[R] Identifying sequences

[Mike Marchywka]

----------------------------------------
> Date: Wed, 1 Jun 2011 17:12:29 +0200
> From: cjpauw at gmail.com
> To: r-help at r-project.org
> Subject: Re: [R] Identifying sequences
>
> Thanks to David, Thierry and Jonathan for your help.
> I have been able to put this function together
>
> a=1:10
> b=20:30
> c=40:50
> x=c(a,b,c)
>
> seq.matrix <- function(x){
> lower<- x[which(diff(x) != 1)]
> upper <- x[which(diff(x) != 1)+1]
> extremities <- c(1,lower, upper,x[length(x)])
> m <- data.frame(matrix(extremities[order(extremities)],ncol=2,byrow=TRUE,dimnames=list(rows=paste("group",1:(length(lower)+1),sep=""),cols=c("lower","upper"))))
> m$length=m$upper-m$lower+1
> m
> }
>
> s.m=seq.matrix(x)
> s.m
> lower upper length
> group1 1 10 10
> group2 20 30 11
> group3 40 50 11
>
> One can then make a test to see if a certain value (say 9) falls
> within one of the groups and use that to find the group name or lower
> or upper border

As I understand, you are looking for large derivatives
or approx discontinuity against smooth signal.
This seems like a natural application for wavelets,
try the haar wavelet and use package 
wavelets,

> library(wavelets)
> f=wt.filter(c(-1,1),modwt=T)
> z<-modwt(X=as.numeric(x),filter=f,n.levels=1)
> z at W
$W1
      [,1]
 [1,]   49
 [2,]   -1
 [3,]   -1
 [4,]   -1
 [5,]   -1
 [6,]   -1
 [7,]   -1
 [8,]   -1
 [9,]   -1
[10,]   -1
[11,]  -10
[12,]   -1
[13,]   -1
[14,]   -1
[15,]   -1
[16,]   -1
[17,]   -1
[18,]   -1
[19,]   -1
[20,]   -1
[21,]   -1
[22,]  -10
[23,]   -1
[24,]   -1
[25,]   -1
[26,]   -1
[27,]   -1
[28,]   -1
[29,]   -1
[30,]   -1
[31,]   -1
[32,]   -1







>
> s.m.test=function(s.m,i){which(s.m[,1] > s.m.test(s.m,i=9)
> [1] 1
> e.g.
> row.names(s.m)[s.m.test(s.m,i=9)]
> [1] "group1"
>
> Cheers
> Christiaan
>
> On 1 June 2011 14:31, Jonathan Daily wrote:
> >
> > I am assuming in this case that you are looking for continuity along
> > integers, so if you expect noninteger values this will not work.
> >
> > You can get the index of where breaks can be found in your example using
> >
> > which(diff(x) > 1)
> >
> > On Wed, Jun 1, 2011 at 6:27 AM, christiaan pauw wrote:
> > > Hallo Everybody
> > >
> > > Consider the following vector
> > >
> > > a=1:10
> > > b=20:30
> > > c=40:50
> > > x=c(a,b,c)
> > >
> > > I need a function that can tell me that there are three set of continuos
> > > sequences and that the first is from 1:10, the second from 20:30 and the
> > > third from 40:50. In other words: a,b, and c.
> > >
> > > regards
> > > Christiaan
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > ______________________________________________
> > > 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.
> > >
> >
> >
> >
> > --
> > ===============================================
> > Jon Daily
> > Technician
> > ===============================================
> > #!/usr/bin/env outside
> > # It's great, trust me.
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.