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
遵循Aaron Maenpaa提出的类Java枚举实现,我得出了以下结论。我们的想法是使它具有通用性和可解析性。
class Enum:
#'''
#Java like implementation for enums.
#
#Usage:
#class Tool(Enum): name = 'Tool'
#Tool.DRILL = Tool.register('drill')
#Tool.HAMMER = Tool.register('hammer')
#Tool.WRENCH = Tool.register('wrench')
#'''
name = 'Enum' # Enum name
_reg = dict([]) # Enum registered values
@classmethod
def register(cls, value):
#'''
#Registers a new value in this enum.
#
#@param value: New enum value.
#
#@return: New value wrapper instance.
#'''
inst = cls(value)
cls._reg[value] = inst
return inst
@classmethod
def parse(cls, value):
#'''
#Parses a value, returning the enum instance.
#
#@param value: Enum value.
#
#@return: Value corresp instance.
#'''
return cls._reg.get(value)
def __init__(self, value):
#'''
#Constructor (only for internal use).
#'''
self.value = value
def __str__(self):
#'''
#str() overload.
#'''
return self.value
def __repr__(self):
#'''
#repr() overload.
#'''
return "<" + self.name + ": " + self.value + ">"
其他回答
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)
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。
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。
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!