How do I test if int value exists in Python Enum without using try/catch?

You could use Enum.__members__ – an ordered dictionary mapping names to members:

In [12]: 'Apple' in Fruit.__members__
Out[12]: True

In [13]: 'Grape' in Fruit.__members__
Out[13]: False

Just check whether it’s in Enum. _value2member_map_

In[15]: Fruit._value2member_map_
Out[15]: {4: <Fruit.Apple: 4>, 5: <Fruit.Orange: 5>, 6: <Fruit.Pear: 6>}

In[16]: 6 in Fruit._value2member_map_
Out[16]: True

In[17]: 7 in Fruit._value2member_map_
Out[17]: False

Building on what Reda Maachi started:

6 in Fruit.__members__.values() 

returns True

7 in Fruit.__members__.values()  

returns False

Originally, Don’t.

If you are using Enum, you can test for enum with

     if isinstance(key, Fruit):

But otherwise, try.. is the pythonic way to test for enum. Indeed, for any break in the duck-typing paradigm.

The correct, and pythonic, way of testing for an int in an IntEnum is to give it a go and to catch a ValueError if there’s a failure.

Many of the solutions proposed above are actively deprecated and will be disallowed by 3.8 ( "DeprecationWarning: using non-Enums in containment checks will raise TypeError in Python 3.8" )

From Python 3.8 – Coerce it into a list.

After 3.8 or so, you can test by coercion of the enum into a list as follows (using IntEnum)

from enum import IntEnum
class Fruit(IntEnum):
     Kiwi = 2
     Apple = 4
     Orange = 5
     Pear = 6

for x in range(8):
     print(f'{x}, {x in list(Fruit)}')

Which will print as follows

0, False
1, False
2, True
3, False
4, True
5, True
6, True
7, False

An EAFP version of the answer:

try: 
    Fruit(val)
    return True
except ValueError:
    return False

You could use __members__ special attribute to iterate over members:

from enum import Enum

class Fruit(Enum):
    Apple = 4
    Orange = 5
    Pear = 6

    @staticmethod
    def has_value(item):
        return item in [v.value for v in Fruit.__members__.values()]

There’s another one liner solution nobody has mentioned yet:

is_value_in_fruit = any(f.value == value_to_check for f in Fruit)

Also, if you use IntEnum instead of Enum, (class Fruit(IntEnum)) you can just do this

is_value_in_fruit = any(f == value_to_check for f in Fruit) 

IntEnum + __members__

You could use IntEnum and __members__ to achieve required behaviour:

from enum import IntEnum

class Fruit(IntEnum):
    Apple = 4
    Orange = 5
    Pear = 6

>>> 6 in Fruit.__members__.values()
True
>>> 7 in Fruit.__members__.values()
False

Enum + list comprehension + .value

If you must/want stick to Enum, you can do:

>>> 6 in [f.value for f in Fruit]
True
>>> 7 in [f.value for f in Fruit]
False

EAPF + ValueError

Or you can use easier to ask for forgiveness than permission method:

try:
    Fruit(x)
except ValueError:
    return False
else:
    return True

There is a way to have all the enums be able to check if an item is present:

import enum 

class MyEnumMeta(enum.EnumMeta): 
    def __contains__(cls, item): 
        return isinstance(item, cls) or item in [v.value for v in cls.__members__.values()] 

class MyEnum(enum.Enum, metaclass=MyEnumMeta): 
   FOO = "foo" 
   BAR = "bar"

Now you can do an easy check:

>>> "foo" in MyEnum
True

It can even be made simpler if all the enum’s values will always be the same type — for example strings:

import enum 
 
class MyEnumMeta(enum.EnumMeta):  
    def __contains__(cls, item): 
        return item in cls.__members__.values()

class MyEnum(str, enum.Enum, metaclass=MyEnumMeta): 
    FOO = "foo" 
    BAR = "bar"

Edit: As per the comments by @MestreLion, I’ve updated the first example to include the isinstance check; without it the test fails if we’re checking an already instantiated enum value.

Edit: Yet another version, technically the most correct one:

import enum 

class MyEnumMeta(enum.EnumMeta): 
    def __contains__(cls, item): 
        try:
            cls(item)
        except ValueError:
            return False
        else:
            return True

class MyEnum(enum.Enum, metaclass=MyEnumMeta): 
   FOO = "foo" 
   BAR = "bar"

If the enum has many members, this approach can be faster because it doesn’t make a new list and stops walking the enum when the given value is found:

any(x.value == 5 for x in Fruit)  # True
any(x.value == 7 for x in Fruit)  # False

how about this?

from enum import Enum

class Fruit(Enum):
    Apple = 4
    Orange = 5
    Pear = 6

has_apples = 4 in [n.value for n in Fruit]

This would let you also do:

has_apples = "Apple" in [n.name for n in Fruit]

I just convert an IntEnum to a list and test it normally:

from enum import IntEnum
class Foo(IntEnum):
    ONE = 1
    TWO = 2
    THREE = 3

print(1 in list(Foo))
True
print(4 in list(Foo))
False

class MyEnumMixin:

    raw_values = None  # for IDE autocomplete

    def __new__(cls, value):
        if 'raw_values' not in cls.__dict__:
            cls.raw_values = set()
        cls.raw_values.add(value)
        if cls.__bases__[0] is MyEnumMixin:
            member = object().__new__(cls)
        else:
            member = super().__new__(cls, value)
        member._value_ = value
        return member



class MyEnum(MyEnumMixin, Enum):
    FOO = 1
    BAR = 'bar'

print(1 in MyEnum.raw_values)
print('bar' in MyEnum.raw_values)



class MyStrEnumMixin(MyEnumMixin, str):
    pass

class MyStrEnum(MyStrEnumMixin, Enum):
    FOO = 'foo'
    BAR = 'bar'

print('foo' in MyStrEnum.raw_values)
print('bar' in MyStrEnum.raw_values)