Re: removing sublist . Again and Different
- To: mathgroup at smc.vnet.net
- Subject: [mg56327] Re: removing sublist . Again and Different
- From: Peter Pein <petsie at arcor.de>
- Date: Fri, 22 Apr 2005 06:22:56 -0400 (EDT)
- References: <d47tgv$575$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
giampi1960 wrote:
> hello i read borges2003xx at yahoo.it meassage :
>
>
>
>>i'm a newbie. How implement the _faster functon_ which removes in a
>>list every element that are a subelement.
>>which means
>>f[x]:= {..,{1,2,3,4},..,{2,3},..} removes {2,3}.
>>thanx a lot.
>>giorgio borghi
>
>
>
> i ask your help for faster way to do the opposite
> f[x]:= {..,{1,2,3,4},..,{2,3},..} removes {1,2,3,4}.
>
> best regards
>
> giampiero
>
remSupersets[ls_] :=
Cases[Transpose[{ls,
Not /@ Or @@@ Table[
If[i == j, False, Complement[ls[[j]], ls[[i]]] == {}],
{i, Length[ls]}, {j, Length[ls]}]
}], {x_, True} :> x]
lst = {{1, 4, 2}, {1, 2}, {1, 2, 3, 4}, {4, 4}, {2, 3}, {1, 2, 3}};
remSupersets[lst]
--> {{1, 2}, {4, 4}, {2, 3}}
--
Peter Pein
Berlin