Use variable in R substitute

Question

I have an expression stored in a variable

a <- expression(10 + x + y)

I want to use substitute to fill the expression with x = 2

substitute(a, list(x=2))

But this returns a and a evaluates to expression(10 + x + y)

Ideally a would evaluate to expression(12 + y) (or (10 + 2 + y))

Is there any way to implement this behavior while using an expression stored in the variable a (mandated by other parts of my project)?

Maybe worth pointing out that `substitute(expression(10 + x + y), list(x = 2))` works as desired, so the question is how to use substitute on an expression stored in a variable. (Implied by your title, but nice to be explicit.) — Gregor Thomas, Dec 12 '17 at 19:35

score 3 · Accepted Answer · answered Dec 12 '17 at 19:36

3

Use do.call. substitute won't descend into some objects but if you use a[[1]] here then it will work.

a <- expression(10 + x + y)
do.call("substitute", list(a[[1]], list(x = 2)))
## 10 + 2 + y

answered Dec 12 '17 at 19:36

G. Grothendieck

score 2 · Answer 2 · answered Dec 12 '17 at 19:37

2

You could do that with pryr if you can use an alternative (i.e. quote instead of expression):

a <- quote(10 + x + y)

library(pryr)
substitute_q(a, list(x=2))
#10 + 2 + y

answered Dec 12 '17 at 19:37

LyzandeR

Andrii · Answer 3 · 2017-12-12T19:51:31.453

0

Maybe this one snippet can be useful

  x <- 1; y <- 2
  a <- expression(10 + x + y)
  eval(a)

edited Dec 12 '17 at 19:51

answered Dec 12 '17 at 19:45

Andrii

3 Answers3