Using Python to implement a complex formula calculator function

Give a section of the following line, more complex operation formula, write your own code calculation 1-2.99 * ((60.2-30 + ( -40/5) * (9-2*5/3 + 7/3*99/4*2998 +10.5 * 568/14))-( -4*3)/(16-3* 2)  ideas: Step1. Need to check legality first, detect the illegal character step2 with no letters and other non-operations. Format formulas, remove spaces, replace (--,+-,-+,++) with double operators such as (+,-,-,+ Step3. First calculate () the multiplication, add and subtract, and then calculate the last non-contained () multiplication, plus and minus

#Check the legality of the formula, there are no illegal charactersdefCheck (s):if  notS.count ('(') = = S.count (')'):        Print('Please check that parentheses are not closed')    elifRe.findall ('[[Email protected]#$]','s') is  notNone:Print('Please check that the formula contains illegal symbols')    elifRe.findall ('[A-za-z]','s') is  notNone:Print('Please check that the formula contains letters')    elifS >= u'\u4e00'  andS <= u'\u9fa5':        Print('Please check that the formula contains Chinese')#formatting formulas, removing spaces, redefining double symbolsdefformats (s): s= S.replace (' ',"') s= S.replace ('++','+') s= S.replace ('--','+') s= S.replace ('+-|-+','-') s= S.replace ('*+','*') s= S.replace ('/+','/')    returnS

Take the number and the method with the decimal point, 2 kinds of results are the same, pay attention to experience
ret = Re.findall (R ' [\d\.] + ', ' NUMBRT 1.58 ABC 123 ')
Print (ret)  # [' 1.58 ', ' 123 ']
ret = Re.findall (' \d\.? \d* ', ' NUMBRT 1.58 ABC 123 ')
Print (ret)  # [' 1.58 ', ' 123 ']

defCheng_chu (s):#handling multiplication with a minus signs =formats (s) r= Re.compile (r'[\d\.] +[\*/]-? [\d\.] +')     whileRe.search (R'[\*/]', s): Ma=Re.search (R, s). Group ()#print (MA)Li = Re.findall (r'(-?[ \d\.] +|\*|/)', MA)ifLI[1] = ='*': Result= str (float (li[0]) * FLOAT (li[2]))        Else: Result= str (float (li[0])/float (li[2])) s= S.replace (MA, result, 1)    returnsdefJia_jian (s):#processing add and subtract, becoming an array, full pluss =formats (s) Li= Re.findall (r'([\d\.] +|\+|-)', s) nums=0 forIinchRange (len (LI)):ifLi[i] = ='-': Li[i]='+'Li[i+ 1] = float (li[i + 1]) *-1 forIinchLi:ifi = ='+': I=0 Nums= Nums +float (i)returnStr (nums)defSimple (x):#handle without parentheses.    returnJia_jian (Cheng_chu (x))defJisuan (x):#Handle the parentheses     while '(' inchX:reg= Re.compile (r'\([^\(\)]+\)') Ma=Re.search (Reg, X). Group () result= Simple (ma[1:-1]) x= X.replace (MA, result, 1)    returnSimple (x)

' 1-2.99 * ((60.2-30 + ( -40/5) * (9-2*5/3 + 7/3*99/4*2998 +10.5 * 568/14))-( -4*3)/(16-3*2) ) '. Replace (' ")print(' your results:')  , Jisuan (ss))print('eval calculation result:', eval (ss))

The following test phase is strange:

If Jisuan (ss) = = eval (ss):
   Print (' Calculate right, you're great yo @[email protected] ')
Else
   Print (' not calculated correctly, please try again! ‘)

# Here is strange, if not str (), direct comparison, it will show ' calculation is incorrect ', is there a decimal point behind it?
If STR (Jisuan (ss)) = = str (eval (ss)):
   Print (' Calculate right, you're great yo @[email protected] ')
Else
   Print (' not calculated correctly, please try again! ‘)

Ask the big guy to point out.

Using Python to implement a complex formula calculator function