Split a vector by NA's - is there a better solution then a loop ?

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

Another option could be:

split(x, replace(cumsum(is.na(x)), is.na(x), -1))[-1]

On Thu, Apr 29, 2010 at 1:27 PM, Henrique Dallazuanna <wwwhsd at gmail.com> wrote:

Another option could be:

split(x, replace(cumsum(is.na(x)), is.na(x), -1))[-1]

One thing none of the solutions so far do (except I haven't tried
Tal's original code) is insert an empty group between adjacent NA
values, for example in:

x = c(1,2,3,NA,NA,4,5,6)

split(x, replace(cumsum(is.na(x)), is.na(x), -1))[-1]

$`0`
[1] 1 2 3

$`2`
[1] 4 5 6

Maybe this never happens in Tal's case, or it's not what he wanted
anyway, but I thought I'd point it out!

y <- rle(!is.na(x))
split(x, rep( cumsum(y$val)*y$val, y$len ) )[-1]

Barry