Question

i am having a problem with scope. i would like to have a function operate on the result of the builtin function dir(). so, as a simple example, i want to

>>> print (dir())
>>>  .... bunch of stuff ...

but, i want define a function which operates on this results of dir(). however, if try this like so

>>> def print_dir(dummy_var=None):
        print dir()
>>> print_dir()
>>> ['dummy_var']

i cant access the namespace of what called print_dir which i understand is a scope thing and makes sense, but is there a way to access the namespace from which a function is called? or do i have to pass the result of dir() to the function? like.

>>> act_on_namespace(dir())

the specific goal in mind is something like this,

def act_on_namespace()
    for k in dir():
        try:
            print(eval(k).__module__)
        except:
            pass

Answer 1

No, there is no straightforward way to manipulate the calling namespace like that. The simplest way is to do what you already suggested: call act_on_namespace(dir()). If you really want to get crazy, you could look at the inspect module, which allows you to look through the call stack, but that is probably not a good idea unless what you’re doing is writing some kind of debugging tool.