Tom McCallum wrote:
> Hi Everyone,
>
> I have a question about for loops. If you have something like:
>
> f <- function(x) {
> y <- rep(NA,10);
> for( i in 1:10 ) {
> if ( i > 3 ) {
> if ( is.na(y[i-3]) == FALSE ) {
> # some calculation F which depends on one or
> more of the previously
> generated values in the series
> y[i] = y[i-1]+x[i];
> } else {
> y[i] <- x[i];
> }
> }
> }
> y
> }
>
> e.g.
>
>> f(c(1,2,3,4,5,6,7,8,9,10,11,12));
> [1] NA NA NA 4 5 6 13 21 30 40
>
> is there a faster way to process this than with a 'for' loop? I have
> looked at lapply as well but I have read that lapply is no faster than a
> for loop and for my particular application it is easier to use a for loop.
> Also I have seen 'rle' which I think may help me but am not sure as I have
> only just come across it, any ideas?
Hi Tom,
In the general case, you need a loop in order to propagate calculations
and their results across a vector.
In _your_ particular case however, it seems that all you are doing is a
cumulative sum on x (at least this is what's happening for i >= 6).
So you could do:
f2 <- function(x)
{
offset <- 3
start_propagate_at <- 6
y_length <- 10
init_range <- (offset+1):start_propagate_at
y <- rep(NA, offset)
y[init_range] <- x[init_range]
y[start_propagate_at:y_length] <- cumsum(x[start_propagate_at:y_length])
y
}
and it will return the same thing as your function 'f' (at least when 'x'
doesn't
contain NAs) but it's not faster :-/
IMO, using sapply for propagating calculations across a vector is not
appropriate
because:
(1) It requires special care. For example, this:
> x <- 1:10
> sapply(2:length(x), function(i) {x[i] <- x[i-1]+x[i]})
doesn't work because the 'x' symbol on the left side of the <- in the
anonymous function doesn't refer to the 'x' symbol defined in the global
environment. So you need to use tricks like this:
> sapply(2:length(x),
function(i) {x[i] <- x[i-1]+x[i]; assign("x", x,
envir=.GlobalEnv); x[i]})
(2) Because of this kind of tricks, then it is _very_ slow (about 10 times
slower or more than a 'for' loop).
Cheers,
H.
>
> Many thanks
>
> Tom
>
>
>
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