similar to: Documenting R package with Rd file

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

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

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,]

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: >

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

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,

The matrix equivalent of x <- ... v <- ... x[length(x)+1] <- v is m <- ... r <- ... m <- rbind(m, r) or m <- ... k <- ... m <- cbind(m, c) A vector or matrix so constructed never has "holes" in it. It's better to think of CONSTRUCTING vectors and matrices rather than INITIALISING them, because always being fully defined is important. It

Hmm. You raised an interesting point. Actually I am not having problems with aod per se?-it is just a supporting package I need while using old R. The essential package I need, maxLik, simply works better under R-3.0.3, for reason I do not understand?specifically the numerical gradients of the likelihood function are not evaluated as accurately in newer versions of R in my experience, which is why

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

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

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

mydata[, -grep("^yr",colnames(mydata))] On Sat, Jan 14, 2023 at 8:57 AM Steven T. Yen <styen at ntu.edu.tw> wrote: > 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]

Hi Steven Which optimisation algorithms in maxLik work better under R-3.0.3 than under the current version of R? /Arne On Thu, 8 Oct 2020 at 21:05, Steven Yen <styen at ntu.edu.tw> wrote: > > Hmm. You raised an interesting point. Actually I am not having problems with aod per se?-it is just a supporting package I need while using old R. The essential package I need, maxLik, simply

"It would be really really helpful to have a clearer idea of what you are trying to do." Amen! But in R, "constructing" objects by extending them piece by piece is generally very inefficient (e.g. https://r-craft.org/growing-objects-and-loop-memory-pre-allocation/), although sometimes?/often? unavoidable (hence the relevance of your comment above). R generally prefers to take

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: grepl("^yr", colnames(mydata)) will tell you which 'colnames' start with "yr". If you'd rather you use globbing:

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

?load Read this carefully. Pay attention to its instructions re: overwriting existing objects. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Jan 16, 2018 at 12:43 AM, Steven Yen <styen at ntu.edu.tw> wrote: >

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

Steven, Just want to add a few things to what people wrote. In base R, the methods mentioned will let you make a copy of your original DF that is missing the items you are selecting that match your pattern. That is fine. For some purposes, you want to keep the original data.frame and remove a column within it. You can do that in several ways but the simplest is something where you sat the

You rang sir? library(tidyverse) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat1 <- data.frame(xx , yr1, yr2, y3) dat1 %>% select(!starts_with("yr")) or for something a bit more exotic as I have been trying to learn a bit about the "data.table package library(data.table) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat2 <- data.table(xx , yr1, yr2, yr3) dat2[, !names(dat2)