Hi:

Does this work for you?

dat <- as.data.frame(dat)
lapply(split(dat, dat$V1), function(x) split(x, x$V2))

The result contains two components for 100 and 200, and subcomponents
within each component.

HTH,
Dennis

On Tue, Mar 30, 2010 at 12:20 AM, Megh <megh700004@yahoo.com> wrote:

>
> I need to split a given matrix in a sequential order. Let my matrix is :
>
> > dat <- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10,
> > T), sample(seq(20, 200, by=20), 10, T)); dat
>      [,1] [,2] [,3]
>  [1,]  200  100   80
>  [2,]  100  180   80
>  [3,]  200  150  180
>  [4,]  200   50  140
>  [5,]  100  150   60
>  [6,]  100   50   60
>  [7,]  100  100  100
>  [8,]  200  150  100
>  [9,]  100   50  120
> [10,]  200   50  180
>
> Now I need to split above matrix according to unique numbers in the 2nd
> column. Therefore I have following :
>
> > dat1 <- dat[which(dat[,1] == unique(dat[,1])[1]),]
> > dat2 <- dat[-which(dat[,1] == unique(dat[,1])[1]),]; dat1; dat2
>     [,1] [,2] [,3]
> [1,]  200  100   80
> [2,]  200  150  180
> [3,]  200   50  140
> [4,]  200  150  100
> [5,]  200   50  180
>     [,1] [,2] [,3]
> [1,]  100  180   80
> [2,]  100  150   60
> [3,]  100   50   60
> [4,]  100  100  100
> [5,]  100   50  120
>
> Now each of dat1 and dat2 needs to be splited according to the it's 2nd
> column i.e.
>
> > dat11 <- dat1[which(dat1[,2] == unique(dat1[,2])[1]),]
> > dat12 <- dat1[which(dat1[,2] == unique(dat1[,2])[2]),]
> > dat13 <- dat1[which(dat1[,2] == unique(dat1[,2])[3]),]; dat11; dat12;
> > dat13
> [1] 200 100  80
>     [,1] [,2] [,3]
> [1,]  200  150  180
> [2,]  200  150  100
>     [,1] [,2] [,3]
> [1,]  200   50  140
> [2,]  200   50  180
>
> similarly for dat2..............
>
> This kind of sequential spliting would continue for
> (no_of_cols_of_ogirinal_matrix -1) times. It would be greate if again I can
> put all those matrices within a "list" object for further calculations.
>
> Therefore you see if the original matrix is of small_size then that can be
> handled manually. However for a moderately large matrix that task would be
> very clumbersome. Therefore I am looking for some mechanized way to do that
> for an arbitrary matrix.
>
> Can anyone here help me on this regard?
>
> Thank you so much for your kind attention.
>
> --
> View this message in context:
> http://n4.nabble.com/Need-help-to-split-a-given-matrix-is-a-sequential-way-tp1744803p1744803.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> 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.
>

	[[alternative HTML version deleted]]

