similar to: grep

Yes. Any of the following worked. The pipe greater than (|>) is neat! Thanks. > v<-goprobit.p$est > names(v) |> grep("somewhat|very", x = _) ?[1]? 6? 7? 8? 9 10 11 12 13 28 29 30 31 32 33 34 35 50 51 52 53 54 55 56 57 > v |> names() |> grep("somewhat|very", x = _) ?[1]? 6? 7? 8? 9 10 11 12 13 28 29 30 31 32 33 34 35 50 51 52 53 54 55 56 57 >

Now I've found another way to make it work. All I need is to pick up the names in the column (x.1.age...). > v<-pr(goprobit.p); v Maximum-Likelihood Estimates weighted = FALSE iterations = 5 logLik = -14160.75 finalHessian = TRUE Covariance matrix is Robust Number of parameters = 66 Sample size = 17922 est se t p g sig x.1.age 0.0341 0.0138 2.4766 0.0133 -3.8835e-04 ** x.1.sleep

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

Could not get "which" to work, but my grep worked. Thanks. > which(grep("very|somewhat",names(goprobit.p$est))) Error in which(grep("very|somewhat", names(goprobit.p$est))) : argument to 'which' is not logical > grep("very|somewhat",names(goprobit.p$est)) [1] 6 7 8 9 10 11 12 13 28 29 30 31 32 33 34 35 50 51 52 53 54 55 56 57 On 7/12/2024

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

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

OK. The fact it's in a function is making things clearer. Are you trying to update the values of an object from within the function, and have them available outside the function. I don't speak functional programming articulately enough but basically v <- 1 funA <- function() { v <- v+1 } funA() cat (v) # 1 You either return the v from the function so funB <- function() {

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

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

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

? Fri, 4 Oct 2024 20:28:01 +0800 Steven Yen <styen at ntu.edu.tw> ?????: > Suppose I have two vectors, x and y. Is there a way > to do the covariance matrix with ?apply?. There is no covariance matrix for just two samples (vectors) 'x' and 'y'. You can only get one covariance value for these. If you had a pair of vectors of _random variates_, the situation would be

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]

"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 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,]

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:

?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

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,