问题描述
我正在尝试使用指数函数拟合数据
import numpy as np
def exponentional(k,alpha,k0,c):
return k0 * np.exp(k *-alpha) + c
我从scipy.optimize使用了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,'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()
结果:
它看起来不太合适。
哪里错了?
解决方法
最后,我没有使用curve_fit
。我使用了https://mathworld.wolfram.com/LeastSquaresFittingExponential.html
同样,我也需要根据幂律来拟合其他数据。
#%%
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,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)
第一个图形用于指数拟合,第二个图形用于幂律。 我认为结果令人信服。