fast way to filter a set?

[fortepianissimo]

But this created two lists (keys and dummy)? Compared to the filter()
method, which created a list and a new set, maybe your suggestion is
still a bit faster...

--- Skip Montanaro <skip at pobox.com> wrote:
> 
>     fortepianissimo> I know I can do things like
>     fortepianissimo> s=Set(range(1,11))
>     fortepianissimo> s=Set(filter(lambda x:x%2==0,s))
> 
>     fortepianissimo> But this seems a bit slow since filter returns a
> list
>     fortepianissimo> which then must be converted back to a set. Any
> tips?
> 
> The only thing which comes to mind is:
> 
>     >>> s = sets.Set(range(1,11))
>     >>> s
>     Set([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
>     >>> keys = list(s)
>     >>> dummy = [s.discard(x) for x in keys if x%2]
>     >>> s
>     Set([2, 4, 6, 8, 10])
> 
> You still create a list, but don't create a second set.
> 
> Skip


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