How to fit a weibull distribution to data using python?

Question

I am looking to find the best fit weibull parameters to a set of data using Python 3.4.

import scipy.stats as ss
list1 = []
list2 = []
for x in range(0, 10):
    list1.append(ss.exponweib.pdf(x, a=1, c=2.09, scale=10.895, loc=0))
    list2.append(ss.weibull_min.pdf(x, c=2.09, loc=0, scale=10.895))
    if list1[x]-list2[x] < .000000001:
        list1[x]=list2[x]

if list1 == list2:
    print("true")

print(ss.distributions.weibull_min.fit(list1, floc=0))
print(ss.distributions.weibull_min.fit(list1, loc=0))
print(ss.distributions.weibull_min.fit(list1, floc=0))
print(ss.distributions.exponweib.fit(list1, 1,1))
print(ss.distributions.exponweib.fit(list1, floc=0, f0=1))
print(ss.distributions.exponweib.fit(list1, floc=0, a=1, f0=1))

Everything that I have tried doesn't yield the input parameters and I can't figure out why.

The output of this code is:

true
(2.8971366871403661, 0, 0.065615284314998634)
(0.71134622938358294, 0.014105558832066645, 0.076662586739229072)
(2.8971366871403661, 0, 0.065615284314998634)
(0.27753056922336583, 3.1962672780921197, -3.4788071110631162e-27, 0.077986010645321888)
(1, 2.8971366871403661, 0, 0.065615284314998634)
(1, 2.8971366871403661, 0, 0.065615284314998634)

None of which are the correct input parameters. (2.09 and 10.895.) Any help is appreciated. Thanks.

Warren Weckesser · Accepted Answer · 2016-01-21 10:50:52Z

6

The first argument to the fit() method is a sample of values from the distribution to be fit (not PDF values). So you should use the rvs() method to generate your data, not the pdf() method.

Here's a simple example where I generate a sample of 250 values from the exponweib distribution, and then use fit() on that sample. I'll assume that when I fit the data, I know that that the shape parameter a must be 1 and the loc parameter must be 0:

In [178]: from scipy.stats import exponweib

In [179]: sample = exponweib.rvs(a=1, c=2.09, scale=10.895, loc=0, size=250)

In [180]: exponweib.fit(sample, floc=0, fa=1)
Out[180]: (1, 2.0822583185068915, 0, 10.946962241403902)

answered Jan 21, 2016 at 10:50

Warren Weckesser

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