[R] Modifying the embed-results

[Gabor Grothendieck]

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 <attenka at utu.fi> 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 <attenka at utu.fi> 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 at 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.
> > >
> >
>