'Where is the NoneType located in Python 3.x?
In Python 3, I would like to check whether value is either string or None.
One way to do this is
assert type(value) in { str, NoneType }
But where is NoneType located in Python?
Without any import, using NoneType produces NameError: name 'NoneType' is not defined.
Solution 1:[1]
types.NoneType is being reintroduced in Python 3.10.
Improved Modules
Reintroduced the
types.EllipsisType,types.NoneTypeandtypes.NotImplementedTypeclasses, providing a new set of types readily interpretable by type checkers. (Contributed by Bas van Beek in bpo-41810.)
The discussion about the change was motivated by a need for types.EllipsisType, leading to types.NoneType also being added for consistency.
Solution 2:[2]
You can use type(None) to get the type object, but you want to use isinstance() here, not type() in {...}:
assert isinstance(value, (str, type(None)))
The NoneType object is not otherwise exposed anywhere in Python versions older than 3.10*.
I'd not use type checking for that at all really, I'd use:
assert value is None or isinstance(value, str)
as None is a singleton (very much on purpose) and NoneType explicitly forbids subclassing anyway:
>>> type(None)() is None
True
>>> class NoneSubclass(type(None)):
... pass
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: type 'NoneType' is not an acceptable base type
* As of Python 3.10, you could use types.NoneType, to be consistent with the other singleton types being added to the types module.
Solution 3:[3]
Please use type(None). You can use python shell to check like in the below function in which I use type(None) in order to change from None to NoneType.
def to_unicode(value):
'''change value to unicode'''
try:
if isinstance(value, (str,type(None))):
return value
if not isinstance(value, bytes):
raise TypeError("Expected bytes, unicode, or None; got %r" % type(value))
return value.decode("utf-8")
except UnicodeDecodeError:
return repr(value)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | |
| Solution 3 | sam |
