similar to: Define replacement functions

No. I fixed that, the NAMESPACE file now contains: S3method("[[<-", mylist) S3method("$<-", mylist) It still does not work. I also created a print method (print.mylist) which did work out of the box, regardless of being in the NAMESPACE file or not. Could it be somehow in here (also in my NAMESPACE file): exportPattern("^[[:alpha:]]+") Or could it be that

I have created a function to do something: i <- factor(sample(c("A", "B", "C", NA), 793, rep=T, prob=c(8, 7, 5, 1))) k <- factor(sample(c("X", "Y", "Z", NA), 793, rep=T, prob=c(12, 7, 9, 1))) mytable <- function(x){ xtb <- x btx <- x # do more with x, not relevant here cat("The table has been created,

On Tue, 14 May 2019 at 12:31, Pavel Krivitsky <pavel at uow.edu.au> wrote: > > > Note that disabling name-based dispatch implies two things: 1) the > > inability to override your method by defining gen.formula in the > > global environment, and 2) another package can break yours (i.e., > > internal calls to gen()) by registering an S3 method for gen() after >

### example:start v <- sample(rnorm(200), 100, replace=T) k <- rep.int(c("locA", "locB", "locC", "locD"), 25) tapply(v, k, summary) ### example:end ... (hopefully) produces 4 summaries of v according to k group membership. How can I transform the output into a nice table with the croups as columns and the interesting statistics as lines? Thx,

CCing r-devel. On Tue, 14 May 2019 at 02:11, Pavel Krivitsky <pavel at uow.edu.au> wrote: > > Dear All, > > I've run into this while updating a package with unfortunately named > legacy functions. It seems like something that might be worth changing > in R, and I want to get a sense of whether this is a problem before > submitting a report to the Bugzilla. > >

Dear All, In R 3.6.3 (and earlier), method dispatch used to work for methods stored in local environments that are attached to the search path. For example: myfun <- function(y) { out <- list(y=y) class(out) <- "myclass" return(out) } print.myclass <- function(x, ...) print(formatC(x$y, format="f", digits=5)) myfun(1:4) # prints: [1]

tl;dr: S3 lookup no longer works in custom non-namespace environments as of R 3.6.1. Is this a bug? I am implementing S3 dispatch for generic methods in environments that are not packages. I am trying to emulate the R package namespace mechanism by having a ?namespace? environment that defines generics and methods, but only exposes the generics themselves, not the methods. To make S3 lookup work

A Likert scale may have produced counts of answers per category. According to theory I may expect equality over the categories. A statistical test shall reveal the actual equality in my sample. When applying a chi square test with increasing number of repetitions (simulate.p.value) over a fixed sample, the p-value decreases dramatically (looks as if converge to zero). (1) Why? (2) (If

On Sat, 18 May 2019 at 23:34, Pavel Krivitsky <pavel at uow.edu.au> wrote: > > > The issue here is that you are registering a non-standard name > > (.gen.formula) for that generic and then defining what would be the > > standard name (gen.formula) for... what purpose? IMHO, this is a bad > > practice and should be avoided. > > The situation initially arose

I posted on this topic to r-help, but never got a sufficient answer, so I'm reposting here. [Env: R 2.11.1, Win Xp, using Eclipse/StatET] In a package I'm working on, I want to create as.matrix() and as.array() methods for a particular kind of object (log odds ratios). These are returned in a loddsratio object as the $coefficients component, a vector, but really reflect an underlying

A list contains several matrices. Over all matrices (list elements) I'd like to access one matrix cell: m <- matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3])) l <- list(m1=m, m2=m*2, m3=m*3) l[[3]] # works l[[3]][1:2, ] # works l[[1:3]][1, 1] # does not work How can I slice all C-c combinations in the list? S?ren -- S?ren Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag,

Hi, y is nominal (3 categories), x1 to 3 is scale. What I want is a regression, showing the probability to fall in one of the three categories of y according to the x. How can I perform such a regression in R? Thanks for your help S?ren

Hello, I read the help as well as the examples, but I can not figure out why the following code does not produce the *given* row names, "x" and "y": x <- 1:20 y <- 21:40 rbind( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) Could you please help? Thank you S?ren

Hello, how can I return the name of a variable, say "a$b", from a function? fun <- function(x){ return(substitute(x)); } a <- data.frame(b=1:10); fun(a$b) ... returns a$b, but this is a type language, thus I can't use it as a character string, can I? How? Thanks for help, S?ren

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 <-

Hi, I'm pretty new to R. I have an object (say a list) and I I have a function that I call on various columns in that list (excuse terminology if it's wrong/ambiguous). Imagine its like this (actual values are unimportant) and called mylist: >mylist A B 1 5 2 5 3 6 4 8 5 0 I have a function: foo = function(param){ #modify list A or B values depending on

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

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?

Hello, after the creation of a data.frame I like to add NAs as follows: n <- 743; x <- runif(n, 1, 7); Y <- runif(n, 1, 7); Ag6 <- runif(n, 1, 7); df <- data.frame(x, Y, Ag6); # a list with positions: v <- apply(df, 2, function(x) sample(n, sample(1:ceiling(5*n/100), 1), repl=F)); # a loop too much? for (i in 1:length(df)){ df[unlist(v[i]), i] <- NA; } summary(df); This

Hi I have a list : mylist <- list( a = NULL, b = 1, c = 2 ) > mylist[1] $a NULL > is.null(mylist[1]) [1] FALSE > is.null(mylist$a) [1] TRUE why? I need to use mylist[1]