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我想优化 SIR 模型的拟合。如果我只用 60 个数据点拟合 SIR 模型，我会得到一个“好”的结果。“好”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，也许基于所有数据点？

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：

使用随机 ydata 拟合模型

4

1 回答 1

8

我正在将我的评论转换为成熟的答案。

问题是由于模型设置不正确而引起的。为了简化微分方程，我将dS(t)/dt和分别dI(t)/dt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

通过不正确地建立微分方程，变化率，即从y(t)到y(t+dt)的变化是错误的。因此，您不仅获得了不正确积分的 I(t)，而且还进一步将其除以 N（或 k，正如您所说的那样），使其更加错误。

我们知道这些特定方程的耦合系统要求 S(t) + I(t) + R(t) = N，其中 N 是总体常数。从你声明初始条件的方式来看，我们推断 N 为 1。注意这也与max(ydata)小于 1 的 一致。

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

此外，你处理的方式k真的很可疑。您的数据似乎已经标准化，但您将其乘以 0.1 倍。如您所见，k = 1./sum(ydata), 与总体常数无关。通过I0 = ydata[0] * k将 I(t) 除以k，您可以有效地缩小数据规模，只是为了以后放大它们。无论总体常数是多少，这几乎都将 I(t) 限制在 0-1 范围内。

您可以通过简单地设置所有初始条件和未知参数并查看结果来验证您的模型是否错误odeint()。您会注意到 S(0)、I(0) 和 R(0) 可能与您给它们的值不对应，这是做错事的标志k。但是要发现有缺陷的动力学演化，您必须简单地查看您的模型。

在此处输入图像描述

一个 hacky 解决方案是设置k = 1.0. 一切正常，因为乘法和除法没有效果，即使你在技术上仍然在做错误的计算。但是，如果您的总体常数应该与 1 不同，那么一切都会中断。所以，为了彻底，

  • 手动设置k为总体常数，除非您也尝试适应 S0、I0 和/或 R0，否则您应该知道这一点。

  • S在模型中为和写出正确的变化率I。

  • 摆脱np.divide(array, k)您的任何计算，并且

  • k从参数中删除fitFunc()并且不要将其附加到param_init列表中。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，优化求解器会尝试为其找到最佳值，即使您最终没有在任何地方使用它来影响您的计算。

用curve_fit()解决同样的问题

如果要进行最小二乘拟合，可以使用curve_fit()，它在内部调用最小二乘方法。您仍然需要为拟合创建一个包装函数，该函数必须针对各种 beta 和 gamma 值对系统进行数值积分，但您不必手动进行任何 SSE 计算。

curve_fit()还将返回协方差矩阵，您可以使用它来估计拟合变量的置信区间。可以在此处找到有关从协方差矩阵计算置信区间的进一步相关讨论。

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求解器 (Levenburg-Marquardt) 的初始搜索，该求解器尝试了一些值，beta并gamma在积分期间导致数值溢出。但是，它应该尽快稳定到更合理的值。如果您尝试不同的求解器minimize()，您会注意到类似的警告。

于 2015-12-22T22:12:22.817 回答