similar to: getting the "name" of an object

Wouldn't that change how simplify='array' is handled? > str(sapply(1:3, function(x)diag(x,5,2), simplify="array")) int [1:5, 1:2, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=TRUE)) int [1:10, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=FALSE)) List of 3 $ : int [1:5, 1:2] 1 0 0 0 0 0 1 0 0

You?re right, it sure does. My suggestion causes it to fail when simplify = ?array? From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:11 PM To: Doran, Harold <HDoran at air.org> Cc: r-help at r-project.org Subject: Re: [R] Possible Improvement to sapply Wouldn't that change how simplify='array' is handled? > str(sapply(1:3,

On 03/13/2018 09:23 AM, Doran, Harold wrote: > While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function > > if (!identical(simplify, FALSE) && length(answer)) > > This seems superfluous to me,

While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function if (!identical(simplify, FALSE) && length(answer)) This seems superfluous to me, in particular this part: !identical(simplify, FALSE) The preceding

Martin In terms of context of the actual problem, sapply is called millions of times because the work involves scoring individual students who took a test. A score for student A is generated and then student B and such and there are millions of students. The psychometric process of scoring students is complex and our code makes use of sapply many times for each student. The toy example used

I have a list that is made of lists of varying length. I wish to create a new vector that contains the last element of each list. So far I have used sapply to determine the length of each list, but I'm stymied at the part where I index the list to make a new vector containing only the last item of each list mylist =

Could your code use vapply instead of sapply? vapply forces you to declare the type and dimensions of FUN's output and stops if any call to FUN does not match the declaration. It can use much less memory and time than sapply because it fills in the output array as it goes instead of calling lapply() and seeing how it could be simplified. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue,

Hi there During execution of sapply I want to extract the number of times the function given to supply has been executed. I came up with: mylist <- list(a=3,b=6,c=9) sapply(mylist,function(x)as.numeric(gsub("[^0-9]","",deparse(substitute(x))))) This works fine, but looks quite ugly. I'm sure that there's a more elegant way to do this. Any suggestion? Christian

Quite possibly, and I?ll look into that. Aside from the work I was doing, however, I wonder if there is a way such that sapply could avoid the overhead of having to call the identical function to determine the conditional path. From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:14 PM To: Doran, Harold <HDoran at air.org> Cc: Martin Morgan <martin.morgan

Hi all, I would like to iterate through a list with named elements and access the names within an lapply / sapply call. One way to do this is iterate through the names and index the list with the name. Is there a way to iterate through the list elements themselves and access the element names within in the function? For example, mylist <-

Hi, I am trying to install pdftools package on R 3.3.1 (Ubuntu 16.04), but the following issue occurs: ------------------------- ANTICONF ERROR --------------------------- Configuration failed because poppler-cpp was not found. Try installing: * deb: libpoppler-cpp-dev (Debian, Ubuntu, etc) * rpm: poppler-cpp-devel (Fedora, CentOS, RHEL) * csw: poppler_dev (Solaris) * brew: poppler (Mac OSX)

Excuse me for what I'm sure is a stupid beginner's question, but I've given up trying to find the answer to this question from the help, RSiteSearch, or any of the usual places. I have a list that looks like this: >myList $first [1] "--" "18" "8" "32" $second [1] "--" "--" "40" "54" I want a

Hello, I have a number of files output1.dat, output2.dat, ... , output20.dat, each of which monitors several variables over a fixed number of timepoints. From this I want to create a data frame which contains the mean value between all files, for each timepoint and each variable. The code below works, but it seems like I should be able to do the second part without a for loop. I played

Hi Francois, Thanks for your quick response. Actually, I had already done that... sudo apt-get install libpoppler-cpp-dev Reading package lists... Done Building dependency tree Reading state information... Done libpoppler-cpp-dev is already the newest version (0.41.0-0ubuntu1.1). 0 upgraded, 0 newly installed, 0 to remove and 107 not upgraded. Therefore, I assume I have this installed. Best

Hi It seems, I just miss something. I defined treshold <- function(pred) { if (pred < 0.5) pred <- 0 else pred <- 1 return(pred) } and want to use apply it on a vector sapply(mylist[,,3],threshold) but I get: Error in match.fun(FUN) : Object "threshold" not found thanks for help cheers chris -- Christoph Lehmann <christoph.lehmann at gmx.ch>

I've been doing some consulting with students who seem to come to R from SAS. They are usually pre-occupied with do loops and it is tough to persuade them to trust R lists rather than keeping 100s of named matrices floating around. Often it happens that there is a list with lots of matrices or data frames in it and we need to "stack those together". I thought it would be a simple

Dear all, I think that my question is very simple but I failed to solve it. I have a list which elements are matrices like this: >mylist [[1]] [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 [[2]] [,1] [,2] [,3] [1,] 7 9 11 [2,] 8 10 12 I'd like to create a matrix M<-mylist[[1]]+mylist[[2]] [,1] [,2] [,3] [1,] 8 12 16 [2,] 10 14 18

I am trying to use apply and a list to supply names to a set of tables I want to generate. Below is an example that I hope mimics the larger original problem. EXAMPLE aa <- c( 2,2,1,1,2) bb <- c(5,6,6,7,4) aan <- c("yes", "no") bbn <- c("a", "b", "c", "d") mynames <- c("abby", "billy") mylist <-

FYI, in R devel (to become 3.5.0), there's isFALSE() which will cut some corners compared to identical(): > microbenchmark::microbenchmark(identical(FALSE, FALSE), isFALSE(FALSE)) Unit: nanoseconds expr min lq mean median uq max neval identical(FALSE, FALSE) 984 1138 1694.13 1218.0 1337.5 13584 100 isFALSE(FALSE) 713 761 1133.53 809.5 871.5

Hello guys, I am working on a matrix which looks like this one: > initialMatrix <- > rbind(cbind(rep("A",3),seq(1,3)),cbind(rep("B",4),seq(1,4)),cbind(rep("C",3),seq(1,3))) > initialMatrix [,1] [,2] [1,] "A" "1" [2,] "A" "2" [3,] "A" "3" [4,] "B" "1" [5,]