how do I define a function which is equivalent to `deparse(substitute(x))`?
Dear R-Help,
I was going to ask Jeff to read the entire works of William
Shakespeare to learn why his reply was not helpful to me...
Then I realized that the answer, as always, lies within...
desub <- function(y) {
e1=substitute(y, environment())
e2=do.call(substitute,list(e1), env=parent.frame())
deparse(e2)
}
Sorry to trouble the list; other solutions still welcome.
Cheers,
Frederick
On Sun, Dec 11, 2016 at 12:46:23AM -0800, Jeff Newmiller wrote:
No. Read Hadley Wickham's "Advanced R" to learn why not. -- Sent from my phone. Please excuse my brevity. On December 10, 2016 10:24:49 PM PST, frederik at ofb.net wrote:
Dear R-Help, I asked this question on StackOverflow, http://stackoverflow.com/questions/41083293/in-r-how-do-i-define-a-function-which-is-equivalent-to-deparsesubstitutex but thought perhaps R-help would be more appropriate. I want to write a function in R which grabs the name of a variable from the context of its caller's caller. I think the problem I have is best understood by asking how to compose `deparse` and `substitute`. You can see that a naive composition does not work: # a compose operator
> `%c%` = function(x,y)function(...)x(y(...))
# a naive attempt to combine deparse and substitute
> desub = deparse %c% substitute
> f=function(foo) { message(desub(foo)) }
> f(log)
foo # this is how it is supposed to work
> g=function(foo) { message(deparse(substitute(foo))) }
> g(log)
log Is there a way I can define a function `desub` so that `desub(x)` has the same value as `deparse(substitute(x))` in every context? Thank you, Frederick Eaton
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