接受输入（专业）并返回输出（饮料类型）的写函数[关闭]

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关闭29分钟前。

此代码适用于所有输入，但如果输入无效，则应返回“啤酒”。我不确定是否有办法在字典中包含一个变量，该变量允许python中所有其他未分配的输入（例如“ else”语句），因此在末尾尝试了“ else”语句，但是此语法无效，因为之前没有“ if”语句。最后，如果输入不在列表中，那么我做了一个笨拙的添加，然后返回啤酒，但是这不起作用。

def get_drink_by_profession(param):
    p_to_d = param.title()
    return{
        "Jabroni":"Patron Tequila",
        "School Counselor":"Anything with Alcohol",
        "Programmer":"Hipster Craft Beer",
        "Bike Gang Member":"Moonshine" ,
        "Politician":"Your tax dollars",
        "Rapper":"Cristal",
        }[p_to_d]
    if param not in p_to_d:
        return "Beer"

Answer 1

您想进行安全获取并将“啤酒”作为默认值，如果不存在密钥refer here

def get_drink_by_profession(param):
    p_to_d = param.title()
    return{
        "Jabroni":"Patron Tequila",
        "School Counselor":"Anything with Alcohol",
        "Programmer":"Hipster Craft Beer",
        "Bike Gang Member":"Moonshine" ,
        "Politician":"Your tax dollars",
        "Rapper":"Cristal",
        }.get(p_to_d, "Beer")

Answer 2

您可以在大型return语句之前访问p_to_d，因此您要做的就是更改其顺序。

def get_drink_by_profession(param):
    p_to_d = param.title()
    if param not in p_to_d:
        return "Beer"    
    return{
        "Jabroni":"Patron Tequila",
        "School Counselor":"Anything with Alcohol",
        "Programmer":"Hipster Craft Beer",
        "Bike Gang Member":"Moonshine" ,
        "Politician":"Your tax dollars",
        "Rapper":"Cristal",
        }[p_to_d]