na.locf fill NAs up to maxgap even if gap > maxgap, with groups

Question

I've seen a solution to this, but can't get it to work for groups (Fill NA in a time series only to a limited number), and thought there must be a neater way to do this also?

Say I have the following dt:

dt <- data.table(ID = c(rep("A", 10), rep("B", 10)), Price = c(seq(1, 10, 1), seq(11, 20, 1)))
dt[c(1:2, 5:10), 2] <- NA 
dt[c(11:13, 15:19) ,2] <- NA 
dt
    ID Price
 1:  A    NA
 2:  A    NA
 3:  A     3
 4:  A     4
 5:  A    NA
 6:  A    NA
 7:  A    NA
 8:  A    NA
 9:  A    NA
10:  A    NA
11:  B    NA
12:  B    NA
13:  B    NA
14:  B    14
15:  B    NA
16:  B    NA
17:  B    NA
18:  B    NA
19:  B    NA
20:  B    20

What I would like to do, is to fill NAs both forward and back from the most recent non-NA value, but only up to a maximum of two rows forward or back.

I also need it to be done by group (ID).

I have tried using na.locf/na.approx with maxgap = x etc, but it does not fill NAs where the gap between non-NA values is greater than maxgap. Whereas I want to fill these forward and back even if the gap between non-NA values is greater than maxgap, but only by two rows.

The final result should looks something like:

    ID Price Price_Fill
 1:  A    NA          3
 2:  A    NA          3
 3:  A     3          3
 4:  A     4          4
 5:  A    NA          4
 6:  A    NA          4
 7:  A    NA         NA
 8:  A    NA         NA
 9:  A    NA         NA
10:  A    NA         NA
11:  B    NA         NA
12:  B    NA         14
13:  B    NA         14
14:  B    14         14
15:  B    NA         14
16:  B    NA         14
17:  B    NA         NA
18:  B    NA         20
19:  B    NA         20
20:  B    20         20

In reality, my data set is massive, and I want to be able to fill NAs forward and back for up to 672 rows, but no more, by group.

Thanks!

Answer 1

For the example showed, we group by 'ID', get the shift of 'Price' with n = 0:2, and type as 'lead' to create 3 temporary columns, get the pmax from this, use the output to do the shift with type = 'lag' (by default it is 'lag') and same n, get the pmin and assign it as 'Price_Fill'

dt[, Price_Fill := do.call(pmin, c(shift(do.call(pmax, c(shift(Price, n = 0:2, 
                  type = "lead"), na.rm=TRUE)), n= 0:2), na.rm = TRUE)) , by = ID]
dt
#    ID Price Price_Fill
#1:  A    NA          3
#2:  A    NA          3
#3:  A     3          3
#4:  A     4          4
#5:  A    NA          4
#6:  A    NA          4
#7:  A    NA         NA
#8:  A    NA         NA
#9:  A    NA         NA
#10: A    NA         NA
#11: B    NA         NA
#12: B    NA         14
#13: B    NA         14
#14: B    14         14
#15: B    NA         14
#16: B    NA         14
#17: B    NA         NA
#18: B    NA         20
#19: B    NA         20
#20: B    20         20

---

A more general approach would be to do the pmin/pmax on .I as the 'Price' can be different and not the sequence number as showed in the OP's post.

i1 <- dt[,  do.call(pmin, c(shift(do.call(pmax, c(shift(NA^(is.na(Price))* 
    .I, n = 0:2, type = "lead"), na.rm = TRUE)), n = 0:2), na.rm = TRUE)), ID]$V1

dt$Price_Fill <  dt$Price[i1]
dt$Price_Fill
#[1]  3  3  3  4  4  4 NA NA NA NA NA 14 14 14 14 14 NA 20 20 20

i.e. suppose we change the 'Price', it will be different

dt$Price[3] <- 10
dt$Price[14] <- 7
dt$Price_Fill <- dt$Price[i1]
dt$Price_Fill
#[1] 10 10 10  4  4  4 NA NA NA NA NA  7  7  7  7  7 NA 20 20 20