Convert String to Enum in Python

I wonder what's the correct way of converting (deserializing) a string to a Python's Enum class. Seems like getattr(YourEnumType, str) does the job, but I'm not sure if it's safe enough.

Just to be more specific, I would like to convert a 'debug'string to an Enum object like this:

class BuildType(Enum):
    debug = 200
    release = 400
0

8 Answers

This functionality is already built in to Enum [1]:

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

The member names are case sensitive, so if user-input is being converted you need to make sure case matches:

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

[1] Official docs: Enum programmatic access

8

Another alternative (especially useful if your strings don't map 1-1 to your enum cases) is to add a staticmethod to your Enum, e.g.:

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

    @staticmethod
    def from_str(label):
        if label in ('single', 'singleSelect'):
            return QuestionType.SINGLE_SELECT
        elif label in ('multi', 'multiSelect'):
            return QuestionType.MULTI_SELECT
        else:
            raise NotImplementedError

Then you can do question_type = QuestionType.from_str('singleSelect')

2

My Java-like solution to the problem. Hope it helps someone...

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`
1
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)

Or do you need to convert string to known 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
3

An improvement to the answer of @rogueleaderr :

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
4

Since MyEnum['dontexist'] will result in error KeyError: 'dontexist', you might like to fail silently (eg. return None). In such case you can use the following static method:

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')

Change your class signature to this:

class BuildType(str, Enum):
1

I just want to notify this does not work in python 3.6

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

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

You will have to give the data as a tuple like this

MyEnum(('aaa',))

EDIT: This turns out to be false. Credits to a commenter for pointing out my mistake

2

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Elena Rostova

Elena Rostova

Lead Health, Wellness & Medical Journalist

Elena Rostova holds a Master's degree in Public Health Journalism. She covers groundbreaking medical research, holistic wellness trends, mental health awareness, and nutritional science.

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