Thanks!
I see, that do.call-function is used often in R-algorithms... Looking over
some extra do.call-examples seems useful. This tail-function is also new for
me.
Is there some reason to use seq(along = VECTOR) instead of 1:length(VECTOR)?
Atte
> You can replace the for with lapply like this:
>
> VECTOR <- c(0, 3, 6, 3, 11, 2, 11, 4, 3, 4, 7, 7, 6, 4, 8)
> f <- function(i) unique(tail(VECTOR, length(VECTOR)-i+1))[1:5]
> out <- do.call(rbind, lapply(seq(along = VECTOR), f))
> na.omit(rbind(rep(NA, 5), out))
>
> Note that a matrix with zero rows is returned in the case
> that VECTOR has zero length and in the case that VECTOR
> has fewer than 5 unique elements.
>
>
> On 8/26/06, Atte Tenkanen <[EMAIL PROTECTED]> wrote:
> > Again my example was't very clear: there were not enough same
> numbers in the VECTOR.
> > What I need is something like this:
> >
> > VECTOR<-c(0,3,6,3,11,2,11,4,3,4,7,7,6,4,8)
> > MATRIX<-c()
> > for(i in 1:length(VECTOR)){
> > v<-(unique(VECTOR[i:length(VECTOR)])[1:5])
> > MATRIX<-rbind(MATRIX,v)
> > }
> >
> > MATRIX<-na.omit(MATRIX)
> > MATRIX
> >
> > > data.frame(MATRIX, row.names=NULL)
> > X1 X2 X3 X4 X5
> > 1 0 3 6 11 2
> > 2 3 6 11 2 4
> > 3 6 3 11 2 4
> > 4 3 11 2 4 7
> > 5 11 2 4 3 7
> > 6 2 11 4 3 7
> > 7 11 4 3 7 6
> > 8 4 3 7 6 8
> > 9 3 4 7 6 8
> >
> > So, there are no duplicates in rows.
> > VECTOR is always scanned forward as long as the number of items
> (here 5) becomes full.
> >
> > Atte
> >
> > > Try:
> > >
> > > embed(VECTOR, 5)[,5:1]
> > >
> > > On 8/25/06, Atte Tenkanen <[EMAIL PROTECTED]> wrote:
> > > > Hi,
> > > >
> > > > Here is a vector and the result from the embed-command:
> > > >
> > > > VECTOR=c(0,3,6,3,11,2,4,3,7,6,4,5,10,2,3,5,8)
> > > >
> > > > > embed(VECTOR, dimension=5)
> > > > [,1] [,2] [,3] [,4] [,5]
> > > > [1,] 11 3 6 3 0
> > > > [2,] 2 11 3 6 3
> > > > [3,] 4 2 11 3 6
> > > > [4,] 3 4 2 11 3
> > > > [5,] 7 3 4 2 11
> > > > [6,] 6 7 3 4 2
> > > > [7,] 4 6 7 3 4
> > > > [8,] 5 4 6 7 3
> > > > [9,] 10 5 4 6 7
> > > > [10,] 2 10 5 4 6
> > > > [11,] 3 2 10 5 4
> > > > [12,] 5 3 2 10 5
> > > > [13,] 8 5 3 2 10
> > > >
> > > > Is there a way to little modify the algorithm so that the result
> > > looks> like this:
> > > >
> > > > [1] 0 3 6 11 2 <- beginning from the first number of the
> VECTOR> > > [1] 3 6 11 2 4 <- beginning from the second number
> of the
> > > VECTOR etc
> > > > [1] 6 3 11 2 4
> > > > [1] 3 11 2 4 7
> > > > [1] 11 2 4 3 7
> > > > [1] 2 4 3 7 6
> > > > [1] 4 3 7 6 5
> > > > [1] 3 7 6 4 5
> > > > [1] 7 6 4 5 10
> > > > [1] 6 4 5 10 2
> > > > [1] 4 5 10 2 3
> > > > [1] 5 10 2 3 8
> > > > [1] 10 2 3 5 8
> > > >
> > > > Every row consists of next five unique(!) member of the
> VECTOR. I
> > > made> this example result with a time consuming algorithm which
> > > uses for-loops
> > > > and whiles.
> > > >
> > > > How to do this better??
> > > >
> > > > Thanks in advance!
> > > >
> > > > Atte Tenkanen
> > > > University of Turku
> > > >
> > > > ______________________________________________
> > > > R-help@stat.math.ethz.ch 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.
> > > >
> > >
> >
>
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