Question

我有一个字典，其中包含通过Web应用程序提供的一组键值：我想处理用户提供的公式，如：（（值1 +值3）/ VALUE4）* 100

最简单的方法是使公式计算出匹配值与字典中的值？

考虑这个例子：

#!/usr/bin/python
values={'value1':10,'value2':1245,'value3':674365,'value4':65432,'value5':131}
formula=raw_input('Please enter formula:')

如果我提供公式'（（value1 + value3）/ value4）* 100'，我如何将value1等映射到字典中的map1并计算结果？

干杯，

Answer 1

eval可用于执行恶意代码。

您相信您的用户吗？如果是这样，您可以将values作为全局字典传递给eval使用。因此，eval可以直接评估用户公式，而无需任何其他字符串操作：

values={'value1':10,'value2':1245,'value3':674365,'value4':65432,'value5':131}
formula=raw_input('Please enter formula:')
values=eval(formula,values)
print(values)

如果您不信任潜在用户，可以使用pyparsing：以下是Paul McGuire的numeric expression parser, fourFn.py，包含在一个类中以便于使用。

utils_parse_numeric.py：

from __future__ import division
from pyparsing import (Literal,CaselessLiteral,Word,Combine,Group,Optional,
                       ZeroOrMore,Forward,nums,alphas,oneOf)
import math
import operator   

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()
        addop  = plus | minus
        multop = mult | div
        expop = Literal( "^" )
        pi    = CaselessLiteral( "PI" )
        expr = Forward()
        atom = ((Optional(oneOf("- +")) +
                 (pi|e|fnumber|ident+lpar+expr+rpar).setParseAction(self.pushFirst))
                | Optional(oneOf("- +")) + Group(lpar+expr+rpar)
                ).setParseAction(self.pushUMinus)       
        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 ) )
        self.bnf = expr
        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}
    def evaluateStack(self, s ):
        op = s.pop()
        if op == 'unary -':
            return -self.evaluateStack( s )
        if op in "+-*/^":
            op2 = self.evaluateStack( s )
            op1 = self.evaluateStack( s )
            return self.opn[op]( op1, op2 )
        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[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

然后你的脚本可以这样做：

import re
import utils_parse_numeric as upn

my_dict={
    'number1':54,
    'number2':1234,
    'number3':778,
    'number25':2109}

这使用re模块将my_dict [“numberXXX”]替换为'numberXXX'：

def callback(match):
    num=match.group(1)
    key='number{0}'.format(num)
    val=my_dict[key]
    return str(val)

astr='((number1+number3)/number2)*100'
astr=re.sub('number(\d+)',callback,astr)

以下是NumericStringParser如何用于安全地评估数值表达式：

nsp=upn.NumericStringParser()
result=nsp.eval(astr)
print(result)

这比使用eval更安全。所有无效表达式都会引发pyparsing.ParseException。

Answer 2

验证公式和数字值（通过正则表达式后）后，您可以执行以下操作：

arr = {'num1':4, 'num2':5, 'num3':7}
formula = '(num1+num2)*num3'

for key, val in arr.items():
    formula = formula.replace(key, str(val))

res = eval(formula)
print res

Answer 3

感谢所有输入。我个人发现lacopo的答案最适合我的情况。

以下是对解决方案的粗略概念：

import sys

values={'one':10,'two':1245,'three':674365,'four':65432,'five':131}
print str(values)

formula=raw_input('Please enter formula:')

for key, val in values.items():
        formula = formula.replace(key, str(val))

whitelist=[ '+','-','/','*','^','(',')' ]

to_evaluate=re.findall('\D',formula)
to_evaluate=list(set(to_evaluate))

for element in to_evaluate:
        if not element in whitelist:
                print "Formula contains an invalid character: "+str(element)
                sys.exit(1)


print eval(formula)

如何处理用户提供的公式？

3 个答案: