1. matrices are stored columnwise so R is better at column-wise operations
than row-wise.
2. Here is one way to do it (although I am not sure its better than the
index approach):
row.apply <- function(f, a, b)
t(mapply(f, as.data.frame(t(a)), as.data.frame(t(b))))
3. The code for the example in this post could be simplified to:
first.1 <- apply(cbind(goldstandard, 1), 1, which.max)
ifelse(col(newtest) > first.1, NA, newtest)
4. given that both examples did not inherently need row by row operations
I wonder if that is the wrong generalization in the first place?
On 8/10/07, Johannes Hüsing <[EMAIL PROTECTED]> wrote:
> [Apologies to Gabor, who I sent a personal copy of the reply
> erroneously instead of posting to List directly]
>
> [...]
> > Perhaps what you really intend is to
> > take the average over those elements in each row of the first matrix
> which correspond to 1's in the second in the corresponding
> > row of the second. In that case its just:
> >
> > rowSums(newtest * goldstandard) / rowSums(goldstandard)
> >
>
> Thank you for clearing my thoughts about the particular example.
> My question was a bit more general though, as I have different
> functions which are applied row-wise to multiple matrices. An
> example that sets all values of a row of matrix A to NA after the
> first occurrence of TRUE in matrix B.
>
> fillfrom <- function(applvec, testvec=NULL) {
> if (is.null(testvec)) testvec <- applvec
> if (length(testvec) != length(applvec)) {
> stop("applvec and testvec have to be of same length!")
> } else if(any(testvec, na.rm=TRUE)) {
> applvec[min(which(testvec)) : length(applvec)] <- NA
> }
> applvec
> }
>
> fillafter <- function(applvec, testvec=NULL) {
> if (is.null(testvec)) testvec <- applvec
> fillfrom(applvec, c(FALSE, testvec[-length(testvec)]))
> }
>
> numtest <- 6
> numsubj <- 20
>
> newtest <- array(rbinom(numtest*numsubj, 1, .5),
> dim=c(numsubj, numtest))
> goldstandard <- array(rbinom(numtest*numsubj, 1, .5),
> dim=c(numsubj, numtest))
>
> newtest.NA <- t(sapply(1:nrow(newtest), function(i) {
> fillafter(newtest[i,], goldstandard[i,]==1)}))
>
> My general question is if R provides some syntactic sugar
> for the awkward sapply(1:nrow(A)) expression. Maybe in this
> case there is also a way to bypass the apply mechanism and
> my way of thinking about the problem has to be adapted. But
> as the *apply calls are galore in R, I feel this is a standard
> way of dealing with vectors and matrices.
>
>
>
>
>
> --
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>
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