我非常喜欢Alec Thomas的解决方案(http://stackoverflow.com/a/1695250):

def enum(**enums):
    '''simple constant "enums"'''
    return type('Enum', (object,), enums)

它看起来优雅而简洁，但它只是一个创建具有指定属性的类的函数。

通过对函数进行一些修改，我们可以让它表现得更像“枚举”:

注意:我通过尝试重现 pygtk的新样式'enums'的行为(如gtk . messagtype . warning)

def enum_base(t, **enums):
    '''enums with a base class'''
    T = type('Enum', (t,), {})
    for key,val in enums.items():
        setattr(T, key, T(val))

    return T

这将基于指定类型创建枚举。除了像前面的函数那样提供属性访问外，它的行为与您期望Enum对类型的行为一样。它还继承了基类。

例如，整数enum:

>>> Numbers = enum_base(int, ONE=1, TWO=2, THREE=3)
>>> Numbers.ONE
1
>>> x = Numbers.TWO
>>> 10 + x
12
>>> type(Numbers)
<type 'type'>
>>> type(Numbers.ONE)
<class 'Enum'>
>>> isinstance(x, Numbers)
True

用这个方法还可以做一件有趣的事情，那就是通过覆盖内置方法来定制特定的行为:

def enum_repr(t, **enums):
    '''enums with a base class and repr() output'''
    class Enum(t):
        def __repr__(self):
            return '<enum {0} of type Enum({1})>'.format(self._name, t.__name__)

    for key,val in enums.items():
        i = Enum(val)
        i._name = key
        setattr(Enum, key, i)

    return Enum



>>> Numbers = enum_repr(int, ONE=1, TWO=2, THREE=3)
>>> repr(Numbers.ONE)
'<enum ONE of type Enum(int)>'
>>> str(Numbers.ONE)
'1'

如果你需要数值，这是最快的方法:

dog, cat, rabbit = range(3)

在Python 3中。X你也可以在最后添加一个星号占位符，它将吸收范围内所有剩余的值，以防你不介意浪费内存和无法计数:

dog, cat, rabbit, horse, *_ = range(100)

从Python 3.4开始，正式支持枚举。您可以在Python 3.4文档页面上找到文档和示例。

枚举是使用类语法创建的，这使得它们很容易 读和写。中描述了另一种创建方法 功能的API。定义枚举，子类Enum如下所示:

from enum import Enum
class Color(Enum):
     red = 1
     green = 2
     blue = 3

我非常喜欢Alec Thomas的解决方案(http://stackoverflow.com/a/1695250):

def enum(**enums):
    '''simple constant "enums"'''
    return type('Enum', (object,), enums)

它看起来优雅而简洁，但它只是一个创建具有指定属性的类的函数。

通过对函数进行一些修改，我们可以让它表现得更像“枚举”:

注意:我通过尝试重现 pygtk的新样式'enums'的行为(如gtk . messagtype . warning)

def enum_base(t, **enums):
    '''enums with a base class'''
    T = type('Enum', (t,), {})
    for key,val in enums.items():
        setattr(T, key, T(val))

    return T

这将基于指定类型创建枚举。除了像前面的函数那样提供属性访问外，它的行为与您期望Enum对类型的行为一样。它还继承了基类。

例如，整数enum:

>>> Numbers = enum_base(int, ONE=1, TWO=2, THREE=3)
>>> Numbers.ONE
1
>>> x = Numbers.TWO
>>> 10 + x
12
>>> type(Numbers)
<type 'type'>
>>> type(Numbers.ONE)
<class 'Enum'>
>>> isinstance(x, Numbers)
True

用这个方法还可以做一件有趣的事情，那就是通过覆盖内置方法来定制特定的行为:

def enum_repr(t, **enums):
    '''enums with a base class and repr() output'''
    class Enum(t):
        def __repr__(self):
            return '<enum {0} of type Enum({1})>'.format(self._name, t.__name__)

    for key,val in enums.items():
        i = Enum(val)
        i._name = key
        setattr(Enum, key, i)

    return Enum



>>> Numbers = enum_repr(int, ONE=1, TWO=2, THREE=3)
>>> repr(Numbers.ONE)
'<enum ONE of type Enum(int)>'
>>> str(Numbers.ONE)
'1'

我使用元类来实现枚举(在我的想法中，它是一个const)。代码如下:

class ConstMeta(type):
    '''
    Metaclass for some class that store constants
    '''
    def __init__(cls, name, bases, dct):
        '''
        init class instance
        '''
        def static_attrs():
            '''
            @rtype: (static_attrs, static_val_set)
            @return: Static attributes in dict format and static value set
            '''
            import types
            attrs = {}
            val_set = set()
            #Maybe more
            filter_names = set(['__doc__', '__init__', '__metaclass__', '__module__', '__main__'])
            for key, value in dct.iteritems():
                if type(value) != types.FunctionType and key not in filter_names:
                    if len(value) != 2:
                        raise NotImplementedError('not support for values that is not 2 elements!')
                    #Check value[0] duplication.
                    if value[0] not in val_set:
                        val_set.add(value[0])
                    else:
                        raise KeyError("%s 's key: %s is duplicated!" % (dict([(key, value)]), value[0]))
                    attrs[key] = value
            return attrs, val_set

        attrs, val_set = static_attrs()
        #Set STATIC_ATTRS to class instance so that can reuse
        setattr(cls, 'STATIC_ATTRS', attrs)
        setattr(cls, 'static_val_set', val_set)
        super(ConstMeta, cls).__init__(name, bases, dct)

    def __getattribute__(cls, name):
        '''
        Rewrite the special function so as to get correct attribute value
        '''
        static_attrs = object.__getattribute__(cls, 'STATIC_ATTRS')
        if name in static_attrs:
            return static_attrs[name][0]
        return object.__getattribute__(cls, name)

    def static_values(cls):
        '''
        Put values in static attribute into a list, use the function to validate value.
        @return: Set of values
        '''
        return cls.static_val_set

    def __getitem__(cls, key):
        '''
        Rewrite to make syntax SomeConstClass[key] works, and return desc string of related static value.
        @return: Desc string of related static value
        '''
        for k, v in cls.STATIC_ATTRS.iteritems():
            if v[0] == key:
                return v[1]
        raise KeyError('Key: %s does not exists in %s !' % (str(key), repr(cls)))


class Const(object):
    '''
    Base class for constant class.

    @usage:

    Definition: (must inherit from Const class!
        >>> class SomeConst(Const):
        >>>   STATUS_NAME_1 = (1, 'desc for the status1')
        >>>   STATUS_NAME_2 = (2, 'desc for the status2')

    Invoke(base upper SomeConst class):
    1) SomeConst.STATUS_NAME_1 returns 1
    2) SomeConst[1] returns 'desc for the status1'
    3) SomeConst.STATIC_ATTRS returns {'STATUS_NAME_1': (1, 'desc for the status1'), 'STATUS_NAME_2': (2, 'desc for the status2')}
    4) SomeConst.static_values() returns set([1, 2])

    Attention:
    SomeCosnt's value 1, 2 can not be duplicated!
    If WrongConst is like this, it will raise KeyError:
    class WrongConst(Const):
        STATUS_NAME_1 = (1, 'desc for the status1')
        STATUS_NAME_2 = (1, 'desc for the status2')
    '''
    __metaclass__ = ConstMeta
##################################################################
#Const Base Class ends
##################################################################


def main():
    class STATUS(Const):
        ERROR = (-3, '??')
        OK = (0, '??')

    print STATUS.ERROR
    print STATUS.static_values()
    print STATUS.STATIC_ATTRS

    #Usage sample:
    user_input = 1
    #Validate input:
    print user_input in STATUS.static_values()
    #Template render like:
    print '<select>'
    for key, value in STATUS.STATIC_ATTRS.items():
        print '<option value="%s">%s</option>' % (value[0], value[1])
    print '</select>'


if __name__ == '__main__':
    main()

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!