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Hello, I have a problem with getting items from ListCtrl. Although my ListCtrl is in LC_REPORT mode, it has columns set up, every cell is set with set_item and everything displays fine, this code: x = myList.get_item(row,col).get_text() always returns an empty string. For example: myList.set_item(0,0,"Hello") x = myList.get_item(0,0).get_text() now x equals "" ! I

Hello, does anyone know of a quick way to create a data frame "like" another, but with more rows? What I'd like to do is this: if mydata is a data.frame like a b c 1 TRUE yes 2 FALSE no 3 TRUE yes I'd like to get mydata2 with the same column names and column types, but without the values and with more rows. All I could think of was to manually do

I have a set of character vectors of uneven length that I have stored in a list. I can easily enough get any column of them using lapply but what I want is to be able to create a matrix of them. Other than some kind of brute force looping approach I have drawn a blank. Would somebody please suggest something? Thanks Example. mylist <- list(aa=c("cat","peach" ),

Dear all, I have a question with regard to the use of hclust. I would like to be able to specify my own distance matrix instead of asking R to compute the distance matrix for me. It is computationally easier for me this way. My question is: How can I get hclust to accept this? Thanks, Ranjan -- *************************************************************************** Ranjan

Thanks! Yes, however, this seems a bit wasteful. Just wondering if there are other, more efficient options possible. Best wishes, Ranjan On Thu, 29 Mar 2018 22:20:19 -0400 Boris Steipe <boris.steipe at utoronto.ca> wrote: > If one is equal to the reverse of another, keep only one of the pair. > > B. > > > > > On Mar 29, 2018, at 9:48 PM, Ranjan Maitra

Dear friends, Is there a way to read data files written in lisp into R? Here is the file: https://archive.ics.uci.edu/ml/machine-learning-databases/university/university.data I would like to read it into R. Any suggestions? Thanks very much in advance for pointers on this and best wishes, Ranjan -- Important Notice: This mailbox is ignored: e-mails are set to be deleted on receipt. Please

Dear list, I have been using the Rmath library for quite a while: in the current instance, I am calling dnt (non-central t density function) repeatedly for several million. When the argument is small, I get the warning message: full precision was not achieved in 'pnt' which is nothing unexpected. (The density calls pnt, if you look at the function dnt.) However, to have this happen a

Jeff, I wanted to let you know that your function is faster than generating the directional circular permutations and weeding. Here is the time for n = 10. I compared with just doing the permutations, there is no point in proceeding further with the weeding since it is slower at the start itself. system.time(directionless_circular_permutations(10)) user system elapsed 1.576 0.000

I don't know if this is more efficient than enumerating with distinct directions and weeding... it seems kind of heavyweight to me: ####### library(gtools) directionless_circular_permutations <- function( n ) { v <- seq.int( n-1 ) ix <- combinations( n-1, 2 ) jx <- permutations( n-3, n-3 ) x <- lapply( seq.int( nrow( ix ) ) , function( i ) {

Dear friends, Sorry for this somewhat generically titled posting but I had a question with using contrasts in a manova context. So here is my question: Suppose I am interested in doing inference on \beta in the case of the model given by: Y = X %*% \beta + e where Y is a n x p matrix of observations, X is a n x m design matrix, \beta is m x p matrix of parameters, and e is a

https://bugzilla.mindrot.org/show_bug.cgi?id=3202 Bug ID: 3202 Summary: Ed25519 key on HSM is not getting listed in ssh-add -l command Product: Portable OpenSSH Version: 8.2p1 Hardware: ARM64 OS: Linux Status: NEW Severity: enhancement Priority: P5 Component: ssh-add

Dear friends, I would like to get all possible arrangements of n objects listed 1:n on a circle. Now this is easy to do in R. Keep the last spot fixed at n and fill in the rest using permuations(n-1, n-1) from the gtools package. However, what if clockwise or counterclockwise arrangements are the same? I know that half of the above (n - 1)! arrangements are redundant. Is there an easy way to

1) Definitely yes. They are on CRAN. Just type install.packages( c("dplyr", "tidyr" ) ) at the R console. 2) Don't know, but most likely the answer is yes. Since all R packages on Linux are compiled by R when installed, you either need to activate your virtual machine, compile the packages, and save the state for future use, or you have to build the package library on

Hello all, I''m just getting started with rails. I''m trying to install it on a CentOS 4.2 (i.e. Red Hat Enterprise 4.2) x86_64 system. The OS install is brand new. Ruby version is: # ruby --version ruby 1.8.1 (2003-12-25) [x86_64-linux-gnu] Gem: # gem --version 0.8.11 When I attempt to install rails, I get this: # gem install rails --include-dependencies Attempting

Hello I tried to define replacement functions for the class "mylist". When I test them in an active R session, they work -- however, when I put them into a package, they don't. Why and how to fix? make_my_list <- function( x, y ) { return(structure(list(x, y, class="mylist"))) } mylist <- make_my_list(1:4, letters[3:7]) mylist mylist[['x']] <- 4:6

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]

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 <-

Dear colleagues, I have an image which I can display in the greyscale using image. On this image, for some pixels, which I know, I want to display their activity based on a third measure. One way to do that would be to color these differently, and use an opacity measure to display the third measure. An example of what I am trying to do is at:

Hi r-users, I have a matrix B and a list of 3x3 matrices (mylist). I want to calculate the quantiles in the list using each of the value of B as probabilities. The codes I wrote are: B <- matrix (runif(12, 0, 1), 3, 4) mylist <- lapply(mylist, function(x) {matrix (rnorm(9), 3, 3)}) for (i in 1:length(B)) { quant <- lapply (mylist, quantile, probs=B[i]) } But quant

Dear all, I think that my question is very simple but I failed to solve it. I have a list which elements are matrices like this: >mylist [[1]] [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 [[2]] [,1] [,2] [,3] [1,] 7 9 11 [2,] 8 10 12 I'd like to create a matrix M<-mylist[[1]]+mylist[[2]] [,1] [,2] [,3] [1,] 8 12 16 [2,] 10 14 18