Murta
Posts:
86
Registered:
7/5/10


Re: Group and Replace itens sequence in a list
Posted:
Aug 25, 2012 4:34 AM


Tks for Bob and Ulrich! I choosed to work with this combination of both
SplitAt[list_,spliter_]:=Module[{l}, l=(list//.{s___,Sequence@@spliter,r___}>{{s},{r}})/.{}>Sequence[]; Cases[l,_?VectorQ,Infinity] ]
On Friday, August 24, 2012 6:07:23 AM UTC3, Bob Hanlon wrote: > Clear[f] > > > > f[list_?VectorQ] := Cases[ > > (list //. {s___, a, b, a, r___} > {{s}, {r}}) /. > > {} > > > Sequence[], _?VectorQ, Infinity] > > > > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12}] > > > > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}} > > > > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, > > 11, 12}] > > > > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}} > > > > f[{1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, 11, 12, > > a, b, a}] > > > > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}} > > > > f[{a, b, a, 1, 2, 3, a, b, a, 4, 5, 6, a, b, c, 7, 8, 9, a, b, a, 10, > > 11, 12, a, b, a}] > > > > {{1, 2, 3}, {4, 5, 6, a, b, c, 7, 8, 9}, {10, 11, 12}} > > > > > > Bob Hanlon > > > > > > On Thu, Aug 23, 2012 at 2:52 AM, Murta <rodrigomurtax@gmail.com> wrote: > > > Hi All > > > > > > I have a simple problem that is: > > > > > > l={1,2,3,a,b,a,4,5,6,a,b,c,7,8,9,a,b,a,10,11,12} > > > > > > I want to replace all a,b,a sequence by X to get: > > > > > > l={1,2,3,X,4,5,6,a,b,7,8,9,X,10,11,12} > > > > > > Then I want to group it by X intervals as > > > l={{1,2,3},{4,5,6,a,b,7,8,9},{10,11,12}} > > > > > > If I don't need to put the intermediate X, even better! > > > I think the with pattern, RaplaceAll and DeleteCases I can do It. Some clue? > > > Tks > > > Murta > > > > > > > > >

