Is it possible to fix parameters while fitting distributions in SciPy? For example, this code:
import scipy.stats as st
xx = st.expon.rvs(size=100)
print st.expon.fit(xx, loc=0)
results in non-zero location (loc).
When some parameter is provided to the fit function it is considered as an initial guess. And if it is provided to the constructor (st.expon(loc=0)) the distribution becomes "frozen" and can not be used for fitting.
Warren Weckesser
To fix loc, use the argument floc:
print st.expon.fit(xx, floc=0)
E.g.
In [33]: import scipy.stats as st
In [34]: xx = st.expon.rvs(size=100)
In [35]: print st.expon.fit(xx, floc=0)
(0, 0.77853895325584932)
Some related questions:
来源：https://stackoverflow.com/questions/21610034/fitting-distribution-with-fixed-parameters-in-scipy