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 <[email protected]> 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. > > ______________________________________________ > [email protected] 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]] ______________________________________________ [email protected] 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.
