Question

我只想简单列出各种咖啡的清单，但会收到一条错误消息，指出该清单未定义。引用类变量时，是否必须在构造函数中使用self？

我尝试将return语句更改为返回self.coffelist.append(name)，但是又收到另一个错误：'Function' object has no attribute 'append'。

class coffe:

    coffelist = []

    def __init__(self,name,origin,price):
        self.name = name
        self.origin = origin
        self.price = price
        return (self.coffelist.append(self.name))

    def coffelist(self):
        print(coffelist)


c1=coffe("blackcoffe","tanz",55)
c2=coffe("fineroasted","ken",60)

Answer 1

这是因为您将一种方法命名为coffelist。

Answer 2

我认为这表明了如何做自己想做的。我还修改了您的代码以遵循PEP 8 - Style Guide for Python Code并更正了一些拼写错误的单词。

class Coffee:  # Class names should Capitalized.

    coffeelist = []  # Class attribute to track instance names.

    def __init__(self,name,origin,price):
        self.name = name
        self.origin = origin
        self.price = price
        self.coffeelist.append(self.name)

    def print_coffeelist(self):
        print(self.coffeelist)


c1 = Coffee("blackcoffee", "tanz", 55)
c1.print_coffeelist()  # -> ['blackcoffee']
c2 = Coffee("fineroasted", "ken", 60)
c1.print_coffeelist()  # -> ['blackcoffee', 'fineroasted']

# Can also access attribute directly through the class:
print(Coffee.coffeelist)  # -> ['blackcoffee', 'fineroasted']

Answer 3

是的，这正是我想要的！我不确定..我以为您可以在return语句中同时做两件事，都返回append。我猜想python的分配时间非常灵活，有时甚至不是。谢谢

如何将列表用作类变量，以便将实例对象（参数）附加到列表？

3 个答案: