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 >