使用从 json 键定义的抽象方法创建 ABC

Question

假设我有一个 json 文件，如下所示：

{
    "foo": ["hi", "there"],
    "bar": ["nothing"]
}

我想创建一个抽象基础class（ABC），其中抽象方法的名称是上面json的键，即：

from abc import ABCMeta, abstractmethod

class MyABC(metaclass=ABCMeta):

    @abstractmethod
    def foo(self):
        pass

    @abstractmethod
    def bar(self):
        pass

问题是 json 文件实际上有很多密钥。我想知道有没有这样的方法：

import json

with open("the_json.json") as f:
    the_json = json.load(f)

class MyABC(metaclass=ABCMeta):

    # for k in the_json.keys():
    #     create abstract method k

感谢评论中的建议，但不知何故没有按预期工作。这是我尝试过的：

class MyABC(metaclass=ABCMeta):
    pass

def f(self):
    pass

setattr(MyABC, "foo", abstractmethod(f))
# I also tried
# setattr(MyABC, "foo", abstractmethod(lambda self: ...))

# Try to define another class that inherits MyABC
class MyClass(MyABC):
    pass

c = MyClass()
# Now this should trigger TypeError but it doesn't
# I can even call c.foo() without getting any errors

Answer 1

这可能有效：

from abc import ABCMeta, abstractmethod

with open("the_json.json") as f:
    the_json = json.load(f)

class MyABC(metaclass=ABCMeta):

    def func(self):
        pass

    for k in the_json:
        locals()[k] = abstractmethod(func)

# Delete attribute "func" is a must
# Otherwise it becomes an additional abstract method in MyABC
delattr(MyABC, "func")
delattr(MyABC, "f")
delattr(MyABC, "k")

class MyClass(MyABC):
    pass

MyClass()
# TypeError: Can't instantiate abstract class MyClass with abstract methods bar, foo

如果您尝试实例化 MyABC 或未实现抽象方法的 MyABC 的子类，它将正确地抛出错误。

使用从 json 键定义的抽象方法创建 ABC

Create an ABC with abstract methods defined from json keys

python

json

python-3.x

abc