Hi all,
I have some rough code to sample consecutive integers with length
according to a vector of lengths
#sample space (representing positions)
pos<-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths<-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the sampled
integers need to be consecutive. Im hoping somebody knows a faster way
of doing it than I have. ATM its way to slow on large vectors.
samplePos<-function(l){
start.pos<-sample(pos,1)
end.pos<-start.pos+l-1
posies<-start.pos:end.pos
posies
}
s.start<-c()
newPos<-function(a){
rp<-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp<-samplePos(a)}
pos<-setdiff(pos,rp)
rp[1]
}
newps<-c()
newps<-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample
space is quite large so I dont think there would be too many rejections.
Many thanks,
Chris
On Thu, 28 Aug 2008, Chris Oldmeadow wrote:> Hi all, > > I have some rough code to sample consecutive integers with length according > to a vector of lengths > > #sample space (representing positions) > pos<-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20) > > #sample lengths > lengths<-c(2,3,2) > > From these two vectors I need a vector of sampled positions. > > > the sampling is without replacement, making things tough as the sampled > integers need to be consecutive. Im hoping somebody knows a faster way of > doing it than I have. ATM its way to slow on large vectors. > > > samplePos<-function(l){ > start.pos<-sample(pos,1) > end.pos<-start.pos+l-1 > posies<-start.pos:end.pos > posies > } > > s.start<-c() > > > newPos<-function(a){ > rp<-samplePos(a) > #test sampled range is consecutive, if not resample > if (length(rp) != rp[a]+1 -rp[1]){rp<-samplePos(a)} > pos<-setdiff(pos,rp) > rp[1] > } > > newps<-c() > newps<-unlist(lapply(lengths,newPos)) > > I think the bottleneck may be on the setdiff() function - the sample space is > quite large so I dont think there would be too many rejections.The bottleneck is in the formulation of the sampling scheme. This is a simple combinatorics problem. There are 3360 possible values ( prod(16:14) ) for the start positions of the three elements, and you can form a bijection between 1:3360 and the individual samples. If the number of possible sample is small enough, it would be most efficient to sample from the corresponding integer vector and then translate it to the corresponding sample. For larger values where the number of possible samples become a challenge for 32-bit integer arithmetic, I expect this approach would be preferred: Permute length ( pos ) - sum ( lengths ) + length( lengths ) distinct (consecutively labelled) elements: elz <- sample( length ( pos ) - sum ( lengths ) + length( lengths ) ) Take the lengths of the original objects to be z.lens <- rep( 1, length( elz ) ) z.lens[ seq(along = lengths ) ] <- lengths (i.e. objects longer than 1 appear first) Determine the start positions of the objects as if they were laid down consecutively according to the permutation: start <- head( cumsum( c(0, z.lens[ elz ]) ) + 1 , -1 ) Find the start positions of just those with lengths greater than 1 gt.1 <- match( seq(along=lengths) , elz ) Report the start positions start[ gt.1 ] --- If length( pos ) is large, you can rewrite the above to simply sample the positions (in the ordering) of the objects with lengths greater than 1. You will have to revise the calculation of start and gt.1 in that case. HTH, Chuck >> > > Many thanks, > Chris > > ______________________________________________ > R-help at 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. >Charles C. Berry (858) 534-2098 Dept of Family/Preventive Medicine E mailto:cberry at tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
Charles C. Berry wrote:> On Thu, 28 Aug 2008, Chris Oldmeadow wrote: > >> Hi all, >> >> I have some rough code to sample consecutive integers with length >> according to a vector of lengths >> >> #sample space (representing positions) >> pos<-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20) >> >> #sample lengths >> lengths<-c(2,3,2) >> >> From these two vectors I need a vector of sampled positions. >> >> >> the sampling is without replacement, making things tough as the >> sampled integers need to be consecutive. Im hoping somebody knows a >> faster way of doing it than I have. ATM its way to slow on large >> vectors. >> >> >> samplePos<-function(l){ >> start.pos<-sample(pos,1) >> end.pos<-start.pos+l-1 >> posies<-start.pos:end.pos >> posies >> } >> >> s.start<-c() >> >> >> newPos<-function(a){ >> rp<-samplePos(a) >> #test sampled range is consecutive, if not resample >> if (length(rp) != rp[a]+1 -rp[1]){rp<-samplePos(a)} >> pos<-setdiff(pos,rp) >> rp[1] >> } >> >> newps<-c() >> newps<-unlist(lapply(lengths,newPos)) >> >> I think the bottleneck may be on the setdiff() function - the sample >> space is quite large so I dont think there would be too many rejections. > > > The bottleneck is in the formulation of the sampling scheme. > > This is a simple combinatorics problem. There are 3360 possible values > ( prod(16:14) ) for the start positions of the three elements, and you > can form a bijection between 1:3360 and the individual samples. If the > number of possible sample is small enough, it would be most efficient > to sample from the corresponding integer vector and then translate it > to the corresponding sample. For larger values where the number of > possible samples become a challenge for 32-bit integer arithmetic, I > expect this approach would be preferred: > > Permute length ( pos ) - sum ( lengths ) + length( lengths ) distinct > (consecutively labelled) elements: > > elz <- sample( length ( pos ) - sum ( lengths ) + length( lengths ) ) > > > Take the lengths of the original objects to be > > z.lens <- rep( 1, length( elz ) ) > z.lens[ seq(along = lengths ) ] <- lengths > > (i.e. objects longer than 1 appear first) > > Determine the start positions of the objects as if they were laid down > consecutively according to the permutation: > > start <- head( cumsum( c(0, z.lens[ elz ]) ) + 1 , -1 ) > > Find the start positions of just those with lengths greater than 1 > > gt.1 <- match( seq(along=lengths) , elz ) > > Report the start positions > > start[ gt.1 ] > > --- > > If length( pos ) is large, you can rewrite the above to simply sample > the positions (in the ordering) of the objects with lengths greater > than 1. You will have to revise the calculation of start and gt.1 in > that case. > > HTH, > > Chuck > > > > >> >> >> Many thanks, >> Chris >> >> ______________________________________________ >> R-help at 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. >> > > Charles C. Berry (858) 534-2098 > Dept of Family/Preventive > Medicine > E mailto:cberry at tajo.ucsd.edu UC San Diego > http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego > 92093-0901 > > >Thanks! that worked perfectly. Chris
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