[R] fast way to create composite matrix based on mixed indices?
toth.denes at ttk.mta.hu
Thu Sep 17 23:22:56 CEST 2015
you could use matrix indexing. Here is a possible solution, which could
be optimized further (probably).
# The old matrix
(old.mat <- matrix(1:30,nrow=3,byrow=TRUE))
# matrix of indices
index <- matrix(c(1,1,1,4,
# expected result
new.mat <- matrix(c(1:4,25:30,11:13,4:8,29:30),
# column indices
ind <- mapply(seq, index[, 3], index[,4],
SIMPLIFY = FALSE, USE.NAMES = FALSE)
ind_len <- vapply(ind, length, integer(1))
ind <- unlist(ind)
# old indices
old.ind <- cbind(rep(index[,2], ind_len), ind)
# new indices
new.ind <- cbind(rep(index[,1], ind_len), ind)
# create the new matrix
result <- matrix(NA_integer_, max(index[,1]), max(index[,4]))
# fill the new matrix
result[new.ind] <- old.mat[old.ind]
# check the results
On 09/17/2015 10:36 PM, Matthew Keller wrote:
> HI all,
> Sorry for the title here but I find this difficult to describe succinctly.
> Here's the problem.
> I want to create a new matrix where each row is a composite of an old
> matrix, but where the row & column indexes of the old matrix change for
> different parts of the new matrix. For example, the second row of new
> matrix (which has , e.g., 10 columns) might be columns 1 to 3 of row 2 of
> old matrix, columns 4 to 8 of row 1 of old matrix, and columns 9 to 10 of
> row 3 of old matrix.
> Here's an example in code:
> #The old matrix
> (old.mat <- matrix(1:30,nrow=3,byrow=TRUE))
> #matrix of indices to create the new matrix from the old one.
> #The 1st column gives the row number of the new matrix
> #the 2nd gives the row of the old matrix that we're going to copy into the
> new matrix
> #the 3rd gives the starting column of the old matrix for the row in col 2
> #the 4th gives the end column of the old matrix for the row in col 2
> index <- matrix(c(1,1,1,4,
> I will be given old.mat and index and want to create new.mat from them.
> I want to create a new.matrix of two rows that looks like this:
> new.mat <- matrix(c(1:4,25:30,11:13,4:8,29:30),byrow=TRUE,nrow=2)
> So here, the first row of new.mat is columns 1 to 4 of row 1 of the old.mat
> and columns 5 to 10 of row 3 of old.mat.
> new.mat and old.mat will always have the same number of columns but the
> number of rows could differ.
> I could accomplish this in a loop, but the real problem is quite large
> (new.mat might have 1e8 elements), and so a for loop would be prohibitively
> I may resort to unix tools and use a shell script, but wanted to first see
> if this is doable in R in a fast way.
> Thanks in advance!
More information about the R-help