similar to: Stacking matrix columns

You could also do dim(x) <- c(length(x), 1) On Sat, Aug 5, 2023, 20:12 Steven Yen <styen at ntu.edu.tw> wrote: > I wish to stack columns of a matrix into one column. The following > matrix command does it. Any other ways? Thanks. > > > x<-matrix(1:20,5,4) > > x > [,1] [,2] [,3] [,4] > [1,] 1 6 11 16 > [2,] 2 7 12 17 > [3,]

Stacking columns of a matrix is a standard operation in multilinear algebra, usually written as the operator vec(). I checked to see if there is an R package that deals with multilinear algebra. I found rTensor, which has a function vec(). So, yet another way to accomplish what you want would be: > library(rTensor) > vec(as.tensor(x)) Eric On Sun, Aug 6, 2023 at 5:05?AM Bert Gunter

Eric, I am not sure your solution is particularly economical albeit it works for arbitrary arrays of any dimension, presumably. But it seems to involve converting a matrix to a tensor just to undo it back to a vector. Other solutions offered here, simply manipulate the dim attribute of the data structure. Of course, the OP may have uses in mind which the package might make easier. We often get

Or just dim(x) <- NULL. (as matrices in base R are just vectors with a dim attribute stored in column major order) ergo: > x [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 > x<- 1:20 ## a vector > is.matrix(x) [1] FALSE > dim(x) <- c(5,4) > is.matrix(x) [1] TRUE > attributes(x) $dim [1] 5 4 > ## in painful and unnecessary detail as dim() should

Avi, I was not trying to provide the most economical solution. I was trying to anticipate that people (either the OP or others searching for how to stack columns of a matrix) might be motivated by calculations in multilinear algebra, in which case they might be interested in the rTensor package. On Sun, Aug 6, 2023 at 6:16?PM <avi.e.gross at gmail.com> wrote: > > Eric, > > I

I wish to stack columns of a matrix into one column. The following matrix command does it. Any other ways? Thanks. > x<-matrix(1:20,5,4) > x ???? [,1] [,2] [,3] [,4] [1,]??? 1??? 6?? 11?? 16 [2,]??? 2??? 7?? 12?? 17 [3,]??? 3??? 8?? 13?? 18 [4,]??? 4??? 9?? 14?? 19 [5,]??? 5?? 10?? 15?? 20 > matrix(x,ncol=1) ????? [,1] ?[1,]??? 1 ?[2,]??? 2 ?[3,]??? 3 ?[4,]??? 4 ?[5,]??? 5

I updated to latest RStudio (RStudio-2023.03.0-386.exe) but R would not run. Error message: Error Starting R The R session failed to start. RSTUDIO VERSION RStudio 2023.03.0+386 "Cherry Blossom " (3c53477a, 2023-03-09) for Windows [No error available] I also tried RStudio 2022.12.0+353 --- same problem. I then tried another older version of RStudio (not sure version as I changed file

You could declare a matrix much larger than you intend to use. This works with a few megabytes of data. It is not very efficient, so scaling up may become a problem. m22 <- matrix(NA, 1:600000, ncol=6) It does not work to add a new column to the matrix, as in you get an error if you try m22[ , 7] but convert to data frame and add a column m23 <- data.frame(m22) m23$x7 <- 12 The only

"Or use <<- assignment I think. (I usually return, but return can only return one object and I think you want two or more" You can return any number of objects by putting them in a list and returning the list. Use of "<<-" is rarely a good idea in R. -- Bert On Fri, Aug 9, 2024 at 1:53?AM CALUM POLWART <polc1410 at gmail.com> wrote: > > OK. The fact

x["V2"] is more efficient than using drop=FALSE, and perfectly normal syntax (data frames are lists of columns). I would ignore the naysayers, or put a comment in if you want to accelerate their uptake. As I understand it, one of the main reasons tibbles exist is because of drop=TRUE. List-slice (single-dimension) indexing works equally well with both standard and tibble types of data

I just like the subroutine to spit out results (Mean, Std.dev, etc.) and also be able to access the results for further processing, i.e., v$Mean v$Std.dev On 3/26/2024 11:24 AM, Richard O'Keefe wrote: > Not clear what you mean by "saved". > If you call a function and the result is printed, the result is > remembered for a wee while in > the variable .Last.value, so

How can I have both printout and saved results at the same time. The subroutine first return "out" and the printout gets printed, but not saved. I then run the "invisible" line. Results got saved and accessible but no printout. How can I have both printout and also have the results saved? Thank you! > dstat4 <- function(data,digits=3){ +?? Mean??? <-

Thanks to all. Very helpful. Steven from iPhone > On Jan 14, 2023, at 3:08 PM, Andrew Simmons <akwsimmo at gmail.com> wrote: > > ?You'll want to use grep() or grepl(). By default, grep() uses extended > regular expressions to find matches, but you can also use perl regular > expressions and globbing (after converting to a regular expression). > For example: >

I need to read csv files repeatedly, named data1.csv, data2.csv,? data24.csv, 24 altogether. That is, data<-read.csv(?data1.csv?) ? data<-read.csv(?data24.csv?) ? Is there a way to do this in a loop? Thank you. Steven from iPhone [[alternative HTML version deleted]]

Your desire is not unusual among novices... but it is really not a good idea for your function to be making those decisions. Look at how R does things: The lm function prints nothing... it returns an object containing the result of a linear regression. If you happen to call it directly from the R command prompt and don't assign it to a variable, then the command interpreter notices that

Thanks. Will try it. Have not tried it but I think the following may work: out$results<-NULL out$results$ei<-ap out$results$vi<-vap All I need is printing by returning out (unless I turn it off). And, retrieve ap and vap as needed as shown above. Guess I need to read more about invisible. On 8/11/2024 10:09 PM, Rui Barradas wrote: > ?s 09:51 de 11/08/2024, Steven Yen escreveu:

I have a data frame containing variables "yr3",...,"yr28". How do I remove them with a wild card----something similar to "del yr*" in Windows/doc? Thank you. > colnames(mydata) ? [1] "year"?????? "weight"???? "confeduc"?? "confothr" "college" ? [6] ... ?[41] "yr3"??????? "yr4"???????

x <- numeric(0) for (...) { x[length(x)+1] <- ... } works. You can build a matrix by building a vector one element at a time this way, and then reshaping it at the end. That only works if you don't need it to be a matrix at all times. Another approach is to build a list of rows. It's not a matrix, but a list of rows can be a *ragged* matrix with rows of varying length. On Wed,

Just FYI, the R interpreter typically saves the last value returned briefly in a variable called .Last.value that can be accessed before you do anything else. > sin(.5) [1] 0.4794255 > temp <- .Last.value > print(temp) [1] 0.4794255 > sin(.666) [1] 0.6178457 > .Last.value [1] 0.6178457 > temp [1] 0.4794255 > invisible(sin(0.2)) > .Last.value [1] 0.1986693 So perhaps if

I have trouble documenting an R package. In my .Rd file (sixth line below), I have uhat<-m%*%y but when the package is built (successfully), the matrix multiplication part does not show up in the documentation. The line become (missing %*% y) uhat<-m === \examples{ x<-c(1,2,3,4,5) y<-c(1,1,2,2,4) x<-cbind(1,x) m<-mmat(x) uhat<-m%*%y dstat(uhat) } ?-- styen at ntu.edu.tw