search for: sappli

I am discovering sapply! :-) Could you please help me with a very elementary question? Here is what I know. The following two programs generate the same answer. --------------------------------+---------------------------------------- Loops version | sapply version --------------------------------+----------------------------------------

Hello everyone, I am attempting to write a function to count the number of non-missing values of each column in a data frame using the sapply function. I have the following code which is receiving the error message below. > n.valid<-sapply(data1,sum(!is.na)) Error in !is.na : invalid argument type Ultimately, I would like for this to be 1 conponent in a larger function that will produce

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

Hi, I bet this one has be asked before, but doing sapply(x, FUN=as.character) where 'x' is a vector, then the result "should [] be simplified to a vector" according to ?sapply, correct? However, > x <- 1:10 > sapply(x, FUN=as.character) [1] "1" "2" "3" "4" "5" "6" "7" "8"

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

Hi, I use sapply very frequently, but I have recently noticed a behavior of sapply which I don't understand and have never seen before. Basically, sapply returns what looks like a matrix, says it a matrix, and appears to let me do matrix things (like transpose). But it is also a list and behaves like a list when I subset it, not a vector (so I can't sort a row for instance). I

I am grateful to Andy Liaw, Douglas Grove, Brian Ripley, Tony Plate, Dirk Eddelbuettel and Sundar Dorai-Raj all of whom got together and drilled sense into my skull. I would like to take some effort into explaining what the question was, that I was grappling with, and the (nice) R way of solving the question. My apologies: I am still a victim of too many years of writing C, so I'm a bit dense

Hi, I have this code here and try to use sapply code.  But I got error message that I don't really understand to correct. bt   <- c(24.96874, 19.67861, 23.51001, 19.86868); round(bt,2) alp  <- c(2.724234, 3.914649, 3.229146, 3.120719); round(alp,2) bt_alp <- data.frame(bt,alp) sapply(bt_alp, function(bt,alp) ((bt_m/bt)^alp), bt_m = min(bt)) > sapply(bt_alp, function(bt,alp)

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

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 <-

Dear All, Suppose I have a program as below: Outside is a loop for simulation (with random generated data), inside there are several sapply()'s (10~100) over the data and something else, but these sapply's have to be sequential. And each sapply do not involve very intensive calculation (a few seconds only). So the outside loop takes minutes to finish one iteration. I guess the better way

Hi. I'm having difficulty following the advice given in help(Recall). Consider the two following toy functions: f1 <- function(n){ if(length(n)>1){return(sapply(n,f1))} matrix(n,n,n) } f2 <- function(n){ if(length(n)>1){return(sapply(n,Recall))} matrix(n,n,n) } f1() works as desired (that is, f(1:3), say, gives me a three element list whose i-th element is an

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,

Can't seem to get the code below working. It gets stuck on line 24 inside the function hm; comments show the line in question. The function hm is called by sapply and is at the bottom of the code. Other stuff above line 24 works correctly including the first couple of lines of the function hm. Should I be using a different apply function or am I doing something wrong with xmlTreeParse ?

I am puzzled by the scope rules that apply with sapply. If I want to modify a vector with sapply I tried... N <- 10 vec <- vector(mode="numeric", length=N) test <- function(i){ vec[i] <- i } sapply(1:N, test) vec but it not work. How can this be done? Worik [[alternative HTML version deleted]]

Hello everyone, I have the following call to sapply() and error message. Is the most efficient way to deal with this to make sum(!is.na(x)) a function in a separate line prior to this call? If not, please advise. N.Valid=sapply(x,sum(!is.na(x))) Error in match.fun(FUN) : 'sum(!is.na(x))' is not a function, character or symbol Thanks! Dan [[alternative HTML version deleted]]

sapply() stems from S / S+ times and hence has a long tradition. In spite of that I think that it should be enhanced... As the subject mentions, sapply() produces a matrix in cases where the list components of the lapply(.) results are of the same length (and ...). However, it unfortunately "stops there". E.g., if you *nest* two sapply() calls where the inner one produces a matrix, very

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,

Dear All, applying some function within a with() function I wanted to use also sapply() and get() to form a data.frame, but did not succede. Below is a simplified example. It is possible to use sapply() within a with() function, it is also possible to use get() within a with() function, but when I try to use get within sapply within with I arrive at "Error in get(x, envir, mode, inherits) :

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