我选择使用元类方法来解决这个问题。
from enum import EnumMeta
class MetaClsEnumJoin(EnumMeta):
"""
Metaclass that creates a new `enum.Enum` from multiple existing Enums.
@code
from enum import Enum
ENUMA = Enum('ENUMA', {'a': 1, 'b': 2})
ENUMB = Enum('ENUMB', {'c': 3, 'd': 4})
class ENUMJOINED(Enum, metaclass=MetaClsEnumJoin, enums=(ENUMA, ENUMB)):
pass
print(ENUMJOINED.a)
print(ENUMJOINED.b)
print(ENUMJOINED.c)
print(ENUMJOINED.d)
@endcode
"""
@classmethod
def __prepare__(metacls, name, bases, enums=None, **kargs):
"""
Generates the class's namespace.
@param enums Iterable of `enum.Enum` classes to include in the new class. Conflicts will
be resolved by overriding existing values defined by Enums earlier in the iterable with
values defined by Enums later in the iterable.
"""
#kargs = {"myArg1": 1, "myArg2": 2}
if enums is None:
raise ValueError('Class keyword argument `enums` must be defined to use this metaclass.')
ret = super().__prepare__(name, bases, **kargs)
for enm in enums:
for item in enm:
ret[item.name] = item.value #Throws `TypeError` if conflict.
return ret
def __new__(metacls, name, bases, namespace, **kargs):
return super().__new__(metacls, name, bases, namespace)
#DO NOT send "**kargs" to "type.__new__". It won't catch them and
#you'll get a "TypeError: type() takes 1 or 3 arguments" exception.
def __init__(cls, name, bases, namespace, **kargs):
super().__init__(name, bases, namespace)
#DO NOT send "**kargs" to "type.__init__" in Python 3.5 and older. You'll get a
#"TypeError: type.__init__() takes no keyword arguments" exception.
这个元类可以这样使用：
>>> from enum import Enum
>>>
>>> ENUMA = Enum('ENUMA', {'a': 1, 'b': 2})
>>> ENUMB = Enum('ENUMB', {'c': 3, 'd': 4})
>>> class ENUMJOINED(Enum, metaclass=MetaClsEnumJoin, enums=(ENUMA, ENUMB)):
... e = 5
... f = 6
...
>>> print(repr(ENUMJOINED.a))
<ENUMJOINED.a: 1>
>>> print(repr(ENUMJOINED.b))
<ENUMJOINED.b: 2>
>>> print(repr(ENUMJOINED.c))
<ENUMJOINED.c: 3>
>>> print(repr(ENUMJOINED.d))
<ENUMJOINED.d: 4>
>>> print(repr(ENUMJOINED.e))
<ENUMJOINED.e: 5>
>>> print(repr(ENUMJOINED.f))
<ENUMJOINED.f: 6>
此方法使用与源Enums 相同的名称-值对创建一个新的Enum，但生成的Enum 成员仍然是唯一的。名称和值将是相同的，但它们将无法按照 Python 的 Enum 类设计的精神与其起源进行直接比较：
>>> ENUMA.b.name == ENUMJOINED.b.name
True
>>> ENUMA.b.value == ENUMJOINED.b.value
True
>>> ENUMA.b == ENUMJOINED.b
False
>>> ENUMA.b is ENUMJOINED.b
False
>>>
注意在命名空间冲突时会发生什么：
>>> ENUMC = Enum('ENUMA', {'a': 1, 'b': 2})
>>> ENUMD = Enum('ENUMB', {'a': 3})
>>> class ENUMJOINEDCONFLICT(Enum, metaclass=MetaClsEnumJoin, enums=(ENUMC, ENUMD)):
... pass
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 19, in __prepare__
File "C:\Users\jcrwfrd\AppData\Local\Programs\Python\Python37\lib\enum.py", line 100, in __setitem__
raise TypeError('Attempted to reuse key: %r' % key)
TypeError: Attempted to reuse key: 'a'
>>>
这是由于基本 enum.EnumMeta.__prepare__ 返回一个特殊的 enum._EnumDict 而不是典型的 dict 对象，该对象在键分配时表现不同。您可能希望通过用try-except TypeError 将其包围来抑制此错误消息，或者可能有一种方法可以在调用super().__prepare__(...) 之前修改命名空间。