search for: vappli

Hi all Sorry for the confusing title. I noticed the following inconsistency: If i have a function that returns a named list with 2 (or more) elements, then using 'vapply' retains the names of the elements: > vapply(1:3, function(x) list("foo" = "bar", "hello" = "world"), > vector("list", 2)) [,1] [,2] [,3] foo

The documentation of ?lapply includes: > lapply and vapply are primitive functions. However, both evaluate to FALSE in `is.primitive()`: is.primitive(vapply) #FALSE is.primitive(lapply) #FALSE It appears that they are not primitives and that the documentation might be outdated. Thank you for your time and work. Cole Miller P.S. During research, my favorite `help()` is

Hi, Both vapply() and sapply() support the 'USE.NAMES' argument. According to the man page: USE.NAMES: logical; if ?TRUE? and if ?X? is character, use ?X? as ?names? for the result unless it had names already. But if 'X' has names already and 'USE.NAMES' is FALSE, it's not clear what will happen to the names. Are they going to propagate to the result

vapply <- function(X, FUN, FUN.VALUE, ..., USE.NAMES = TRUE) { FUN <- match.fun(FUN) if(!is.vector(X) || is.object(X)) X <- as.list(X) .Internal(vapply(X, FUN, FUN.VALUE, USE.NAMES)) } This is an implementor question. Basically, what happened to the '...' args in the call to the .Internal? cf lapply:, where the ... is passed. lapply <- function (X, FUN, ...)

The following code seems to contain an inconsistency in the behavior of vapply(). Am I missing something here? ## This function assumes v is a 3d vector, beta a scalar. f3d <- function(v,beta) { v+beta } ## This expression applies f3d to a vector of scalars, and ## specifies the template 'array(10,3)' for the return value. dat <- vapply(seq(0,1,length=10), function(beta) {

Changing the vapply() behavior makes sense in principle. I analyzed the CRAN code base using the R parser and found 143 instances of calling vapply with USE.NAMES=FALSE. These would need to be inspected to understand the consequences of the change. For reference: /AzureML/R/datasets.R:226 /BBmisc/R/toRangeStr.R:33 /DBI/R/DBDriver.R:205 /Kmisc/R/str_rev.R:37 /Matrix/R/diagMatrix.R:98

On 12/16/2014 08:20 PM, Mick Jordan wrote: > vapply <- function(X, FUN, FUN.VALUE, ..., USE.NAMES = TRUE) > { > FUN <- match.fun(FUN) > if(!is.vector(X) || is.object(X)) X <- as.list(X) > .Internal(vapply(X, FUN, FUN.VALUE, USE.NAMES)) > } > > This is an implementor question. Basically, what happened to the '...' args in > the call to the

> -----Original Message----- > From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Dean Attali > If i have a function that returns a named list with 2 (or more) elements, > then using 'vapply' retains the names of the elements: > .... > But if the function only returns one element, then the name "foo" is lost vapply _always simplifies_

Quote---------------- > If i have a function that returns a named list with 2 (or more) elements, > then using 'vapply' retains the names of the elements: > .... > But if the function only returns one element, then the name "foo" is lost vapply _always simplifies_ according to the documentation. In the first case (function return value contains more than one

>>>>> Herv? Pag?s <hpages at fredhutch.org> >>>>> on Mon, 8 Feb 2016 10:48:50 -0800 writes: > Hi, > Both vapply() and sapply() support the 'USE.NAMES' argument. According > to the man page: > USE.NAMES: logical; if ?TRUE? and if ?X? is character, use ?X? as > ?names? for the result unless it had names already.

Friends I clearly donot understand how sapply and vapply work. What I have is a function that returns a matrix with an indeterminate number of rows (some times zero) but a constant number of columns. I cannot reliably use an apply function to assemble the matrices into a matrix. I am not sure it is possible. I can demonstrate the core of my confusion with this simple code. A.f <-

On 1/20/20 12:33 PM, Martin Maechler wrote: > > It's really something that should be discussed (possibly not > here, .. but then I've started it here ...). > > The NEWS for R 3.0.0 contain (in NEW FEATURES) : > > * Functions rbinom(), rgeom(), rhyper(), rpois(), rnbinom(), > rsignrank() and rwilcox() now return integer (not double) > vectors.

The algorithm does make a differece. You can use Kahan's summation algorithm (https://en.wikipedia.org/wiki/Kahan_summation_algorithm) to reduce the error compared to the naive summation algorithm. E.g., in R code: naiveSum <- function(x) { s <- 0.0 for(xi in x) s <- s + xi s } kahanSum <- function (x) { s <- 0.0 c <- 0.0 # running compensation for lost

I just got a new Windows laptop (i7, 10th generation CPU), installed 'Windows Subsystem for Linux 2' and then installed Ubuntu 20.04 and used 'apt-get install' to install packages that the R build seems to require. In particular, I am using gcc version 9.3.0. The build went without a hitch but the tests showed that deparse(1e-16) produced "1.00000000000000e-16" instead

Hi all, This is the first time I'm writing to R-devel, and this time I'm just asking for the purpose for a certain line of code in stats::as.hclust.dendrogram, which comes up as I'm trying to fix dendextend. The line in question is at line 128 of dendrogram.R in R-3.3.3, at stats::as.hclust.dendrogram: stopifnot(length(s) == 2L, all( vapply(s, is.integer, NA) )) Is there any

? Wed, 12 Jun 2024 23:42:18 +0000 "Levine, Michael" <mlevins at purdue.edu> ?????: > f.int1 <- function(x,y) {Vectorize (function(y) f(c(x,y),H=H,j=j))} Vectorize returns a callable function(y), so wrapping it in a function(x,y) will not work. Since you'd like to integrate over y, we can perform the same transformation manually using vapply: # A version of f(...) that

Hello, I've been dealing with a set of values that contain time stamps and part of my summary needs to look at just weekend data. In trying to limit the data I've found a large difference in performance in the way I index a data frame. I've constructed a minimal example here to try to explain my observation. is.weekend <- function(x) { tm <-

Dear All, I have trouble generizising some code. > index <- 0 > sapply(list(c(1,2,3),c(1,2),c(1)),function(x){x[max(length(x)-index,0)]}) Will yield a wished for vector like so: [1] 3 2 1 But in this case (trying to select te second to last element in each vector of the list) > index <- 1 > sapply(list(c(1,2,3),c(1,2),c(1)),function(x){x[max(length(x)-index,0)]}) I end up

There turn out to be a few more things to fix. One problem is easy to solve: vapply() needs a third argument specifying the type of the return value. (Can we have unit tests for tab completion?) The other problem is harder: `comps` defaults to an empty string, and you can't have a symbol consisting of an empty string, because this value is internally reserved for missing function arguments.

On Wed, 18 Nov 2020 at 10:26, Tomas Kalibera <tomas.kalibera at gmail.com> wrote: > > On 11/17/20 9:34 PM, Bill Dunlap wrote: > > I just got a new Windows laptop (i7, 10th generation CPU), installed > > 'Windows Subsystem for Linux 2' and then installed Ubuntu 20.04 and > > used 'apt-get install' to install packages that the R build seems > > to