Default scope of variables

[Joshua Landau]

On 4 July 2013 05:36, Chris Angelico <rosuav at gmail.com> wrote:
> On Thu, Jul 4, 2013 at 2:30 PM, Joshua Landau
> <joshua.landau.ws at gmail.com> wrote:
>> That said, I'm not too convinced. Personally, the proper way to do
>> what you are talking about is creating a new closure. Like:
>>
>> for i in range(100):
>>     with new_scope():
>>         for i in range(100):
>>             func(i)
>>     func(i) # Using i from original loop
>>
>> But it's not like Python'll ever support that.
>>
>
> def foo():
>   for i in range(3):
>     print("outer",i)
>     def inner():
>       for i in range(4):
>         print("inner",i)
>     inner()
>     print("outer",i)
>
> That works, but you then have to declare all your nonlocals, and it
> hardly reads well.

Stealing concepts shamelessly from
http://www.slideshare.net/r1chardj0n3s/dont-do-this-24000445, you can
do this:

import inspect
from contextlib import contextmanager

@contextmanager
def scope(namespace=None):
    old_names = inspect.currentframe().f_back.f_back.f_locals.copy()

    yield

    names = inspect.currentframe().f_back.f_back.f_locals

    if namespace is not None:
        new_names = {k:v for k, v in names.items() if k not in
old_names and v is not namespace}
        namespace.update(**new_names)

    names.clear()
    names.update(old_names)

So you *can* do:

>>> for i in range(3):
...     with scope():
...         for i in range(3):
...             print("Inner:", i)
...     print("Outer", i)
Inner: 0
Inner: 1
Inner: 2
Outer 0
Inner: 0
Inner: 1
Inner: 2
Outer 1
Inner: 0
Inner: 1
Inner: 2
Outer 2

:)

If you pass scope() a dictionary, all the new variables will get added to it.