使用_member_names_快速简单的结果，如果它只是名称，即。

Color._member_names_

还有_member_map_，它返回元素的有序字典。这个函数返回一个集合。OrderedDict，因此您可以使用color . _member_map_items()和color . _member_map_values()。如。

return list(map(lambda x: x.value, Color._member_map_.values()))

将返回颜色的所有有效值

你可以使用IntEnum:

from enum import IntEnum

class Color(IntEnum):
   RED = 1
   BLUE = 2


print(int(Color.RED))   # prints 1

获取整型数组的列表:

enum_list = list(map(int, Color))
print(enum_list) # prints [1, 2]

基于@Jeff的回答，重构为使用类方法，以便您可以对任何枚举重用相同的代码:

from enum import Enum

class ExtendedEnum(Enum):

    @classmethod
    def list(cls):
        return list(map(lambda c: c.value, cls))

class OperationType(ExtendedEnum):
    CREATE = 'CREATE'
    STATUS = 'STATUS'
    EXPAND = 'EXPAND'
    DELETE = 'DELETE'

print(OperationType.list())

生产:

['CREATE', 'STATUS', 'EXPAND', 'DELETE']

使用__members__类方法:

class RoleNames(str, Enum):
    AGENT = "agent"
    USER = "user"
    PRIMARY_USER = "primary_user"
    SUPER_USER = "super_user"
    
    @classmethod
    def list_roles(cls):
        role_names = [member.value for role, member in cls.__members__.items()]
        return role_names

>>> role_names = RoleNames.list_roles()
>>> print(role_names)

或者如果你有多个Enum类并且想要抽象类方法:

class BaseEnum(Enum):
    @classmethod
    def list_roles(cls):
        role_names = [member.value for role, member in cls.__members__.items()]
        return role_names


class RoleNames(str, BaseEnum):    
    AGENT = "agent"
    USER = "user"
    PRIMARY_USER = "primary_user"
    SUPER_USER = "super_user"
    

class PermissionNames(str, BaseEnum):
    READ = "updated_at"
    WRITE = "sort_by"
    READ_WRITE = "sort_order"

您可以执行以下操作:

[e.value for e in Color]

一种方法是获取_value2member_map_属性的键值:

class Color(Enum):
    RED = 1
    BLUE = 2

list(Color._value2member_map_.keys())
# [1, 2]