python 计算器的（正则匹配+递归）

经过2天的长时间的战斗，python计算器终于完成了。

 import re
 val="1-2*((60-30*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2))"
 def chen(arg):
     f = arg
 #    print (f)
     if "*" in arg or "/" in arg:
         '''如果包含*号或者除好则进如该方法'''
         b2_2 = re.search("\d+\.?\d*[*/]+-?\d+\.?\d*", arg)
         m = b2_2.group()
         k = re.split("([*/])", m, 1)
         kn = k[0] + '\\' + k[1] + k[2]
         if k[1] == '*':
            new_a = float(k[0]) * float(k[2])
            f = re.sub(kn, str(new_a), arg)
            new_z = chen(f)
            return new_z
         elif k[1] == '/':
            new_a = float(k[0]) / float(k[2])
            f = re.sub(kn, str(new_a), arg)
            new_z = chen(f)
            return new_z
     elif re.match("-+\d+\.?\d*[+-]+\d+\.?\d*",arg):
         '''如果是负数的减法则进如该方法'''
         m = re.search("-+\d+\.?\d*[+-]+\d+\.?\d*", arg).group()
         k = re.split("([+-])", m, 1)
         kt =re.split("([+-])", m, 2)
         kn = k[0] + '\\' + k[1] + k[2]
         aa = "%s%s" %(kt[1],kt[2])
         if kt[3] == '-':
             new_a = float(aa) - float(kt[4])
             f = re.sub(kn, str(new_a), arg)
             new_z = chen(f)
             return new_z
     elif re.search("\d+\.?\d*[+-]+\d+\.?\d*", arg):
         '''如果是 加号和减号则进如该方法'''
 #        b2_2 = re.search("\d+\.*\d*[+-]+\d+\.*\d*", arg)
         b2_2 = re.search("\d+\.?\d*[+-]+\d+\.?\d*", arg)
         m = b2_2.group()
         k = re.split("([+-])", m, 1)
         kn = k[0] + '\\' + k[1] + k[2]
        # print (k)
         if k[1] == '+':
             new_a = float(k[0]) + float(k[2])
             f = re.sub(kn, str(new_a), arg)
             new_z = chen(f)
             return new_z
         elif k[1] == '-':
             new_a = float(k[0]) - float(k[2])
             f = re.sub(kn, str(new_a), arg)
             new_z = chen(f)
             return new_z
     return f

 #b = re.split("\(([^()]+)\)",val,1)
 while True:
     b = re.split("\(([^()]+)\)", val, 1)
     if len(b) == 3:
     #    print (b)
         a1,a2,a3=b
         chen1=chen(a2)
         val = "%s%s%s" %(a1,chen1,a3)
     else:
         val=chen(val)
         print (val)
         break