similar to: print.<strorageMode>() not called when autoprinting

It also is a problem with storage.modes "integer" and "complex": 3.6.0> print.integer <- function(x,...) "integer vector" 3.6.0> 1:10 [1] 1 2 3 4 5 6 7 8 9 10 3.6.0> print(1:10) [1] "integer vector" 3.6.0> 3.6.0> print.complex <- function(x, ...) "complex vector" 3.6.0> 1+2i [1] 1+2i 3.6.0>

Letting a user supply the autoprint function would be nice also. In a way you can already do that, using addTaskCallback(), but that doesn't let you suppress the standard autoprinting. Having the default autoprinting do its own style of method dispatch doesn't seem right. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, May 21, 2019 at 10:50 AM Lionel Henry <lionel at

>>>>> William Dunlap >>>>> on Tue, 21 May 2019 12:11:45 -0700 writes: > Letting a user supply the autoprint function would be nice also. In a way > you can already do that, using addTaskCallback(), but that doesn't let you > suppress the standard autoprinting. > Having the default autoprinting do its own style of method dispatch

FWIW it was the intention of the patch to make printing of unclassed functions consistent with other base types. This was documented in the "patch 3" section: https://bugs.r-project.org/bugzilla/show_bug.cgi?id=17398 I think we need a general way to customise auto-printing for base types and even classed objects as that'd be useful for both users and IDEs. However S3 dispatch may

Hi Martin, > On 22 May 2019, at 03:50, Martin Maechler <maechler at stat.math.ethz.ch> wrote: > > I'm pretty sure that all teaching and documentation about S and R > has suggested that print(f) and auto-printing should result in > the same output _ AFAIR also for S4 objects I agree with the principle that autoprint and print() should be equivalent for users. However

Hi, Peter et al.: On 2017-01-09 4:24 AM, peter dalgaard wrote: > On 09 Jan 2017, at 10:53 , Spencer Graves <spencer.graves at prodsyse.com> wrote: > >> # Define an object of class 'dum' >> k <- 1 >> class(k) <- 'dum' >> str(k) # as expected >> >> # Define print.dum >> print.dum <- function(x, ...) >>

Hi, All: I'm having trouble getting deparse(substitute(x)) inside print.generic to consistently I'm having trouble getting a print.something to work consistently. Consider the following toy example: # Define an object of class 'dum' k <- 1 class(k) <- 'dum' str(k) # as expected # Define print.dum print.dum <- function(x, ...)

You could replace your 'depth' argument with one that shows where in the original data you are at: leaf.func <- function(data, where) { if(is.null(data)) stop("Null data at ", deparse(where)) return(mean(data)) } visit.level <- function(data, where = integer()) { if (length(where) == 2) { return(leaf.func(data, where)) } else

Hello, I'm writing a program that takes a tree in input (nested lists) and returns a copy of it replacing the leaves with something else (eg: a computation done on the original leaves). In the example below, the tree is composed by countries and cities, and the leaves (children of the cities) are vectors of numbers. The program takes this tree and replaces the vectors at the bottom by their

Dear R Helpers, I am missing something very elementary here, and I don't seem to get it from the help pages of the ave, seq and seq_along functions, so I wonder if you could offer a quick help. To use an example from an earlier post on this list, I have a dataframe of this kind: dat = data.frame(name = rep(c("Mary", "Sam", "John"), c(3,2,4))) dat$freq =

Dear folks: Let?s suppose I want a function to print return the name of the object passed to it. > myname <- function(object) {out<-deparse(substitute(object)); out} This works fine on a single object: > O1 <-c(1:4) > myname(O1) [1] "O1" However it does not work if you use lapply to pass it the same object from a list: > O2 <-c(1:4) > object.list <-

Buenas Tengo en un string guardado lo siguiente: > parametros [1] "ntree=10" "ntree=30" "ntree=50" "ntree=100" "ntree=200" Con un bucle for quiero ir metiendolo en el modelo, pero no se muy bien como hacerlo, ya que con deparse no me funciona, con get tampoco (obvio, no es un objeto), y no se muy bien como hacerlo de manera dinamica

Hello there, I am not quite sure how to interpret the output of Rprof (in the following the output I was staring at). I was poking around the web a little bit for documentation but without much success. I guess if I want to figure out what takes so long in my code the 2nd table $by.total and the total.pct column (pct = percent) is the most helpful. What does it mean that [ or [.data.frame is

Dear List, I'm struggling with the signature writing cbind/rbind functions for a S4 class. First of all, I'm very happy that it is now possible to dispatch on ... I follow the example for "paste" in ?dotMethods, which works as far as this: ### start example setClass ("cbtest", representation = representation (data = "data.frame"),

%=>% would have precendence ('order of operations') problems also. A + B %=>% C is equivalent to A + ( B %=>% C) and I don't think that is what you want. as.list(quote(A + B %=>% C)) shows the first branch in the parse tree. The following function, str.language, shows the entire parse tree, as in > str.language(quote(A + B %=>% C)) `quote(A + B %=>%

On 09 Jan 2017, at 10:53 , Spencer Graves <spencer.graves at prodsyse.com> wrote: > # Define an object of class 'dum' > k <- 1 > class(k) <- 'dum' > str(k) # as expected > > # Define print.dum > print.dum <- function(x, ...) > deparse(substitute(x)) > > print(k) # Prints "k" as expected > #####**** THE FOLLOWING PRINTS

Hello List I have a list question. I'm doing some data wrangling for a colleague and I have nested list in the following format: structure(list(MU10 = structure(c(0.80527905920989, 0.4350488707836, 0.455195366623, 0.565174432205497, 0.208180556861924), .Names = c("MU.16", "MU.19", "MU.21", "mean", "sd")), MU11 =

A private reply by Martin made me realize that I was wrong about stopifnot(exprs=TRUE) . It actually works fine. I apologize. What I tried and was failed was stopifnot(exprs=T) . Error in exprs[[1]] : object of type 'symbol' is not subsettable The shortcut assert <- function(exprs) stopifnot(exprs = exprs) mentioned in "Warning" section of the documentation similarly fails

Dear R-devel, I have encountered a crash-inducing scenario and would like to enquire as to whether this would be considered a bug. To reproduce the crash: X <- sample(letters, 3000, TRUE) D <- data.frame(X, 1:3000, X, X, X, X, X) D$X1.3000 <- paste0("GSM", D) The reason why I'm not sure if this would be considered a bug is because I typed this by accident, when what I

I put the idea below into a function that gives nicer looking results. Here's the new code: dupnames <- function(path = ".") { Rfiles <- pkgload:::find_code(path) allnames <- data.frame(names=character(), filename=character(), line = numeric()) result <- NULL for (f in Rfiles) { exprs <- parse(f, keep.source = TRUE) locs <-