data.table assignment by reference using lapply and also returning the rest of the columns

Question

I want to be able to apply a function over a subset of columns, and return those columns that have been manipulated along with the rest of the data columns that weren't touched. Is there a way to do this with data.table. I wasn't able to figure out the syntax.

In this example I have NAs and want to overwrite them with something else for a few different columns. I need a way to also return other columns that weren't touched.

library(data.table)

# make data set
a <- sample(c(letters[1:5], NA), 50, replace=TRUE)
b <- sample(c(LETTERS[1:5], NA), 50, replace=TRUE)
c <- sample(runif(50))

x <- data.table(a,b,c)

# function to apply to a single column
overwriteNA <- function(vec, new="") ifelse(is.na(vec), new, vec)

# Only returns .SDcols but would like to also include rest of columns in data.table
x[, lapply(.SD, overwriteNA), .SDcols=c("a", "b")]

# Need something along these lines
x[, `:=` lapply(.SD, overwriteNA), .SDcols=c("a", "b")]

David Arenburg · Accepted Answer · 2016-07-21T05:09:22.790

9

Try

x[,  c("a", "b") := lapply(.SD, overwriteNA), .SDcols = c("a", "b")]

Edit:

Per OPs additional request.

myCols <- c("a", "b")  
x[, (myCols) := lapply(.SD, overwriteNA), .SDcols = myCols]

edited Jul 21 '16 at 05:09

answered May 18 '14 at 21:32

David Arenburg

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Thanks. Any idea how to do so with a character vector of column names? This doesn't seem to work. `myCols = c("a", "b")`; `x[, myCols := lapply(.SD, overwriteNA), .SDcols=myCols]` – iAmAMutt May 18 '14 at 22:10
Why do you use eval() ? – skan Jul 20 '16 at 22:50
1

@skan Because you need to tell `data.table` to evaluate `myCols`. Otherwise it will create a new column called `myCols`. Though you can evaluate it with just parenthesis – David Arenburg Jul 21 '16 at 05:08
Hello, @DavidArenburg , and what's the difference between using (myCols) and .(myCols) ? , I've seen the latter option, with a dot, in some examples. – skan Jul 21 '16 at 18:29
@skan `.()` is short for `list()` – David Arenburg Jul 21 '16 at 18:34
Then here does the same? – skan Jul 21 '16 at 19:08
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@skan, no, `list` isn't evaluating the expression within it. So you can't use it in this case – David Arenburg Jul 21 '16 at 19:09

data.table assignment by reference using lapply and also returning the rest of the columns

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