similar to: extract index during execution of sapply

Dear all I want to generate a list like this: a <- data.frame(1:10) attr(a,'myattribute') <- 'something' b <- data.frame(11:20) attr(b,'myattribute') <- 'anything' mylist <- list(a,b) Is there a way to place the dataframes into the list giving them the attribute at the same time? I don't want to create all the dataframes in my workspace first.

Hi all! I would like to import a matlab or gauss code to R. Could you help me? Bye, Sebasti?n. 2007/6/23, r-help-request en stat.math.ethz.ch <r-help-request en stat.math.ethz.ch>: > Send R-help mailing list submissions to > r-help en stat.math.ethz.ch > > To subscribe or unsubscribe via the World Wide Web, visit >

Hello I am using R 2.2.0 with Windows XP. I've got a five element list object, each element containing two dataframes of equivalent size. > str(mylist) List of 1 $ data1:List of 2 ..$ data1a :`data.frame': 77 obs. of 63 variables: .. ..$ var1 : num [1:77] 0.41375 0.00056 1.43040 1.43528 0.61730 ... .. ..$ var2 : num [1:77] 1.154 1.686 0.673 0.800 0.760 ... ..

Dear all I imported a Stata .dta file with the read.dta-function from the foreign-package. The dataframe's dimensions are > dim(d.apc) [1] 15806 1300 Importing needs up to 15 min and calculations with these data are rather slow (although I subset the data before starting analyses). My questions are: 1. Has someone experiences importing Stata files (alternatives to read.dta) ? 2.

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 <-

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

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,

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 there I want to pass arguments (i.e. the response variable and the subset argument) in a self-made function to glm. Here is one way I can do this: f.myglm <- function(y,subfact,subval) { glm(d.mydata[,y]~d.mydata[,'x1'],family=binomial,subset=d.mydata[,subfact]==subval) } > str(d.mydata) `data.frame': 15806 obs. of 3 variables: $ y : Factor w/ 2 levels

Dear R users, Let's say I have a list with components being 'm' matrices (as exemplified in the "mylist" object below). Now, I'd like to subset this list based on an index vector, which will partition each matrix 'm' in 2 sub-matrices. My questions are: 1. Is there an elegant way to have the results shown in mylist2 for an arbitrary number of matrices in mylist?

Hi, I'm have a list of integer vectors and I want to perform an outer() like operation on the list. As an example, take the following list: mylist <- list(1:5,3:9,8:12) A simple example of the kind of thing I want to do is to find the sum of the shared numbers between each vector to give a result like: result <- array(c(15,12,0,12,42,17,0,17,50), dim=c(3,3)) Two for() loops is the

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 =

Hi R users I like to extract (or collect) a numeric element with a name from a list. Is there any way to extract just a numeric element without the name attached to the element. For example, >mylist Mantel-Haenszel chi-squared test with continuity correction data: table(mydata[, x]) Mantel-Haenszel X-squared = 8.3832, df = 1, p-value = 0.003787 alternative hypothesis: true

Hello I tried to define replacement functions for the class "mylist". When I test them in an active R session, they work -- however, when I put them into a package, they don't. Why and how to fix? make_my_list <- function( x, y ) { return(structure(list(x, y, class="mylist"))) } mylist <- make_my_list(1:4, letters[3:7]) mylist mylist[['x']] <- 4:6

Hello! # I have a list with several data frames: mylist<-list(data.frame(a=1:2,b=2:3), data.frame(a=3:4,b=5:6),data.frame(a=7:8,b=9:10)) (mylist) # I want to grab only one specific column from each list element neededcolumns<-c(1,2,0) # number of the column I need from each element of the list # Below, I am doing it using a loop: newlist<-NULL for(i in 1:length(mylist) ) {

I can define a list containing NULL elements: > myList <- list("aaa",NULL,TRUE) > names(myList) <- c("first","second","third") > myList $first [1] "aaa" $second NULL $third [1] TRUE > length(myList) [1] 3 However, if I assign NULL to any of the list element then such element is deleted from the list: > myList$second <-