将自定义公式转换为python函数

Question

考虑一下我们有以下输入

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

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

是否有任何python库可以解析公式并将其转换为python函数表示.

例如.

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

我目前正在研究类似的东西

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))))))]))

然后进一步处理这个树.

你知道任何更清洁的替代解决方案吗？非常感谢任何帮助或见解!

Answer 1

您可以使用该re模块通过正则表达式模式匹配和相对直接的文本替换来执行您想要的操作。

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

输出：

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