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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.

Vladimir
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1 Answers1

10

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)

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Warren Weckesser
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