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MathGroup Archive 2005

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Re: the faster way to find repeated sublists

  • To: mathgroup at smc.vnet.net
  • Subject: [mg60958] Re: [mg60923] the faster way to find repeated sublists
  • From: Pratik Desai <pdesai1 at umbc.edu>
  • Date: Wed, 5 Oct 2005 02:27:57 -0400 (EDT)
  • References: <200510040524.BAA17873@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

giampiero wrote:

>i'm newbie
>with a stupid problem
>
>a function for find repetead sublist in faster way
>
>ex
>f[{1,2,1,2,1,2},{1,2}]-> True cause {1,2} is repated three times
>f[{1,2,3,1,2,3},{1,2,3}]-> True cause {1,2,3} is repeated two times
>f[{1,2,1,2,3},{1,2}]->False cause after two {1,2} there is another
>symbol.
>
>True if the second list in containes many times exactly in first list
>False otherwise.
>
>
>bye everyone and sorry for my stupidity.
>
>giampiero
>
>  
>
Hi
Here something I adapted from an example in  help on Count for your problem


Clear[ww,pat]
ww = {1,2,1,2,1,2}
pat = {1, 2}
If[Count[NestList[Rest, ww,
    Length[ww]], Append[pat, ___]] == Length[ww]/Length[pat], Print[
        "True"], Print["False"]]
 >>True

Clear[ww,pat]
ww = {1, 2, 3, 1, 2, 3}
pat = {1, 2, 3}
If[Count[NestList[Rest, ww, Length[ww]], Append[
  pat, ___]] == Length[ww]/Length[pat], Print["True"], Print["False"]]
 >>True




Clear[ww,pat]
ww = {1, 2, 1, 2, 3}
pat = {1, 2}
If[Count[NestList[Rest, ww,
    Length[ww]], Append[pat, ___]] == Length[ww]/Length[pat], Print[
        "True"], Print["False"]]
 >>False


Hope this helps

Pratik  .


-- 
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134



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