How to create functors?

[Rami Chowdhury]

> The example I gave earlier is a bit contrived, the real example
> fundamentally requires a lambda since I am actually passing in local
> variables into the functions the lambda is wrapping. Example:
>
> def MyFunction():
>   localVariable = 20
>   CreateTask( lambda: SomeOtherFunction( localVariable ) ) # CreateTask
> () executes the functor internally
>
> This is more or less like the real scenario I'm working with.

You could define a local function ('closure') inside the function:

def MyFunction():
	localVariable = 20
	def UseLocalVariable():
		SomeOtherFunction(localVariable)
	CreateTask(UseLocalVariable)

I would suggest this is much better for readability.

On Tue, 18 Aug 2009 13:51:19 -0700, Robert Dailey <rcdailey at gmail.com>  
wrote:

> On Aug 18, 3:45 pm, "Rami Chowdhury" <rami.chowdh... at gmail.com> wrote:
>> Lambda expressions are, I believe, syntactically limited to a single  
>> expression -- no statements, like 'print' is in Python 2.x.
>>
>> If you are strongly against just defining a function, you might have to  
>>  
>> use a trick to get around it -- this page  
>> (http://p-nand-q.com/python/stupid_lambda_tricks.html) has some  
>> suggestions.
>>
>> On Tue, 18 Aug 2009 13:32:55 -0700, Robert Dailey <rcdai... at gmail.com>  
>> wrote:
>>
>>
>>
>>
>>
>> > On Aug 18, 3:31 pm, Duncan Booth <duncan.bo... at invalid.invalid> wrote:
>> >> Robert Dailey <rcdai... at gmail.com> wrote:
>> >> > Hello,
>>
>> >> > I want to simply wrap a function up into an object so it can be  
>> called
>> >> > with no parameters. The parameters that it would otherwise have  
>> taken
>> >> > are already filled in. Like so:
>>
>> >> >       print1 = lambda: print( "Foobar" )
>> >> >       print1()
>>
>> >> > However, the above code fails with:
>>
>> >> >   File "C:\IT\work\distro_test\distribute_radix.py", line 286
>> >> >     print1 = lambda: print( "Foobar" )
>> >> >                          ^
>> >> > SyntaxError: invalid syntax
>>
>> >> > How can I get this working?
>>
>> >> def print1():
>> >>     print "Foobar"
>>
>> >> It looks like in your version of Python "print" isn't a function. It  
>>  
>> >> always
>> >> helps if you say the exact version you are using in your question as  
>> the
>> >> exact answer you need may vary.
>>
>> > I'm using Python 2.6. And using the legacy syntax in the lambda does
>> > not work either. I want to avoid using a def if possible. Thanks.
>>
>> --
>> Rami Chowdhury
>> "Never attribute to malice that which can be attributed to stupidity"  
>> --  
>> Hanlon's Razor
>> 408-597-7068 (US) / 07875-841-046 (UK) / 0189-245544 (BD)
>
> The example I gave earlier is a bit contrived, the real example
> fundamentally requires a lambda since I am actually passing in local
> variables into the functions the lambda is wrapping. Example:
>
> def MyFunction():
>   localVariable = 20
>   CreateTask( lambda: SomeOtherFunction( localVariable ) ) # CreateTask
> () executes the functor internally
>
> This is more or less like the real scenario I'm working with. There
> are other (more personal) reasons why I prefer to avoid 'def' in this
> case. I want to keep the functor as central to the code that needs it
> as possible to improve code readability.
>
> Thanks for the help everyone. I guess in Python 3.0 the print()
> function will not require the import from __future__ to work in this
> particular case?



-- 
Rami Chowdhury
"Never attribute to malice that which can be attributed to stupidity" --  
Hanlon's Razor
408-597-7068 (US) / 07875-841-046 (UK) / 0189-245544 (BD)