python：如何验证用户定义的条件？

Question

我正在努力解决的是测试预定义条件，该条件采用用户提供的参数，如下例所示：

cond = "if ( 1 is yes and 2 is no ) or ( 1 is yes and 2 is no )"  
cond2 = "if (3 is no or 1 is no )"  
vars = []  

lst = cond.split()  
lst += cond2.split()  

for l in lst:  
   if l.isdigit():  
      if l not in vars:  
         vars.append(l)  
# ... sort 
# ... read user answers => x = no, y = no, y = yes 
# ... replace numbers with input (yes or no)

# ... finally I have 

cond = "if ( no is yes and no is no ) or ( no is yes and no is no )"
cond2 = "if (yes is no or no is no )" 

首先，这是正确的方法吗？
其次，如果 True 或 False ，如何验证上述条件？

先感谢您。

Answer 1

使用 Python 的语言服务 解析和编译字符串，然后执行生成的 AST。

Answer 2

用户的所有答案加在一起可用于形成唯一的二进制数（仅由零和一组成）。您可以为每个条件创建一个表（列表），该表（列表）由每个位置中答案的每个可能组合索引，存储给定条件（表示为 lambda 函数）对于该集合的值。

设置这些表后，您可以通过查找由给定答案组合索引的相应表中的值来确定任何条件是否为真。以下是如何设置的示例。

NUM_ANSWERS = 4
NUM_COMBOS = 2**NUM_ANSWERS
NO,YES = 'no','yes'

def index(answers):
    """ Convert a list of yes/no answers into binary number. """
    binstr = ''.join([('1' if a is 'yes' else '0') for a in answers])
    return int(binstr, 2)

def answers(index):
    """ Convert binary value of number into list of yes/no answers. """
    masks = [2**p for p in range(NUM_ANSWERS-1, -1, -1)]
    bits = [((index & m) / m) for m in masks]
    return [[NO,YES][b] for b in bits]

# condition expressions
cond_expr1 = lambda a1,a2,a3,a4: a1 is YES and a2 is NO  # a3,a4 ignored
cond_expr2 = lambda a1,a2,a3,a4: (
                ( a1 is YES and a2 is NO ) or ( a3 is YES and a4 is NO )
             )
# build tables for each condition
cond1 = []
cond2 = []
for i in range(NUM_COMBOS):
    ans_combo = answers(i)
    cond1.append( cond_expr1(*ans_combo) )
    cond2.append( cond_expr2(*ans_combo) )

# once tables are built, you can lookup the corresponding conditional
print cond1[ index(['yes', 'no', 'no', 'yes']) ]  # True
print cond2[ index(['yes', 'no', 'yes', 'no']) ]  # True

Answer 3

因此，在根据 Ignacio 的提示进行了一些阅读之后，我想我在进行一些字符串修改后就得到了它。但是，仍然不太确定这是否是正确的方法。
因此，我的条件变量定义如下

cond = """if ( 'no' is 'yes' and 'no' is 'no' ):
   result.append[1]
else:
   result.append[0]
"""

此外，我创建变量来存储条件结果以便稍后对其进行评估

result = []

我在字符串上运行 exec

exec(cond)

最后，我可以评估结果。

if result[0] == 1
   print "Condition was met"
else: 
   print "Condition wasn't met"

任何想法或评论都受到高度赞赏。