Python 将递归计算器重构为迭代计算器
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(比如Excel公式),然后在另一个屏幕上输入(单元格)数据 在将字段链接到它们的值之后,我得到了一个如下的公式Python 将递归计算器重构为迭代计算器,python,recursion,iteration,pyparsing,Python,Recursion,Iteration,Pyparsing,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(比如Excel公式),然后在另一个屏幕上输入(单元格)数据 在将字段链接到它们的值之后,我得到了一个如下的公式 SUM(1,2,4) 可以任意深,例如 SUM(1,SUM(5,DIFFERENCE(6,DIVISION(8,10),7),4),2) 让我头疼的一个公式是用户在我们系统中输入的一个更复杂的公式: ROUND(MULTIPLICATION(DIVISION(ROUND(SUM(MULTIPLICAT
SUM(1,2,4)
可以任意深,例如
SUM(1,SUM(5,DIFFERENCE(6,DIVISION(8,10),7),4),2)
让我头疼的一个公式是用户在我们系统中输入的一个更复杂的公式:
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,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,我在pyparsing中遇到了递归限制
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
以及公式的解析器:
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
class FormulaParser():
定义初始化(self,公式=None):
self.formula=公式
#文法
#最终结果
expr=Forward()
公式=前进()
#计算关键字
计算关键字=lambda名称:关键字(名称)
计算=[calc_关键字(calc)用于计算类型中的计算]
calc_name=Group(reduce(lambda y,z:y|z,[x代表计算中的x])。setResultsName('calculation_name'))
#象征
oparen,cparen,逗号,点,减号=映射(文字,,(),.-)
dnumber=组合(可选(减)+字(nums)+可选(点+字(nums)))
#可能的公式
expr=Group(formula).setResultsName('formulas',listalMatches=True)| Group(dnumber).setResultsName(
“dnumbers”,listAllMatches=True)
exprs=expr+ZeroOrMore(逗号+expr)
#完整公式
公式我不确定这是否能帮助您在解析堆栈中递归。如果您只想对表达式求值,那么您可以使用解析操作处理所有事情,并在解析时进行求值。请参阅my mods中嵌入的注释,以查看您提供的源代码:
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()
sample=“”四舍五入(乘法(除法)(四舍五入(和)(乘法(50.00000,50.00000),乘法(50.00000,50.00000,2),乘法(50.00000,50.00000)),-2),300),差(差(41,7),0.3)),0“
从pyparsing导入*
计算\u TYPES=“四舍五入乘法除法和差”.split()
函数映射={
“圆形”:lambda参数:圆形(参数[0]),
“乘法”:lambda args:args[0]*args[1],
“除法”:lambda args:args[0]/args[1],
“总和”:lambda args:args[0]+args[1],
“差异”:lambda args:args[0]-args[1],
}
类FormulaParser():
定义初始化(self,公式=None):
self.formula=公式
#文法
#最终结果
expr=Forward()
公式=前进()
#计算关键字
计算关键字=lambda名称:关键字(名称)
计算=[calc_关键字(calc)用于计算类型中的计算]
calc_name=Group(reduce(lambda y,z:y|z,[x代表计算中的x])。setResultsName('calculation_name'))
#创建匹配的更简单方法首先是计算
#另外,保存结果名称,以便在将小图元组合成较大图元时使用
计算名称=匹配优先(计算)
#象征
oparen,cparen,逗号,点,减号=映射(文字,,(),.-)
#dnumber=组合(可选(减)+字(nums)+可选(点+字(nums)))
#要点-使用此解析操作在解析时将数字转换为浮点数
dnumber=Regex(r'-?\d+(\.\d+)).setParseAction(lambda-toks:float(toks[0]))
#可能的公式
#expr=Group(formula).setResultsName('formulas',listAllMatches=True)|
#组(dnumber).setResultsName('dnumbers',listAllMatches=True)
#exprs=expr+ZeroOrMore(逗号+expr)
#完整公式
#谢谢。这是一个很好的解决方案。最后,我决定不把担忧混为一谈。我想得太多了。我将解析器简化为我感兴趣的标记,并使用一个简单的堆栈实现来进行计算。