On 4 July 2013 05:36, Chris Angelico <ros...@gmail.com> wrote:
> On Thu, Jul 4, 2013 at 2:30 PM, Joshua Landau
> <joshua.landau...@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.
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