similar to: Bugs due to naive copying of list elements

I think the discussion of this issue has gotten more complicated than necessary. First, there really is a bug. You can see this also by the fact that delayed warning messages are wrong. For instance, in R-3.1.2: > lapply(c(-1,2,-1),sqrt) [[1]] [1] NaN [[2]] [1] 1.414214 [[3]] [1] NaN Warning messages: 1: In FUN(c(-1, 2, -1)[[3L]], ...) : NaNs produced 2: In

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

Hi, There are quiet a few different 'apply' functions, such as lapply, sapply and many more. I'm very familiar with the 'Apply' function in Mathematica. Can somebody point me a summary of all the 'apply' functions in R. Also, I'm curious that what 'l' and 's' (and other prefixes) stand for in 'lapply' and 'sapply' Regards, Peng

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

Hi,   it seems a simple problem, but I can not find a clear way. I have a list: lst=list(m=c('a','b','c'),n=c('c','a'),l=c('a','bc')) > lst $m [1] "a" "b" "c" $n [1] "c" "a" $l [1] "a"  "bc" how can I get list elements that include a given subset? for example, for given

Hello. Please forgive me if this problem has already been posted (and solved) by someone else ... I can't find it anywhere though it seems so very basic. Here it is: I have a list comprised of several matrices, each of which has two columns. > list [[1]] [,1] [,2] [1,] 1 3 [2,] 2 4 [[2]] [,1] [,2] [1,] 5 7 [2,] 6 8 [[3]] [,1] [,2]

The following looks like an 'eapply' bug to me: t/subtest> e <- new.env() t/subtest> e$tempo <- quote( 1+'hi') t/subtest> lapply( ls( e), function( x) length( get( x,e))) [[1]] [1] 3 # seems reasonable-- e$tempo is a 'call' object of length 3 t/subtest> eapply( e, length) Error in 1 + "hi" : non-numeric argument to binary operator

Hi, I have 3 lists, x, y, z and I'd like to perform a calculation over all the lists simultaneously. If it were a single list I could use lapply, but for more than one list I'm using a for loop. Is there an idiom that would let me use something like lapply, but the function specified to lappy would have access to an element from each list? (In Python, I would have used for a,b,c in

In an earlier post, a person wanted to divide each of the rows of rawdata by the row vector sens so he did below but didn't like it and asked if there was a better solution. rawdata <- data.frame(rbind(c(1,2,2), c(4,5,6))) sens <- c(2,4,6) temp <- t(rawdata)/sens temp <- t(temp) print(temp) Gabor sent three other solutions and I understood 2 of them but not the

Dear R developers, Reviewing my code, I have realized that about 80% of the time in the lapply I need to access the names of the objects inside the loop. In such cases I iterate over indexes or names: lapply(names(x), ... [i]), lapply(seq_along(x), ... x[[i]] ... names(x)[i] ), or for(i in seq_along(x)) ... which is rather inconvenient. How about an argument to lapply which would specify the

eapply() works on most environments, but not on baseenv(). For example, > x <- 1 > eapply(globalenv(), function(x) x) $x [1] 1 > eapply(baseenv(), function(x) x) list() I'm probably not going to have time to work on this before 2.3.0, but I don't think it's really urgent; if no one else fixes it first I'll do it after the release. Duncan Murdoch

David, If you are referring to the solution that would be: rapply(list(test), eval, envir = fenv) I thought I explained in the question that the above code does not work. It does not throw an error, but the behavior is no different (at least in the output or result). Using the above code still results in the x object not being stored in fenv on 3.1.2. Dayne On Wed, Jul 15, 2015 at 4:40 PM,

On Jul 15, 2015, at 12:51 PM, William Dunlap wrote: > I think rapply() was changed to act like lapply() in this respect. > When I looked at the source of the difference, it was that typeof() returned 'language' in 3.2.1, while it returned 'list' in the earlier version of R. The first check in rapply's code in both version was: if (typeof(object) != "list")

In 3.1.2 eval does not store the result of the bquote-generated call in the given environment. Interestingly, in 3.2.1 eval does store the result of the bquote-generated call in the given environment. In other words if I run the given example with eval rather than evalq, on 3.1.2 "x" is never stored in "fenv," but it is when I run the same code on 3.2.1. However, the given

Dear R People: I want to get the class of all of the objects in my directory. I was trying: do.call(class,list=ls()) but got an "unused argument error". I'm sure it's simple, but I'm just not seeing it. Any help would be much appreciated. Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston -

Hi, I have a question about applying a function recursively through a list. Suppose I have a list where the different elements have different levels of recursion: > test.list<-list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f"), "C"=c("g", "h", "i")), +

Dear colleagues, I've just come across a problem with the following command which is a part of the "metaOverview.R" code file provided as an monography- accompanying file at http://www.bioconductor.org/docs/mogr/metadata: ################################## R> hasChr <- eapply(GOTERM, function(x) + x[grep("chromosome", Term(x))]) Error in

hi all members, please, i need you help... now a i´m working with veronoi polygons in a area with projections, but i need cut the polygons left. On other words, i need cut the polygons in the worked area. R help say that use the command voronoi.findrejectsites, but in this command i need put the numbers, any way...this command not cut!! do you can help me? Thank you for help me! José Bustos

Hello there, any ideas on how to save all the objects on my workspace inside a list object? For example, say my workspace is as follows ls() [1] "x" "y" "z" and suppose I want to put these objects inside a list object, say object.list <- list() without having to explicitly write down their names as in object.list$x = x object.list$y = y object.list$z = z Is

Is there a function in R that corresponds to the function ``map'' of perl? It could be called like: vector.a <- map( vector.b, FUN, args.for.FUN ) It should execute for each element ele.b of vector.b: FUN( vector.b, args.for.FUN) It should return a vector (or data.frame) of the results of the calls of FUN. It nearly works using: apply( data.frame( vector.b ), 1, FUN,