我正在尝试将一些数据拟合到对数正态分布,并使用优化的参数生成随机对数正态分布 . 经过一番搜索,我发现了一些解决方案,但没有人说服:
solution1 using the fit function:
import numpy as np
from scipy.stats import lognorm
mydata = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,8,8,8,8,8,9,9,9,10,10,11,12,13,14,14,15,19,19,21,23,25,27,28,30,31,36,41,45,48,52,55,60,68,75,86,118,159,207,354]
shape, loc, scale = lognorm.fit(mydata)
rnd_log = lognorm.rvs (shape, loc=loc, scale=scale, size=100)
or Solution 2 using mu and sigma from original data:
import numpy as np
from scipy.stats import lognorm
mydata = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,8,8,8,8,8,9,9,9,10,10,11,12,13,14,14,15,19,19,21,23,25,27,28,30,31,36,41,45,48,52,55,60,68,75,86,118,159,207,354]
mu = np.mean([np.log(i) for i in mydata])
sigma = np.std([np.log(i) for i in mydata])
distr = lognorm(mu, sigma)
rnd_log = distr.rvs (size=100)
这些解决方案都不合适:
import pylab
pylab.plot(sorted(mydata, reverse=True), 'ro')
pylab.plot(sorted(rnd_log, reverse=True), 'bx')
我不确定我是否理解使用发行版的方式,或者我是否遗漏了其他内容......
我虽然在这里找到了解决方案:Does anyone have example code of using scipy.stats.distributions?但是我无法从我的数据中得到形状...我是否在使用fit函数时遗漏了一些东西?
谢谢
EDIT:
this is an example in order to understand better my problem:
print 'solution 1:'
means = []
stdes = []
distr = lognorm(mu, sigma)
for _ in xrange(1000):
rnd_log = distr.rvs (size=100)
means.append (np.mean([np.log(i) for i in rnd_log]))
stdes.append (np.std ([np.log(i) for i in rnd_log]))
print 'observed mean:',mu , 'mean simulated mean:', np.mean (means)
print 'observed std :',sigma, 'mean simulated std :', np.mean (stdes)
print '\nsolution 2:'
means = []
stdes = []
shape, loc, scale = lognorm.fit(mydata)
for _ in xrange(1000):
rnd_log = lognorm.rvs (shape, loc=loc, scale=scale, size=100)
means.append (np.mean([np.log(i) for i in rnd_log]))
stdes.append (np.std ([np.log(i) for i in rnd_log]))
print 'observed mean:',mu , 'mean simulated mean:', np.mean (means)
print 'observed std :',sigma, 'mean simulated std :', np.mean (stdes)
the result is:
solution 1:
observed mean: 1.82562655734 mean simulated mean: 1.18929982267
observed std : 1.39003773799 mean simulated std : 0.88985924363
solution 2:
observed mean: 1.82562655734 mean simulated mean: 4.50608084668
observed std : 1.39003773799 mean simulated std : 5.44206119499
while, if I do the same in R:
mydata <- c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,8,8,8,8,8,9,9,9,10,10,11,12,13,14,14,15,19,19,21,23,25,27,28,30,31,36,41,45,48,52,55,60,68,75,86,118,159,207,354)
meanlog <- mean(log(mydata))
sdlog <- sd(log(mydata))
means <- c()
stdes <- c()
for (i in 1:1000){
rnd.log <- rlnorm(length(mydata), meanlog, sdlog)
means <- c(means, mean(log(rnd.log)))
stdes <- c(stdes, sd(log(rnd.log)))
}
print (paste('observed mean:',meanlog,'mean simulated mean:',mean(means),sep=' '))
print (paste('observed std :',sdlog ,'mean simulated std :',mean(stdes),sep=' '))
我得到:
[1] "observed mean: 1.82562655733507 mean simulated mean: 1.82307191072317"
[1] "observed std : 1.39704049131865 mean simulated std : 1.39736545866904"
这更接近,所以我猜我在使用scipy时做错了...
1 回答
scipy中的对数正态分布参数化与通常的方法略有不同 . 请参阅scipy.stats.lognorm docs,尤其是"Notes"部分 .
以下是如何获得您期望的结果(请注意,我们在拟合时将位置保持为0):
现在您可以生成样本并确认您的期望: