动态评估Python中的简单布尔逻辑

Question

我有一些动态生成的布尔逻辑表达式,如:

• (A或B)和(C或D)
• A或(A和B)
• 一个
• empty - 评估为True

占位符被布尔代替.我是不是该,

1. 转换这些信息像Python表达式True or (True or False)和eval它？
2. 创建一个二叉树,其中一个节点是一个bool或Conjunction/ Disjunction对象并递归评估它？
3. 将其转换为嵌套的S表达式并使用Lisp解析器？
4. 别的什么？

建议欢迎.

Answer 1

这是一个小的(可能是74行,包括空白)模块,我在大约一个半小时内构建(再加上几乎一小时的重构):

str_to_token = {'True':True,
                'False':False,
                'and':lambda left, right: left and right,
                'or':lambda left, right: left or right,
                '(':'(',
                ')':')'}

empty_res = True


def create_token_lst(s, str_to_token=str_to_token):
    """create token list:
    'True or False' -> [True, lambda..., False]"""
    s = s.replace('(', ' ( ')
    s = s.replace(')', ' ) ')

    return [str_to_token[it] for it in s.split()]


def find(lst, what, start=0):
    return [i for i,it in enumerate(lst) if it == what and i >= start]


def parens(token_lst):
    """returns:
        (bool)parens_exist, left_paren_pos, right_paren_pos
    """
    left_lst = find(token_lst, '(')

    if not left_lst:
        return False, -1, -1

    left = left_lst[-1]

    #can not occur earlier, hence there are args and op.
    right = find(token_lst, ')', left + 4)[0]

    return True, left, right


def bool_eval(token_lst):
    """token_lst has length 3 and format: [left_arg, operator, right_arg]
    operator(left_arg, right_arg) is returned"""
    return token_lst[1](token_lst[0], token_lst[2])


def formatted_bool_eval(token_lst, empty_res=empty_res):
    """eval a formatted (i.e. of the form 'ToFa(ToF)') string"""
    if not token_lst:
        return empty_res

    if len(token_lst) == 1:
        return token_lst[0]

    has_parens, l_paren, r_paren = parens(token_lst)

    if not has_parens:
        return bool_eval(token_lst)

    token_lst[l_paren:r_paren + 1] = [bool_eval(token_lst[l_paren+1:r_paren])]

    return formatted_bool_eval(token_lst, bool_eval)


def nested_bool_eval(s):
    """The actual 'eval' routine,
    if 's' is empty, 'True' is returned,
    otherwise 's' is evaluated according to parentheses nesting.
    The format assumed:
        [1] 'LEFT OPERATOR RIGHT',
        where LEFT and RIGHT are either:
                True or False or '(' [1] ')' (subexpression in parentheses)
    """
    return formatted_bool_eval(create_token_lst(s))

简单的测试给出:

>>> print nested_bool_eval('')
True
>>> print nested_bool_eval('False')
False
>>> print nested_bool_eval('True or False')
True
>>> print nested_bool_eval('True and False')
False
>>> print nested_bool_eval('(True or False) and (True or False)')
True
>>> print nested_bool_eval('(True or False) and (True and False)')
False
>>> print nested_bool_eval('(True or False) or (True and False)')
True
>>> print nested_bool_eval('(True and False) or (True and False)')
False
>>> print nested_bool_eval('(True and False) or (True and (True or False))')
True

[可能部分偏离主题]

注意,您可以轻松地配置您使用的令牌(操作数和操作符)与穷人依赖注入方式(token_to_char=token_to_char和朋友)同时拥有多个不同的评估者(只需重置"默认注入") "全局变量会让你只有一个行为."

例如:

def fuzzy_bool_eval(s):
    """as normal, but:
    - an argument 'Maybe' may be :)) present
    - algebra is:
    [one of 'True', 'False', 'Maybe'] [one of 'or', 'and'] 'Maybe' -> 'Maybe'
    """
    Maybe = 'Maybe' # just an object with nice __str__

    def or_op(left, right):
        return (Maybe if Maybe in [left, right] else (left or right))

    def and_op(left, right):
        args = [left, right]

        if Maybe in args:
            if True in args:
                return Maybe # Maybe and True -> Maybe
            else:
                return False # Maybe and False -> False

        return left and right

    str_to_token = {'True':True,
                    'False':False,
                    'Maybe':Maybe,
                    'and':and_op,
                    'or':or_op,
                    '(':'(',
                    ')':')'}

    token_lst = create_token_lst(s, str_to_token=str_to_token)

    return formatted_bool_eval(token_lst)

得到:

>>> print fuzzy_bool_eval('')
True
>>> print fuzzy_bool_eval('Maybe')
Maybe
>>> print fuzzy_bool_eval('True or False')
True
>>> print fuzzy_bool_eval('True or Maybe')
Maybe
>>> print fuzzy_bool_eval('False or (False and Maybe)')
False

Answer 2

编写一个可以处理此问题的评估程序,例如使用pyparsing,应该不难.您只需要处理一些操作(和,或,和分组？),因此您应该能够自己解析和评估它.

您不需要显式地形成二叉树来评估表达式.

Answer 3

如果你使用你关心的locals和globals设置dicts,那么你应该能够安全地将它们与表达式一起传递给eval().

Answer 4

使用SymPy逻辑模块听起来像小菜一碟。他们甚至在文档上都有一个示例：http : //docs.sympy.org/0.7.1/modules/logic.html