使用 data.table 按组删除特定列中具有前导缺失值的行

Question

I have a data.table like this:

DT <- data.table(id = c(rep("a", 3), rep("b", 3)),
                 col1 = c(NA,1,2,NA,3,NA), col2 = c(NA,NA,5,NA,NA,NA))
   id col1 col2
1:  a   NA   NA
2:  a    1   NA
3:  a    2    5
4:  b   NA   NA
5:  b    3   NA
6:  b   NA   NA

For each id, I would like to remove rows with leading NA s in 'col1' using zoo::na.trim . Here's the result I'm expecting:

   id col1 col2
1:  a    1   NA
2:  a    2    5
3:  b    3   NA
4:  b   NA   NA

Here's what I have tried so far. This indeed removes leading NA in 'col1', but it omits 'col2' from the result:

DT[ , na.trim(col1), by = id]
   id V1
1:  a  1
2:  a  2
3:  b  3

This is also not working:

DT[ , .SD[na.trim(col1)], by = id]
   id col1 col2
1:  a   NA   NA
2:  a    1   NA
3:  b   NA   NA

Answer 1

A possible solution without using the zoo -package:

DT[DT[, .I[!!cumsum(!is.na(col1))], by = id]$V1]

you get:

   id col1 col2
1:  a    1   NA
2:  a    2    5
3:  b    3   NA
4:  b   NA   NA

What this does:

• With DT[, .I[!!cumsum(!is.na(col1))], id]$V1 you create a vector of rownumbers to keep. By using !!cumsum(!is.na(col1)) you make sure that only the leading missing values of col1 are omitted.
• Next you use that vector to subset the data.table.
• !!cumsum(!is.na(col1)) does the same as cumsum(!is.na(col1))!=0 . Using !! converts all number higher than zero to TRUE and all zeros to FALSE .
• .I isn't necessarily needed, you can also use: DT[DT[, !!cumsum(!is.na(col1)), by = id]$V1] which subsets the data.table with a logical vector.

---

Two alternatives with cummax by @lmo from the comments:

# alternative 1:
DT[DT[, !!(cummax(!is.na(col1))), by = id]$V1]

# alternative 2:
DT[as.logical(DT[, cummax(!is.na(col1)), by = id]$V1)]

Another alternative by @jogo:

DT[, .SD[!!cumsum(!is.na(col1))], by = id]

Another alternative by @Frank:

DT[, .SD[ rleid(col1) > 1L | !is.na(col1) ], by = id]

Answer 2

na.trim would be used like this with data.table. See ?na.trim for more info on its arguments.

DT[, na.trim(.SD, sides = "left", is.na = "all"), by = id]

giving:

   id col1 col2
1:  a    1   NA
2:  a    2    5
3:  b    3   NA
4:  b   NA   NA

ADDED:

In comment poster clarified that only column 1 NAs should be operated on by na.trim . In that case append a column of row numbers, .I, and after involing na.trim subset using those row numbers.

DT[DT[, na.trim(data.table(col1, .I), "left"), by = id]$.I, ]

Answer 3

We can use 1:.N >= which.max(...) to subset the required rows

> DT[, .SD[1:.N >= which.max(!is.na(col1))], id]
   id col1 col2
1:  a    1   NA
2:  a    2    5
3:  b    3   NA
4:  b   NA   NA