我该如何使用pyparsing从布尔表达式生成元素列表?

3 投票
1 回答
633 浏览
提问于 2025-04-17 12:58

我有一堆布尔表达式,这些表达式代表可以组合在一起形成更大物体的物理对象。它们看起来像这样:((A and B) or C)。这个对象可以通过A和B的组合来表示,也可以单独用C来表示。我想生成一个字符串的列表列表,用来创建这个对象。在这个例子中,我想要的是[[A,B], [C]]。

我觉得Pyparsing挺有意思的,所以决定用它来解决这个问题。在几次失败的尝试后,我决定改编网站上的fourFn.py示例。这是我目前的进展:

from pyparsing import Literal, CaselessLiteral, Word, Combine, \
    Group, Optional, ZeroOrMore, Forward, alphanums

exprStack = []

def myAnd(op1, op2):
    if isinstance(op1, str):
        return([op1, op2])
    else:
    return op1.append(op2)

def myOr(op1, op2):
    if isinstance(op1, str):
        return([[op1], [op2]])
    else:
        return op1.append([op2])

def pushFirst(strg, loc, toks):
    exprStack.append(toks[0])

bnf = None
def BNF():
    """
    boolop  :: 'and' | 'or'
    gene    :: alphanum
    atom    :: gene | '(' expr ')'
    """
    global bnf
    if not bnf:
        element = Word(alphanums)
        andop  = Literal( "and" )
        orop = Literal( "or" )
        lpar  = Literal( "(" ).suppress()
        rpar  = Literal( ")" ).suppress()
        boolop  = andop | orop

        expr = Forward()
        atom = ((element | lpar + expr + rpar).setParseAction(pushFirst) | (lpar + expr.suppress() + rpar))
        expr << atom + ZeroOrMore((boolop + expr).setParseAction(pushFirst))

        bnf = expr
    return bnf

# map operator symbols to corresponding arithmetic operations
fn  = {"or": myOr,
       "and": myAnd}

def evaluateStack( s ):
    op = s.pop()
    if op in fn:
        op2 = evaluateStack(s)
        op1 = evaluateStack(s)
        return fn[op](op1, op2)
    else:
        return op

if __name__ == "__main__":

    def test(s, expVal):
        global exprStack
        exprStack = []
        results = BNF().parseString(s)
        val = evaluateStack(exprStack[:])
        if val == expVal:
            print s, "=", val, results, "=>", exprStack
        else:
            print "!!! "+s, val, "!=", expVal, results, "=>", exprStack

    test("((A and B) or C)", [['A','B'], ['C']])
    test("(A and B) or C", [['A','B'], ['C']])
    test("(A or B) and C", [['A', 'C'], ['B', 'C']])
    test("A and B", ['A', 'B'])
    test("A or B", [['A'], ['B']])

前面三个测试在这里失败了,只返回每个括号内表达式的第一个元素。A会被多次推入栈中。看起来我修改fourFn.py的方式破坏了我的脚本处理这些组合的能力。有没有更好的方法来解决这个问题呢?

编辑 喝了一杯咖啡后,我意识到我遇到的问题其实很容易解决。我的新and和or函数如下:

def myAnd(op1, op2):
    if isinstance(op1, str) and isinstance(op2, str):
        newlist = [op1, op2]

    elif isinstance(op1, str):
        newlist = [op1]
        newlist.append(op2)

    elif isinstance(op2, str):
        newlist = op1
        newlist.append(op2)

    else:
        newlist = [op1.append(item) for item in op2]
    return newlist

def myOr(op1, op2):
    if isinstance(op1, str) and isinstance(op2, str):
        newlist = [[op1], [op2]]
        r
    elif isinstance(op1, str):
        newlist = [op1]
        newlist.append([op2])

    elif isinstance(op2, str):
        newlist = [op1]
        newlist.append([op2])

    else:
        newlist = [op1, [op2]]
    return newlist1

解析器的构建方式如下:

expr = Forward()
atom = element.setParseAction(pushFirst) | (lpar + expr + rpar)
expr << atom + ZeroOrMore((boolop + expr).setParseAction(pushFirst))

一个新的、更有趣的问题是如何处理像这样(A or B)和C的情况。结果应该是[[A, C], [B, C]]。有没有一种典型的pyparsing方法来处理这个问题呢?

1 个回答

0

为了将来参考,这里有一种方法,它适用于我的测试案例,但与上面提到的抽象语法树(AST)方法有所不同:

from pyparsing import Literal, Word, Optional, \
    Group, ZeroOrMore, Forward, alphanums
import ffparser, sys

exprStack = []

def myAnd(op1, op2):
    if isinstance(op1, str) and isinstance(op2, str):
        newlist = [[op1, op2]]

    elif isinstance(op1, str):
        newlist = op2
        [item.insert(0, op1) for item in newlist]

    elif isinstance(op2, str):
        newlist = op1
        [item.append(op2) for item in op1]

    else:
        newlist = [op1.append(item) for item in op2]

    return newlist

def myOr(op1, op2):
   if isinstance(op1, str) and isinstance(op2, str):
        newlist = [[op1], [op2]]

    elif isinstance(op1, str):
        newlist = op2
        newlist.insert(0, [op1])

    elif isinstance(op2, str):
        newlist = op1
        newlist.append([op2])

    else:
        newlist = []
        [newlist.append(item) for item in op1]
        [newlist.append(item) for item in op2]

    return newlist

def pushFirst(strg, loc, toks):
    exprStack.append(toks[0])

bnf = None
def BNF():
    """
    boolop  :: 'and' | 'or'
    gene    :: alphanum
    atom    :: gene | '(' expr ')'
    """
    global bnf
    if not bnf:
        element = Word(alphanums)
        andop  = Literal( "and" )
        orop = Literal( "or" )
        lpar  = Literal( "(" ).suppress()
    rpar  = Literal( ")" ).suppress()
    boolop  = andop | orop

    expr = Forward()
    atom = element.setParseAction(pushFirst) | (Optional(lpar) + expr + Optional(rpar))
    expr << atom + ZeroOrMore((boolop + expr).setParseAction(pushFirst))

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