如何在Python中实现可订阅类(可订阅类,而不是可订阅对象)？

Question

要实现可订阅对象很简单,只需__getitem__在此对象的类定义中实现即可.
但是现在我想实现一个可订阅的类.例如,我想实现此代码:

class Fruit(object):
    Apple = 0
    Pear = 1
    Banana = 2
    #________________________________ 
    #/ Some other definitions,         \
    #\ make class 'Fruit' subscriptable. /
    # -------------------------------- 
    #        \   ^__^
    #         \  (oo)\_______
    #            (__)\       )\/\
    #                ||----w |
    #                ||     ||

print Fruit['Apple'], Fruit['Banana']
#Output: 0 2

我知道getattr可以做同样的事情,但我觉得下标访问更优雅.

Answer 1

似乎通过改变元类来工作.对于Python 2:

class GetAttr(type):
    def __getitem__(cls, x):
        return getattr(cls, x)

class Fruit(object):
    __metaclass__ = GetAttr

    Apple = 0
    Pear = 1
    Banana = 2

print Fruit['Apple'], Fruit['Banana']
# output: 0 2

在Python 3上,您应该直接使用Enum:

import enum

class Fruit(enum.Enum):
    Apple = 0
    Pear = 1
    Banana = 2

print(Fruit['Apple'], Fruit['Banana'])
# Output: Fruit.Apple, Fruit.Banana
print(Fruit['Apple'].value, Fruit['Banana'].value)
# Output: 0 2

Answer 2

在课堂上添加以下内容:

class Fruit(object):
     def __init__(self):
         self.Fruits = {"Apple": 0, "Pear": 1, "Banana": 2}
     def __getitem__(self, item):
         return self.Fruits[item]

Answer 3

我确实认为您是在询问订阅类而不是类的实例。

这是我对这个问题的回答：“如何在Python中创建可下标的类？”

class Subscriptable:
    def __class_getitem__(cls, item):
        return cls._get_child_dict()[item]

    @classmethod
    def _get_child_dict(cls):
        return {k: v for k, v in cls.__dict__.items() if not k.startswith('_')}


class Fruits(Subscriptable):
    Apple = 0
    Pear = 1
    Banana = 2

>>> Fruits['Apple']
    0
>>> Fruits['Pear']
    1

Answer 4

扩展@LuisKleinwort 的答案，如果您想对所有类属性执行此操作：

fruits_dict = {'apple':0, 'banana':1}

class Fruits(object):
    def __init__(self, args):
        for k in args:
            setattr(self, k, args[k])
            
    def __getitem__(self, item):
        return getattr(self, item)

fruits = Fruits(fruits_dict)
print(fruits.apple)
print(fruits['apple'])