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！！！