Damien,
You don't give an example of what your data frame looks like or what you
want the new column to look like (given that example data), but I created
an example data frame for z, and wrote a few lines of code to add a new
column. Check it out and see if it comes close to doing what you want.
first - function(x) c(1, 1-(x[-1]==x[-length(x)]))
n - 25
z - data.frame(flagFoehn3_durr=sample(0:2, n, TRUE), Guetsch=sample(0:2,
n, TRUE))
z$newColumn - cumsum(first(z$flagFoehn3_durr==1) z$flagFoehn3_durr==1)
z$newColumn[z$flagFoehn3_durr!=1] - 0
Jean
On Tue, Jan 8, 2013 at 12:33 PM, Damien Pilloud damien.pill...@gmail.comwrote:
Dear R-helper,
I am working on a very large data frame and I am trying to add a new column
and write in it with certain conditions. I have try to use this code with
the data frame p :
ID = 0
p[,newColumn]-
ifelse (p$flagFoehn3_durr == 1,
ifelse(p$Guetsch == 0,
ID - ID ++
,
ID
)
,
0
)
What I am trying to do is to increment the ID when p$Guetsch == 0 and to
put this result in the column. The problem is that ID does not increment
itself.
An other way is to use a loop for like this example :
ID = 0
for (s in 1:(nrow(z))){
z[s,newColumn]-
if (z$flagFoehn3_durr[s] == 1){
if(z$flagFoehn3_durr[s-1] == 0){
ID -ID+1
}else{
ID
}
}else{
0
}
}
This work perfectly, but the problem is that it will take me more than a
month to run it.
Is there a way to increment with the first code I used or a way of running
the second code faster (I have more than 1 million rows)
Thanks!
Cheers,
Damien
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and provide commented, minimal, self-contained, reproducible code.
