fast way to filter a set?

[fortepianissimo]

Actually I like your idea most (out of other iterator approach) - now
if we could do an in-place set filtering... (maybe have a different
version of set.discard() which takes a lambda expression?)

--- Skip Montanaro <skip at pobox.com> wrote:
> 
>     fortepianissimo> But this created two lists (keys and dummy)?
> Compared
>     fortepianissimo> to the filter() method, which created a list and
> a new
>     fortepianissimo> set, maybe your suggestion is still a bit
> faster...
> 
> You asked for alternatives.  I offered one.  I'll let you do the
> performance
> testing. ;-)
> 
> Skip
> 
>     fortepianissimo> --- 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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