On 3/9/2010 9:48 AM, Bruno Desthuilliers wrote:
John Posner a écrit :
On 3/8/2010 11:55 PM, Gary Herron wrote:
<snip>
The form of import you are using
from helpers import mostRecent
makes a *new* binding to the value in the module that's doing the
import.
<snip>
What you can do, is not make a separate binding, but reach into the
helpers module to get the value there. Like this:
import helpers
print helpers.mostRecent
Gary, are you asserting that in these separate situations:
one.py:
from helpers import mostRecent
x = mostRecent
two.py:
import helpers
x = helpers.mostRecent
... the name "x" will be bound to different objects?
Nope. What he's saying is that one.x and two.x are two different names -
each living in it's own namespace - that happen to be bound to the same
object. Now rebiding one.x to a different object will _of course_ have
no impact on the object two.x is bound to.
Sure -- my bad, Python 101. In this case, the module *helpers* is the
equivalent of a container object, one of whose attributes *mostRecent*
gets bound to a series of immutable string objects.
That's exactly the same as:
one = dict()
two = dict()
# one['x'] and two['x'] refer to the same object
one['x'] = two['x'] = ["foo", "bar"]
print one['x'], two['x'], one['x'] is two['x']
# mutating one['x'], visible in two['x']
# (of course since it's the same object)
one['x'].append("baaz")
print one['x'], two['x'], one['x'] is two['x']
# now rebind one['x']
one['x'] = 42
# obvious result: one['x'] and two['x'] now refer to 2 different objects
print one['x'], two['x'], one['x'] is two['x']
If in doubt about namespaces, think dicts. Namespaces are like dicts -
and are often nothing else that a plain old dict FWIW.
No doubts here, I hope. I even tend to see namespaces where, strictly
speaking, they don't exist. As I said at the end of a recent (and
flame-ridden) thread [1]:
-----------
* A dict is a collection of user-devised names, each of which
is assigned to an object.
* A list/tuple is an interpreter-maintained collection of integer
names (0, 1, 2, ...), each of which is assigned to an object.
-----------
So in my world, in mylist[cnt+1], the expression *cnt+1* is equivalent
to the integer "name" 4 (if cnt==3, that is), so that
mylist[cnt+1] = obj.subobj.attr
... is just a NAME2 = NAME1 kind of assignment statement, binding an
additional name to an object that already has a name.
Tx,
John
[1] http://mail.python.org/pipermail/python-list/2010-February/1236318.html
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