fast way to filter a set?

[Fortepianissimo]

I know I can do things like

s=Set(range(1,1 1))
s=Set(filter(la mbda x:x%2==0,s))

But this seems a bit slow since filter returns a list which then must
be converted back to a set. Any tips? Thanks!

[Nicola Mingotti]

Re: fast way to filter a set?

fortepianissimo @yahoo.com.tw (Fortepianissim o) writes:
[color=blue]
> I know I can do things like
>
> s=Set(range(1,1 1))
> s=Set(filter(la mbda x:x%2==0,s))
>
> But this seems a bit slow since filter returns a list which then must
> be converted back to a set. Any tips? Thanks![/color]

Can you ?
I tried python2.3 and 2.2 and it replies
: "NameError: name 'Set' is not defined" .

There is a PEP but it seems it's steel 'under consideration'.

http://www.python.org/peps/pep-0218.html

May be , if this proposal is accepted you
one day could write :
{ x for x in S if x%2 == 0 } # S be a set .

bye.

[Skip Montanaro]

Re: fast way to filter a set?


fortepianissimo > I know I can do things like
fortepianissimo > s=Set(range(1,1 1))
fortepianissimo > s=Set(filter(la mbda 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:
[color=blue][color=green][color=darkred]
>>> s = sets.Set(range( 1,11))
>>> s[/color][/color][/color]
Set([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])[color=blue][color=green][color=darkred]
>>> keys = list(s)
>>> dummy = [s.discard(x) for x in keys if x%2]
>>> s[/color][/color][/color]
Set([2, 4, 6, 8, 10])

You still create a list, but don't create a second set.

Skip

[Peter Otten]

Re: fast way to filter a set?

Nicola Mingotti wrote:
[color=blue]
> fortepianissimo @yahoo.com.tw (Fortepianissim o) writes:
>[color=green]
>> I know I can do things like
>>
>> s=Set(range(1,1 1))
>> s=Set(filter(la mbda x:x%2==0,s))
>>
>> But this seems a bit slow since filter returns a list which then must
>> be converted back to a set. Any tips? Thanks![/color]
>
> Can you ?
> I tried python2.3 and 2.2 and it replies
> : "NameError: name 'Set' is not defined" .[/color]

In 2.3 it works if you do

from sets import Set

first.

Peter

[Peter Otten]

Re: fast way to filter a set?

Fortepianissimo wrote:
[color=blue]
> I know I can do things like
>
> s=Set(range(1,1 1))
> s=Set(filter(la mbda x:x%2==0,s))
>
> But this seems a bit slow since filter returns a list which then must
> be converted back to a set. Any tips? Thanks![/color]

The Set constructor accepts any iterable, so you can do it all with
iterators instead of temporary lists:

from sets import Set
from itertools import ifilter

s = Set(range(1, 11))
print Set(ifilter(lam bda x: x % 2 == 0, s))

Peter