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