Is this what you want? I was not clear on your algorithm, but is looks
like you wanted descending values:
> testx <-
+ function(n, verbose = FALSE)
+ {
+ mat <- cbind(optA = sample(n, n, TRUE), perm = seq(n))
+ if (verbose){
+ cat("***starting matrix\n")
+ print(mat)
+ }
+ # iterate till done
+ while(nrow(mat) > 1){
+ high <- diff(mat[, 'optA']) > 0
+ if (!any(high)) break # done
+ # find which one to delete
+ delete <- which.max(high) + 1L
+ mat <- mat[-delete, ]
+ }
+ mat
+ }
> testx(20, verbose = TRUE)
***starting matrix
optA perm
[1,] 13 1
[2,] 12 2
[3,] 7 3
[4,] 10 4
[5,] 11 5
[6,] 4 6
[7,] 11 7
[8,] 2 8
[9,] 6 9
[10,] 5 10
[11,] 6 11
[12,] 18 12
[13,] 9 13
[14,] 16 14
[15,] 18 15
[16,] 9 16
[17,] 2 17
[18,] 7 18
[19,] 15 19
[20,] 7 20
optA perm
[1,] 13 1
[2,] 12 2
[3,] 7 3
[4,] 4 6
[5,] 2 8
[6,] 2 17
>
On Fri, May 31, 2013 at 2:02 PM, Laz <lmra...@ufl.edu> wrote:
> Dear R Users,
>
> I created a function which returns a value every time it's run (A
> simplified toy example is attached on this mail).
>
> For example spat(a,b,c,d) # run the first time and it gives you ans1=
> 10, perm=1 ;
> run the second time and gives you ans2= 7, perm=2 etc
> I store both the result and the iteration on a matrix called vector
> with columns:1==ans, column2==permutation
>
> The rule I want to implement is: compare between ans1 and ans2. If
> ans1>ans2, keep both ans1 and ans2. if ans1<ans2, drop the whole row of
> the second permutation (that is drop both ans2 and perm2 but continue
> counting all permutations).
> Re-run the function for the 3rd time and repeat comparison between the
> value of the last run and the current value obtained.
> Return matrix ans with the saved ans and their permutations and another
> full matrix with all results including the dropped ans and their
> permutation numbers.
>
> I need to repeat this process 1000 times but I am getting stuck below.
> See attached R code.
> The code below works only for the first 6 permutations. suppose I want
> 1000 permutations, where do I keep the loop?
>
>
> Example: Not a perfect code though it somehow works:
> testx<-function(perm){
> optA<-c()
> #while(perm<=2){
> for (k in 1:perm){
> #repeat {
> optA[k]<-sample(1:1000,1,replace=TRUE)
> perm[k]<-k
> }
> mat2<-as.matrix(cbind(optA=optA,perm=perm))
> result<-mat2
> lenm<-nrow(result)
> if(result[1,1]<=result[2,1]) result<-result[1,]
> return(list(mat2=mat2,result=result))
> #}
> if(perm>2){
> mat2<-as.matrix(cbind(optA=optA,perm=perm))
> result<-mat2
> lenm<-nrow(result)
> if(result[1,1]<=result[2,1]) result<-result[1,]
> if(result[lenm-1,1]<=result[lenm,1]) result<-result[-lenm,]
> if(result[(lenm-2),1]<=result[(lenm-1),1])
> result<-result[-(lenm-1),]
> if(result[(lenm-3),1]<=result[(lenm-2),1])
> result<-result[-(lenm-2),]
> if(result[(lenm-4),1]<=result[(lenm-3),1])
> result<-result[-(lenm-3),]
> if(result[(lenm-5),1]<=result[(lenm-4),1])
> result<-result[-(lenm-4),]
> return(list(mat2=mat2,result=result))
> }
> }
> ## Now calling my function below:
>
> >testx(perm=6)
>
>
>
> $mat2
> optA perm
> [1,] 272 1
> [2,] 858 2
> [3,] 834 3
> [4,] 862 4
> [5,] 650 5
> [6,] 405 6
>
> $result
> optA perm
> 272 1
>
>
> > testx(perm=2)
> $mat2
> optA perm
> [1,] 398 1
> [2,] 816 2
>
> $result
> optA perm
> 398 1
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
