similar to: Modifying parsed code

Hi, I'm trying to get ...as a list of unevaluated arguments, ie. substitute(list(...)) gives me an unevaluated list of the arguments, but I want a list of the unevaluated arguments. My attempts so far: (function(...) substitute(...))(a=1, b=a) # Only returns first (function(...) substitute(list(...)))(a=1, b=a) # Unevaluated list, not list of unevaluated (function(...)

Since you're already using bang-bang ;) library(rlang) dots1 <- function(...) as.list(substitute(list(...)))[-1L] dots2 <- function(...) as.list(substitute(...())) dots3 <- function(...) match.call(expand.dots = FALSE)[["..."]] dots4 <- function(...) exprs(...) bench::mark( dots1(1+2, "a", rnorm(3), stop("bang!")), dots2(1+2, "a",

Consider the following example: # substitute a with b in the indicated function. Seems to work. > z <- substitute( function()a+1, list(a=quote(b)) ) > z function() b + 1 # z is an object of class call so use eval # to turn it into an object of class expression; however, # when z is evaluated, the variable a returns. > eval(z) function()a+1 Why did a suddenly reappear again

I am looking for a way to extract the name of a variable that has been passed into a function for example foo <-function(x){ write.csv(x, file = paste(NAME(x), "csv", sep =".")) } is there a function "NAME" that would let the calls foo(bar) write the file bar.csv and foo(stuff) write the file stuff.csv Robert [[alternative HTML version deleted]]

Hi all, It's possible to capture the expression associated with a promise (using substitute). Is there any way to capture the environment associated with a promise? Similarly, is there any way to tell if something is a promise without accidentally evaluating it? Thanks! Hadley -- RStudio / Rice University http://had.co.nz/

Hi all, I think there's a small buglet in quote: str(quote()) # Error in quote() : 0 arguments passed to 'quote' which requires 1 str(quote(expr = )) # symbol I bring this up because this seems like the most natural way of capturing the "missing" symbol with pure R code, compared to substitute() or bquote() or formals(plot)$x Hadley -- Chief Scientist, RStudio

I am sure it is just me not understanding how R works, but could somebody explain why curve(cos(x)) works and curve(expression(cos(x)) does not? I have done some investigating and here is what I found. If I comment out the line of curve indicated below, both calls work fine. function (expr, from, to, n = 101, add = FALSE, type = "l", ylab = NULL, log = NULL, xlim =

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

> lm2 <- function(...) lm(...) > lm2(mpg ~ wt, data=mtcars) Call: lm(formula = ..1, data = ..2) Coefficients: (Intercept) wt 37.285 -5.344 > lm2(mpg ~ wt, weights=cyl, data=mtcars) Error in eval(expr, envir, enclos) : ..2 used in an incorrect context, no ... to look in Can anyone explain why this is happening? (Obviously this is a manufactured example, but it

> On Jun 8, 2018, at 1:49 PM, Hadley Wickham <h.wickham at gmail.com> wrote: > > Hmmm, yes, there must be some special case in the C code to avoid > recycling a length-1 logical vector: Here is a version that (I think) handles Herve's issue of arrays having one or more 0 dimensions. subset_ROW <- function(x,i) { dims <- dim(x) index_list <-

Hi all, Is there a better to way to subset the ROWs (in the sense of NROW) of an vector, matrix, data frame or array than this? subset_ROW <- function(x, i) { nd <- length(dim(x)) if (nd <= 1L) { x[i] } else { dims <- rep(list(quote(expr = )), nd - 1L) do.call(`[`, c(list(quote(x), quote(i)), dims, list(drop = FALSE))) } } subset_ROW(1:10, 4:6) #> [1] 4 5 6

Hi all, What's the official name for the value that represents a missing argument? e.g. formals(plot)$x str(formals(plot)$x) deparse(formals(plot)$x) is.symbol(formals(plot)$x) What's the correct way to create an object like this? (for example if you are manipulating the formals of a function to add an argument with no default value, as in http://stackoverflow.com/questions/3892580/).

Hi all, Does there already exist a version of substitute that evaluates it's first argument? (i.e. it accepts an already quoted expression). This seems like something that's pretty handy, but I haven't found any existing function to do it: substitute_e <- function(expr, env) { eval(substitute(substitute(expr, env), list(expr = expr))) } f <- quote(x + y + z) substitute(f,

Interestingly, as.list(substitute(...())) also works. On Sun, Aug 12, 2018 at 1:16 PM, Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > On 12/08/2018 4:00 PM, Henrik Bengtsson wrote: >> >> Hi. For any number of *known* arguments, we can do: >> >> one <- function(a) list(a = substitute(a)) >> two <- function(a, b) list(a = substitute(a), b =

Michael Lawrence (as last in long series of posters)... > Yes, it would bind the language object to the environment, like an > R-level promise (but "promise" of course refers specifically to just > _lazy_ evaluation). > > For the uqs() thing, expanding calls like that is somewhat orthogonal > to NSE. It would be nice in general to be able to write something like >

As instructed, I have spent a long time searching the web for an answer to this question. I am trying to use Sweave to produce lecture slides, and have the problem that I can't control the formatting of my R source. Setting options(width), as recommended in this forum, works fine on the R _output_, but seems to have unpredictable effects on the echoing of the source code. If I try setting

>After off list discussions with Jonathan Carrol and with >Michael Lawrence I think it's doable, unambiguous, >and even imo pretty intuitive for an "unquote" operator. For those of us who are not CS/Lisp mavens, what is an "unquote" operator? Can you expression quoting and unquoting in R syntax and show a few examples where is is useful, intuitive, and fits in to

Although I owe thanks to Ramus and Ivan, I still do not know how to write and "advanced" function. My most recent try (after looking at the material Ramus and Ivan set) still does not work. I am trying to run the lm function on two different formulae: 1) y~x, 2) y~x+z Any corrections would be appreciated! Thank you, John doit <- function(x){ ds <- deparse(substitute(x))

Your example x = 5 exp = parse(text="f(uq(x)) + y +z") # expression: f(uq(x)) +y + z do_unquote(expr) # -> the language object f(5) + y + z could be done with the following wrapper for bquote my_do_unquote <- function(language, envir = parent.frame()) { if (is.expression(language)) { # bquote does not go into expressions, only calls

I suspect this will have suboptimal performance since the TRUEs will get recycled. (Maybe there is, or could be, ALTREP, support for recycling) Hadley On Fri, Jun 8, 2018 at 10:16 AM, Berry, Charles <ccberry at ucsd.edu> wrote: > > >> On Jun 8, 2018, at 8:45 AM, Hadley Wickham <h.wickham at gmail.com> wrote: >> >> Hi all, >> >> Is there a better to