# [R] Split a vector by NA's - is there a better solution then a loop ?

Romain Francois romain.francois at dbmail.com
Thu Apr 29 09:56:25 CEST 2010

```Maybe this :

> foo <- function( x ){
+   idx <- 1 + cumsum( is.na( x ) )
+   not.na <- ! is.na( x )
+   split( x[not.na], idx[not.na] )
+ }
> foo( x )
\$`1`
[1] 2 1 2

\$`2`
[1] 1 1 2

\$`3`
[1] 4 5 2 3

Romain

Le 29/04/10 09:42, Tal Galili a écrit :
>
> 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

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Romain Francois
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```

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