Question

我的问题涉及统计和python，我是两者的初学者。我正在运行模拟，并且对于自变量（X）的每个值，我为因变量（Y）生成1000个值。我所做的是，我计算了X的每个值的平均Y，并使用scipy.optimize.curve_fit拟合这些平均值。曲线非常适合，但我想绘制置信区间。我不确定我所做的是否正确或者我想做什么都可以完成，但我的问题是如何从curve_fit产生的协方差矩阵中获得置信区间。代码首先从文件中读取平均值，然后只使用curve_fit。

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit


def readTDvsTx(L, B, P, fileformat):
    # L should be '_Fixed_' or '_'
    TD = []
    infile = open(fileformat.format(L, B, P), 'r')
    infile.readline()  # To remove header
    for line in infile:
        l = line.split()  # each line contains TxR followed by CD followed by TD
        if eval(l[0]) >= 70 and eval(l[0]) <=190:
            td = eval(l[2])
            TD.append(td)
    infile.close()
    tdArray = np.array(TD)

    return tdArray


def rec(x, a, b):
    return a * (1 / (x**2)) + b



fileformat = 'Densities_file{}BS{}_PRNTS{}.txt'
txR = np.array(range(70, 200, 20))
parents = np.array(range(1,6))
disc_p1 = readTDvsTx('_Fixed_', 5, 1, fileformat)


popt, pcov = curve_fit(rec, txR, disc_p1)


plt.plot(txR, rec(txR, popt[0], popt[1]), 'r-')
plt.plot(txR, disc_p1, '.')

print(popt)
plt.show()

以下是合适的结果：

Answer 1

这是一个快速错误的答案：您可以将a和b参数的协方差矩阵中的误差近似为其对角线的平方根：np.sqrt(np.diagonal(pcov))。然后可以使用参数不确定性来绘制置信区间。

答案是错误的，因为在将数据与模型拟合之前，您需要估算平均disc_p1点上的错误。平均时，您已经丢失了有关人口分布的信息，导致curve_fit认为您提供的y点是绝对且无可争议的。这可能会导致低估您的参数错误。

要估算平均Y值的不确定性，您需要估算它们的色散度量并将其传递给curve_fit，同时说明您的误差是绝对的。下面是一个如何对随机数据集执行此操作的示例，其中每个点包含从正态分布中提取的1000个样本。

from scipy.optimize import curve_fit
import matplotlib.pylab as plt
import numpy as np

# model function
func = lambda x, a, b: a * (1 / (x**2)) + b 

# approximating OP points
n_ypoints = 7 
x_data = np.linspace(70, 190, n_ypoints)

# approximating the original scatter in Y-data
n_nested_points = 1000
point_errors = 50
y_data = [func(x, 4e6, -100) + np.random.normal(x, point_errors,
          n_nested_points) for x in x_data]

# averages and dispersion of data
y_means = np.array(y_data).mean(axis = 1)
y_spread = np.array(y_data).std(axis = 1)

best_fit_ab, covar = curve_fit(func, x_data, y_means,
                               sigma = y_spread,
                               absolute_sigma = True)
sigma_ab = np.sqrt(np.diagonal(covar))

from uncertainties import ufloat
a = ufloat(best_fit_ab[0], sigma_ab[0])
b = ufloat(best_fit_ab[1], sigma_ab[1])
text_res = "Best fit parameters:\na = {}\nb = {}".format(a, b)
print(text_res)

# plotting the unaveraged data
flier_kwargs = dict(marker = 'o', markerfacecolor = 'silver',
                    markersize = 3, alpha=0.7)
line_kwargs = dict(color = 'k', linewidth = 1)
bp = plt.boxplot(y_data, positions = x_data,
                 capprops = line_kwargs,
                 boxprops = line_kwargs,
                 whiskerprops = line_kwargs,
                 medianprops = line_kwargs,
                 flierprops = flier_kwargs,
                 widths = 5,
                 manage_xticks = False)
# plotting the averaged data with calculated dispersion
#plt.scatter(x_data, y_means, facecolor = 'silver', alpha = 1)
#plt.errorbar(x_data, y_means, y_spread, fmt = 'none', ecolor = 'black')

# plotting the model
hires_x = np.linspace(50, 190, 100)
plt.plot(hires_x, func(hires_x, *best_fit_ab), 'black')
bound_upper = func(hires_x, *(best_fit_ab + sigma_ab))
bound_lower = func(hires_x, *(best_fit_ab - sigma_ab))
# plotting the confidence intervals
plt.fill_between(hires_x, bound_lower, bound_upper,
                 color = 'black', alpha = 0.15)
plt.text(140, 800, text_res)
plt.xlim(40, 200)
plt.ylim(0, 1000)
plt.show()

修改如果您没有考虑数据点上的固有错误，那么使用＆＃34; qiuck和错误＆＃34;我之前提到过的案例。然后可以使用协方差矩阵的对角线条目的平方根来计算置信区间。但请注意，由于我们已经放弃了不确定性，因此置信区间缩小了：

from scipy.optimize import curve_fit import matplotlib.pylab as plt import numpy as np func = lambda x, a, b: a * (1 / (x**2)) + b n_ypoints = 7 x_data = np.linspace(70, 190, n_ypoints) y_data = np.array([786.31, 487.27, 341.78, 265.49, 224.76, 208.04, 200.22]) best_fit_ab, covar = curve_fit(func, x_data, y_data) sigma_ab = np.sqrt(np.diagonal(covar)) # an easy way to properly format parameter errors from uncertainties import ufloat a = ufloat(best_fit_ab[0], sigma_ab[0]) b = ufloat(best_fit_ab[1], sigma_ab[1]) text_res = "Best fit parameters:\na = {}\nb = {}".format(a, b) print(text_res) plt.scatter(x_data, y_data, facecolor = 'silver', edgecolor = 'k', s = 10, alpha = 1) # plotting the model hires_x = np.linspace(50, 200, 100) plt.plot(hires_x, func(hires_x, *best_fit_ab), 'black') bound_upper = func(hires_x, *(best_fit_ab + sigma_ab)) bound_lower = func(hires_x, *(best_fit_ab - sigma_ab)) # plotting the confidence intervals plt.fill_between(hires_x, bound_lower, bound_upper, color = 'black', alpha = 0.15) plt.text(140, 630, text_res) plt.xlim(60, 200) plt.ylim(0, 800) plt.show()

如果您不确定是否包含绝对错误或如何在您的情况下估算它们，那么最好在Cross Validated寻求建议，因为Stack Overflow主要用于讨论关于回归方法的实现，而不是讨论基础统计数据。

如何从curve_fit获得置信区间

1 个答案: