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
>