使用CURVE_FIT在Python中拟合Lognormal分布

Question

我有一个x的假设y函数,试图找到/拟合一个对数正态分布曲线,该曲线最好地塑造数据.我正在使用curve_fit函数并且能够适合正态分布,但曲线看起来并不优化.

下面给出y和x数据点,其中y = f(x).

y_axis = [0.00032425299473065838, 0.00063714106162861229, 0.00027009331177605913, 0.00096672396877715144, 0.002388766809835889, 0.0042233337680543182, 0.0053072824980722137, 0.0061291327849408699, 0.0064555344006149871, 0.0065601228278316746, 0.0052574034010282218, 0.0057924488798939255, 0.0048154093097913355, 0.0048619350036057446, 0.0048154093097913355, 0.0045114840997070331, 0.0034906838696562147, 0.0040069911024866456, 0.0027766995669134334, 0.0016595801819374015, 0.0012182145074882836, 0.00098231827111984341, 0.00098231827111984363, 0.0012863691645616997, 0.0012395921040321833, 0.00093554121059032721, 0.0012629806342969417, 0.0010057068013846018, 0.0006081017868837127, 0.00032743942370661445, 4.6777060529516312e-05, 7.0165590794274467e-05, 7.0165590794274467e-05, 4.6777060529516745e-05]

y轴是在x轴时间仓中发生的事件的概率:

x_axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0]

使用excel和lognormal方法,我能够更好地适应我的数据.当我尝试在python中使用lognormal时,拟合不起作用,我做错了.

下面是我用于拟合正态分布的代码,这似乎是我唯一能够适应python的代码(很难相信):

#fitting distributino on top of savitzky-golay
%matplotlib inline
import matplotlib
import matplotlib.pyplot as plt
import pandas as pd
import scipy
import scipy.stats
import numpy as np
from scipy.stats import gamma, lognorm, halflogistic, foldcauchy
from scipy.optimize import curve_fit

matplotlib.rcParams['figure.figsize'] = (16.0, 12.0)
matplotlib.style.use('ggplot')
# results from savgol
x_axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0,     13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0]
y_axis = [0.00032425299473065838, 0.00063714106162861229, 0.00027009331177605913, 0.00096672396877715144, 0.002388766809835889, 0.0042233337680543182, 0.0053072824980722137, 0.0061291327849408699, 0.0064555344006149871, 0.0065601228278316746, 0.0052574034010282218, 0.0057924488798939255, 0.0048154093097913355, 0.0048619350036057446, 0.0048154093097913355, 0.0045114840997070331, 0.0034906838696562147, 0.0040069911024866456, 0.0027766995669134334, 0.0016595801819374015, 0.0012182145074882836, 0.00098231827111984341, 0.00098231827111984363, 0.0012863691645616997, 0.0012395921040321833, 0.00093554121059032721, 0.0012629806342969417, 0.0010057068013846018, 0.0006081017868837127, 0.00032743942370661445, 4.6777060529516312e-05, 7.0165590794274467e-05, 7.0165590794274467e-05, 4.6777060529516745e-05]

## y_axis values must be normalised
sum_ys = sum(y_axis)

# normalize to 1
y_axis = [_/sum_ys for _ in y_axis]

# def gamma_f(x, a, loc, scale):
#     return gamma.pdf(x, a, loc, scale)

def norm_f(x, loc, scale):
#     print 'loc: ', loc, 'scale: ', scale, "\n"
    return norm.pdf(x, loc, scale)

fitting = norm_f

# param_bounds = ([-np.inf,0,-np.inf],[np.inf,2,np.inf])
result = curve_fit(fitting, x_axis, y_axis)
result_mod = result

# mod scale
# results_adj  = [result_mod[0][0]*.75, result_mod[0][1]*.85]

plt.plot(x_axis, y_axis, 'ro')
plt.bar(x_axis, y_axis, 1, alpha=0.75)
plt.plot(x_axis, [fitting(_, *result[0]) for _ in x_axis], 'b-')
plt.axis([0,35,0,.1])

# convert back into probability
y_norm_fit = [fitting(_, *result[0]) for _ in x_axis]
y_fit = [_*sum_ys for _ in y_norm_fit]
print list(y_fit)

plt.show()

我试图得到两个问题的答案:

1. 这是我从正态分布曲线得到的最佳拟合吗？我怎样才能改善我的健康？

正态分布结果:

2. 如何将对数正态分布拟合到此数据中,还是可以使用更好的分布？

我正在玩对数正态分布曲线调整μ和sigma,看起来有可能更好的拟合.我不明白我在python中得到类似结果我做错了什么.

Answer 1

实际上，伽玛分布可能很适合@Glen_b 提出的。我使用第二个定义 \alpha 和 \beta 。

注意：我用于快速拟合的技巧是计算均值和方差，对于典型的二参数分布，它足以恢复参数并快速了解它是否适​​合。

代码

import math
from scipy.misc import comb

import matplotlib.pyplot as plt

y_axis = [0.00032425299473065838, 0.00063714106162861229, 0.00027009331177605913, 0.00096672396877715144, 0.002388766809835889, 0.0042233337680543182, 0.0053072824980722137, 0.0061291327849408699, 0.0064555344006149871, 0.0065601228278316746, 0.0052574034010282218, 0.0057924488798939255, 0.0048154093097913355, 0.0048619350036057446, 0.0048154093097913355, 0.0045114840997070331, 0.0034906838696562147, 0.0040069911024866456, 0.0027766995669134334, 0.0016595801819374015, 0.0012182145074882836, 0.00098231827111984341, 0.00098231827111984363, 0.0012863691645616997, 0.0012395921040321833, 0.00093554121059032721, 0.0012629806342969417, 0.0010057068013846018, 0.0006081017868837127, 0.00032743942370661445, 4.6777060529516312e-05, 7.0165590794274467e-05, 7.0165590794274467e-05, 4.6777060529516745e-05]
x_axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0]

## y_axis values must be normalised
sum_ys = sum(y_axis)

# normalize to 1
y_axis = [_/sum_ys for _ in y_axis]

m = 0.0
for k in range(0, len(x_axis)):
    m += y_axis[k] * x_axis[k]

v = 0.0
for k in range(0, len(x_axis)):
    t = (x_axis[k] - m)
    v += y_axis[k] * t * t

print(m, v)

b = m/v
a = m * b

print(a, b)

z = []
for k in range(0, len(x_axis)):
    q = b**a * x_axis[k]**(a-1.0) * math.exp( - b*x_axis[k] ) / math.gamma(a)
    z.append(q)

plt.plot(x_axis, y_axis, 'ro')
plt.plot(x_axis, z, 'b*')
plt.axis([0, 35, 0, .1])
plt.show()