not for me...

lanx@nc10-ubuntu:~$ python
Python 2.5.2 (r252:60911, Jan 20 2010, 23:16:55) 
[GCC 4.3.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> def rubyyielder(gen):
...     def wrapped_gen(block):
...         for elem in gen():
...             block(elem)
...     return wrapped_gen
... 
>>> @rubyyielder
... def test():
...     print("In test")
...     yield 1
...     print("back in test")
...     yield 2
... 
>>> test(lambda a: print("You are in block %s" % a))
  File "<stdin>", line 1
    test(lambda a: print("You are in block %s" % a))
                       ^
SyntaxError: invalid syntax
[download]

I suppose the lamda syntax has more restrictions...

> If you used a non-keyword such as "send" instead of "yield" you could do Evil and avoid the first decorator. This is left as an exercise.

Simple, I can just port the semantic of my OP and let send execute the callback which is passed to @test.¹

Update

¹) at second thought this would require a possibility to access the arguments of the caller. I suppose the caller is an object where arguments are accessible.

Update

deleted undiplomatic irony =)

Ah, you're using an old version where print was still a statement not a function; Python is finicky about how statements are used. v2.6-v2.7 allows "from __future__ import print_function", and in v3.0+ it is a function. If you really must use a lambda in v2.5- then "import sys" and use "sys.stdout.write()"... no you mustn't.

-T. "We now return you to your regularly scheduled Perl"

> Python is finicky

indeed!

> -T. "We now return you to your regularly scheduled Perl"

agreed! =)

Cheers Rolf

( addicted to the Perl Programming Language)