A replacement for lambda
stefan.rank at ofai.at
Sat Jul 30 11:08:28 CEST 2005
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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