Carry last Factor observation forward and backward in group of rows in R

Question

Suppose I have dataset which looks like

ID Name 
1  JAY
1  
1  JAY
2  LAY
2  LAY
2  
3  NA
3  KAY
3  

I want to fill the rows with missing values(empty or with NA) based on the observation already available in the group. So the resultant data frame will look like

ID Name 
1  JAY
1  JAY
1  JAY
2  LAY
2  LAY
2  LAY
3  KAY
3  KAY
3  KAY

I tried using na.locf but it didnt work for non numeric value.

DF1 = setDT(DF)[,  N := na.locf(na.locf(Name(NA_real_^!Name),na.rm=FALSE), fromLast=TRUE, na.rm=FALSE), ID][is.na(N), N := 0]

Answer 1

One option is after grouping by 'ID', subset the 'Name' that are not NA and not a blank (nzchar(Name)), get the last observation (tail(...)) and assign (:=) it to 'Name'.

setDT(DF)[, Name := tail(Name[!is.na(Name) & nzchar(Name)], 1), by = ID]
DF
#   ID Name
#1:  1  JAY
#2:  1  JAY
#3:  1  JAY
#4:  2  LAY
#5:  2  LAY
#6:  2  LAY
#7:  3  KAY
#8:  3  KAY
#9:  3  KAY

If the 'Name' column is factor change the nzchar(Name) to nzchar(as.character(Name))

---

Or specify the logical vector in the 'i' and assign (:=) the last observation (Name[.N]) to 'Name' after grouping by 'ID'

setDT(DF)[!is.na(Name) & nzchar(Name), Name := Name[.N], ID]

Note: For the second solution to work, the 'Name' should be character class.

Answer 2

Solution in base R (using split and do.call(bind, ...). Assume d contains your dataframe:

tmp <- lapply(split(d, d$ID), function(x) { 
    # Explanation:
    # decreasing = TRUE so that empty strings are at the end
    # na.last = NA so that NA's are omitted
    x$Name <- sort(x$Name, decreasing = TRUE, na.last = NA)[1];
    return(x);
})

d.new <- do.call(rbind, tmp);

print(d.new);
ID Name
1.1  1  JAY
1.2  1  JAY
1.3  1  JAY
2.4  2  LAY
2.5  2  LAY
2.6  2  LAY
3.7  3  KAY
3.8  3  KAY
3.9  3  KAY