CODEWITHSUNDEEP
r
jangorecki
jangorecki

Reputation: 16697

Substitute on quoted arguments

Looking for a way to use substitute on quoted language objects as expression.
substitute expects to get lazy expression in expr.
To goal is to substitute .expr in expr.template which is a language objects generated dynamically based on metadata. 

## data
expr = quote(x <- f(10))
expr.template = quote(g(.expr, flag = TRUE))

## expected output
quote(g(x <- f(10), flag = TRUE))
#g(x <- f(10), flag = TRUE)

## current workaround
h = function(expr, expr.template){
    eval(substitute(
        substitute(
            .expr.template,
            list(.expr = expr)
        ),
        list(.expr.template = expr.template)
    ))
}
h(expr = expr, expr.template = expr.template)
#g(x <- f(10), flag = TRUE)

So I would be surprised if there wouldn't be any more canonical way to deal with it. Base R solution preferred.

Upvotes: 4

Views: 89

Answers (1)

G. Grothendieck
G. Grothendieck

Reputation: 269371

Use do.call:

do.call("substitute", list(expr.template, list(.expr = expr)))
## g(x <- f(10), flag = TRUE)

Upvotes: 6

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