R: Confusion with apply() vs for loop

Question

I know that I should avoid for-loops, but I\'m not exactly sure how to do what I want to do with an apply function.

Here is a slightly simplified model of what I\'

Answer 1

This question touches several points that are made in 'The R Inferno' http://www.burns-stat.com/pages/Tutor/R_inferno.pdf

There are some loops you should avoid, but not all of them. And using an apply function is more hiding the loop than avoiding it. This example seems like a good choice to leave in a 'for' loop.

Growing objects is generally bad form -- it can be extremely inefficient in some cases. If you are going to have a blanket rule, then "not growing objects" is a better one than "avoid loops".

You can create a list with the final length by:

result <- vector("list", ncol(g))
for(i in 1:ncol(g)) {
    # stuff
    result[[i]] <- #results
}

In some circumstances you might think the command:

window<-5

means give me a logical vector stating which values of 'window' are less than -5.

Spaces are good to use, mostly not to confuse humans, but to get the meaning directly above not to confuse R.

Answer 2

Using an apply function to do your regression is mostly a matter of preference in this case; it can handle some of the bookkeeping for you (and so possibly prevent errors) but won't speed up the code.

I would suggest using vectorized functions though to compute your first's and last's, though, perhaps something like:

window <- 5
ng <- 15 #or ncol(g)
xy <- data.frame(first = pmax( (1:ng) - window, 1 ), 
                  last = pmin( (1:ng) + window, ng) )

Or be even smarter with

xy <- data.frame(first= c(rep(1, window), 1:(ng-window) ), 
                 last = c((window+1):ng, rep(ng, window)) )

Then you could use this in a for loop like this:

results <- list()
for(i in 1:nrow(xy)) {
  results[[i]] <- xy$first[i] : xy$last[i]
}
results

or with lapply like this:

results <- lapply(1:nrow(xy), function(i) {
  xy$first[i] : xy$last[i]
})

where in both cases I just return the sequence between first and list; you would substitute with your actual regression code.