How do I detect whether a Python variable is a function?


I have a variable, x, and I want to know whether it is pointing to a function or not.

I had hoped I could do something like:

>>> isinstance(x, function)

But that gives me:

Traceback (most recent call last):
  File "<stdin>", line 1, in ?
NameError: name 'function' is not defined

The reason I picked that is because

>>> type(x)
<type 'function'>
2/24/2018 11:38:54 AM

Accepted Answer

If this is for Python 2.x or for Python 3.2+, you can also use callable(). It used to be deprecated, but is now undeprecated, so you can use it again. You can read the discussion here: You can do this with:


If this is for Python 3.x but before 3.2, check if the object has a __call__ attribute. You can do this with:

hasattr(obj, '__call__')

The oft-suggested types.FunctionTypes approach is not correct because it fails to cover many cases that you would presumably want it to pass, like with builtins:

>>> isinstance(open, types.FunctionType)

>>> callable(open)

The proper way to check properties of duck-typed objects is to ask them if they quack, not to see if they fit in a duck-sized container. Don't use types.FunctionType unless you have a very specific idea of what a function is.

9/2/2016 8:48:14 AM

Builtin types that don't have constructors in the built-in namespace (e.g. functions, generators, methods) are in the types module. You can use types.FunctionType in an isinstance call.

In [1]: import types
In [2]: types.FunctionType
Out[2]: <type 'function'>
In [3]: def f(): pass
In [4]: isinstance(f, types.FunctionType)
Out[4]: True
In [5]: isinstance(lambda x : None, types.FunctionType)
Out[5]: True

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