There is an equation of exponential truncated power law in the article of Gonzalez1 below:

$P(r_g) = (r_g + r_g^0)^{-\beta_r} e^{-r_g/K}$

To solve this problem, we asked it on stackoverflow, and Michael suggested to have a look at scipy.optimize.curve_fit.

## Toy Data

import numpy as np
rg = np.array([ 20.7863444 ,   9.40547933,   8.70934714,   8.62690145,
7.16978087,   7.02575052,   6.45280959,   6.44755478,
5.16630287,   5.16092884,   5.15618737,   5.05610068,
4.87023561,   4.66753197,   4.41807645,   4.2635671 ,
3.54454372,   2.7087178 ,   2.39016885,   1.9483156 ,
1.78393238,   1.75432688,   1.12789787,   1.02098332,
0.92653501,   0.32586582,   0.1514813 ,   0.09722761,
0.        ,   0.        ])

# calculate P(rg)
rg = sorted(rg, reverse=True)
rg = np.array(rg)
prg = np.arange(len(rg)) / float(len(rg)-1)


## Solution

def func(rg, rg0, beta, K):
return (rg + rg0) ** (-beta) * np.exp(-rg / K)

from scipy import optimize
popt, pcov = optimize.curve_fit(func, rg, prg, p0=[1.8, 0.15, 5])
print popt
print pcov


[  1.04303608e+03   3.02058550e-03   4.85784945e+00]
[[  1.38243336e+18  -6.14278286e+11  -1.14784675e+11]
[ -6.14278286e+11   2.72951900e+05   5.10040746e+04]
[ -1.14784675e+11   5.10040746e+04   9.53072925e+03]]

%matplotlib inline
import matplotlib.pyplot as plt

plt.plot(rg, prg, 'bs', alpha = 0.3)
plt.plot(rg, (rg+popt[0])**-(popt[1])*np.exp(-rg/popt[2]), 'r-' )
plt.yscale('log')
plt.xscale('log')
plt.xlabel('$r_g$', fontsize = 20)
plt.ylabel('$P(r_g)$', fontsize = 20)
plt.show()


# 参考文献

1. Gonzalez, M. C., Hidalgo, C. A., & Barabasi, A. L. (2008). Understanding individual human mobility patterns. Nature, 453(7196), 779-782.

Tags:

Categories:

Updated: