我尝试使用pyparsing来解析逻辑表达式,例如这些
x
FALSE
NOT x
(x + y <= 5) AND (y >= 10) OR NOT (z < 100 OR w)
(A=True OR NOT (G < 8) => S = J) => ((P = A) AND not (P = 1) AND (B = O)) => (S = T)
((P = T) AND NOT (K =J) AND (B = F)) => (S = O) AND
((P = T) OR (k and b => (8 + z <= 10)) AND NOT (a + 9 <= F)) => (7 = a + z)
我在下面写的代码似乎工作正常 - 但它很慢(例如上面的最后一个例子需要几秒钟)。我是否以某种低效的方式构造语法?应该使用递归而不是operatorPrecedence?有没有办法加快速度?
identifier = Group(Word(alphas, alphanums + "_") + Optional("'"))
num = Regex(r"[+-]?\d+(:?\.\d*)?(:?[eE][+-]?\d+)?")
operator = Regex(">=|<=|!=|>|<|=")
operand = identifier | num
aexpr = operatorPrecedence(operand,
[('*',2,opAssoc.LEFT,),
('+',2,opAssoc.LEFT,),
(operator,2,opAssoc.LEFT,)
])
op_prec = [(CaselessLiteral('not'),1,opAssoc.RIGHT,),
(CaselessLiteral('and'),2,opAssoc.LEFT ,),
(CaselessLiteral('or'), 2,opAssoc.LEFT ,),
('=>', 2,opAssoc.LEFT ,),
]
sentence = operatorPrecedence(aexpr,op_prec)
return sentence
答案 0 :(得分:6)
我遇到了同样的问题。在此处找到解决方案(parserElement.enablePackrat()
):http://pyparsing.wikispaces.com/share/view/26068641?replyId=26084853
现在立即解析以下代码(之前60秒)
ParserElement.enablePackrat()
integer = Word(nums).setParseAction(lambda t:int(t[0]))('int')
operand = integer | variable('var')
# Left precedence
eq = Literal("==")('eq')
gt = Literal(">")('gt')
gtEq = Literal(">=")('gtEq')
lt = Literal("<")('lt')
ltEq = Literal("<=")('ltEq')
notEq = Literal("!=")('notEq')
mult = oneOf('* /')('mult')
plus = oneOf('+ -')('plus')
_and = oneOf('&& and')('and')
_or = oneOf('|| or')('or')
# Right precedence
sign = oneOf('+ -')('sign')
negation = Literal('!')('negation')
# Operator groups per presedence
right_op = negation | sign
# Highest precedence
left_op_1 = mult
left_op_2 = plus
left_op_3 = gtEq | ltEq | lt | gt
left_op_4 = eq | notEq
left_op_5 = _and
left_op_6 = _or
# Lowest precedence
condition = operatorPrecedence( operand, [
(right_op, 1, opAssoc.RIGHT),
(left_op_1, 2, opAssoc.LEFT),
(left_op_2, 2, opAssoc.LEFT),
(left_op_3, 2, opAssoc.LEFT),
(left_op_4, 2, opAssoc.LEFT),
(left_op_5, 2, opAssoc.LEFT),
(left_op_6, 2, opAssoc.LEFT)
]
)('computation')
答案 1 :(得分:4)
我把你的代码放到一个小程序中
from sys import argv
from pyparsing import *
def parsit(aexpr):
identifier = Group(Word(alphas, alphanums + "_") + Optional("'"))
num = Regex(r"[+-]?\d+(:?\.\d*)?(:?[eE][+-]?\d+)?")
operator = Regex(">=|<=|!=|>|<|=")
operand = identifier | num
aexpr = operatorPrecedence(operand,
[('*',2,opAssoc.LEFT,),
('+',2,opAssoc.LEFT,),
(operator,2,opAssoc.LEFT,)
])
op_prec = [(CaselessLiteral('not'),1,opAssoc.RIGHT,),
(CaselessLiteral('and'),2,opAssoc.LEFT ,),
(CaselessLiteral('or'), 2,opAssoc.LEFT ,),
('=>', 2,opAssoc.LEFT ,),
]
sentence = operatorPrecedence(aexpr,op_prec)
return sentence
def demo02(arg):
sent = parsit(arg)
print arg, ":", sent.parseString(arg)
def demo01():
for arg in ["x", "FALSE", "NOT x",
"(x + y <= 5) AND (y >= 10) OR NOT (z < 100 OR w)",
"(A=True OR NOT (G < 8) => S = J) => ((P = A) AND not (P = 1) AND (B = O)) => (S = T)",
"((P = T) AND NOT (K =J) AND (B = F)) => (S = O) AND ((P = T) OR (k and b => (8 + z <= 10)) AND NOT (a + 9 <= F)) => (7 = a + z)"
]:
demo02(arg)
if len(argv) <= 1:
demo01()
else:
for arg in argv[1:]:
demo02(arg)
并浏览cProfile
$ python -m cProfile pyparsetest.py
您会发现许多 parseImpl
来电,但在输出中间有
2906500/8 26.374 0.000 72.667 9.083 pyparsing.py:913(_parseNoCache)
212752/300 1.045 0.000 72.608 0.242 pyparsing.py:985(tryParse)
72.667
总计72
计算时间{/ 1}}。
因此,我冒昧地猜测“缓存”会提供良好的杠杆。
仅仅启用http://pyparsing-public.wikispaces.com/FAQs并没有帮助。我添加了行
import pyparsing
pyparsing.usePackrat = True
并且运行时是相同的。
Number-Regex对我来说也很好 - 我觉得非常标准。例如,用
替换它#num = Regex(r"[+-]?\d+(:?\.\d*)?(:?[eE][+-]?\d+)?")
num = Regex(r"8|1|10|100|5")
也没有帮助。在我的简单变体中没有“空匹配”,我猜这可能是个问题 - 但似乎没有。
最后一次尝试是用以下内容查看结果解析器:
....
sentence = operatorPrecedence(aexpr,op_prec)
print sentence
return sentence
....
而且......哇......太长了!
嗯,不使用你的第一个operatorPrecedence
很多更快,但是对于算术不再有效。
因此,我冒昧地猜测,是的,尝试更多地分离两种表达式(布尔和算术)。也许这会改善它。我也会研究它,它也让我感兴趣。