How to access the NoneType type?

Question

For example looking at the type of None, we can see that it has NoneType:

>>> type(None)
NoneType

However a NameError results when trying to access the NoneType:

>>> NoneType
NameError: name 'NoneType' is not defined

How can the NoneType be accessed?

Asked By: Greg

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Source

Answer 1

Well, in Python 2, or 3.10+, you can import it from the types module:

from types import NoneType

but it’s not actually implemented there or anything. types.py just does NoneType = type(None). You might as well just use type(None) directly.

For some reason, they decided to take the types.NoneType name out in Python 3, before putting it back in 3.10. On versions where the name doesn’t exist, just use type(None).

If type(None) shows NoneType for you, rather than something like <class 'NoneType'>, you’re probably on some nonstandard interpreter setup, such as IPython. It’d usually show up as something like <class 'NoneType'>, making it clearer that you can’t just type NoneType and get the type.

If the reason you want this is for type annotations, use None, not types.NoneType:

def returns_none() -> None:
    return None

Static type checkers special-case this use of None.

Answered By: user2357112

Answer 2

As suggested by @dawg in the comments, you can do

if (type(some_object).__name__ == "NoneType"):
  # Do some
  pass

You can also do

NoneType = type(None)    
isinstance(some_object, NoneType)

Answered By: hungryWolf

How to access the NoneType type?

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