data.table：组内存在值的计算条件

Question

With some data

library(data.table); set.seed(42)
dat <- data.table(id=1:5, group=c(1,1,1,2,2), time=c(1,2,3,1,2), val=runif(5))
> dat
   id group time       val
1:  1     1    1 0.9148060
2:  2     1    2 0.9370754
3:  3     1    3 0.2861395
4:  4     2    1 0.8304476
5:  5     2    2 0.6417455

and I would like to apply some calculation, say val*2 , to time point 2 only to those groups for which there is no third time point. The expected output is therefore

> res
   id group time       val
1:  1     1    1 0.9148060
2:  2     1    2 0.9370754
3:  3     1    3 0.2861395
4:  4     2    1 0.8304476
5:  5     2    2 1.2834910

where the value of time 2 in group 2 was changed. I suspected it is something along the lines of

dat[,val:=val[max(time)==2]*2, by=group]

but this would not work. Because I want to apply the calculation to a different time point than the one I am subsetting on, I felt this can't be done in i but I would not know how to do it instead.

Answer 1

Based on my previous answer (before edit) and on @Axeman's one, you could do the following

dat[, val2 := if(max(time) == 2) ifelse(time==2, 2*val, val) else val, group][]
##     id group time       val      val2
##  1:  1     1    1 0.9148060 0.9148060
##  2:  2     1    2 0.9370754 0.9370754
##  3:  3     1    3 0.2861395 0.2861395
##  4:  4     2    1 0.8304476 0.8304476
##  5:  5     2    2 0.6417455 1.2834910

and replace the 2*val by any function you would like.

Answer 2

Like this:

dat[, val := val*(1 + (time==2 & max(time)==2)), by=group]
##    id group time       val
## 1:  1     1    1 0.9148060
## 2:  2     1    2 0.9370754
## 3:  3     1    3 0.2861395
## 4:  4     2    1 0.8304476
## 5:  5     2    2 1.2834910

Answer 3

The data is sorted by time, so we can join on the last row per group and edit iff it meets the criterion:

dat[.(unique(group)), on=.(group), mult="last", 
  val := if (time == 2) val*2 else val
, by=.EACHI]

We can use if / else because mult="last" (and nomatch=NA ) guarantees that time has a length of 1. (This contrasts with the other two answers where the full time vector for each group is handled.)