我試圖將一些數據擬合到對數正態分佈,並由此使用優化參數生成隨機對數正態分佈。 一些搜索後,我找到了一些解決方案,但沒有說服力:使用觀測數據的形狀生成隨機對數正態分佈
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)
或解決方案2使用mu和sigma從原始數據:
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)
使用擬合函數
解決方法1
這些解決方案都不太適合:
import pylab
pylab.plot(sorted(mydata, reverse=True), 'ro')
pylab.plot(sorted(rnd_log, reverse=True), 'bx')
我不知道我的理解以及使用分配的方式,或者如果我失去了一些東西別的......
我雖然在這裏找到解決方案:Does anyone have example code of using scipy.stats.distributions? 但我無法得到的形狀從我的數據...我錯過了使用適合函數的東西?
感謝
編輯:
這是一個例子,以便更好地瞭解我的問題:
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)
結果是:
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
同時,如果我做同樣的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"
是更近了,所以我想我使用SciPy的時候做錯了什麼。 ..
什麼這是mydata數組?爲了擬合,我期望看到x值和y值......應該如何解釋這個數組? – Tanriol 2012-01-02 19:14:51
你有沒有看過[關於對數正態分佈參數估計的許多論文](http://scholar.google.com/scholar?q=lognormal+parameter+estimation&hl=zh-CN&as_sdt=0&as_vis=1&oi=scholart)? – 2012-01-02 19:32:43
好的,我很抱歉,我想我的問題還不夠清楚。我編輯它。 – fransua 2012-01-03 10:25:06