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MathGroup Archive 2003

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Re: Need a nice way to do this

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40289] Re: [mg40279] Need a nice way to do this
  • From: "Mihajlo Vanevic" <mvane at eunet.yu>
  • Date: Sun, 30 Mar 2003 04:08:04 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

First rename a->1 b->2 (with s/.{a->1, b->2})
and try this (a 'more general version')

In[]:=
s = {1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1};

In[]:=
t = Table[0, {Max[s]}];
sumt = 0;
g = (sumt++ - t[[#]]++) & /@ s



Regards,				 
         Mihajlo Vanevic
         mvane at EUnet.yu
         2003-03-29

**************************************************************
*    At 2003-03-29, 05:19:00 
*        Steve Gray, stevebg at adelphia.net  wrote:
**************************************************************
>	Given a list consisting of only two distinct values, such as
>s={a,b,b,a,a,a,b,a,b,a,a}, I want to derive a list of equal length
>g={0,1,1,2,2,2,4,3,5,4,4}. The rule is: for each position
>1<=p<=Length[s], look at list s and set g[[p]] to the number of
>elements in s to the left of p which are not equal to s[[p]].
>	In a more general version, which I do  not need now, s would
>not be restricted to only two distinct values.
>
>	Thank you for any ideas, including other applications where
>this particular calculation is used. The current application is an
>unusual conjecture in geometry.
**************************************************************







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