pyboolean

Parse infix boolean expressions to postfix (RPN), evaluate and generate truth tables.

Compatibilities

• Python 3.x
• Any Operating System

Installation

pyboolean is published on PyPi, so you only need to run the following command:

$ pip install pyboolean

Usage

Creating a new boolean expression:

from pyboolean import BoolExpr

bool_expr = BoolExpr('1 or 0')

Printing the formatted version of the expression:

print(bool_expr)
# Output: "1 + 0"

Evaluating our boolean expression:

print(bool_expr.eval())
# Output: "1"

A boolean expression can also include variables:

bool_expr = BoolExpr('1 . m + ( 0 + ! n ) . m')

In order to evaluate an expression with variables, the eval() function can take arguments to replace them:

print(bool_expr.eval(1, 0))
# Output: "1"

For expressions with variables, one can generate a truth table like so:

print(bool_expr.truthtable())
# Output:
#╔════════════╗
#║  m  n  ┃ O ║
#║━━━━━━━━╋━━━║
#║  0  0  ┃ 0 ║
#║────────╂───║
#║  0  1  ┃ 0 ║
#║────────╂───║
#║  1  0  ┃ 1 ║
#║────────╂───║
#║  1  1  ┃ 1 ║
#╚════════════╝

Or if you need to manipulate the raw truth table data, generate a dictionary:

print(bool_expr.truthdict())
# Output: "{('0', '0'): '0', ('0', '1'): '0', ('1', '0'): '1', ('1', '1'): '1'}"

Finally, a note on formatting. When creating an expression, all spaces are ignored and many different versions of the operator symbols are accepted, so one can be very 'creative' with the input and not need to worry:

bool_expr = BoolExpr('p+1and   ¬m.(1∧p ) ∨notxand~0')
print(bool_expr)
# Output: "p + 1 . ! m . ( 1 . p ) + ! x . ! 0"

Full list of accepted operator symbols:

AND:

and . & ∧

OR:

or + | ∨

NOT:

not ! ~ ¬