CODEWITHSUNDEEP
rdata.tablerlangquasiquotes
JBGruber
JBGruber

Reputation: 12450

Quasiquotation with data.table

I'm trying to wrap my head around quasiquotation so that I could use it together with a data.table call. Here is an example:

library(data.table)
library(rlang)
dt <- data.table(col1 = 1:10, col2 = 11:20)

dt[, col1]

If I wanted to wrap this into function, how would I do this? I tried:

foo <- function(dt, col) {
  col <- quo(col)

  expr(dt[, !!col1])
}

foo(dt, col1)

But get Error in enexpr(expr) : object 'col1' not found. I assume I'm missing some steps as data.table evaluates this differently than dplyr.

Upvotes: 4

Views: 698

Answers (2)

Etienne Kintzler
Etienne Kintzler

Reputation: 682

You can use deparse and substitute and use the argument with=FALSE as in :

foo <- function(dt, col){
  col_str = deparse(substitute(col))
  dt[, col_str, with = F]
}

or you can use eval and substitute and use the default data.table argument with=TRUE as in :

foo <- function(dt, col){
  col_symb = substitute(col)
  dt[, eval(col_symb)] # by default: with=TRUE
}

In both cases, substitute will get the name of the argument that you pass to the parameter col. In the first case deparse will convert this name as a string, thus enabling us to select it from the data.table using with = FALSE. In the second case we evaluate (using eval) the name of the argument in the context of the data.table.

Upvotes: 1

MrFlick
MrFlick

Reputation: 206516

You want to capture the column name as a symbol with

col <- ensym(col)

rather than quo() and then use 

expr(dt[, !!col])

(not col1 which doesn't exist there) but that will just return an expression. If you want to evaluated it, you'd need 

eval_tidy(expr(dt[, !!col]))

But really the quasinotation stuff works best in the tidyverse and not with data.table functions natively. The "data.table" way might be more like something in this existing question: Pass column name in data.table using variable. data.table very much prefers strings to symbols.

Upvotes: 6

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