similar to: Intercepts in linear models.

I have factors with levels ``Unit", "Achieved", and "Scholarship"; I wish to replace these with "U", "A", and "S". So I do fff <- factor(fff,labels=c("U","A","S")) This works as long as all of the levels are actually present in the factor. But if ``Scholarship'' is absent (as if often is) then

I want to do something like: TH <- sprintf("%1.1f",c(0.3,0.5,0.7,0.9,1)) plot(1:10) legend("bottomright",pch=1:5,legend=parse(text=paste("theta ==",TH))) Notice that the final "1" comes out in the legend as just plain "1" and NOT as "1.0" although TH is [1] "0.3" "0.5" "0.7" "0.9"

Hi Rolf, this topic is probably already saturated, but here is a tidyverse solution: ``` library(purrr) x <- list( ? `1` = c(7, 13, 1, 4, 10), ? `2` = c(2, 5,? 14, 8, 11), ? `3` = c(6, 9, 15, 12, 3) ) x |> ? pmap(~ c(..1, ..2, ..3)) |> ? reduce(c) #> [1]? 7? 2? 6 13? 5? 9? 1 14 15? 4? 8 12 10 11? 3 ``` Here, we map over the elements of the list in parallel (hence pmap),

Dear R users, I want a help to write an algorithm for swapping rows and columns in a matrix thanks in advance [[alternative HTML version deleted]]

For reasons best known only to myself ( :-) ) I wish to create a data frame with 0 rows and 9 columns. The best I've been able to come up with is: junk <- as.data.frame(matrix(0,nrow=0,ncol=9)) Is there a sexier way? cheers, Rolf ###################################################################### Attention:\ This e-mail message is privileged and confid...{{dropped:9}}

Sorry to append, but I just realised that of course ``` x |> ? pmap(c) |> ? reduce(c) |> ? unname() ``` also works and is a general solution in case your list has more than three elements. Here, we map in parallel over all elements of the list, always combining the current set of elements into a vector, and then reduce the resulting list into a vector by combining the elements

I am interested in whether the slopes in a linear model are different from 0. I.e. I would like to obtain the slope estimates, and their standard errors, ``relative to 0'' for each group, rather than relative to some baseline. Explicitly I would like to write/represent the model as y = a_i + b_i*x + E i = 1, ..., K, where x is a continuous variate and i indexes groups (levels of a

I have (toy example): x <- list(`1` = c(7, 13, 1, 4, 10), `2` = c(2, 5, 14, 8, 11), `3` = c(6, 9, 15, 12, 3)) and f <- factor(rep(1:3,5)) I want to create a vector v of length 15 such that the entries of v, corresponding to level l of f are the entries of x[[l]]. I.e. I want v to equal c(7, 2, 6, 13, 5, 9, 1, 14, 15, 4, 8, 12, 10, 11, 3) I can create v

I see a book coming: "666 ways to do the same thing in R ranked by sexiness." Kidding aside, if you look under the covers of some of the functions we are using, we may find we are taking steps back as some of them use others and perhaps more functionality than we need. But for a new reader , looking at many approaches may open up other ways and ideas and see the problem space as quite

Avi I fear this was all a huge social experiment. Testing if a post titled "sexy way" would increase engagement... On Sat, 28 Sep 2024, 07:21 , <avi.e.gross at gmail.com> wrote: > I see a book coming: > "666 ways to do the same thing in R ranked by sexiness." > > Kidding aside, if you look under the covers of some of the functions we > are

Admit it, Rolf. Haven't you wondered if S, in a more private way, is sexier than R? OK, kidding aside, we have talked this to death. Just FYI, the conversation was stimulating for some of us and I have continued on my own and located functions I see as useful in the stringi and stringr packages to make my silly version ever less silly! LOL! -----Original Message----- From: Rolf Turner

I wish to pass a vector ``y'', some of whose entries are NAs to a fortran subroutine which I am dynamically loading and calling by means of .Fortran(). The subroutine runs through the vector entry by entry; obviously I want to have it do one thing if y[i] is present and a different thing if it is missing. The way I am thinking of proceeding is along the xlines of: ymiss <- is.na(y)

Dimitris Rizopoulos wrote: > look at ?rowMeans; you can also use "apply(mat, 1, mean)" but > rowMeans() is better. By my reading of the question, this is not what Ezhil wants. He said: ``I have a 992 x 74 matrix. I would like to form a new matrix by averaging each 4 rows from the original one.'' I.e. he wants (I think) the first row of the new matrix to be the

A colleague of mine is trying to use nls() to effect an optimization, and is encountering a scoping problem. I should know how to solve it for him but .... well, I just don't. I also had a quick scrounge of the archives --- I know I've seen this topic addressed before --- but I couldn't track it down. So here's a toy example that demonstrates the problem: hhh <-

>>>>> Chris Evans via R-help >>>>> on Fri, 27 Sep 2024 12:20:47 +0200 writes: > Oh glorious!? Thanks Duncan. > Fortune cookie nomination! I don't disagree with the nomination -- thank you, Duncan! However, please note that I'm sure Rolf's was challenged / question was ment to work correctly for all factors `f` with levels

Is there a way of calculating the derivative of a function returned by splinefun()? Such a function is a cubic spline, whence it has a calculable derivative, but is there a (simple) way of getting at it? One workaround that I have thought of is to take a fine grid of points, evaluate the function returned by splinefun() at these points, put an interpolating spline through these points using

Suppose that I have a function fff which calls a couple of functions foo and bar, and that foo takes optional arguments a and b, and bar takes optional arguments u and v. Is there an elegant way of passing ***both*** sets of optional arguments via a ``...'' argument for fff? I'd like to be able to have a structure like: fff(x,y,z,...) { . a1 <- foo(x,y,...) b1

I'm experiencing a new and annoying phenomenon which seems to consist of an unfortunate interaction between R-1.7.0 and netscape version 7. When I invoke help.start(), a netscape window duly appears with the browser pointed at the file .../R/doc/html/index.html as one would hope and expect. However if I then ask for help on a function, e.g. > help(glm) the help does NOT get displayed

I would like to centre titles for pairs of plots in a 3-x-2 array. Each row of the array corresponds to a calendar year and I would like to have the year value centred between the two plots in the row, and just above their upper edges. I have attached an example in "demo.pdf" showing roughly what I want. I managed to produce the example using

I had such good luck with my previous question to r-help, (a few minutes ago) that I thought I would try again with the following query: > Suppose I have > > xNm <- "gamma" > > I would like to be able to do > > plot(1:10,xlab = <something involving xNm">) > > and get the x axis label to be the Greek letter gamma > (rather than the