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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: [mg40296] Re: [mg40279] Need a nice way to do this
  • From: Dr Bob <majort at cox-internet.com>
  • Date: Sun, 30 Mar 2003 04:09:47 -0500 (EST)
  • References: <200303291019.FAA04099@smc.vnet.net>
  • Reply-to: majort at cox-internet.com
  • Sender: owner-wri-mathgroup at wolfram.com

leftCount[s_List] := Block[{count},
    count[any_] := 0;
    Range@Length@s - Rest@FoldList[++count[#2] &, s, s]]
leftCount[s]

{0, 1, 1, 2, 2, 2, 4, 3, 5, 4, 4}

or

leftCount[s_List] := Block[{count, n = 0},
    count[any_] := 0;
    Rest@FoldList[++n - (++count[#2]) &, s, s]]
leftCount[s]

{0, 1, 1, 2, 2, 2, 4, 3, 5, 4, 4}

Bobby

On Sat, 29 Mar 2003 05:19:39 -0500 (EST), 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.
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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