Where can I find .join() in the docs

[Steve Holden]

"Geiger Ho" <s997659 at ee.cuhk.edu.hk> wrote ...
> Hi all,
>
>   I would like to know how could I modify the following code so that it
> can be a function in a module? I've tried. But the interpreter warns me
> that no __dic__ attribute in __builtins__.
>
>   Also, how can I know how many namespaces I create in an environment and
> how can I modify the code so that it can show what is in a specific
> namespace.
>
>   Thanks in advance.
>
> Regards,
> Geiger
>
>
> On Thu, 30 May 2002, Duncan Booth wrote:
>
> > You might find this useful as it shows you what is really in builtins,
not
> > what is documented:
> > >>> def showNamespace(namespace=None):
> >         '''Summarise namespace contents'''
> >         if namespace is None: namespace = __builtins__.__dict__
> >         names = {}
> >         # Build a dictionary mapping type to list of names
> >         for k, v in namespace.items():
> >             names.setdefault(type(v), []).append(k)
> >         # Extract typename and name lists into a list
> >         items = [(t.__name__, v) for t, v in names.items()]
> >         items.sort() # Sort by typename
> >         # Print out the variables categorised by type
> >         for k, v in items:
> >             print "%s:" % k, ", ".join(v)
> >             print
> >

Well, you could try typing in the code you quoted, rather than misspelling
__dict__ as __dic__. Surely the error message gave you *some* idea where to
look?

To give a meaningful answer to your second question is diffcult. Each object
has its own namespace, as does each function execution, as does each module.
I'm not sure *why* you might want to to this, unless you are simply (!)
writing a debugger, or trying to improve your command of Python.

what's-in-a-namespace-ly y'rs  - steve
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Steve Holden                                 http://www.holdenweb.com/
Python Web Programming                http://pydish.holdenweb.com/pwp/
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