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Last active Aug 29, 2015

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python + lea: probabilisitc programming

A script that replicates all examples in my blog post using Python and Lea to do some probabilistic programming: Lea post

Run all the examples

    $ python lea_ex01.py

Or,

    $ chmod u+x lea_ex01.py
    $ ./lea_ex01.py

That's it, give it a try!

#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=utf-8
#
# Copyright © 2015 Christopher C. Strelioff <chris.strelioff@gmail.com>
#
# Distributed under terms of the MIT license.
"""
An example of probabilistic programming using Lea.
Blog post: http://chrisstrelioff.ws/sandbox/2015/05/04/probabilistic_programming_with_python_and_lea.html
"""
from __future__ import division, print_function
from lea import Lea
# define coin
coin = Lea.fromValFreqs(('H', 1),
('T', 1))
print('Coin Distribution',
coin,
sep='\n')
# define six-sided die
die6 = Lea.fromValFreqs(('1', 1),
('2', 1),
('3', 1),
('4', 1),
('5', 1),
('6', 1))
print('Six-sided Die Distribution',
die6,
sep='\n')
# define four-side die
die4 = Lea.fromValFreqs(('1', 1),
('2', 1),
('3', 1),
('4', 1))
print('Four-sided Die Distribution',
die4,
sep='\n')
# construct Scenario 1
scenario1 = Lea.buildCPT((coin == 'H', die6),
(coin == 'T', die6))
print('Scenario 1',
scenario1,
sep='\n')
# construct Scenario 2
scenario2 = Lea.buildCPT((coin == 'H', die6),
(coin == 'T', die4))
print('Scenario 2',
scenario2,
sep='\n')
# get joint probs for all events
# -- scenario 1
joint_prob1 = Lea.cprod(coin, scenario1)
print('Scenario 1',
'* Joint Probabilities',
joint_prob1,
sep='\n')
# get joint probs for all events
# -- scenario 2
joint_prob2 = Lea.cprod(coin, scenario2)
print('Scenario 2',
'* Joint Probabilities',
joint_prob2,
sep='\n')
# prob coin given D=6, scenario 1
print("Scenario 1 -> P(C|D=6)",
coin.given(scenario1 == '6'),
sep='\n')
# prob coin given D=6, scenario 2
print("Scenario 2 -> P(C|D=6)",
coin.given(scenario2 == '6'),
sep='\n')
# prob coin given D=4, scenario 1
print("Scenario 1 -> P(C|D=4)",
coin.given(scenario1 == '4'),
sep='\n')
# prob coin given D=4, scenario 2
print("Scenario 2 -> P(C|D=4)",
coin.given(scenario2 == '4'),
sep='\n')
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