pyparsing

Question

所以这是从 fourFn.py:

中提取的解析器

from pyparsing import (
    Literal,
    Word,
    Group,
    Forward,
    alphas,
    alphanums,
    Regex,
    ParseException,
    CaselessKeyword,
    Suppress,
    delimitedList
)
import math
import operator

exprStack = []


def push_first(toks):
    exprStack.append(toks[0])


def push_unary_minus(toks):
    for t in toks:
        if t == "-":
            exprStack.append("unary -")
        else:
            break


bnf = None


def BNF():
    """
    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 ]*
    """
    global bnf
    if not bnf:
        # use CaselessKeyword for e and pi, to avoid accidentally matching
        # functions that start with 'e' or 'pi' (such as 'exp'); Keyword
        # and CaselessKeyword only match whole words
        e = CaselessKeyword("E")
        pi = CaselessKeyword("PI")
        # fnumber = Combine(Word("+-"+nums, nums) +
        #                    Optional("." + Optional(Word(nums))) +
        #                    Optional(e + Word("+-"+nums, nums)))
        # or use provided pyparsing_common.number, but convert back to str:
        # fnumber = ppc.number().addParseAction(lambda t: str(t[0]))
        fnumber = Regex(r"[+-]?\d+(?:\.\d*)?(?:[eE][+-]?\d+)?")
        ident = Word(alphas, alphanums + "_$")

        plus, minus, mult, div = map(Literal, "+-*/")
        lpar, rpar = map(Suppress, "()")
        addop = plus | minus
        multop = mult | div
        expop = Literal("^")

        expr = Forward()
        expr_list = delimitedList(Group(expr))
        # add parse action that replaces the function identifier with a (name, number of args) tuple
        def insert_fn_argcount_tuple(t):
            fn = t.pop(0)
            num_args = len(t[0])
            t.insert(0, (fn, num_args))

        fn_call = (ident + lpar - Group(expr_list) + rpar).setParseAction(
            insert_fn_argcount_tuple
        )
        atom = (
            addop[...]
            + (
                (fn_call | pi | e | fnumber | ident).setParseAction(push_first)
                | Group(lpar + expr + rpar)
            )
        ).setParseAction(push_unary_minus)

        # 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 + (expop + factor).setParseAction(push_first)[...]
        term = factor + (multop + factor).setParseAction(push_first)[...]
        bnf <<= term + (addop + term).setParseAction(push_first)[...]
    return bnf


# map operator symbols to corresponding arithmetic operations
epsilon = 1e-12
opn = {
    "+": operator.add,
    "-": operator.sub,
    "*": operator.mul,
    "/": operator.truediv,
    "^": operator.pow,
}

fn = {
    "sin": math.sin,
    "cos": math.cos,
    "tan": math.tan,
    "exp": math.exp,
    "abs": abs,
    "trunc": int,
    "round": round,
    "sgn": lambda a: -1 if a < -epsilon else 1 if a > epsilon else 0,
    # functionsl with multiple arguments
    "multiply": lambda a, b: a * b,
    "hypot": math.hypot,
    # functions with a variable number of arguments
    "all": lambda *a: all(a),
}


def evaluate_stack(s):
    op, num_args = s.pop(), 0
    if isinstance(op, tuple):
        op, num_args = op
    if op == "unary -":
        return -evaluate_stack(s)
    if op in "+-*/^":
        # note: operands are pushed onto the stack in reverse order
        op2 = evaluate_stack(s)
        op1 = evaluate_stack(s)
        return opn[op](op1, op2)
    elif op == "PI":
        return math.pi  # 3.1415926535
    elif op == "E":
        return math.e  # 2.718281828
    elif op in fn:
        # note: args are pushed onto the stack in reverse order
        args = reversed([evaluate_stack(s) for _ in range(num_args)])
        return fn[op](*args)
    elif op[0].isalpha():
        raise Exception("invalid identifier '%s'" % op)
    else:
        # try to evaluate as int first, then as float if int fails
        try:
            return int(op)
        except ValueError:
            return float(op)

此代码适用于小型计算：

>>> import fourFn
>>> fourFn.BNF().parseString("9+9", parseAll=True)
>>> val = fourFn.evaluate_stack(fourFn.exprStack[:])
>>> print(val)
18
>>>

但是当我尝试像 "9^9^9" 这样的大计算时，程序在尝试计算 val 时冻结。

>>> import fourFn
>>> fourFn.BNF().parseString("9^9^9", parseAll=True)
>>> val = fourFn.evaluate_stack(fourFn.exprStack[:])

是否可以让评估停止并在它变得太大时抛出错误？

Answer 1

这很可能是不可能的，因为它 Python 很慢，而不是你的口译员。您正在尝试计算 9 ** 387420489，这是一个巨大的数字（并且 Python 的整数实际上是无界的，因此它将继续计算该整数，直到运行超出记忆）。由于 Python 一次只能运行一个线程，并且整数求幂是在 C 中实现的，因此纯 Python 代码无法知道解释器正在“卡住”计算某些东西。

您可以设置一些其他代码来监视此代码的执行，如果它没有在 n 秒内终止，则将其终止，但这可能有点矫枉过正。顺便说一句，我已经计算了这个数字 5 分钟了，仍然没有答案。

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