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

def enum( *names ):

    '''
    Makes enum.
    Usage:
        E = enum( 'YOUR', 'KEYS', 'HERE' )
        print( E.HERE )
    '''

    class Enum():
        pass
    for index, name in enumerate( names ):
        setattr( Enum, name, index )
    return Enum

下面是亚历克·托马斯解决方案的一个变种:

def enum(*args, **kwargs):
    return type('Enum', (), dict((y, x) for x, y in enumerate(args), **kwargs)) 

x = enum('POOH', 'TIGGER', 'EEYORE', 'ROO', 'PIGLET', 'RABBIT', 'OWL')
assert x.POOH == 0
assert x.TIGGER == 1

这是我见过的最好的一个:“Python中的First Class Enums”

http://code.activestate.com/recipes/413486/

It gives you a class, and the class contains all the enums. The enums can be compared to each other, but don't have any particular value; you can't use them as an integer value. (I resisted this at first because I am used to C enums, which are integer values. But if you can't use it as an integer, you can't use it as an integer by mistake so overall I think it is a win.) Each enum is a unique value. You can print enums, you can iterate over them, you can test that an enum value is "in" the enum. It's pretty complete and slick.

编辑(cfi):上面的链接不兼容Python 3。下面是我将枚举.py移植到Python 3的端口:

def cmp(a,b):
   if a < b: return -1
   if b < a: return 1
   return 0


def Enum(*names):
   ##assert names, "Empty enums are not supported" # <- Don't like empty enums? Uncomment!

   class EnumClass(object):
      __slots__ = names
      def __iter__(self):        return iter(constants)
      def __len__(self):         return len(constants)
      def __getitem__(self, i):  return constants[i]
      def __repr__(self):        return 'Enum' + str(names)
      def __str__(self):         return 'enum ' + str(constants)

   class EnumValue(object):
      __slots__ = ('__value')
      def __init__(self, value): self.__value = value
      Value = property(lambda self: self.__value)
      EnumType = property(lambda self: EnumType)
      def __hash__(self):        return hash(self.__value)
      def __cmp__(self, other):
         # C fans might want to remove the following assertion
         # to make all enums comparable by ordinal value {;))
         assert self.EnumType is other.EnumType, "Only values from the same enum are comparable"
         return cmp(self.__value, other.__value)
      def __lt__(self, other):   return self.__cmp__(other) < 0
      def __eq__(self, other):   return self.__cmp__(other) == 0
      def __invert__(self):      return constants[maximum - self.__value]
      def __nonzero__(self):     return bool(self.__value)
      def __repr__(self):        return str(names[self.__value])

   maximum = len(names) - 1
   constants = [None] * len(names)
   for i, each in enumerate(names):
      val = EnumValue(i)
      setattr(EnumClass, each, val)
      constants[i] = val
   constants = tuple(constants)
   EnumType = EnumClass()
   return EnumType


if __name__ == '__main__':
   print( '\n*** Enum Demo ***')
   print( '--- Days of week ---')
   Days = Enum('Mo', 'Tu', 'We', 'Th', 'Fr', 'Sa', 'Su')
   print( Days)
   print( Days.Mo)
   print( Days.Fr)
   print( Days.Mo < Days.Fr)
   print( list(Days))
   for each in Days:
      print( 'Day:', each)
   print( '--- Yes/No ---')
   Confirmation = Enum('No', 'Yes')
   answer = Confirmation.No
   print( 'Your answer is not', ~answer)

虽然最初的全会提案PEP 354在几年前被否决，但它不断被提出。本打算将某种enum添加到3.2，但它被推回到3.3，然后被遗忘。现在有一个PEP 435打算包含在Python 3.4中。PEP 435的参考实现是flufl.enum。

截至2013年4月，似乎有一个普遍的共识，即应该在3.4的标准库中添加一些东西——只要人们能够就“一些东西”应该是什么达成一致。这是最难的部分。看看从这里和这里开始的线程，以及2013年初的六个其他线程。

与此同时，每次出现这个问题时，PyPI、ActiveState等上都会出现大量的新设计和实现，所以如果你不喜欢FLUFL设计，可以尝试一下PyPI搜索。

来自PyPI的枚举包提供了枚举的健壮实现。之前的回答提到了PEP 354;这一提议遭到拒绝，但该提议得到了执行 http://pypi.python.org/pypi/enum。

用法简单而优雅:

>>> from enum import Enum
>>> Colors = Enum('red', 'blue', 'green')
>>> shirt_color = Colors.green
>>> shirt_color = Colors[2]
>>> shirt_color > Colors.red
True
>>> shirt_color.index
2
>>> str(shirt_color)
'green'