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Topic: Group and Replace itens sequence in a list
Replies: 10   Last Post: Aug 29, 2012 1:14 AM

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Murta

Posts: 86
Registered: 7/5/10
Re: Group and Replace itens sequence in a list
Posted: Aug 25, 2012 4:34 AM
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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 UTC-3, 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
>
> >
>
> >
>
> >




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