how to use 'which' inside of 'apply'?
I think something like this should do it at a huge speed up, though I'd advise you check it to make sure it does exactly what you want: there's also nothing to guarantee that something beats the threshold, so that might make the whole thing fall apart (though I don't think it will) # Sample data df = data.frame(x = sample(5, 15,T), y = sample(5, 15, T), z = sample(5, 15,T), w = (1:5)/2 + 0.5, th = (1:5)/2, doy = rep(0,15)) wd <- which(df[,1:4] > df[,5], arr.ind = TRUE) # identify all elements that beat the threshold value by their indices wd <- wd[!duplicated(wd[,1]),] # select only the first appearance of each "row" value in wd -- this keeps the earliest column beating the threshold wd <- wd[order(wd[,"row"]),] # sort them by row df$doy = (wd[,"col"]-1)*16 + 1 # The column transform you used. Hope this helps, Michael
On Mon, Oct 17, 2011 at 1:03 PM, Nathan Piekielek <npiekielek at gmail.com> wrote:
Hello R-community,
I am trying to populate a column (doy) in a large dataset with the first
column number that exceeds the value in another column (thold) using the
'apply' function.
Sample data:
? ? pt D1 D17 D33 D49 D65 D81 D97 D113 ? D129 ? D145 ? ? ?D161 ? ? ?D177
D193 ? D209 ? D225 ? D241 ? D257
1 39177 ?0 ? 0 ? 0 ? 0 ? 0 ? 0 ? 0 ? ?0 0.4336 0.4754 0.5340667 0.5927334
0.6514 0.6966 0.5900 0.5583 0.5676
2 39178 ?0 ? 0 ? 0 ? 0 ? 0 ? 0 ? 0 ? ?0 0.3420 0.4543 0.5397666 0.6252333
0.7107 0.7123 0.5591 0.4617 0.4206
3 39164 ?0 ? 0 ? 0 ? 0 ? 0 ? 0 ? 0 ? ?0 0.4830 0.4943 0.5740333 0.6537667
0.7335 0.6255 0.6228 0.5255 0.5436
4 39143 ?0 ? 0 ? 0 ? 0 ? 0 ? 0 ? 0 ? ?0 0.3088 0.3753 0.4466000 0.5179000
0.5892 0.6468 0.4794 0.4411 0.4307
5 39144 ?0 ? 0 ? 0 ? 0 ? 0 ? 0 ? 0 ? ?0 0.3390 0.4152 0.5147000 0.6142000
0.7137 0.6914 0.6381 0.5704 0.5619
6 39146 ?0 ? 0 ? 0 ? 0 ? 0 ? 0 ? 0 ? ?0 0.4232 0.4442 0.5084000 0.5726000
0.6368 0.5896 0.4703 0.4936 0.5353
? ?D273 ? ?D289 ? D305 ? ?D321 D337 D353 ? ?thold doy
1 0.4682 0.35115 0.2341 0.11705 ? ?0 ? ?0 0.406825 ? 0
2 0.3867 0.25780 0.1289 0.00000 ? ?0 ? ?0 0.420600 ? 0
3 0.5541 0.46195 0.3698 0.18490 ? ?0 ? ?0 0.459200 ? 0
4 0.3632 0.34355 0.3239 0.00000 ? ?0 ? ?0 0.477800 ? 0
5 0.5347 0.49760 0.4605 0.00000 ? ?0 ? ?0 0.526350 ? 0
6 0.4067 0.39685 0.3870 0.00000 ? ?0 ? ?0 0.511900 ? 0
For the first record in above example I would expect doy = 129.
I can achieve this with the following loop, but it takes several days to run
and there must be a more efficient solution:
for (i in (1:152000)) {
t=which(data[i,2:24]>data[i,25])
r=min(t)
data[i,26]=(r-1)*16+1
}
How do I write this using 'apply' or another function that will be more
efficient?
I have tried the following:
data$doy=apply(which(data[,2:24]>data[,25]),1,min)
Which returns the following error message:
"Error in apply(which(new[, 2:24] > new[, 25]), 1, min) :
?dim(X) must have a positive length"
Any help would be much appreciated.
Nathan
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