Re: How to get current module object
On 19 feb, 03:33, Alex [EMAIL PROTECTED] wrote: GabrielGenellinawrote: En Mon, 18 Feb 2008 14:49:02 -0200, Alex [EMAIL PROTECTED] escribió: That's what I've been searching for, thanks. By the way, I know it might be trivial question... but function and class namespaces have __name__ attribute too. Why is global one always returned? I don't understand the question (even with the later correction namespaces-objects) There's no question anymore, I just failed to distinguish function local variables (which don't include __name__) and function object's attributes Why do you want to get the module object? globals() returns the module namespace, its __dict__, perhaps its only useful attribute... To pass it as a parameter to a function (in another module), so it can work with several modules (plugins for main program) in a similar manner. The function could receive a namespace to work with (a dictionary). Then you just call it with globals() == the namespace of the calling module. Yes, but access to module seems more verbose: module_dict['x']() xxx Instead of just: module.x() xxx You could write a wrapper class on the client side, but I guess it's easier to pass the module object directly, as you said earlier. Anyway, this class would fake attribute access for dictionary entries: py class Idx2Attr(object): ... def __init__(self, d): self.__dict__[None] = d ... def __getattr__(self, name): ... try: return self.__dict__[None][name] ... except KeyError: raise NameError, name ... def __setattr__(self, name, value): ... self.__dict__[None][name]=value ... py import htmllib py dir(htmllib) ['AS_IS', 'HTMLParseError', 'HTMLParser', '__all__', '__buil tins__', '__doc__', '__file__', '__name__', 'sgmllib', 'test '] py mod = Idx2Attr(htmllib.__dict__) # htmllib.__dict__ is what you get if you use globals() inside htmllib py mod.__file__ 'C:\\APPS\\PYTHON25\\lib\\htmllib.pyc' py mod.foo = 3 py htmllib.foo 3 ([None] is to avoid name collisions to some extent) -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: How to get current module object
Gabriel Genellina wrote:
En Sun, 17 Feb 2008 16:25:44 -0200, Alex [EMAIL PROTECTED] escribi�:
Can I get reference to module object of current module (from which the
code is currently executed)? I know __import__('filename') should
probably do that, but the call contains redundant information (filename,
which needs to be updated), and it'll perform unnecessary search in
loaded modules list.
It shouldn't be a real problem (filename can probably be extracted from
the traceback anyway), but I wonder if there is more direct and less
verbose way.
sys.modules[__name__]
That's what I've been searching for, thanks. By the way, I know it might
be trivial question... but function and class namespaces have __name__
attribute too. Why is global one always returned?
Why do you want to get the module object? globals() returns the module
namespace, its __dict__, perhaps its only useful attribute...
To pass it as a parameter to a function (in another module), so it can
work with several modules (plugins for main program) in a similar manner.
--
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Re: How to get current module object
Alex wrote: function and class namespaces have __name__ attribute too I was wrong - these were function and class _objects'_ namespaces -- http://mail.python.org/mailman/listinfo/python-list
Re: How to get current module object
En Mon, 18 Feb 2008 14:49:02 -0200, Alex [EMAIL PROTECTED] escribió: Gabriel Genellina wrote: That's what I've been searching for, thanks. By the way, I know it might be trivial question... but function and class namespaces have __name__ attribute too. Why is global one always returned? I don't understand the question (even with the later correction namespaces-objects) Why do you want to get the module object? globals() returns the module namespace, its __dict__, perhaps its only useful attribute... To pass it as a parameter to a function (in another module), so it can work with several modules (plugins for main program) in a similar manner. The function could receive a namespace to work with (a dictionary). Then you just call it with globals() == the namespace of the calling module. -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: How to get current module object
Gabriel Genellina wrote: En Mon, 18 Feb 2008 14:49:02 -0200, Alex [EMAIL PROTECTED] escribió: That's what I've been searching for, thanks. By the way, I know it might be trivial question... but function and class namespaces have __name__ attribute too. Why is global one always returned? I don't understand the question (even with the later correction namespaces-objects) There's no question anymore, I just failed to distinguish function local variables (which don't include __name__) and function object's attributes Why do you want to get the module object? globals() returns the module namespace, its __dict__, perhaps its only useful attribute... To pass it as a parameter to a function (in another module), so it can work with several modules (plugins for main program) in a similar manner. The function could receive a namespace to work with (a dictionary). Then you just call it with globals() == the namespace of the calling module. Yes, but access to module seems more verbose: module_dict['x']() xxx Instead of just: module.x() xxx -- http://mail.python.org/mailman/listinfo/python-list
How to get current module object
Can I get reference to module object of current module (from which the
code is currently executed)? I know __import__('filename') should
probably do that, but the call contains redundant information (filename,
which needs to be updated), and it'll perform unnecessary search in
loaded modules list.
It shouldn't be a real problem (filename can probably be extracted from
the traceback anyway), but I wonder if there is more direct and less
verbose way.
Thanks
--
http://mail.python.org/mailman/listinfo/python-list
Re: How to get current module object
On Feb 18, 5:25 am, Alex [EMAIL PROTECTED] wrote:
Can I get reference to module object of current module (from which the
code is currently executed)? I know __import__('filename') should
probably do that, but the call contains redundant information (filename,
which needs to be updated), and it'll perform unnecessary search in
loaded modules list.
It shouldn't be a real problem (filename can probably be extracted from
the traceback anyway), but I wonder if there is more direct and less
verbose way.
Try this:
C:\junktype whoami.py
def showme():
import sys
modname = globals()['__name__']
print repr(modname)
module = sys.modules[modname]
print repr(module)
print dir(module)
C:\junkpython
Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit
(Intel)] on win32
Type help, copyright, credits or license for more information.
import whoami
whoami.showme()
'whoami'
module 'whoami' from 'whoami.py'
['__builtins__', '__doc__', '__file__', '__name__', 'showme']
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Re: How to get current module object
En Sun, 17 Feb 2008 16:25:44 -0200, Alex [EMAIL PROTECTED] escribi�:
Can I get reference to module object of current module (from which the
code is currently executed)? I know __import__('filename') should
probably do that, but the call contains redundant information (filename,
which needs to be updated), and it'll perform unnecessary search in
loaded modules list.
It shouldn't be a real problem (filename can probably be extracted from
the traceback anyway), but I wonder if there is more direct and less
verbose way.
sys.modules[__name__]
Why do you want to get the module object? globals() returns the module
namespace, its __dict__, perhaps its only useful attribute...
--
Gabriel Genellina
--
http://mail.python.org/mailman/listinfo/python-list
