[R] fast way to create composite matrix based on mixed indices?
Matthew Keller
mckellercran at gmail.com
Thu Sep 17 22:36:50 CEST 2015
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,
1,3,5,10,
2,2,1,3,
2,1,4,8,
2,3,9,10),
nrow=5,byrow=TRUE,
dimnames=list(NULL,c('new.mat.row','old.mat.row','old.mat.col.start','old.mat.col.end')))
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
slow.
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!
Matt
--
Matthew C Keller
Asst. Professor of Psychology
University of Colorado at Boulder
www.matthewckeller.com
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