为什么枚举必须是整数?不幸的是，在不改变Python语言的情况下，我想不出任何好看的构造来生成它，所以我将使用字符串:

class Enumerator(object):
    def __init__(self, name):
        self.name = name

    def __eq__(self, other):
        if self.name == other:
            return True
        return self is other

    def __ne__(self, other):
        if self.name != other:
            return False
        return self is other

    def __repr__(self):
        return 'Enumerator({0})'.format(self.name)

    def __str__(self):
        return self.name

class Enum(object):
    def __init__(self, *enumerators):
        for e in enumerators:
            setattr(self, e, Enumerator(e))
    def __getitem__(self, key):
        return getattr(self, key)

也许现在更好的是，为了配置文件或其他远程输入，我们可以自然地对字符串进行测试。

例子:

class Cow(object):
    State = Enum(
        'standing',
        'walking',
        'eating',
        'mooing',
        'sleeping',
        'dead',
        'dying'
    )
    state = State.standing

In [1]: from enum import Enum

In [2]: c = Cow()

In [3]: c2 = Cow()

In [4]: c.state, c2.state
Out[4]: (Enumerator(standing), Enumerator(standing))

In [5]: c.state == c2.state
Out[5]: True

In [6]: c.State.mooing
Out[6]: Enumerator(mooing)

In [7]: c.State['mooing']
Out[7]: Enumerator(mooing)

In [8]: c.state = Cow.State.dead

In [9]: c.state == c2.state
Out[9]: False

In [10]: c.state == Cow.State.dead
Out[10]: True

In [11]: c.state == 'dead'
Out[11]: True

In [12]: c.state == Cow.State['dead']
Out[11]: True

另一个非常简单的Python枚举实现，使用namedtuple:

from collections import namedtuple

def enum(*keys):
    return namedtuple('Enum', keys)(*keys)

MyEnum = enum('FOO', 'BAR', 'BAZ')

,或者

# With sequential number values
def enum(*keys):
    return namedtuple('Enum', keys)(*range(len(keys)))

# From a dict / keyword args
def enum(**kwargs):
    return namedtuple('Enum', kwargs.keys())(*kwargs.values())




# Example for dictionary param:
values = {"Salad": 20, "Carrot": 99, "Tomato": "No i'm not"} 
Vegetables= enum(**values)

# >>> print(Vegetables.Tomato)        'No i'm not'


# Example for keyworded params: 
Fruits = enum(Apple="Steve Jobs", Peach=1, Banana=2)

# >>> print(Fruits.Apple)             'Steve Jobs'

就像上面子类设置的方法一样，这允许:

'FOO' in MyEnum
other = MyEnum.FOO
assert other == MyEnum.FOO

但是具有更大的灵活性，因为它可以有不同的键和值。这允许

MyEnum.FOO < MyEnum.BAR

如果使用填充连续数字值的版本，则按预期操作。

Alec Thomas简洁回答的一个变体(支持获取枚举值的名称):

class EnumBase(type):
    def __init__(self, name, base, fields):
        super(EnumBase, self).__init__(name, base, fields)
        self.__mapping = dict((v, k) for k, v in fields.iteritems())
    def __getitem__(self, val):
        return self.__mapping[val]

def enum(*seq, **named):
    enums = dict(zip(seq, range(len(seq))), **named)
    return EnumBase('Enum', (), enums)

Numbers = enum(ONE=1, TWO=2, THREE='three')
print Numbers.TWO
print Numbers[Numbers.ONE]
print Numbers[2]
print Numbers['three']

我喜欢在Python中这样定义枚举:

class Animal:
  class Dog: pass
  class Cat: pass

x = Animal.Dog

这比使用整数更有漏洞，因为你不必担心确保整数是唯一的(例如，如果你说Dog = 1和Cat = 1，你就完蛋了)。

它比使用字符串更防bug，因为你不必担心拼写错误(例如。 x == "猫"无声失败，但x ==动物。Catt是一个运行时异常)。

附录: 你甚至可以通过让Dog和Cat继承一个具有正确元类的符号类来增强这个解决方案:

class SymbolClass(type):
    def __repr__(self): return self.__qualname__
    def __str__(self): return self.__name__

class Symbol(metaclass=SymbolClass): pass


class Animal:
    class Dog(Symbol): pass
    class Cat(Symbol): pass

然后，如果你使用这些值来索引一个字典，请求它的表示将使它们看起来很漂亮:

>>> mydict = {Animal.Dog: 'Wan Wan', Animal.Cat: 'Nyaa'}
>>> mydict
{Animal.Dog: 'Wan Wan', Animal.Cat: 'Nyaa'}

def M_add_class_attribs(attribs):
    def foo(name, bases, dict_):
        for v, k in attribs:
            dict_[k] = v
        return type(name, bases, dict_)
    return foo

def enum(*names):
    class Foo(object):
        __metaclass__ = M_add_class_attribs(enumerate(names))
        def __setattr__(self, name, value):  # this makes it read-only
            raise NotImplementedError
    return Foo()

像这样使用它:

Animal = enum('DOG', 'CAT')
Animal.DOG # returns 0
Animal.CAT # returns 1
Animal.DOG = 2 # raises NotImplementedError

如果你只想要唯一的符号，不关心值，替换这行:

__metaclass__ = M_add_class_attribs(enumerate(names))

用这个:

__metaclass__ = M_add_class_attribs((object(), name) for name in names)

对于旧的Python 2.x

def enum(*sequential, **named):
    enums = dict(zip(sequential, [object() for _ in range(len(sequential))]), **named)
    return type('Enum', (), enums)

如果你命名它，是你的问题，但如果不创建对象而不是值允许你这样做:

>>> DOG = enum('BARK', 'WALK', 'SIT')
>>> CAT = enum('MEOW', 'WALK', 'SIT')
>>> DOG.WALK == CAT.WALK
False

当使用这里的其他实现时(在我的例子中使用命名实例时)，必须确保永远不要尝试比较来自不同枚举的对象。这里有一个可能的陷阱:

>>> DOG = enum('BARK'=1, 'WALK'=2, 'SIT'=3)
>>> CAT = enum('WALK'=1, 'SIT'=2)
>>> pet1_state = DOG.BARK
>>> pet2_state = CAT.WALK
>>> pet1_state == pet2_state
True

呵!