如何在data.table中添加延迟并导致每个观察结果中的更多变量排除NA？

Question

I have a data.table similar to this:

library(data.table)
mydt <- data.table(id = LETTERS[1:6], x = 1:6, y = 2:3) 
> mydt
   id x y
1:  A 1 2
2:  B 2 3
3:  C 3 2
4:  D 4 3
5:  E 5 2
6:  F 6 3

I would like to replace the value columns with adding the lag and lead to each observation (ie x[-1] + x + x[1] ). I can do something like this with the amazing shift() feature.

cols <- c('x', 'y')
mydt[
    ,
    (cols) := shift(.SD, 1) + .SD + shift(.SD, 1, type = 'lead'),
    .SDcols = cols
][]
   id  x  y
1:  A NA NA
2:  B  6  7
3:  C  9  8
4:  D 12  7
5:  E 15  8
6:  F NA NA

But this introduces NAs for rows where there is no lead/lag value. How can I modify the calculation to use the available two values only for these rows (like na.rm = TRUE )? So that the output would be

   id  x  y
1:  A  3  5
2:  B  6  7
3:  C  9  8
4:  D 12  7
5:  E 15  8
6:  F 11  5

I tried using sum(..., na.rm = TRUE) instead of the + operator but that gives error: Error in sum(shift(.SD, 1), .SD, shift(.SD, 1, type = "lead"), na.rm = TRUE) : invalid 'type' (list) of argument .

I also tried the following but that apparently gives something else as a result.

mydt[
    ,
    (cols) := lapply(
        .SD, 
        function(x) sum(shift(x, 1), x, shift(x, 1, type = 'lead'), na.rm = TRUE)
    ),
    .SDcols = cols
][]
   id   x  y
1:  A 126 90
2:  B 126 90
3:  C 126 90
4:  D 126 90
5:  E 126 90
6:  F 126 90

Answer 1

As @akrun and @DavidArenburg pointed out, the shift function has a fill parameter which solves the issue.

cols <- c('total_open', 'total_send')
mydt[
    ,
    (cols) := shift(.SD, 1, fill = 0) + .SD + shift(.SD, 1, type = 'lead', fill = 0),
    .SDcols = cols
][]
   id  x y
1:  A  3 5
2:  B  6 7
3:  C  9 8
4:  D 12 7
5:  E 15 8
6:  F 11 5