我想为如下所示的 ODE 拟合一条曲线-
dA/dt = k1*profit + k2
我有观察到的变量 A
和 profit
的时间序列,我想获得 k1
和 的最优值>k2
在 python 中使用曲线拟合技术。我能够为此编写下面的代码,但解决方案不太合适,或者我的方法可能是错误的。
import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import odeint
def fitfunc(t, k1, k2):
'Function that returns A computed from an ODE for k1 and k2'
def myode(area, t):
profit = get_profit(t)
return k1*profit + k2
A0 = 200000 #initial value of A
out = odeint(myode, A0, t)
return out[:,0]
k_fit, kcov = curve_fit(fitfunc, time_span, obs_A) #time span from 1999-2019 and obs_A is the observed values of A
modeled_A = fitfunc(time_span, k_fit[0], k_fit[1])
20年期间的利润和obs_A数据为:
profit = [ 7.65976374e+06, -6.13172279e+06, 1.03946093e+07, 2.59937877e+06,
-7.88358386e+06, -1.38918115e+04, -3.13403157e+06, -4.74348806e+06,
1.87296164e+07, 4.13680709e+07, -1.77191198e+07, 2.39249499e+06,
1.38521564e+07, 6.52548348e+07, -5.78102494e+07, -5.72469988e+07,
-5.99056006e+06, -1.72424523e+07, 1.78509987e+07, 9.27860105e+06,
-9.96709853e+06]
obs_A = [200000., 165000., 150000., 180000., 190000., 195000., 200000.,
165000., 280000., 235000., 250000., 250000., 250000., 295000.,
295000., 285000., 245000., 315000., 235000., 245000., 305000.]
time_span = np.arange(1999,2020)
这里的 get_profit
是一个输出给定 t
利润值的函数,它是通过对观察到的 profit
数据进行插值而创建的,如下面-
profit_fun = interp1d(t, profit.values, 1, fill_value="extrapolate")
def get_profit(t):
return profit_fin(t)
我不确定如何在此处使用 profit
变量,因为它会在每个时间步发生变化。我的方法正确吗?
最佳答案
(根据要求,这是代码)
首先,进行设置。仅添加了 fitfun2
,它修改了 fitfunc
,删除了对 get_profit
的调用(因此不插入数据)。
import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import odeint
from scipy.interpolate import interp1d
import matplotlib.pyplot as plt
def fitfunc(t, k1, k2): # Original
'Function that returns A computed from an ODE for k1 and k2'
def myode(area, t):
profit = get_profit(t)
return k1*profit + k2
A0 = 20000 #initial value of A
out = odeint(myode, A0, t)
return out[:,0]
def fitfunc2(t, k1, k2): # Modified
'Modified fitfunc, removing the call to `profit_fun`'
def myode(area, t):
return k1*t+k2
A0 = 20000 #initial value of A
out = odeint(myode, A0, t)
return out[:,0]
profit = np.array([ 7.65976374e+06, -6.13172279e+06, 1.03946093e+07, 2.59937877e+06,
-7.88358386e+06, -1.38918115e+04, -3.13403157e+06, -4.74348806e+06,
1.87296164e+07, 4.13680709e+07, -1.77191198e+07, 2.39249499e+06,
1.38521564e+07, 6.52548348e+07, -5.78102494e+07, -5.72469988e+07,
-5.99056006e+06, -1.72424523e+07, 1.78509987e+07, 9.27860105e+06,
-9.96709853e+06])
obs_A = np.array([200000., 165000., 150000., 180000., 190000., 195000., 200000.,
165000., 280000., 235000., 250000., 250000., 250000., 295000.,
295000., 285000., 245000., 315000., 235000., 245000., 305000.])
time_span = np.arange(1999,2020)
profit_fun = interp1d(time_span, profit, 1, fill_value="extrapolate")
def get_profit(t):
return profit_fun(t)
现在,拟合并绘制结果
p0 = (1E-2, 1E4)
k_fit, kcov = curve_fit(fitfunc, time_span, obs_A, p0=p0)
k_fit2, kcov2 = curve_fit(fitfunc2, time_span, obs_A, p0=p0)
modeled_A = fitfunc(time_span, *k_fit)
guess_A = fitfunc(time_span, *p0)
modeled_A2 = fitfunc2(time_span, *k_fit2)
guess_A2 = fitfunc2(time_span, *p0)
plt.plot(time_span, obs_A, marker='o', lw=0, label='data')
plt.plot(time_span, modeled_A, label='model A (original)')
plt.plot(time_span, modeled_A2, label='model A (modified)')
plt.plot(time_span, guess_A, label='initial guess (original)')
plt.plot(time_span, guess_A2, label='initial guess (modified)')
plt.legend()
如前所述,修改参数 k
不会影响原始模型的曲线形状。它看起来仍然有点“逐步”。删除对 get_profit
的调用,曲线变得更加平滑,但我不知道这是不是@Isr729 所期望的。
关于python - 曲线拟合常微分方程,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/73042095/