R help - Apr 2010 - 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




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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 ?crit :
>
> 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
-- 
Romain Francois
Professional R Enthusiast
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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.galili@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.Galili@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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-- 
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Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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