KTY - 4 years ago 433

R Question

I am not very clear about use of

`.SD`

`by`

For instance, does the below snippet mean: 'change all the columns in

`DT`

`A`

`B`

`data.table`

`.SD`

`data.table`

`A`

`B`

`DT = DT[ ,lapply(.SD, as.factor), by=.(A,B)]`

However, I also read that

`by`

`colsum`

`A`

`B`

`A`

`B`

`A,B`

`DT[,lapply(.SD,sum),by=.(A,B)]`

Then how do I do a simple

`colsum`

`A`

`B`

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Answer Source

Just to illustrate the comments above with an example, let's take

```
set.seed(10238)
DT <- data.table(A = rep(1:3, each = 5), B = rep(1:5, 3),
C = sample(15), D = sample(15))
DT
# A B C D
# 1: 1 1 14 11
# 2: 1 2 3 8
# 3: 1 3 15 1
# 4: 1 4 1 14
# 5: 1 5 5 9
# 6: 2 1 7 13
# 7: 2 2 2 12
# 8: 2 3 8 6
# 9: 2 4 9 15
# 10: 2 5 4 3
# 11: 3 1 6 5
# 12: 3 2 12 10
# 13: 3 3 10 4
# 14: 3 4 13 7
# 15: 3 5 11 2
```

Compare the following:

```
#Sum all columns
DT[ , lapply(.SD, sum)]
# A B C D
# 1: 30 45 120 120
#Sum all columns EXCEPT A, grouping BY A
DT[ , lapply(.SD, sum), by = A]
# A B C D
# 1: 1 15 38 43
# 2: 2 15 30 49
# 3: 3 15 52 28
#Sum all columns EXCEPT A
DT[ , lapply(.SD, sum), .SDcols = !"A"]
# B C D
# 1: 45 120 120
#Sum all columns EXCEPT A, grouping BY B
DT[ , lapply(.SD, sum), by = B, .SDcols = !"A"]
# B C D
# 1: 1 27 29
# 2: 2 17 30
# 3: 3 33 11
# 4: 4 23 36
# 5: 5 20 14
```

A few notes:

- You said "does the below snippet... change all the columns in
`DT`

..."

The answer is *no*, and this is very important for `data.table`

. The object returned is a *new* `data.table`

, and all of the columns in `DT`

are exactly as they were before running the code.

- You mentioned wanting to change the column types

Referring to the point above again, note that your code (`DT[ , lapply(.SD, as.factor)]`

) returns a *new* `data.table`

and does not change `DT`

at all. One (**incorrect**) way to do this, which is done with `data.frame`

s in `base`

, is to overwrite the old `data.table`

with the new `data.table`

you've returned, i.e., `DT <- DT[ , lapply(.SD, as.factor)]`

.

This is wasteful because it involves creating copies of `DT`

which can be an efficiency killer when `DT`

is large. The correct `data.table`

approach to this problem is to update the columns by reference using``:=``

, e.g., `DT[ , names(DT) := lapply(.SD, as.factor)]`

, which creates no copies of your data. See `data.table`

's reference semantics vignette for more on this.

- You mentioned comparing efficiency of
`lapply(.SD, sum)`

to that of`colSums`

.`sum`

is internally optimized in`data.table`

; to see this in action, let's beef up your`DT`

a bit and run a benchmark:

Results:

```
library(data.table)
set.seed(12039)
nn <- 1e7; kk <- seq(100L)
DT <- as.data.table(replicate(26, sample(kk, nn, T))
)[ , LETTERS[1:2] := .(sample(100, nn, T),
sample(100, nn, T))]
library(microbenchmark)
microbenchmark(times = 100L,
colsums = colSums(DT[ , !c("A", "B"), with = FALSE]),
lapplys = DT[ , lapply(.SD, sum), .SDcols = !c("A", "B")])
# Unit: milliseconds
# expr min lq mean median uq max neval
# colsums 848.9310 886.6289 906.8105 896.7696 925.4353 997.0001 100
# lapplys 144.5028 145.7165 154.4077 147.5586 153.2286 253.6726 100
```

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