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Groups > comp.lang.python.announce > #2897
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| From | "Pierre Denis" <pie.denis@skynet.be> |
| Newsgroups | comp.lang.python.announce |
| Subject | ANN: Lea 3.0 released |
| Date | Wed, 27 Jun 2018 16:44:29 +0200 |
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Lea 3.0 final is now released!
---> http://pypi.org/project/lea/3.0.0
What is Lea?
------------
Lea is a Python module aiming at working with discrete probability
distributions in an intuitive way.
It allows you modeling a broad range of random phenomena: gambling, weather,
finance, etc. More generally, Lea may be used for any finite set of discrete
values having known probability: numbers, booleans, date/times, symbols, .
Each probability distribution is modeled as a plain object, which can be
named, displayed, queried or processed to produce new probability
distributions.
Lea also provides advanced functions and Probabilistic Programming (PP)
features; these include conditional probabilities, Bayesian networks, joint
probability distributions, Markov chains and symbolic computation.
Lea can be used for AI, machine learning, education, ...
LGPL - Python 2.6+ / Python 3 supported
What's new in Lea 3?
--------------------
Compared to latest version (2.3.5), many things have changed to extend the
usability and openness of the library. To name a few:
* ability to choose between different probability representations: floats,
fractions and decimals
* symbolic computation: Lea can now calculate probability *formula* using
the SymPy library (http://www.sympy.org)
* simpler API and compliance with PEP8 naming convention
* revamped tutorials and examples ->
http://bitbucket.org/piedenis/lea/wiki/Home
* paper on the "Statues" algorithm used in Lea ->
http://arxiv.org/abs/1806.09997
Here is a short sample. A biased coins is flipped with 1/4 chance to be
'head'. Suppose that this coin is thrown 6 times. What is the probability to
get no more than two 'heads'? Here is how you could make this calculation in
Lea, using successively float, fraction and symbolic representations:
print (P(lea.binom(6,1/4) <= 2))
# -> 0.83056640625
print (P(lea.binom(6,'1/4') <= 2))
# -> 1701/2048
print (P(lea.binom(6,'p') <= 2))
# -> (p - 1)**4*(10*p**2 + 4*p + 1))
print (P(lea.binom(6,'p') <= 2).subs('p',1/4))
# -> 0.830566406250000
To learn more...
----------------
Lea 3 on PyPI -> http://pypi.org/project/lea/3.0.0
Lea project page -> http://bitbucket.org/piedenis/lea
Documentation -> http://bitbucket.org/piedenis/lea/wiki/Home
Statues algorithm -> http://arxiv.org/abs/1806.09997
With the hope that Lea can make the Universe less hazardous,
Pierre Denis
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ANN: Lea 3.0 released "Pierre Denis" <pie.denis@skynet.be> - 2018-06-27 16:44 +0200
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