Select columns by class (e.g. numeric) from a data.table

Question

Why doesn't this work with data.table?

It works with data.frame. Is there a way to do this with a data table?

x <- data.table(v1=1:20,v2=1:20,v3=1:20,v4=letters[1:20])
y <- x[ , sapply(x, is.numeric)]

This returns:

v1    v2    v3    v4
TRUE  TRUE  TRUE FALSE

The data.table FAQ vignette covers this with FAQs 1.1 and 1.2 — mnel, Aug 05 '14 at 03:34
Very nearly a duplicate of [select multiple columns in data table](http://stackoverflow.com/questions/13383840/select-multiple-columns-in-data-table-r) — thelatemail, Aug 05 '14 at 04:17
Possible duplicate of [Selecting only numeric columns from a data frame](http://stackoverflow.com/questions/5863097/selecting-only-numeric-columns-from-a-data-frame) — Aramis7d, Apr 04 '16 at 07:07

score 39 · Answer 1 · edited Jul 30 '21 at 18:13

39

From data.table 1.13.0 ".SDcols accepts a function which is used to select the columns of .SD". Thus, simply .SDcols = is.numeric:

x[ , .SD, .SDcols = is.numeric]

edited Jul 30 '21 at 18:13

Henrik

answered Sep 05 '15 at 11:31

Artem Klevtsov

1

Hello Hello, I tried. dt_c[,.SD, .SDcols = sapply(dt_c, as.character)] It's not working. a solution – Mathieu L May 14 '20 at 12:06
1

`s/as.character/is.character/`. – Artem Klevtsov May 14 '20 at 17:05

score 15 · Accepted Answer · answered Aug 05 '14 at 03:23

15

data.table needs the with=FALSE to grab column numbers.

tokeep <- which(sapply(x,is.numeric))
x[ , tokeep, with=FALSE]

answered Aug 05 '14 at 03:23

Mike.Gahan

8

Is the which even needed? - `x[,sapply(x,is.numeric),with=FALSE]` – thelatemail Aug 05 '14 at 04:02

akrun · Answer 3 · 2014-08-05T08:19:08.850

3

You may also try:

 x1 <- x[,Filter(is.numeric, .SD)]
 head(x1,3)
 #   v1 v2 v3
#1:  1  1  1
#2:  2  2  2
#3:  3  3  3

Although, I have to admit that it is slow for bigger datasets.

edited Aug 05 '14 at 08:19

answered Aug 05 '14 at 07:51

akrun

score 0 · Answer 4 · answered Jun 17 '20 at 17:54

0

Similar to @akrun's answer

Filter(is.numeric, x)

answered Jun 17 '20 at 17:54

Jeff Bezos

Eyayaw · Answer 5 · 2020-06-17T18:15:53.570

0

We can write a custom helper calledwhere(), and then we can subset a data.frame/data.table where f is satisfied:

where <- function(x, f) {
  colnames(x)[vapply(x, f, logical(1))]
}

df[, where(df, is.numeric), with = FALSE]

edited Jun 17 '20 at 18:15

answered Jun 17 '20 at 18:09

Eyayaw

5 Answers5