I'm not sure if that is the best way to do it. Probably there is some package with exactly that functionality out there. The following approach might not be the one with the very best performance, but it certainly works and should be fine for small to medium datasets. I would be cautious to apply it for very large datasets (more than a million rows or something like that)
fillNAByPreviousData <- function(column) {
# At first we find out which columns contain NAs
navals <- which(is.na(column))
# and which columns are filled with data.
filledvals <- which(! is.na(column))
# If there would be no NAs following each other, navals-1 would give the
# entries we need. In our case, however, we have to find the last column filled for
# each value of NA. We may do this using the following sapply trick:
fillup <- sapply(navals, function(x) max(filledvals[filledvals < x]))
# And finally replace the NAs with our data.
column[navals] <- column[fillup]
column
}
Here is some example using a test dataset:
set.seed(123)
test <- 1:20
test[floor(runif(5,1, 20))] <- NA
> test
[1] 1 2 3 4 5 NA 7 NA 9 10 11 12 13 14 NA 16 NA NA 19 20
> fillNAByPreviousData(test)
[1] 1 2 3 4 5 5 7 7 9 10 11 12 13 14 14 16 16 16 19 20
