这个功能已经内置在Enum中:

>>> from enum import Enum
>>> class Build(Enum):
...   debug = 200
...   build = 400
... 
>>> Build['debug']
<Build.debug: 200>

成员名是大小写敏感的，所以如果用户输入被转换，你需要确保大小写匹配:

an_enum = input('Which type of build?')
build_type = Build[an_enum.lower()]

这个功能已经内置在Enum中:

>>> from enum import Enum
>>> class Build(Enum):
...   debug = 200
...   build = 400
... 
>>> Build['debug']
<Build.debug: 200>

成员名是大小写敏感的，所以如果用户输入被转换，你需要确保大小写匹配:

an_enum = input('Which type of build?')
build_type = Build[an_enum.lower()]

更改你的类签名为:

class BuildType(str, Enum):

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

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