python - 双曲 Curve_Fit 的预测区间 - SciPy

Question

如何使用 SciPy 的曲线拟合函数获得预测区间/预测带？

更具体地说，如何获得通常用于下降曲线分析的双曲曲线的这些预测带？

如有任何帮助，我们将不胜感激。

import pandas as pd
import numpy as np
from datetime import timedelta
from scipy.optimize import curve_fit

def hyperbolic_equation(t, qi, b, di):
    return qi/((1.0+b*di*t)**(1.0/b))


df1 = pd.DataFrame({ 'cumsum_days': [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],
        'prod': [800, 900, 1200, 700, 600, 
                 550, 500, 650, 625, 600,
                 550, 525, 500, 400, 350]})

qi = max(df1['prod'])

#Hyperbolic curve fit the data to get best fit equation
popt_hyp, pcov_hyp = curve_fit(hyperbolic_equation, df1['cumsum_days'], df1['prod'],bounds=(0, [qi,1,20]))

#Passing t to estimate the coefficients:

def fitted_hyperbolic_equation(t):
    return popt_hyp[0]/((1.0+popt_hyp[1]*popt_hyp[2]*t)**(1.0/popt_hyp[1]))

#Creating future time to predict on:
df2 = pd.DataFrame({ 'future_days': [16,17,18,19,20]})

fitted_hyperbolic_equation(df2.future_days)

16    388.259631
17    368.389649
18    349.754534
19    332.264306
20    315.836485

我有我的 future 值，但如何使用 SciPy 生成置信/预测带 (95%)？任何帮助将不胜感激。

Answer 1

我不确定我是否完全理解，但我认为您是在要求曲线拟合模型的预测值具有不确定性。

我建议使用lmfit(免责声明:我是作者)，因为它提供了执行此类计算的方法。恐怕你的模型和数据不太匹配，不确定性很大

使用lmfit并使用普通的numpy数组而不是pandas数据帧(这些可以使用，但它们在这里会分散注意力 -确实需要 numpy 数组)，您的分析可能如下所示:

import numpy as np
from lmfit import Model
import matplotlib.pyplot as plt

cumsum_days = np.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15])
prod = np.array([800, 900, 1200, 700, 600, 550, 500, 650, 625, 600, 550,
                 525, 500, 400, 350])

# plot data
plt.plot(cumsum_days, prod, 'bo', label='data')

def hyperbolic_equation(t, qi, b, di):
    return qi/((1.0+b*di*t)**(1.0/max(b, 1.e-50)))

# build Model
hmodel = Model(hyperbolic_equation)

# create lmfit Parameters, named from the arguments of `hyperbolic_equation`
# note that you really must provide initial values.
params = hmodel.make_params(qi=1000, b=0.5, di=0.1)

# set bounds on parameters
params['qi'].min=0
params['b'].min=0
params['di'].min=0

# do fit, print resulting parameters
result = hmodel.fit(prod, params, t=cumsum_days)
print(result.fit_report())

# plot best fit: not that great of fit, really
plt.plot(cumsum_days, result.best_fit, 'r--', label='fit')

# calculate the (1 sigma) uncertainty in the predicted model
# and plot that as a confidence band
dprod = result.eval_uncertainty(result.params, sigma=1)   
plt.fill_between(cumsum_days,
                 result.best_fit-dprod,
                 result.best_fit+dprod,
                 color="#AB8888",
                 label='uncertainty band of fit')

# now evaluate the model for other values, predicting future values
future_days = np.array([16,17,18,19,20])
future_prod = result.eval(t=future_days)

plt.plot(future_days, future_prod, 'k--', label='prediction')

# ...and calculate the 1-sigma uncertainty in the future prediction
# for 95% confidence level, you'd want to use `sigma=2` here:
future_dprod = result.eval_uncertainty(t=future_days, sigma=1)

print("### Prediction\n# Day  Prod     Uncertainty")

for day, prod, eps in zip(future_days, future_prod, future_dprod):
    print(" {:.1f}   {:.1f} +/- {:.1f}".format(day, prod, eps))

plt.fill_between(future_days,
                 future_prod-future_dprod,
                 future_prod+future_dprod,
                 color="#ABABAB",
                 label='uncertainty band of prediction')

plt.legend(loc='lower left')
plt.show()

这将打印出结果拟合统计数据和参数值

[[Model]]
    Model(hyperbolic_equation)
[[Fit Statistics]]
    # fitting method   = leastsq
    # function evals   = 21
    # data points      = 15
    # variables        = 3
    chi-square         = 238946.482
    reduced chi-square = 19912.2068
    Akaike info crit   = 151.139170
    Bayesian info crit = 153.263321
[[Variables]]
    qi:  993.608482 +/- 163.710950 (16.48%) (init = 1000)
    b:   0.22855837 +/- 2.07615175 (908.37%) (init = 0.5)
    di:  0.06551315 +/- 0.06250023 (95.40%) (init = 0.1)
[[Correlations]] (unreported correlations are < 0.100)
    C(b, di)  =  0.963
    C(qi, di) =  0.888
    C(qi, b)  =  0.771
### Prediction
# Day  Prod     Uncertainty
 16.0   388.258 +/- 1080.106
 17.0   368.387 +/- 1106.336
 18.0   349.752 +/- 1130.091
 19.0   332.261 +/- 1151.634
 20.0   315.833 +/- 1171.196

并给出如下图:

在您的问题中，您没有通过统计或图形方式检查拟合的质量。真的，你会想要这样做。

您还使用了curve_fit，但没有提供初始值。尽管没有底层拟合例程会支持这一点，并且都需要显式初始值，但 curve_fit 允许在没有警告或理由的情况下这样做，并断言所有起始值都将为 1.0。实际上，您必须提供初始值。