Re: A replacement for lambda

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.

.

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