CODEWITHSUNDEEP
rnon-standard-evaluation
RoyalTS
RoyalTS

Reputation: 10203

get names of quoted list without evaluation

I've got a quoted list

quote(list(orders = .N,
           total_quantity = sum(quantity)))

(that I eventually eval in the j part of a data.table)

What I would like is to extract the names of that list without having to evaluate the expression because outside of the correct environment evaluating the expression will produce an error.

Upvotes: 1

Views: 52

Answers (1)

MrFlick
MrFlick

Reputation: 206243

The list doesn't have any names at that point. It's not even a list. It's a call to the list() function. But that said you can parse that function call and extract name parameter. For example

x <- quote(list(orders = .N,
    total_quantity = sum(quantity)))
names(as.list(x))[-1]
# [1] "orders"         "total_quantity"

That as.list() on the expression turns the function call into a (named) list without evaluation.

Upvotes: 5

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