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MathGroup Archive 2007

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Re: Mathematica 6.0 easier for me ... (small review)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76563] Re: [mg76457] Mathematica 6.0 easier for me ... (small review)
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Thu, 24 May 2007 05:57:31 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

lis = Flatten[Transpose[
    Outer[List, Range[9], 10*Range[4]]], 1]

Since you did not provide an example of the desired output, it is ambiguous as to which pair you want to save.

Split will group the elements

sp = Split[lis, #1[[2]] == #2[[2]] &];

First /@ sp

{{1, 10}, {1, 20}, {1, 30}, {1, 40}}

Last /@ sp

{{9, 10}, {9, 20}, {9, 30}, {9, 40}}

Thread[{Most[Last /@ sp], Rest[First /@ sp]}]

{{{9, 10}, {1, 20}}, {{9, 20}, {1, 30}}, 
   {{9, 30}, {1, 40}}}


Bob Hanlon

---- Paul at desinc.com wrote: 
> 
> 4.  At work, I haven't had to resort to other programs because
> Mathematica wasn't the best choice.  This is extremely nice.  BTW,
> There's a "temporal" advantage with procedural programming that hasn't
> been apparant to me in functional or rule.  Still working on it. Maybe
> someone can help.  If I have
> lis={{1,10},{2,10},...{9,10},{11,20},{12,20}...{19,20}
> 
> How do I use functional and/or rule to determine where the second
> number (lis[[i,2]]) jumped from 10 to 20 to 30 and save the pair.
> Assuming there was noise, I only want to store the first 10->20, then
> look for 20->30 and so on.  So in time, I want my search to change as
> I progress through the list.  Any input appreciated!
> 



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