Convert custom formula to python function [duplicate]

Question

Consider that we have the following input

formula = "(([foo] + [bar]) - ([baz]/2) )"

function_mapping = {
                   "foo" : FooFunction,
                   "bar" : BarFunction,
                   "baz" : BazFunction,  
                  }

Is there any python library that lets me parse the formula and convert it into a python function representation.

eg.

converted_formula = ((FooFunction() + BarFunction() - (BazFunction()/2))

I am currently looking into something like

In [11]: ast = compiler.parse(formula)

In [12]: ast
Out[12]: Module(None, Stmt([Discard(Sub((Add((List([Name('foo')]), List([Name('bar')]))), Div((List([Name('baz')]), Const(2))))))]))

and then process this ast tree further.

Do you know of any cleaner alternate solution? Any help or insight is much appreciated!

Answer 1

You could use the re module to do what you want via regular-expression pattern matching and relatively straight-forward text substitution.

import re

alias_pattern = re.compile(r'''(?:\[(\w+)\])''')

def mapper(mat):
    func_alias = mat.group(1)
    function = function_alias_mapping.get(func_alias)
    if not function:
        raise NameError(func_alias)
    return function.__name__ + '()'

# must be defined before anything can be mapped to them
def FooFunction(): return 15
def BarFunction(): return 30
def BazFunction(): return 6

function_alias_mapping =  dict(foo=FooFunction, bar=BarFunction, baz=BazFunction)
formula = "(([foo] + [bar]) - ([baz]/2))"  # Custom formula.

converted_formula = re.sub(alias_pattern, mapper, formula)
print('converted_formula = "{}"'.format(converted_formula))

# define contexts and function in which to evalute the formula expression
global_context = dict(FooFunction=FooFunction,
                      BarFunction=BarFunction,
                      BazFunction=BazFunction)
local_context = {'__builtins__': None}

function = lambda: eval(converted_formula, global_context, local_context)
print('answer = {}'.format(function()))  # call function

Output:

converted_formula = "((FooFunction() + BarFunction()) - (BazFunction()/2))"
answer = 42