python pyparsing FourFn示例：使用两个或多个参数解析函数

Question

我正在使用pyparsing来编写一个简单的计算器。阅读fourFn示例后（并没有完全理解它！）我决定稍微修改它以满足我的需求。

这是我的代码（大部分来自here）：

def summer(p1, p2):
    return p1 + p2

class NumericStringParser(object):
    '''
    Most of this code comes from the fourFn.py pyparsing example

    '''

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

    def pushUMinus(self, strg, loc, toks):
        if toks and toks[0] == '-':
            self.exprStack.append('unary -')

    def __init__(self):
        """
        expop   :: '^'
        multop  :: '*' | '/'
        addop   :: '+' | '-'
        integer :: ['+' | '-'] '0'..'9'+
        atom    :: PI | E | real | fn '(' expr ')' | '(' expr ')'
        factor  :: atom [ expop factor ]*
        term    :: factor [ multop factor ]*
        expr    :: term [ addop term ]*
        """
        point = Literal(".")
        e = CaselessLiteral("E")

        fnumber = Combine(Word("+-" + nums, nums) +
                          Optional(point + Optional(Word(nums))) +
                          Optional(e + Word("+-" + nums, nums)))
        ident = Word(alphas, alphas + nums + "_$")
        plus = Literal("+")
        minus = Literal("-")
        mult = Literal("*")
        div = Literal("/")
        lpar = Literal("(").suppress()
        rpar = Literal(")").suppress()
        comma = Literal(',').suppress()
        addop = plus | minus
        multop = mult | div
        expop = Literal("^")
        pi = CaselessLiteral("PI")
        expr = Forward()
        atom = ((Optional(oneOf("- +")) +
                 (
                 pi | e | fnumber | ident + lpar + expr + rpar | ident + lpar + expr + comma + expr + rpar).setParseAction(
                     self.pushFirst))
                | Optional(oneOf("- +")) + Group(lpar + expr + rpar)
                ).setParseAction(self.pushUMinus)

        # by defining exponentiation as "atom [ ^ factor ]..." instead of
        # "atom [ ^ atom ]...", we get right-to-left exponents, instead of left-to-right
        # that is, 2^3^2 = 2^(3^2), not (2^3)^2.
        factor = Forward()
        factor << atom + ZeroOrMore((expop + factor).setParseAction(self.pushFirst))
        term = factor + ZeroOrMore((multop + factor).setParseAction(self.pushFirst))
        expr << term + ZeroOrMore((addop + term).setParseAction(self.pushFirst))
        # addop_term = ( addop + term ).setParseAction( self.pushFirst )
        # general_term = term + ZeroOrMore( addop_term ) | OneOrMore( addop_term)
        # expr <<  general_term
        self.bnf = expr
        # map operator symbols to corresponding arithmetic operations
        epsilon = 1e-12
        self.opn = {"+": operator.add,
                    "-": operator.sub,
                    "*": operator.mul,
                    "/": operator.truediv,
                    "^": operator.pow}
        self.fn = {"sin": math.sin,
                   "cos": math.cos,
                   "tan": math.tan,
                   "abs": abs,
                   "trunc": lambda a: int(a),
                   "round": round,
                   "sgn": lambda a: abs(a) > epsilon and cmp(a, 0) or 0}
        self.fn2 = {"sum": summer}

    def evaluateStack(self, s):
        op = s.pop()
        if op == 'unary -':
            res = self.evaluateStack(s)
            return -res
        if op in "+-*/^":
            op2 = self.evaluateStack(s)
            op1 = self.evaluateStack(s)
            res = self.opn[op](op1, op2)
            return res
        elif op == "PI":
            return math.pi  # 3.1415926535
        elif op == "E":
            return math.e  # 2.718281828
        elif op in self.fn:
            return self.fn[op](self.evaluateStack(s))
        elif op in self.fn2:
            op2 = self.evaluateStack(s)
            op1 = self.evaluateStack(s)
            return self.fn2[op](op1, op2)
        elif op[0].isalpha():
            return 0
        else:
            return float(op)

    def eval(self, num_string, parseAll=True):
        self.exprStack = []
        results = self.bnf.parseString(num_string, parseAll)

        val = self.evaluateStack(self.exprStack[:])
        return val


n = NumericStringParser()
print(n.eval('(4.0*sum(3,3))^1'))

正如您所看到的，我刚刚将ident + lpar + expr + comma + expr + rpar添加到atom（我猜这是一个最终评估为数字的单位）并将fn2添加到evaluateStack(self, s)

如果我使用print(n.eval('sum(3,7.0)'))一切正常，结果为10.但是当我使用print(n.eval('1*sum(3,7.0)'))时，结果为30.显示sum的第一个参数被计算两次。< / p>

你能帮我理解缺少的东西吗？或者至少帮助我理解这个问题？

修改 当我在s顶部打印堆栈（evaluateStack(self, s)）时，堆栈包含两次第一个参数：

n.eval('1.0*sum(3+0.0,7)') - ＆gt; ['1.0', '3', '0.0', '+', '3', '0.0', '+', '7', 'sum', '*']