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
另一个非常简单的Python枚举实现,使用namedtuple:
from collections import namedtuple
def enum(*keys):
return namedtuple('Enum', keys)(*keys)
MyEnum = enum('FOO', 'BAR', 'BAZ')
,或者
# With sequential number values
def enum(*keys):
return namedtuple('Enum', keys)(*range(len(keys)))
# From a dict / keyword args
def enum(**kwargs):
return namedtuple('Enum', kwargs.keys())(*kwargs.values())
# Example for dictionary param:
values = {"Salad": 20, "Carrot": 99, "Tomato": "No i'm not"}
Vegetables= enum(**values)
# >>> print(Vegetables.Tomato) 'No i'm not'
# Example for keyworded params:
Fruits = enum(Apple="Steve Jobs", Peach=1, Banana=2)
# >>> print(Fruits.Apple) 'Steve Jobs'
就像上面子类设置的方法一样,这允许:
'FOO' in MyEnum
other = MyEnum.FOO
assert other == MyEnum.FOO
但是具有更大的灵活性,因为它可以有不同的键和值。这允许
MyEnum.FOO < MyEnum.BAR
如果使用填充连续数字值的版本,则按预期操作。
其他回答
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)
最好的解决方案取决于你需要从你的假枚举中得到什么。
简单的枚举:
如果你只需要枚举作为标识不同项目的名称列表,Mark Harrison(上图)的解决方案是很棒的:
Pen, Pencil, Eraser = range(0, 3)
使用范围还允许你设置任何起始值:
Pen, Pencil, Eraser = range(9, 12)
除此之外,如果你还要求项属于某种类型的容器,那么将它们嵌入到一个类中:
class Stationery:
Pen, Pencil, Eraser = range(0, 3)
要使用枚举项,你现在需要使用容器名和项名:
stype = Stationery.Pen
复杂的枚举:
对于枚举的长列表或更复杂的enum使用,这些解决方案是不够的。你可以参考Will Ware在Python Cookbook中发布的关于在Python中模拟枚举的食谱。这里有一个在线版本。
更多信息:
PEP 354: Python中的枚举有关于Python中的枚举建议的有趣细节,以及为什么它被拒绝。
在2013-05-10,Guido同意将PEP 435纳入Python 3.4标准库。这意味着Python终于内置了对枚举的支持!
Python 3.3、3.2、3.1、2.7、2.6、2.5和2.4有一个可用的后端端口。它在Pypi上枚举34。
声明:
>>> from enum import Enum
>>> class Color(Enum):
... red = 1
... green = 2
... blue = 3
表示:
>>> print(Color.red)
Color.red
>>> print(repr(Color.red))
<Color.red: 1>
迭代:
>>> for color in Color:
... print(color)
...
Color.red
Color.green
Color.blue
编程访问:
>>> Color(1)
Color.red
>>> Color['blue']
Color.blue
有关更多信息,请参阅提案。官方文件可能很快就会发布。
这个解决方案是获取定义为列表的枚举类的简单方法(没有更多烦人的整数赋值):
enumeration.py:
import new
def create(class_name, names):
return new.classobj(
class_name, (object,), dict((y, x) for x, y in enumerate(names))
)
example.py:
import enumeration
Colors = enumeration.create('Colors', (
'red',
'orange',
'yellow',
'green',
'blue',
'violet',
))
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。