将字符串转换为Enum子类的对应实例的正确方法是什么?似乎getattr(YourEnumType, str)做的工作,但我不确定它是否足够安全。

举个例子,假设我有一个枚举

class BuildType(Enum):
    debug = 200
    release = 400

给定字符串'调试',我怎么能得到BuildType.debug作为结果?


当前回答

我只是想通知这在python 3.6中不起作用

class MyEnum(Enum):
    a = 'aaa'
    b = 123

print(MyEnum('aaa'), MyEnum(123))

你必须像这样以元组的形式给出数据

MyEnum(('aaa',))

编辑: 事实证明这是错误的。感谢一位指出我错误的评论者

其他回答

我只是想通知这在python 3.6中不起作用

class MyEnum(Enum):
    a = 'aaa'
    b = 123

print(MyEnum('aaa'), MyEnum(123))

你必须像这样以元组的形式给出数据

MyEnum(('aaa',))

编辑: 事实证明这是错误的。感谢一位指出我错误的评论者

由于MyEnum['dontexist']将导致错误KeyError: 'dontexist',您可能喜欢无声地失败(例如。返回None)。在这种情况下,你可以使用下面的静态方法:

class Statuses(enum.Enum):
    Unassigned = 1
    Assigned = 2

    @staticmethod
    def from_str(text):
        statuses = [status for status in dir(
            Statuses) if not status.startswith('_')]
        if text in statuses:
            return getattr(Statuses, text)
        return None


Statuses.from_str('Unassigned')

我的类似java的解决方案。希望它能帮助到某人…

from enum import Enum, auto


class SignInMethod(Enum):
    EMAIL = auto(),
    GOOGLE = auto()

    @classmethod
    def value_of(cls, value):
        for k, v in cls.__members__.items():
            if k == value:
                return v
        else:
            raise ValueError(f"'{cls.__name__}' enum not found for '{value}'")


sim = SignInMethod.value_of('EMAIL')
assert sim == SignInMethod.EMAIL
assert sim.name == 'EMAIL'
assert isinstance(sim, SignInMethod)
# SignInMethod.value_of("invalid sign-in method")  # should raise `ValueError`

对@rogueleaderr的回答进行了改进:

class QuestionType(enum.Enum):
    MULTI_SELECT = "multi"
    SINGLE_SELECT = "single"

    @classmethod
    def from_str(cls, label):
        if label in ('single', 'singleSelect'):
            return cls.SINGLE_SELECT
        elif label in ('multi', 'multiSelect'):
            return cls.MULTI_SELECT
        else:
            raise NotImplementedError
def custom_enum(typename, items_dict):
    class_definition = """
from enum import Enum

class {}(Enum):
    {}""".format(typename, '\n    '.join(['{} = {}'.format(k, v) for k, v in items_dict.items()]))

    namespace = dict(__name__='enum_%s' % typename)
    exec(class_definition, namespace)
    result = namespace[typename]
    result._source = class_definition
    return result

MyEnum = custom_enum('MyEnum', {'a': 123, 'b': 321})
print(MyEnum.a, MyEnum.b)

或者需要将字符串转换为已知Enum?

class MyEnum(Enum):
    a = 'aaa'
    b = 123

print(MyEnum('aaa'), MyEnum(123))

Or:

class BuildType(Enum):
    debug = 200
    release = 400

print(BuildType.__dict__['debug'])

print(eval('BuildType.debug'))
print(type(eval('BuildType.debug')))    
print(eval(BuildType.__name__ + '.debug'))  # for work with code refactoring