我主要是c#开发人员,但我目前正在用Python开发一个项目。

我如何在Python中表示等价的Enum ?


当前回答

我用什么:

class Enum(object):
    def __init__(self, names, separator=None):
        self.names = names.split(separator)
        for value, name in enumerate(self.names):
            setattr(self, name.upper(), value)
    def tuples(self):
        return tuple(enumerate(self.names))

使用方法:

>>> state = Enum('draft published retracted')
>>> state.DRAFT
0
>>> state.RETRACTED
2
>>> state.FOO
Traceback (most recent call last):
   File "<stdin>", line 1, in <module>
AttributeError: 'Enum' object has no attribute 'FOO'
>>> state.tuples()
((0, 'draft'), (1, 'published'), (2, 'retracted'))

这就给出了整数常数,比如状态。PUBLISHED和在Django模型中用作选项的二元组。

其他回答

我喜欢Java枚举,这就是我在Python中的做法:

def enum(clsdef):
    class Enum(object):
        __slots__=tuple([var for var in clsdef.__dict__ if isinstance((getattr(clsdef, var)), tuple) and not var.startswith('__')])

        def __new__(cls, *args, **kwargs):
            if not '_the_instance' in cls.__dict__:
                cls._the_instance = object.__new__(cls, *args, **kwargs)
            return cls._the_instance

        def __init__(self):
            clsdef.values=lambda cls, e=Enum: e.values()
            clsdef.valueOf=lambda cls, n, e=self: e.valueOf(n)
            for ordinal, key in enumerate(self.__class__.__slots__):
                args=getattr(clsdef, key)
                instance=clsdef(*args)
                instance._name=key
                instance._ordinal=ordinal
                setattr(self, key, instance)

        @classmethod
        def values(cls):
            if not hasattr(cls, '_values'):
                cls._values=[getattr(cls, name) for name in cls.__slots__]
            return cls._values

        def valueOf(self, name):
            return getattr(self, name)

        def __repr__(self):
            return ''.join(['<class Enum (', clsdef.__name__, ') at ', str(hex(id(self))), '>'])

    return Enum()

示例使用:

i=2
@enum
class Test(object):
    A=("a",1)
    B=("b",)
    C=("c",2)
    D=tuple()
    E=("e",3)

    while True:
        try:
            F, G, H, I, J, K, L, M, N, O=[tuple() for _ in range(i)]
            break;
        except ValueError:
            i+=1

    def __init__(self, name="default", aparam=0):
        self.name=name
        self.avalue=aparam

所有类变量都定义为元组,就像构造函数一样。到目前为止,还不能使用命名参数。

使用下面的方法。

TYPE = {'EAN13':   u'EAN-13',
        'CODE39':  u'Code 39',
        'CODE128': u'Code 128',
        'i25':     u'Interleaved 2 of 5',}

>>> TYPE.items()
[('EAN13', u'EAN-13'), ('i25', u'Interleaved 2 of 5'), ('CODE39', u'Code 39'), ('CODE128', u'Code 128')]
>>> TYPE.keys()
['EAN13', 'i25', 'CODE39', 'CODE128']
>>> TYPE.values()
[u'EAN-13', u'Interleaved 2 of 5', u'Code 39', u'Code 128']

我用它来选择Django模型,它看起来非常python化。它不是一个真正的Enum,但它完成了这项工作。

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)

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'

我喜欢使用列表或集合作为枚举。例如:

>>> packet_types = ['INIT', 'FINI', 'RECV', 'SEND']
>>> packet_types.index('INIT')
0
>>> packet_types.index('FINI')
1
>>>