Hi Igor.
Here is one way:
DF <- read.table(textConnection("90.1194354
87.94788274
80.34744843
64.06080347
30.40173724
0
0
0
0
0
16.28664495
23.88707926
29.31596091
48.85993485
13.02931596
0
0
0
7.600434311
20.62975027
29.31596091
32.5732899"), header=FALSE)
a <- DF$V1[which(DF$V1!=0)]
indx <- which(DF$V1!=0)
blocks <- cut(1:length(a), breaks=c(0,which(diff(indx)!=1), length(a)))
n_levels <- length(levels(blocks))
l <- vector("list", n_levels)
for(i in 1:n_levels)
{
l[[i]] <- a[blocks==levels(blocks)[i]]
}
l
Andrija
On Thu, Apr 18, 2013 at 11:33 AM, Igor Mintz <igormi...@gmail.com> wrote:
> hello
> i have a very long column of numbers. i want R to make a new column every
> time the value changes from zero.
> example for the column:
> 90.1194354
> 87.94788274
> 80.34744843
> 64.06080347
> 30.40173724
> 0
> 0
> 0
> 0
> 0
> 16.28664495
> 23.88707926
> 29.31596091
> 48.85993485
> 13.02931596
> 0
> 0
> 0
> 7.600434311
> 20.62975027
> 29.31596091
> 32.5732899
>
> for this example i want to get 3 columns.
>
> thanks!
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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and provide commented, minimal, self-contained, reproducible code.
