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MathGroup Archive 2006

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Re: programming problem about elements taken

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72478] Re: [mg72470] programming problem about elements taken
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 31 Dec 2006 05:23:04 -0500 (EST)
  • Reply-to: hanlonr at cox.net

For a sorted output:

choose1[a_?VectorQ, epsilon_?NumericQ]:=Module[
      {sa=Sort[a]},
      Fold[If[#2-Last[#1]>epsilon,
            Append[#1,#2],#1]&,{First[sa]},Rest[sa]]];

To retain the original element order:

choose2[a_?VectorQ, epsilon_?NumericQ]:=Module[
      {sa=Sort[a]},
      Select[a,MemberQ[Fold[If[#2-Last[#1]>epsilon,
                  Append[#1,#2],#1]&,{First[sa]},Rest[sa]], #]&]];

a=Table[Random[],{20}]

{0.67319,0.378445,0.981498,0.579474,0.798173,0.537112,0.884835,0.534093,0.\
0471484,0.0516117,0.988746,0.570592,0.261828,0.632062,0.754747,0.436792,0.\
669037,0.149373,0.802271,0.484141}

choose1[a,0.1]

{0.0471484,0.149373,0.261828,0.378445,0.484141,0.632062,0.754747,0.884835,0.\
988746}

choose2[a,0.1]

{0.378445,0.884835,0.0471484,0.988746,0.261828,0.632062,0.754747,0.149373,0.\
484141}

%%==Sort[%]

True

However, the second method (original order) is much slower for a large list.

a=Table[Random[],{1000}];

Timing[choose1[a,0.1];]

{0.008364 Second,Null}

Timing[choose2[a,0.1];]

{1.23431 Second,Null}


Bob Hanlon

---- Barrow <GRseminar at gmail.com> wrote: 
> Dear all,
>   I have a list of numbers (A),in fact, they are numbers distributed
> over [0,1].  Given parameter \epsilon, I have to choose elements from A
> such that their distances are larger than \epsilon, so the elements are
> "distringuishable". My goal is to find the maximum number of elements
> from A to meet the above "distinct criterion".
>   How to do it with the functions build-in in mathematica ??
>   Thanks in advence.           Sincerely           Barrow
> 


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