In data.table, what is the difference between taking dot(.) and 'c' in j while using .SD? How does it work?

user_8376346 0
Viewed 929

Take DT as mtcars data table.

DT <- as.data.table(mtcars)

While taking multiple arguments in 'j' with .SD, if we use dot(.) before j like below code

DT[ , .(lapply(.SD, sum), .N), by = (cyl) ]

the result comes in vertical order with without the column names.

O/P:

cyl V1 N

1: 6 138.2 7

2: 6 1283.2 7

3: 6 856 7

4: 6 25.1 7

5: 6 21.82 7

6: 6 125.84 7

7: 6 4 7

8: 6 3 7

But when I replace that with the dot(.) in 'j' with 'c' like below,

DT[ , c(lapply(.SD, sum), .N), by = (cyl) ]

the result comes in horizontal order.

O/P:

cyl mpg disp hp drat wt qsec vs am gear carb N

6 138.2 1283.2 856 25.10 21.820 125.84 4 3 27 24 7

4 293.3 1156.5 909 44.78 25.143 210.51 10 8 45 17 11

8 211.4 4943.4 2929 45.21 55.989 234.81 0 2 46 49 14

In another case, without lapply, exactly the opposite happens.

DT[ , c(sum(mpg), .N), by = (cyl) ]

gives the output vertically

O/P:

cyl V1

1: 6 138.2

2: 6 7.0

3: 4 293.3

4: 4 11.0

5: 8 211.4

6: 8 14.0

whereas a dot(.) in 'j' gives the output horizonatlly.

DT[ , .(sum(mpg), .N), by = (cyl) ]

O/P:

cyl V1 N

1: 6 138.2 7

2: 4 293.3 11

3: 8 211.4 14

Why does this happen? Why the result is ordered in such way?

data.table r
1 Answers
DT[,    .(sum(mpg), .N), by = (cyl) ] # equal, creates a list with 2 elements (2 columns)
DT[, list(sum(mpg), .N), by = (cyl) ] # equal, to above

DT[,    c(sum(mpg), .N), by = (cyl) ] # creates a vector of length 2 (equal to 2 rows)

another simplified example.

DT[ ,    .(col1 = 1, col2 = 2), by = (cyl) ]
DT[ , list(col1 = 1, col2 = 2), by = (cyl) ]

DT[ , c(element1 = 1, element2 = 2), by = (cyl) ]

To address your last point,

DT[ , c(element1 = 1, element2 = 2, element3 = list(3)), by = (cyl) ]
DT[ , c(element1 = 1, element2 = 2, element3 = 3      ), by = (cyl) ]

You need to learn more about the c function.

So as lapply (listapply) returns a list the c will add the .N as new LIST ELEMENT in c(lapply(.SD, sum), .N).

So you end up with n list elements and therefore n columns.

Just for fun:

DT[ , c(lapply(.SD, sum), .N), by = (cyl) ]
DT[ , c(sapply(.SD, sum), .N), by = (cyl) ] # sapply will simplify the result into a vector, therefore c() will combine into a vector and you end up with many rows.
user_68529370
Related