How to "test" NoneType in python?

It can also be done with isinstance as per Alex Hall's answer :

>>> NoneType = type(None)
>>> x = None
>>> type(x) == NoneType
True
>>> isinstance(x, NoneType)
True

isinstance is also intuitive but there is the complication that it requires the line

NoneType = type(None)

which isn't needed for types like int and float.

if variable is None:
   ...

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if variable is not None:
   ...

As pointed out by Aaron Hall's comment:

Since you can't subclass NoneType and since None is a singleton, isinstance should not be used to detect None - instead you should do as the accepted answer says, and use is None or is not None.

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Original Answer:

The simplest way however, without the extra line in addition to cardamom's answer is probably:
isinstance(x, type(None))

So how can I question a variable that is a NoneType? I need to use if method

Using isinstance() does not require an is within the if-statement:

if isinstance(x, type(None)): 
    #do stuff

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Additional information
You can also check for multiple types in one isinstance() statement as mentioned in the documentation. Just write the types as a tuple.

isinstance(x, (type(None), bytes))

So how can I question a variable that is a NoneType?

Use is operator, like this

if variable is None:

Why this works?

Since None is the sole singleton object of NoneType in Python, we can use is operator to check if a variable has None in it or not.

Quoting from is docs,

The operators is and is not test for object identity: x is y is true if and only if x and y are the same object. x is not y yields the inverse truth value.

Since there can be only one instance of None, is would be the preferred way to check None.

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Hear it from the horse's mouth

Quoting Python's Coding Style Guidelines - PEP-008 (jointly defined by Guido himself),

Comparisons to singletons like None should always be done with is or is not, never the equality operators.