python写计算器

    #！/usr/bin/env python
    # -*- coding:utf-8 -*-
    import re
    def chu(arg1):  #定义加减
        arg = arg1[0]  #beacuse price is a list ,so index 0
        arg = arg.replace('--', '+').replace('++', '+').replace('-+', '-').replace('+-', '-')  #重点重点重点：就是对负数的一个替换
    #    r = '-9-2-5-75884-1.6666666666666667-80.0-0.6+18'
        li = re.findall("[\-\+]?\d+\.?\d*", arg)        #意思是提取每个数，用findall变成列表形式，然后循环相加，张宇我佩服你
                                                                            # 意思是用findall提取每个值包括值前面的运算符
        num = 0 #定义空值，然后循环列表让每个值相加
        for i in li:
            num += float(i)
        return num
    def multiply(arg):      #definition multip
       while True:
            nu = re.split("(\d+\.?\d*[*\/][\-]?\d+\.?\d*)",arg,1) #['-9-2-5-', '2*5', '/3+7/3*99/4*2998+10*568/14']
            print(nu)
            if len(nu) == 3:
                bef, cen, aft = nu    #split get there price
                nu_cen = re.split("[*/]",cen) #['2', '5']
                nu_bef, nu_af = nu_cen
     #           print(nu_bef,type(nu_af))
                if "*" in cen:     #如果*在中间那个
                    nu_cen = re.split("\*", cen)
     #               print(nu_cen)
                    nu_bef, nu_af = nu_cen
                    sum = float(nu_bef)*float(nu_af)
     #               print(sum)
                    nu = bef + str(sum) + aft #重新组合定义arg形参
                    arg = nu
                    return multiply(arg)  #返回重新定义函数
                elif "/" in cen:
                    nu_cen = re.split("[/]", cen)
                    nu_bef, nu_af = nu_cen
                    sum = float(nu_bef)/float(nu_af)
                    nu = bef + str(sum) + aft
                    arg = nu
                    return multiply(arg)# ['-9-2-5-3.3333333333333335+173134.50000000003+405.7142857142857']
            else:   #这个时候如果不等于3，那就是只剩下加减了执行加减
                return chu(nu)
       #return arg
    #acc = "-9-2-5-244*311-5/3-40*4/2-3/5+6*3"
    #acc = "-9-2-5-75884-5/3-160/2-3/5+18"
    # acc = "-9-2-5-2*5/3+7/3*99/4*2998+10*568/14"
    # ac = acc.strip(" ")
    # a = multiply(ac)
    # print(a)
    # def multiply(arg):
    #     return 1
    origin = "1 - 2 *  ( (60-30 +(-9-2-5-24*11-5/3-40*4/2-3/5+6*3) * (-9-2-5-2*5/3 + 7 /3*9/4*98 +10 * 568/14 )) - (-4*3)/ (16-3*2) )"
    #寻找括号最里面的括号
    while True: #  只要里面还有最里面的括号，就循环
        origin = re.sub(r"\s*","",origin)  #no1 strinp *****space**
        print(origin)
        res = re.split("\(([^()]+)\)",origin,1) #分割括号str得到第一个最里面括号内的值 no2
        if len(res) == 3:    #equal（等于）3，证明have bracket
            before,centor,after = res   #no3
            print(centor)#centor是最里面的字符串，也是计算新函数乘法或除的实参
            r = multiply(centor)    #定义definition function multiply or ride  (no4)
            new_res = before + str(r) + after #重新组合
            origin = new_res   #重新定义origin
            print(origin)
        else:
            final = multiply(origin)
            print(final)
            break
    #this is comments
    """
    acc = "-9-25-75884-5/3-160/2-3/5+18"
    nu = re.split("(\d+\.?\d*[*][\-]?\d+\.?\d*)",acc,1)
    print(type(nu[0]))
    """  

注：凡注释的字段皆为测试所写，切记不可先全部乘法或除法结果不样！！！