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