Another option could be:
split(x, replace(cumsum(is.na(x)), is.na(x), -1))[-1]
On Thu, Apr 29, 2010 at 4:42 AM, Tal Galili <tal.gal...@gmail.com> wrote:
> Hi all,
>
> I would like to have a function like this:
> split.vec.by.NA <- function(x)
>
> That takes a vector like this:
> x <- c(2,1,2,NA,1,1,2,NA,4,5,2,3)
>
> And returns a list of length of 3, each element of the list is the relevant
> segmented vector, like this:
>
> $`1`
> [1] 2 1 2
> $`2`
> [1] 1 1 2
> $`3`
> [1] 4 5 2 3
>
>
> I found how to do it with a loop, but wondered if there is some smarter
> (vectorized) way of doing it.
>
>
>
> Here is the code I used:
>
> x <- c(2,1,2,NA,1,1,2,NA,4,5,2,3)
>
>
> split.vec.by.NA <- function(x)
> {
> # assumes NA are seperating groups of numbers
> #TODO: add code to check for it
>
> number.of.groups <- sum(is.na(x)) + 1
> groups.end.point.locations <- c(which(is.na(x)), length(x)+1) # This will
> be
> all the places with NA's + a nubmer after the ending of the vector
> group.start <- 1
> group.end <- NA
> new.groups.split.id <- x # we will replace all the places of the group
> with
> group ID, excapt for the NA, which will later be replaced by 0
> for(i in seq_len(number.of.groups))
> {
> group.end <- groups.end.point.locations[i]-1
> new.groups.split.id[group.start:group.end] <- i
> group.start <- groups.end.point.locations[i]+1 # make the new group start
> higher for the next loop (at the final loop it won't matter
> }
> new.groups.split.id[is.na(x)] <- 0
> return(split(x, new.groups.split.id)[-1])
> }
>
> split.vec.by.NA(x)
>
>
>
>
> Thanks,
> Tal
>
>
>
>
> ----------------Contact
> Details:-------------------------------------------------------
> Contact me: tal.gal...@gmail.com | 972-52-7275845
> Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
> www.r-statistics.com (English)
>
> ----------------------------------------------------------------------------------------------
>
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>
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>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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