search for: vapply

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 "bar" "bar" "bar" hello "world" &qu...

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#FALSE is.primitive(lapply) #FALSE It appears that they are not primitives and that the documentation might be outdated.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...

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...

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 cal...

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), fun...

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 /D...

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...

> -----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_ according to the documentation. In the first case (function return value contains more than one element, and each ), vapply simpli...

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 element, and each ), vapply simplif...

>>>>> 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. > But if 'X' has names already and 'USE.NAMES&...

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...

...eating most warnings as errors anyway) but for the sake of other useRs who may get bitten, perhaps we should be more explicit that backwards-compatibility won't be preserved under certain use patterns, for example: # works (with warning) in R 3.6.2 but fails (with error) in R-devel: vapply(list(1e9, 1e10), ?????? function(lambda) { ????????? rpois(1L, lambda) ?????? }, ?????? FUN.VALUE = integer(1L) ?????? ) # in R-devel, a little extra work to achieve a warning as before: vapply(list(1e9, 1e10), ?????? function(lambda) { ????????? tmp <...

...) s <- s + xi s } kahanSum <- function (x) { s <- 0.0 c <- 0.0 # running compensation for lost low-order bits for(xi in x) { y <- xi - c t <- s + y # low-order bits of y may be lost here c <- (t - s) - y s <- t } s } > rSum <- vapply(c(1:20,10^(2:7)), function(n) sum(rep(1/7,n)), 0) > rNaiveSum <- vapply(c(1:20,10^(2:7)), function(n) naiveSum(rep(1/7,n)), 0) > rKahanSum <- vapply(c(1:20,10^(2:7)), function(n) kahanSum(rep(1/7,n)), 0) > > table(rSum == rNaiveSum) FALSE TRUE 21 5 > table(rSum == rKah...

...oblem - it gives the same results for the fractional part as bc does. bc: A2425FF7 5E14FC31 A125... standalone: 22425FF7 5E14FC32 R: 22425FF7 5E151800 There are lots of other small numbers with the same problem: > grep(value=TRUE, "0e", vapply((1+(0:10000)/1000)*1e-15, deparse, "")) [1] "8.56000000000000e-15" "8.71700000000000e-15" "8.77800000000000e-15" [4] "8.93500000000000e-15" "9.50800000000000e-15" "9.83800000000000e-15" [7] "9.89900000000000e-15" "...

...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 legitimate possibility that s is a nested list? Currently I have a case where a dendrogram object is breaks at this line, because s is a nested list: >str (s) List of 2 $ : int -779 $ :List of 2 ..$ : int -625 ..$ : int 15 I'm unsure if my dendrogram...

...01") format(tm,"%a") %in% c("Sat","Sun") } use.lapply <- function(data) { data[do.call(rbind,lapply(data$TIME,FUN=is.weekend)),] } use.sapply <- function(data) { data[sapply(data$TIME,FUN=is.weekend),] } use.vapply <- function(data) { data[vapply(data$TIME,FUN=is.weekend,FALSE),] } use.indexing <- function(data) { data[is.weekend(data$TIME),] } And the results of these methods: > names(csv.data) [1] "TIME" "FILE" "RADIAN"...

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 func...

...bc: A2425FF7 5E14FC31 A125... > > standalone: 22425FF7 5E14FC32 > > R: 22425FF7 5E151800 > > > > There are lots of other small numbers with the same problem: > > > > > > > grep(value=TRUE, "0e", > > vapply((1+(0:10000)/1000)*1e-15, deparse, "")) > > [1] "8.56000000000000e-15" "8.71700000000000e-15" "8.77800000000000e-15" > > [4] "8.93500000000000e-15" "9.50800000000000e-15" "9.83800000000000e-15" > > [7] "9.8...

...200", "b2=50", "b3=0.3") when what I wanted was pnum<-c(b1=200, b2=50, b3=0.3) There was a list thread in 2010 that shows how to deal with un-named vectors, but the same lapply solution doesn't seem to work here i.e., pnum<-lapply(pstr, as.numeric) or similar vapply version. The names and = signs seem to mess things up. This is clearly not critical, but it would be nice to know an appropriate transformation. JN