On Fri, Feb 24, 2012 at 08:59:44AM -0800, robertfeldt wrote:
> Hi,
>
> I have R code like so:
>
> num.columns.back.since.last.occurence <- function(m, outcome) {
> nrows <- dim(m)[1];
> ncols <- dim(m)[2];
> res <- matrix(rep.int(0, nrows*ncols), nrow=nrows);
> for(row in 1:nrows) {
> for(col in 2:ncols) {
> res[row,col] <- if(m[row,col-1]==outcome) {0} else
> {1+res[row,col-1]}
> }
> }
> res;
> }
>
> but on the very large matrices I apply this the execution times are a
> problem. I would appreciate any help to rewrite this with more
> "standard"/native R functions to speed things up.
Hi.
If the number of columns is large, so the rows are long, then
the following can be more efficient.
oneRow <- function(x, outcome)
{
n <- length(x)
y <- c(0, cumsum(x[-n] == outcome))
ave(x, y, FUN = function(z) seq.int(along=z) - 1)
}
# random matrix
A <- matrix((runif(49) < 0.2) + 0, nrow=7)
# the required transformation
B <- t(apply(A, 1, oneRow, outcome=1))
# verify
all(num.columns.back.since.last.occurence(A, 1) == B)
[1] TRUE
This solution performs a loop over rows (in apply), so if the
number of rows is large and the number of columns is not,
then a solution, which uses a loop over columns, may be
better.
Hope this helps.
Petr Savicky.
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