changing __call__ on demand

[Stefan Behnel]

Hi!

This somewhat puzzles me:

Python 2.4 (#1, Feb 3 2005, 16:47:05)
[GCC 3.3.4 (pre 3.3.5 20040809)] on linux2
Type "help", "copyright", "credits" or "license" for more information.

.>>> class test(object):
... def __init__(self):
... self.__call__ = self.__call1
... def __call1(self):
... print 1
... def __call__(self):
... print 2
...
.>>> t = test()
.>>> t()
2

If I take out the __call__ method completely and only set it in __init__, I
get a TypeError saying that test is not callable.

I want to use this in order to provide different implementations based on the
object configuration. Calculating the right function to call is non-trivial
and calls are frequent, so I want to change __call__ in order to run the right
function directly.

I know, I could use another level of indirection:

def __call__(self):
self.the_right_method()

and then set the_right_method accordingly, but I find that somewhat
sub-optimal. Is there a way to change __call__ after class creation?

Stefan

[Alan McIntyre]

I tried this:

>>>class test(object):

... def __call1(self):
... print 1

... __call__ = __call1
...
>>>t = test()
>>>t()
1
>>>

Is that what you were looking for?

--
Alan McIntyre
ESRG LLC
http://www.esrgtech.com

[Michael Hoffman]

Stefan Behnel wrote:
> Is there a way to change __call__ after class creation?

__call__, like __getitem__, and __getattr__ is called on the
class object, not the instance object. So, no, not as far as I
am aware, without using metaclass trickery. The simplest option
IMO is to use another level of indirection as you suggest.
--
Michael Hoffman

[Michael Hoffman]

Alan McIntyre wrote:
> >>>class test(object):
> ... def __call1(self):
> ... print 1
> ... __call__ = __call1

> Is that what you were looking for?

That still only allows him to have one call function per class.
--
Michael Hoffman

[Stefan Behnel]

Michael Hoffman schrieb:

> __call__, like __getitem__, and __getattr__ is called on the
> class object, not the instance object. So, no, not as far as I
> am aware, without using metaclass trickery. The simplest option
> IMO is to use another level of indirection as you suggest.

Thanks for the quick answer. I didn't know they were class-level methods. Too
bad. Guess I'll stick with indirection then.

Stefan

[Steven Bethard]

Stefan Behnel wrote:
> Is there a way to change __call__ after class creation?

Check out this thread on the topic:

http://mail.python.org/pipermail/python-list/2004-January/203142.html

Basically, the answer is no -- at least not on a per-instance basis.
You can try something like:

py> class Test(object):
... def __new__(cls):
... cls.__call__ = cls.call1
... return object.__new__(cls)
... def call1(self):
... print 'call1'
... def __call__(self):
... print '__call__'
...
py> Test()()
call1

But then the call method is changed for all instances:

py> class Test(object):
... instances = 0
... def __new__(cls):
... if cls.instances == 1:
... print "setting __call__"
... cls.__call__ = cls.call1
... cls.instances += 1
... return object.__new__(cls)
... def call1(self):
... print 'call1'
... def __call__(self):
... print '__call__'
...
py> t1 = Test()
py> t1()
__call__
py> t2 = Test()
setting __call__
py> t2()
call1
py> t1()
call1

Steve

[Kent Johnson]

Stefan Behnel wrote:
> Hi!
>
> This somewhat puzzles me:
>
> Python 2.4 (#1, Feb 3 2005, 16:47:05)
> [GCC 3.3.4 (pre 3.3.5 20040809)] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
>
> .>>> class test(object):
> ... def __init__(self):
> ... self.__call__ = self.__call1
> ... def __call1(self):
> ... print 1
> ... def __call__(self):
> ... print 2
> ...
> .>>> t = test()
> .>>> t()
> 2
>

It works the way you want if test is an old-style class:
>>> class test:

... def __init__(self):
... self.__call__ = self.__call1
... def __call1(self):
... print 1
... def __call__(self):
... print 2
...

>>> t=test()
>>> t()
1

Kent

[Alan McIntyre]

Thanks; I didn't read close enough. :)

--
Alan McIntyre
ESRG LLC
http://www.esrgtech.com

[Michael Hoffman]

Stefan Behnel wrote:
>
> Thanks for the quick answer. I didn't know they were class-level
> methods. Too bad. Guess I'll stick with indirection then.

Here is one way of doing that indirection I just thought of--have
the class __call__ attribute call on the instance __call__
attribute:

>>> class MyClass(object):
... def __init__(self, func):
... self.__call__ = func
... def __call__(self, *args, **keywds):
... return self.__call__(*args, **keywds)
...
>>> def f1(): return "foo"
...
>>> def f2(x, y): return x+y
...
>>> MyClass(f1)()
'foo'
>>> MyClass(f2)(30, 12)
42

I still can't figure out whether this is elegant, or opaque and
to be avoided. <wink>
--
Michael Hoffman

[Hans Nowak]

Stefan Behnel wrote:
> Hi!
>
> This somewhat puzzles me:
>
> Python 2.4 (#1, Feb 3 2005, 16:47:05)
> [GCC 3.3.4 (pre 3.3.5 20040809)] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
>
> .>>> class test(object):
> ... def __init__(self):
> ... self.__call__ = self.__call1
> ... def __call1(self):
> ... print 1
> ... def __call__(self):
> ... print 2
> ...
> .>>> t = test()
> .>>> t()
> 2
>
> If I take out the __call__ method completely and only set it in
> __init__, I get a TypeError saying that test is not callable.

Note that it works just fine if you don't use a new-style class:

>>> class Test:
... def __init__(self):
... self.__call__ = self.foobar
... def foobar(self, *args, **kwargs):
... print "Called with:", args, kwargs
...
>>> t = Test()
>>> t()
Called with: () {}
>>> t(3, 4)
Called with: (3, 4) {}
>>> t(42, x=0)
Called with: (42,) {'x': 0}

--
Hans Nowak
http://zephyrfalcon.org/

[F. Petitjean]

Le Sun, 13 Feb 2005 13:19:03 -0500, Hans Nowak a écrit :
> Stefan Behnel wrote:
>> Hi!
>>
>> This somewhat puzzles me:
>>
>> Python 2.4 (#1, Feb 3 2005, 16:47:05)
>> [GCC 3.3.4 (pre 3.3.5 20040809)] on linux2
>> Type "help", "copyright", "credits" or "license" for more information.
>>
>> .>>> class test(object):
>> ... def __init__(self):

>> ... self.__call__ = self.__call1 # self.__call__ is bound

>> ... def __call1(self):
>> ... print 1

>> ... def __call__(self): # self.__call__ is rebound

>> ... print 2
>> ...
>> .>>> t = test()
>> .>>> t()
>> 2

2 because the last defined __call__ wins. Try dir(t)

>>
>> If I take out the __call__ method completely and only set it in
>> __init__, I get a TypeError saying that test is not callable.

This seems logical.

>
> Note that it works just fine if you don't use a new-style class:
>
> >>> class Test:
> ... def __init__(self):
> ... self.__call__ = self.foobar
> ... def foobar(self, *args, **kwargs):
> ... print "Called with:", args, kwargs
> ...
> >>> t = Test()
> >>> t()
> Called with: () {}
> >>> t(3, 4)
> Called with: (3, 4) {}
> >>> t(42, x=0)
> Called with: (42,) {'x': 0}

Are you sure that if you add a __call__() method, it will still work
fine ?

Regards
>

[Steven Bethard]

Simple enough to check, isn't it?

py> class C:
... def __init__(self):
... self.__call__ = lambda: "__init__"
... def __call__(self):
... return "__call__"
...
py> C()()
'__init__'

Old-style classes lookup methods like __call__ on the instance[1].
New-style classes look them up on the type:

py> class C:
... def __init__(self):
... self.__iter__ = lambda: iter(["__init__"])
... def __iter__(self):
... return iter(["__call__"])
...
py> list(C())
['__init__']
py> class C(object):
... def __init__(self):
... self.__iter__ = lambda: iter(["__init__"])
... def __iter__(self):
... return iter(["__call__"])
...
py> list(C())
['__call__']

AFAICT, non-magic methods are looked up on instance first:

py> class C:
... def __init__(self):
... self.f = lambda: "__init__"
... def f(self):
... return "__call__"
...
py> C().f()
'__init__'
py> class C(object):
... def __init__(self):
... self.f = lambda: "__init__"
... def f(self):
... return "__call__"
...
py> C().f()
'__init__'

STeVe

[1] Well, they look there first.