similar to: count of factors

Full_Name: Laurent Gautier Version: 0.65.0 OS: Irix 6.5 Submission from: (NULL) (195.110.4.98) the following doesn't give what I expect > test _ factor(ORGMORE[[1]],exclude=c(NA,"NOM")) Warning message: NAs introduced by coercion > levels(test) [1] "CYT" "EXC" "MEM" "NOM" "NUC" "SEC" while this works... >

Dear all, I hope the following problem haven't been discussed here recently (I could not find anything about it the help archive files) : I tried to compile R-0.64.2 from the source files without much success... the last lines from the output of the ./configure command were : ---------------------------------------------------------- R is now configured for mips-sgi-irix6.4 Source

Hi ? This is giving me a headache. I?m trying to do a relatively simple optimization ? actually trying to approximate the output from the Excel Solver function but at roughly 1000x the speed. ? The optimization parameters look like this. The only trouble is that I want to add a constraint that sum(wgt.vect)=1, and I can?t figure out how to do that in optim. Mo.vect <-

Hello, I have a question regarding subsetting of vectors. Here's an example of what I'm trying to do: vect.1 <- c(76,195, 290, 380) vect.2 <- c(63, 95, 133, 170, 215, 253, 285, 299, 325, 375) I would like to subset vect.2 so that it has the same length as vect.1, and its numbers are the first corresponging higher value compared to vect.1. The output should be: final.output =

Folks; I am having a problem with the cv.glm and would appreciate someone shedding some light here. It seems obvious but I cannot get it. I did read the manual, but I could not get more insight. This is a database containing 3363 records and I am trying a cross-validation to understand the process. When using the cv.glm, code below, I get mean of perr1 of 0.2336 and SD of 0.000139. When using a

I'm trying to use Rprof() to identify bottlenecks and speed up a particullary slow section of code which reads in a portion of a tif file and compares each of the values to values of predictors used for model fitting. I've written up an example that anyone can run. Generally temp would be a section of a tif read into a data.frame and used later for other processing. The first portion

Hi Michael, A few comments 1. To add the constraint sum(wgt.vect=1) you would use the method of Lagrange multipliers. What this means is that in addition to the w_i (the components of the weight variables) you would add an additional variable, call it lambda. Then you would modify your optim.fun() function to add the term lambda * (sum(wgt.vect - 1) 2. Are you sure that you have defined

You can't -- at least as I read the docs for ?optim (but I'm pretty ignorant about this, so maybe there's a way to tweak it so you can). See here: https://cran.r-project.org/web/views/Optimization.html for other R optimization capabilities. Also, given your credentials, the r-sig-finance list might be a better place for you to post your query. Cheers, Bert Bert Gunter

Thanks Bert. But everyone on that forum wants to use finance tools rather than general optimization stuff! And I am not optimizing a traditional Markowitz mean-variance problem. Plus, smarter people here. :-) > On May 3, 2018, at 3:01 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > You can't -- at least as I read the docs for ?optim (but I'm pretty > ignorant

On Thu, May 3, 2018 at 2:03 PM, Michael Ashton <m.ashton at enduringinvestments.com> wrote: > Thanks Bert. But everyone on that forum wants to use finance tools rather than general optimization stuff! And I am not optimizing a traditional Markowitz mean-variance problem. Plus, smarter people here. :-) > I'm very confused by these statements. Most of the "finance tools"

Hi, I got caught out by this behaviour in 1.8.0 and I wondered why this happens: I have a list of vectors and was using lapply and grep to remove matched elements that occur in only a subset of the elements of the list: locs <- lapply(locs, function(x){x[- grep("^x", x)]}) The problem is that where the grep finds no matches and hence returns a vector of length 0, all the

Hi, I was trying to get error bars in my matplot. I looked at an earlier thread, and the sample code that I made is: #------------------ library(plotrix) mat1 <- matrix(sample(1:30,10),nrow=5,ncol=2) ses <- matrix(sample(1:3,10,replace=T),nrow=5,ncol=2) vect <- seq(20,100,20) rownames(mat1) <- rownames(ses) <- vect colnames(mat1) <- colnames(ses) <- letters[1:2]

Hi, I search a solution to record data in dynamic structures in R. I have an algorithm that will be executed each step and whose output is an array of doubles with unknown size. The solution I found is to use lists 1) I initialise my list l <- list() 2) and at a step numbered i I conacatain the new tab to the list vect <- algorithm() l <<--c( l , list(stepi=vect)) The problem is

Hi R-users, I have to define a noise level function L and its energy in the various moment of the day by: if time is between 18:00:00 and 23:59:59 then L[j] <- L[j]+5 and W <- 10^((L+5)/10) if time is between 22:00:00 and 05:59:59 ==> L <- L+10 and W <- 10^((L+10)/10) else L=L and W = W Could someone help me to realize this function please? You will find my following

If I have several character vectors, for example: a_c('data.') b_c('_myfunction(') c_c('vect.') d_c(')') and, j_1 I can build a vector using paste: x_paste(a,j,b,c,j,d,sep="") x="data.1_myfunction(vect.1)" I have n numeric vectors (vect.1...vect.n) Then, if I could evaluate the string x, I would calculate the result of my function in

Hi again, Beside R_ParseVector()'s possible inconsistent behavior, R's handling of zero-length named elements does not seem consistent either: ``` > lst <- list() > lst[[""]] <- 1 > names(lst) [1] "" > list("" = 1) Error: attempt to use zero-length variable name ``` Should the parser be made to accept as valid what is otherwise possible

Le lun. 9 d?c. 2019 ? 09:57, Tomas Kalibera <tomas.kalibera at gmail.com> a ?crit : > On 12/9/19 2:54 PM, Laurent Gautier wrote: > > > > Le lun. 9 d?c. 2019 ? 05:43, Tomas Kalibera <tomas.kalibera at gmail.com> a > ?crit : > >> On 12/7/19 10:32 PM, Laurent Gautier wrote: >> >> Thanks for the quick response Tomas. >> >> The same error

Hi, I'm looking for a function which returns the name of the argument object as a string: >vect <- 1:3 >function(vect) "vect" I've looked at 'name', 'names', 'objects', none seem to do exactly what I want. ls/objects comes close, but I can't figure out how to force it to give me only one object name. daver

Thanks for the quick response Tomas. The same error is indeed happening when trying to have a zero-length variable name in an environment. The surprising bit is then "why is this happening during parsing" (that is why are variables assigned to an environment) ? We are otherwise aware that the error is not occurring in the R console, but can be traced to a call to R_ParseVector() in

Dear all, I had to extracts/set elements from/in a matrix. Let say I have two vectors dim1 and dim2 of indices in the respective two dimensions of a matrix: I want to extract all the corresponding elements. I the case of a nxn matrix, dim1 <- 1:n and dim2 <- 1:n would extract the diagonal. I know one way would be to use the functions 'row' and 'col', but the matrixes I