CODEWITHSUNDEEP
rdata.table
cauchy
cauchy

Reputation: 198

Combining multiple lists of variable names in data.table?

I'm trying to pass two separate lists of variable names into a data.table (v1.9.4). It returns the correct columns, but it strips the variable names. This works as expected:

dt <- data.table(a=1:3, b=4:6, c=7:9, d=10:12)
dt
   a b c  d
1: 1 4 7 10
2: 2 5 8 11
3: 3 6 9 12

It also works fine to pass a single list of names:

dt[,list(a,b)]
   a b
1: 1 4
2: 2 5
3: 3 6

But when I need to pass multiple lists, it returns the correct columns but strips the variable names:

dt[,c(list(a,b), list(c,d))]
   V1 V2 V3 V4
1:  1  4  7 10
2:  2  5  8 11
3:  3  6  9 12

Why two lists? I'm using multiple quote()'d lists of variables. I've read FAQ question 1.6, and I know that one workaround is to use a character vector using with=FALSE. But my real use case involves passing a mix of names and expressions to a function, e.g.,

varnames <- quote(list(a,b))
expr <- quote(list(a*b, c+d))
function(dt, varnames, expr) {
  dt[,c(varnames, expr)]
}

And I'd like the "varnames" columns to have their proper names (and they do if you just pass a single list like

dt[,list(a,b,a*b,c+d)]
   a b V3 V4
1: 1 4  4 17
2: 2 5 10 19
3: 3 6 18 21

How can I combine multiple lists in a data.table such that it still returns the proper column names? (I'm not completely sure if this is a data.table issue or if I'm just doing something silly in the way I'm trying to combine lists in R, but c() seems to do what I want.)

Upvotes: 10

Views: 1246

Answers (2)

BrodieG
BrodieG

Reputation: 52637

Another option is to construct the full call ahead of time:

varnames[4:5] <- expr[2:3]  # this results in `list(a, b, a * b, c + d)`
dt[, eval(varnames)]

produces:

   a b V3 V4
1: 1 4  4 17
2: 2 5 10 19
3: 3 6 18 21

More generically, suppose you have a list of quoted lists of expressions:

exprlist <- list(quote(list(a, b)), quote(list(c, c %% a)), quote(list(a + b)))
expr <-  as.call(Reduce(function(x, y) c(as.list(x), as.list(y)[-1]), exprlist))  # @eddi
dt[, eval(expr)]

Upvotes: 4

David Arenburg
David Arenburg

Reputation: 92282

Here's a possible workaround using .SD

varnames <- quote(list(a,b))
expr <- quote(list(a*b, c+d))

myFunc <- function(dt, varnames, expr) {
  dt[, c(.SD[, eval(varnames)], eval(expr))]
}

myFunc(dt, varnames, expr)

#    a b V1 V2
# 1: 1 4  4 17
# 2: 2 5 10 19
# 3: 3 6 18 21

Upvotes: 1

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