Determining a variable's type is NoneType in python

Question

I would like to check if a variable is of the NoneType type. For other types we can do stuff like:

    type([])==list

But for NoneType this simple way is not possible. That is, we cannot say type(None)==NoneType. Is there an alternative way? And why is this possible for some types and not for others? Thank you.

Just use `x is None`. There is no advantage whatsoever to checking the type, as there will never under any circumstances be any object other than `None` of type `NoneType`. — Zero Piraeus, Nov 11 '16 at 17:47
I think the idea is to be able to pass `type(x) == y` for any x,y, not to add a special case for `x is None`. Note: you can also do `x == None`. — Jean-François Fabre, Nov 11 '16 at 17:56

score 37 · Accepted Answer · answered Nov 11 '16 at 17:35

37

NoneType just happens to not automatically be in the global scope. This isn't really a problem.

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

In any case it would be unusual to do a type check. Rather you should test x is None.

answered Nov 11 '16 at 17:35

Alex Hall

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Thanks, I suppose both this solution and the solution of @JeanFrancoisFabre do the job. Do we have some reason to prefer one of them over the other (speed,style,etc.)? – splinter Nov 11 '16 at 17:39
@splinter not that I know of. – Alex Hall Nov 11 '16 at 17:41

score 16 · Answer 2 · answered Nov 11 '16 at 17:35

16

Of course you can do it.

type(None)==None.__class__

True

answered Nov 11 '16 at 17:35

Jean-François Fabre

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Determining a variable's type is NoneType in python

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