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 ?
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
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)
其他回答
Davidg建议使用词典。我想更进一步,使用集合:
months = set('January', 'February', ..., 'December')
现在你可以测试一个值是否与集合中的一个值匹配,如下所示:
if m in months:
但是,像dF,我通常只使用字符串常量来代替枚举。
在答案列表中没有看到这个,这是我想出的一个。它允许使用'in'关键字和len()方法:
class EnumTypeError(TypeError):
pass
class Enum(object):
"""
Minics enum type from different languages
Usage:
Letters = Enum(list('abc'))
a = Letters.a
print(a in Letters) # True
print(54 in Letters) # False
"""
def __init__(self, enums):
if isinstance(enums, dict):
self.__dict__.update(enums)
elif isinstance(enums, list) or isinstance(enums, tuple):
self.__dict__.update(**dict((v,k) for k,v in enumerate(enums)))
else:
raise EnumTypeError
def __contains__(self, key):
return key in self.__dict__.values()
def __len__(self):
return len(self.__dict__.values())
if __name__ == '__main__':
print('Using a dictionary to create Enum:')
Letters = Enum(dict((v,k) for k,v in enumerate(list('abcde'))))
a = Letters.a
print('\tIs a in e?', a in Letters)
print('\tIs 54 in e?', 54 in Letters)
print('\tLength of Letters enum:', len(Letters))
print('\nUsing a list to create Enum:')
Letters = Enum(list('abcde'))
a = Letters.a
print('\tIs a in e?', a in Letters)
print('\tIs 54 in e?', 54 in Letters)
print('\tLength of Letters enum:', len(Letters))
try:
# make sure we raise an exception if we pass an invalid arg
Failure = Enum('This is a Failure')
print('Failure')
except EnumTypeError:
print('Success!')
输出:
Using a dictionary to create Enum:
Is a in e? True
Is 54 in e? False
Length of Letters enum: 5
Using a list to create Enum:
Is a in e? True
Is 54 in e? False
Length of Letters enum: 5
Success!
为了解码二进制文件格式,我曾经需要一个Enum类。我碰巧想要的特性是简洁的枚举定义,通过整数值或字符串自由创建枚举实例的能力,以及有用的表示。这是我最后得出的结论:
>>> class Enum(int):
... def __new__(cls, value):
... if isinstance(value, str):
... return getattr(cls, value)
... elif isinstance(value, int):
... return cls.__index[value]
... def __str__(self): return self.__name
... def __repr__(self): return "%s.%s" % (type(self).__name__, self.__name)
... class __metaclass__(type):
... def __new__(mcls, name, bases, attrs):
... attrs['__slots__'] = ['_Enum__name']
... cls = type.__new__(mcls, name, bases, attrs)
... cls._Enum__index = _index = {}
... for base in reversed(bases):
... if hasattr(base, '_Enum__index'):
... _index.update(base._Enum__index)
... # create all of the instances of the new class
... for attr in attrs.keys():
... value = attrs[attr]
... if isinstance(value, int):
... evalue = int.__new__(cls, value)
... evalue._Enum__name = attr
... _index[value] = evalue
... setattr(cls, attr, evalue)
... return cls
...
使用它的一个异想天开的例子:
>>> class Citrus(Enum):
... Lemon = 1
... Lime = 2
...
>>> Citrus.Lemon
Citrus.Lemon
>>>
>>> Citrus(1)
Citrus.Lemon
>>> Citrus(5)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 6, in __new__
KeyError: 5
>>> class Fruit(Citrus):
... Apple = 3
... Banana = 4
...
>>> Fruit.Apple
Fruit.Apple
>>> Fruit.Lemon
Citrus.Lemon
>>> Fruit(1)
Citrus.Lemon
>>> Fruit(3)
Fruit.Apple
>>> "%d %s %r" % ((Fruit.Apple,)*3)
'3 Apple Fruit.Apple'
>>> Fruit(1) is Citrus.Lemon
True
主要特点:
str(), int()和repr()都会产生最有用的输出,分别是枚举的名称,它的整数值,以及返回枚举的Python表达式。 构造函数返回的枚举值严格限制为预定义值,没有意外的枚举值。 枚举值是单例的;它们可以与之严格比较
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。
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。