Python 中 scipy 的 curve_fit 曲线拟合

Question

我正在尝试用指数函数拟合我的数据

import numpy as np

def exponentional(k, alpha, k0, c):
    return k0 * np.exp(k *-alpha) + c

我使用了 scipy.optimize、

中的 curve_fit

from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
uniq_deg = [2,...,103,..,203,...,307,...,506] 
normalized_deg_dist = [0.99,...,0.43,..0.12,..,0.04,..., 0.01]
           
            
popt, pcov = curve_fit(exponentional, uniq_deg, normalized_deg_dist, 
                       p0 = [1,0.00001,1,1], maxfev = 6000)
           
fig = plt.figure() 
ax = fig.add_subplot(111)
             
ax.semilogy(uniq_deg, normalized_deg_dist, 'bo', label = 'Real data') 
ax.semilogy(uniq_deg,[exponentional(d,*popt) for d in uniq_deg], 'r-', label = 'Fit') 
ax.set_xlabel('Degree' ) 
ax.set_ylabel('1-CDF degree') 
ax.legend(loc='best')
ax.set_title(f'Degree distribution in {city}')
plt.show()

导致：

看起来不太合适。

我哪里错了？

Answer 1

最后没用curve_fit。我使用了 https://mathworld.wolfram.com/LeastSquaresFittingExponential.html

中指数拟合的定义

而且我还需要通过 power-law 拟合一些其他数据，我也这样做了。

    #%%
def fit_powerlaw(xs, ys):
    S_lnx_lny = 0.0
    S_lnx_S_lny = 0.0
    S_lny = 0.0
    S_lnx = 0.0
    S_lnx2 = 0.0
    S_ln_x_2 = 0.0
    n = len(xs)
    for (x,y) in zip(xs, ys):
        S_lnx += np.log(x)
        S_lny += np.log(y)
        S_lnx_lny += np.log(x) * np.log(y)
        S_lnx_S_lny = S_lnx * S_lny
        S_lnx2 += np.power(np.log(x),2)
        S_ln_x_2 = np.power(S_lnx,2)
    #end
    b = (n * S_lnx_lny - S_lnx_S_lny ) / (n * S_lnx2 - S_ln_x_2)
    a = (S_lny - b * S_lnx)  / (n)
    return (np.exp(a), b)
#%%
def fit_exp(xs, ys):
    S_x2_y = 0.0
    S_y_lny = 0.0
    S_x_y = 0.0
    S_x_y_lny = 0.0
    S_y = 0.0
    for (x,y) in zip(xs, ys):
        S_x2_y += x * x * y
        S_y_lny += y * np.log(y)
        S_x_y += x * y
        S_x_y_lny += x * y * np.log(y)
        S_y += y
    #end
    a = (S_x2_y * S_y_lny - S_x_y * S_x_y_lny) / (S_y * S_x2_y - S_x_y * S_x_y)
    b = (S_y * S_x_y_lny - S_x_y * S_y_lny) / (S_y * S_x2_y - S_x_y * S_x_y)
    return (np.exp(a), b)

第一张图表用于指数拟合，第二张图表用于幂律。我觉得这个结果很有说服力。

Python 中 scipy 的 curve_fit 曲线拟合

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