A replacement for lambda

[Stefan Rank]

on 30.07.2005 10:20 Paolino said the following:
> why (x**2 with(x))<(x**3 with(x)) is not taken in consideration?
> 
> If 'with' must be there (and substitue 'lambda:') then at least the 
> syntax is clear.IMO Ruby syntax is also clear.
> 

I am sorry if this has already been proposed (I am sure it has).

Why not substitue python-lambdas with degenerated generator expressions::

   (lambda x: func(x)) == (func(x) for x)

i.e. a one time callable generator expression (missing the `in` part). 
The arguments get passed into the generator, I am sure that can be 
combined with the PEP about passing args and Exceptions into a generator.