R help - Jun 2011 - count length of continues elements in a vector

I am performing a precipitation analysis. data is in the form of daily
precipitation amounts, e.g.

x<- c(4,5,3,0,0,0,2,4,6,4,0,0,0,2,2,0,3,4,1,0,...)

I would like to find the length of the "storm", length of storm would
be
defined as the number of days with continues precipitation. in this case the
returned vector would be:

 (3,4,2,3,...)

I would also like the amount of precipitation associated with each
"storm"
in another variable. in this case the variable would look like:

(12,16,4,8,...)

any suggestions would be appreciated.

thanks
Dave

--
View this message in context:
http://r.789695.n4.nabble.com/count-length-of-continues-elements-in-a-vector-tp3579754p3579754.html
Sent from the R help mailing list archive at Nabble.com.

On Jun 7, 2011, at 9:25 AM, davetracz wrote:
> I am performing a precipitation analysis. data is in the form of daily
> precipitation amounts, e.g.
> 
> x<- c(4,5,3,0,0,0,2,4,6,4,0,0,0,2,2,0,3,4,1,0,...)
> 
> I would like to find the length of the "storm", length of storm
would be
> defined as the number of days with continues precipitation. in this case
the
> returned vector would be:
> 
> (3,4,2,3,...)
> 
> I would also like the amount of precipitation associated with each
"storm"
> in another variable. in this case the variable would look like:
> 
> (12,16,4,8,...)
> 
> any suggestions would be appreciated.
> 
> thanks
> Dave

?rle will be useful here

x <- c(4, 5, 3, 0, 0, 0, 2, 4, 6, 4, 0, 0, 0, 2, 2, 0, 3, 4, 1, 0)

# Get the runs of values in 'x' >= 1
R1 <- rle(x >= 1)
> R1
Run Length Encoding
  lengths: int [1:8] 3 3 4 3 2 1 3 1
  values : logi [1:8] TRUE FALSE TRUE FALSE TRUE FALSE ...


# Split 'x' by the sequences of values >= 1 in 'R1'
# The values will be grouped by integers, including '0s'
# > rep(cumsum(R1$values) * R1$values, R1$lengths)
#  [1] 1 1 1 0 0 0 2 2 2 2 0 0 0 3 3 0 4 4 4 0
# so we want to remove the first '0' group
R2 <- split(x, rep(cumsum(R1$values) * R1$values, R1$lengths))[-1]
 
> R2
$`1`
[1] 4 5 3

$`2`
[1] 2 4 6 4

$`3`
[1] 2 2

$`4`
[1] 3 4 1


Now use normal list processing to get the lengths and sums

> sapply(R2, length)
1 2 3 4 
3 4 2 3 
> sapply(R2, sum)
 1  2  3  4 
12 16  4  8 


HTH,

Marc Schwartz