Call Nested Function in Python
I assume do_this
and do_that
are actually dependent on some argument of foo
, since otherwise you could just move them out of foo
and call them directly.
I suggest reworking the whole thing as a class. Something like this:
class Foo(object):
def __init__(self, x, y):
self.x = x
self.y = y
def do_this(self):
pass
def do_that(self):
pass
def __call__(self):
self.do_this()
self.do_that()
foo = Foo(x, y)
foo()
foo.do_this()
These previous answers, telling you that you can not do this, are of course wrong. This is python, you can do almost anything you want using some magic code magic.
We can take the first constant out of foo's function code, this will be the do_this
function. We can then use this code to create a new function with it.
see https://docs.python.org/2/library/new.html for more info on new and https://docs.python.org/2/library/inspect.html for more info on how to get to internal code.
Warning: it's not because you CAN do this that you SHOULD do this, rethinking the way you have your functions structured is the way to go, but if you want a quick and dirty hack that will probably break in the future, here you go:
import new
myfoo = new.function(foo.func_code.co_consts[1],{})
myfoo(x,y) # hooray we have a new function that does what I want
UPDATE: in python3 you can use the types module with foo.__code__
:
import types
myfoo = types.FunctionType(foo.__code__.co_consts[1], {})
myfoo() # behaves like it is do_this()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: do_this() missing 2 required positional arguments: 'x' and 'y'
There is, you have to make them as an attribute of the function object. But this will work only after the first call of foo
.
def foo(x,y):
def do_this(x,y):
pass
def do_that(x,y):
pass
do_this(x,y)
do_that(x,y)
foo.do_this = do_this
foo.do_that = do_that
return
>>> foo.do_this(1, 2)
AttributeError: 'function' object has no attribute 'do_this'
>>> foo(1, 2)
>>> foo.do_this(1, 2)
>>>