Am I correct assuming that all functions (built-in or user-defined) belong to the same class, but that class doesn't seem to be bound to any variable by default?
How can I check that an object is a function?
I can do this I guess:
def is_function(x):
def tmp()
pass
return type(x) is type(tmp)
It doesn't seem neat, and I'm not even 100% sure it's perfectly correct.
in python2:
callable(fn)
in python3:
isinstance(fn, collections.Callable)
as Callable is an Abstract Base Class, this is equivalent to:
hasattr(fn, '__call__')
callable function was first removed in Python 3.0 and then brought back in Python 3.2 - so you can also use it w/ Python 3 if using a recent version of the interpreter. See docs.python.org/3/library/… for more information.How can I check that an object is a function?
Isn't this same as checking for callables
hasattr(object, '__call__')
and also in python 2.x
callable(object) == True
You can do:
def is_function(x):
import types
return isinstance(x, types.FunctionType) \
or isinstance(x, types.BuiltinFunctionType)
if it is of the same class?
is on type objects, yikes. If you must use explicit typechecks, at least use isinstance(), and preferably ask the inspect module to do it instead.
isinstance - learn the new and improved coding idioms!try:
magicVariable()
except TypeError as e:
print( 'was no function' )
