python - 将递归计算器重构为迭代计算器

Question

我有一个 Django 应用程序，它是一个计算器。用户在一个屏幕上配置任意深度的计算(例如 Excel 公式)，然后在另一个屏幕上输入(单元格)数据。

将字段链接到其值后，我最终得到以下形式的公式

SUM(1,2,4)

可能是任意深度，例如

SUM(1,SUM(5,DIFFERENCE(6,DIVISION(8,10),7),4),2)

让我头疼的公式之一是用户在我们系统中输入的更复杂的公式之一:

ROUND(MULTIPLICATION(DIVISION(ROUND(SUM(MULTIPLICATION(50.00000,50.00000),MULTIPLICATION(50.00000,50.00000,2),MULTIPLICATION(50.00000,50.00000)),-2),300),DIFFERENCE(DIFFERENCE(41,7),0.3)),0)

我使用 pyparsing 来解析公式并提取值和嵌套公式，并递归地执行计算。问题是，由于解析每个嵌套计算，我遇到了 pyparsing 的递归限制。

我的递归代码:

class Calculator:
def __init__(self, formula=None):
    self.formula = formula

def do_calculation(self):
    # parse the formula we receive, which returns arguments in groups of numbers and nested calculations
    # e.g. SUM(MULTIPLICATION(12,11),1,5)
    parsed_formula = FormulaParser(self.formula).get_parsed_formula()

    #calculation name is the outermost level calculation
    calc_name = parsed_formula['calculation_name']
    #don't stomp on built in round
    if 'ROUND' in "".join(calc_name):
        calc_name = ["ROUND_CALCULATION"]
    #don't stomp on if
    if 'IF' == "".join(calc_name):
        calc_name = ['IF_STATEMENT']
    #grab the name of the calculation, will match a function name below
    ex = getattr(self, string.lower("".join(calc_name)))
    calc_arguments = []
    #formulas need to be recursively executed
    formulas = parsed_formula.args.formulas.asList() if len(parsed_formula.args.formulas) else []
    #numbers are just added to the arguments
    dnumbers = parsed_formula.args.dnumbers.asList() if len(parsed_formula.args.dnumbers) else []

    for arg in parsed_formula.args.asList():
        if arg in dnumbers:
            calc_arguments.append(''.join(arg))
        elif arg in formulas:
            new_calc = Calculator(''.join(self.flatten(arg[:])))
            calc_arguments.append(new_calc.do_calculation())

    #execute the calculation with the number arguments
    for idx, arg in enumerate(calc_arguments):
        if isinstance(arg, dict) and arg['rounding']:
            calc_arguments[idx] = arg['result']
    result = ex(*calc_arguments)
    #for rounding, output is special to tell the api to not format to default 5 decimal places
    if 'ROUND' in "".join(calc_name):
        return dict(result=result, rounding=True)
    return result

# function called on nested calculations that may have other nested calculations to flatten to a single level list
@staticmethod
def flatten(expr):
    for i, x in enumerate(expr):
        while isinstance(expr[i], list):
            expr[i:i + 1] = expr[i]
    return expr

以及公式解析器:

class FormulaParser():
def __init__(self, formula=None):
    self.formula = formula

    # grammar
    # end result
    expr = Forward()
    formula = Forward()

    #calculation keywords
    calc_keyword = lambda name: Keyword(name)
    calculations = [calc_keyword(calc) for calc in CALCULATION_TYPES]
    calc_name = Group(reduce(lambda y, z: y | z, [x for x in calculations])).setResultsName('calculation_name')

    #symbols
    oparen, cparen, comma, dot, minus = map(Literal, '(),.-')
    dnumber = Combine(Optional(minus) + Word(nums) + Optional(dot + Word(nums)))

    #possible formulas
    expr = Group(formula).setResultsName('formulas', listAllMatches=True) | Group(dnumber).setResultsName(
        'dnumbers', listAllMatches=True)
    exprs = expr + ZeroOrMore(comma + expr)

    #entire formula
    formula << Combine(calc_name + Group(oparen + exprs + cparen).setResultsName('args'))
    self.parsed_formula = formula

def get_parsed_formula(self):
    if self.formula:
        return self.parsed_formula.parseString(self.formula)

    return None

我已经使用 this SO answer 中的堆栈方法将应用程序的其他递归部分重构为迭代。 .

虽然这可以将字段定义链接到用户输入，但当计算归结为仅参数时，我很难理解如何执行计算，然后将结果传递到下一个堆栈级别，等等.

Answer 1

我不确定这是否能帮助您专注于递归解析堆栈。如果您只想评估表达式，那么您可以使用解析操作处理所有事情，并在解析时进行评估。请参阅我的 mods 中对您提供的源代码的嵌入注释:

sample = """ROUND(MULTIPLICATION(DIVISION(ROUND(SUM(MULTIPLICATION(50.00000,50.00000),MULTIPLICATION(50.00000,50.00000,2),MULTIPLICATION(50.00000,50.00000)),-2),300),DIFFERENCE(DIFFERENCE(41,7),0.3)),0)"""

from pyparsing import *

CALCULATION_TYPES = "ROUND MULTIPLICATION DIVISION SUM DIFFERENCE".split()

functionMap = {
    "ROUND"          : lambda args: round(args[0]),
    "MULTIPLICATION" : lambda args: args[0]*args[1],
    "DIVISION"       : lambda args: args[0]/args[1],
    "SUM"            : lambda args: args[0]+args[1],
    "DIFFERENCE"     : lambda args: args[0]-args[1],
    }

class FormulaParser():
    def __init__(self, formula=None):
        self.formula = formula

        # grammar
        # end result
        expr = Forward()
        formula = Forward()

        #calculation keywords
        calc_keyword = lambda name: Keyword(name)
        calculations = [calc_keyword(calc) for calc in CALCULATION_TYPES]
        calc_name = Group(reduce(lambda y, z: y | z, [x for x in calculations])).setResultsName('calculation_name')

        # a simpler way to create a MatchFirst of all your calculations
        # also, save the results names for when you assemble small elements into larger ones
        calc_name = MatchFirst(calculations)

        #symbols
        oparen, cparen, comma, dot, minus = map(Literal, '(),.-')
        #dnumber = Combine(Optional(minus) + Word(nums) + Optional(dot + Word(nums)))
        # IMPORTANT - convert numbers to floats at parse time with this parse action
        dnumber = Regex(r'-?\d+(\.\d+)?').setParseAction(lambda toks: float(toks[0]))

        #possible formulas
        #expr = Group(formula).setResultsName('formulas', listAllMatches=True) |
        #       Group(dnumber).setResultsName('dnumbers', listAllMatches=True)
        #exprs = expr + ZeroOrMore(comma + expr)

        #entire formula
        #formula << Combine(calc_name + Group(oparen + exprs + cparen).setResultsName('args'))
        #self.parsed_formula = formula

        # define what is allowed for a function arg
        arg_expr = dnumber | formula
        def eval_formula(tokens):
            fn = functionMap[tokens.calculation_name]
            return fn(tokens.args)

        # define overall formula, and add results names here
        formula <<= (calc_name("calculation_name") + oparen 
                                        + Optional(delimitedList(arg_expr))('args') 
                                        + cparen).setParseAction(eval_formula)
        self.parsed_formula = formula


    def get_parsed_formula(self):
        if self.formula:
            return self.parsed_formula.parseString(self.formula)

        return None

fp = FormulaParser(sample)
print fp.get_parsed_formula()