149717', '0.00463557784224762', '0.00689146863803467', '0.010227347567051', '0.0151380084180746', '0.0223233100045688', '0.0327384810150231', '0.0476330618585758', '0.0685260046667727', '0.0970432959143974', '0.134525888779423', '0.181363340075877', '0.236189247803334', '0.295374180276257', '0.353377036130714', '0.404138746080267', '0.442876028839178', '0.467273954573897', '0.477529937494976', '0.475582401936257', '0.464137179474659', '0.445930281787152', '0.423331710456602', '0.39821360956389', '0.371967226561944', '0.345577884704341', '0.319716449520481', '0.294819942458255', '0.271156813453547', '0.24887641905719', '0.228045466022105', '0.208674420183194', '0.190736203926912', '0.174179448652951', '0.158937806544529', '0.144936441326754', '0.132096533873646', '0.120338367115739', '0.10958340819268', '0.099755679236243', '0.0907826241267504', '0.0825956203546979', '0.0751302384111894', '0.0683263295744258', '0.0621279977639921', '0.0564834809370572', '0.0513449852139111', '0.0466684871328814', '0.042413516167789', '0.0385429293775096', '0.035022685071934', '0.0318216204865132', '0.0289112368382048', '0.0262654939162707', '0.0238606155312519', '0.021674906523588', '0.0196885815912485', '0.0178836058829335', '0.0162435470852779', '0.0147534385851646', '0.0133996531928511', '0.0121697868544064', '0.0110525517526551', '0.0100376781867076', '0.00911582462544914', '0.00827849534575178', '0.00751796508841916', '0.00682721019158058', '0.00619984569061827', '0.00563006790443123', '0.00511260205894446', '0.00464265452957236', '0.00421586931435123', '0.00382828837833139', '0.00347631553734708', '0.00315668357532714', '0.00286642431380459', '0.00260284137520731', '0.00236348540287827', '0.00214613152062159', '0.00194875883295343']
ydata = [float(d) for d in ydata]
xdata = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23', '24', '25', '26', '27', '28', '29', '30', '31', '32', '33', '34', '35', '36', '37', '38', '39', '40', '41', '42', '43', '44', '45', '46', '47', '48', '49', '50', '51', '52', '53', '54', '55', '56', '57', '58', '59', '60', '61', '62', '63', '64', '65', '66', '67', '68', '69', '70', '71', '72', '73', '74', '75', '76', '77', '78', '79', '80', '81', '82', '83', '84', '85', '86', '87', '88', '89', '90', '91', '92', '93', '94', '95', '96', '97', '98', '99', '100', '101']
xdata = [float(t) for t in xdata]

from scipy.optimize import minimize
from scipy import integrate
import numpy as np
import pylab as pl

def fitFunc(sir_values, time, beta, gamma, k):
    s = sir_values[0]
    i = sir_values[1]
    r = sir_values[2]

    res = np.zeros((3))
    res[0] = - beta * s * i
    res[1] = beta * s * i - gamma * i
    res[2] = gamma * i
    return res

def lsq(model, xdata, ydata, n):
    """least squares"""
    time_total = xdata
    # original record data
    data_record = ydata
    # normalize train data
    k = 1.0/sum(data_record)
    # init t = 0 values + normalized
    I0 = data_record[0]*k
    S0 = 1 - I0
    R0 = 0 
    N0 = [S0,I0,R0]
    # Set initial parameter values
    param_init = [0.75, 0.75]
    param_init.append(k)
    # fitting
    param = minimize(sse(model, N0, time_total, k, data_record, n), param_init, method="nelder-mead").x
    # get the fitted model
    Nt = integrate.odeint(model, N0, time_total, args=tuple(param))
    # scale out
    Nt = np.divide(Nt, k)
    # Get the second column of data corresponding to I
    return Nt[:,1]

def sse(model, N0, time_total, k, data_record, n):
    """sum of square errors"""
    def result(x):
        Nt = integrate.odeint(model, N0, time_total[:n], args=tuple(x))
        INt = [row[1] for row in Nt]
        INt = np.divide(INt, k)
        difference = data_record[:n] - INt
        # square the difference
        diff = np.dot(difference, difference)
        return diff
    return result

result = lsq(fitFunc, xdata, ydata, 60)

# Plot data and fit
pl.clf()
pl.plot(xdata, ydata, "o")
pl.plot(xdata, result)
pl.show()

我期待这样的事情：

Answer 1

我将评论转换为完全成熟的答案。

问题源于错误地设置模型。为了简化微分方程，我将dS(t)/dt和dI(t)/dt分别称为S和I。

# incorrect
S = -S * I * beta
I = S * I * beta - I * gamma

# correct
S = -S * I * beta / N
I = S * I * beta / N - I * gamma

通过错误地设置微分方程，变化率，即从y（t）到y（t + dt）的变化是错误的。因此，您不仅会获得错误集成的I（t），而且还会将其除以N（或k，就像您所说的那样），从而使其更加错误。

我们知道这些特定方程的耦合系统需要S（t）+ I（t）+ R（t）= N，其中N是总体常数。从您声明初始条件的方式，我们推断N是1.请注意，这也与max(ydata)一致，小于1。

# IO + SO + R0 is always 1 regardless of "value"
I0 = value
S0 = 1 - I0
R0 = 0

此外，你处理k的方式真的很可疑。您的数据似乎已经标准化，但您将其乘以系数0.1。如您所见，k = 1./sum(ydata)与人口常数无关。通过执行I0 = ydata[0] * k并将I（t）除以k，您可以有效地缩小数据，以便稍后进行扩展。无论人口常数是多少，这都将I（t）限制在0-1范围内。

您只需设置所有初始条件和未知参数，并查看odeint()的内容，即可验证模型是否有误。您会注意到S（0），I（0）和R（0）可能与您给出的值不对应，这表示k做错了。但是要发现有缺陷的动态演变，你必须简单地检查你的模型。

一个hacky解决方案是设置k = 1.0。一切都有效，因为乘法和除法没有任何效果，即使你在技术上仍在做错误的计算。但是，如果你的人口常数应该与1不同，那么一切都会破裂。所以，为了彻底，

手动将k设置为总体常数，除非您还尝试适合S0，I0和/或R0，否则您应该知道它。
在S和I的模型中写下正确的更改率。
删除您拥有的任何np.divide(array, k)计算，
从k参数中删除fitFunc()，不要将其附加到param_init列表。虽然最后一个操作是可选的并且不会影响结果，但它在技术上仍然是正确的。这是因为通过传递k，优化解算器会尝试为其找到最佳值，即使您不会在任何地方使用它来影响您的计算。

使用curve_fit（）

解决同样的问题

如果你想进行最小二乘拟合，你可以使用curve_fit()，它在内部调用最小二乘法。您仍然需要为拟合创建一个包装函数，该函数必须在数字上将系统集成到各种beta和gamma值，但您不必手动进行任何SSE计算。

curve_fit()还会返回协方差矩阵，您可以使用该矩阵估算拟合变量的confidence intervals。可以找到关于从协方差矩阵计算置信区间的进一步相关讨论here。

import numpy as np
import matplotlib.pyplot as plt
from scipy import integrate, optimize

ydata = ['1e-06', '1.49920166169172e-06', '2.24595472686361e-06', '3.36377954575331e-06', '5.03793663882291e-06', '7.54533628058909e-06', '1.13006564683911e-05', '1.69249500601052e-05', '2.53483161761933e-05', '3.79636391699325e-05', '5.68567547875179e-05', '8.51509649182741e-05', '0.000127522555808945', '0.000189928392105942', '0.000283447055673738', '0.000423064043409294', '0.000631295993246634', '0.000941024110897193', '0.00140281896645859', '0.00209085569326554', '0.00311449589149717', '0.00463557784224762', '0.00689146863803467', '0.010227347567051', '0.0151380084180746', '0.0223233100045688', '0.0327384810150231', '0.0476330618585758', '0.0685260046667727', '0.0970432959143974', '0.134525888779423', '0.181363340075877', '0.236189247803334', '0.295374180276257', '0.353377036130714', '0.404138746080267', '0.442876028839178', '0.467273954573897', '0.477529937494976', '0.475582401936257', '0.464137179474659', '0.445930281787152', '0.423331710456602', '0.39821360956389', '0.371967226561944', '0.345577884704341', '0.319716449520481', '0.294819942458255', '0.271156813453547', '0.24887641905719', '0.228045466022105', '0.208674420183194', '0.190736203926912', '0.174179448652951', '0.158937806544529', '0.144936441326754', '0.132096533873646', '0.120338367115739', '0.10958340819268', '0.099755679236243', '0.0907826241267504', '0.0825956203546979', '0.0751302384111894', '0.0683263295744258', '0.0621279977639921', '0.0564834809370572', '0.0513449852139111', '0.0466684871328814', '0.042413516167789', '0.0385429293775096', '0.035022685071934', '0.0318216204865132', '0.0289112368382048', '0.0262654939162707', '0.0238606155312519', '0.021674906523588', '0.0196885815912485', '0.0178836058829335', '0.0162435470852779', '0.0147534385851646', '0.0133996531928511', '0.0121697868544064', '0.0110525517526551', '0.0100376781867076', '0.00911582462544914', '0.00827849534575178', '0.00751796508841916', '0.00682721019158058', '0.00619984569061827', '0.00563006790443123', '0.00511260205894446', '0.00464265452957236', '0.00421586931435123', '0.00382828837833139', '0.00347631553734708', '0.00315668357532714', '0.00286642431380459', '0.00260284137520731', '0.00236348540287827', '0.00214613152062159', '0.00194875883295343']
xdata = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23', '24', '25', '26', '27', '28', '29', '30', '31', '32', '33', '34', '35', '36', '37', '38', '39', '40', '41', '42', '43', '44', '45', '46', '47', '48', '49', '50', '51', '52', '53', '54', '55', '56', '57', '58', '59', '60', '61', '62', '63', '64', '65', '66', '67', '68', '69', '70', '71', '72', '73', '74', '75', '76', '77', '78', '79', '80', '81', '82', '83', '84', '85', '86', '87', '88', '89', '90', '91', '92', '93', '94', '95', '96', '97', '98', '99', '100', '101']

ydata = np.array(ydata, dtype=float)
xdata = np.array(xdata, dtype=float)

def sir_model(y, x, beta, gamma):
    S = -beta * y[0] * y[1] / N
    R = gamma * y[1]
    I = -(S + R)
    return S, I, R

def fit_odeint(x, beta, gamma):
    return integrate.odeint(sir_model, (S0, I0, R0), x, args=(beta, gamma))[:,1]

N = 1.0
I0 = ydata[0]
S0 = N - I0
R0 = 0.0

popt, pcov = optimize.curve_fit(fit_odeint, xdata, ydata)
fitted = fit_odeint(xdata, *popt)

plt.plot(xdata, ydata, 'o')
plt.plot(xdata, fitted)
plt.show()

您可能会注意到一些运行时警告，但它们主要是由于最小化解算器（Levenburg-Marquardt）的初始搜索，该搜索器会尝试导致beta和gamma的某些值集成期间数值溢出。但是，它应该尽快适应更合理的价值观。如果你为minimize()尝试不同的解算器，你会注意到类似的警告。

基于最小二乘拟合SIR模型

1 个答案:

使用curve_fit（）