similar to: can a key of a list be a variable

Hi, > > I am a java programmer, an I want to use R from java. > could so help me > thanks for telling me whether I 'm using the good mailing list. > > thanks!! > > > -- Bilel [[alternative HTML version deleted]]

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"

Hello, I was used to count how many times every factor is found the following way : (assuming 'vect' is a vector of factors) tapply(vect,vect,length) but recently I experienced weird results with this In fact, my command line was tapply(ORGMORE[[1]][vect],ORGMORE[[1]][vect],length) where - 'vect' is a vector of indices - 'ORGMORE[[1]]' is a slightly long vector of

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 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, 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

Dear Ruser, Below, the deal (you can copy paste): r=3 ; set.seed(1) v <- matrix(c(rnorm(40)),10,4) for (j in 1:4){ for (i in 1:r){ x <- t(v[,j]^(i)*v[,1:4]^((r-(4-1)):r)) print(x) } } How to reach to " x " " row bind " inside or outside (preferred) of the loop. To obtain at the end something like: > x matrix( , 12*4, 10) Any help or

Hi, I've got a dataset with 7 variables for 8 different species. I'd like to test the null hypothesis of no difference among species for these variables. MANOVA seems like the appropriate test, but since I'm unsure of how well the data fit the assumptions of equal variance/covariance and multivariate normality, I want to use a permutation test. I've been through CRAN looking at

Hi guys, I want to create variable on the fly: for example for (i in 1:10) { cat(paste("VAR",i,sep="")) } Will print VAR1, VAR2 etc up to VAR10. However I want to make these into variables, and then give them a value, for example: vect = c(10:20) for (i in 1:10) { cat(paste("VAR",i,sep="")) = vect[i] } THis doesnt work but I hope it demonstrates

Hello, I'd like to take all possible sub-summands of a vector in the quickest and most efficient way possible. By "sub-summands" I mean for each sub-vector, take its sum. Which is to say: if I had the vector x<-1:4 I'd want the "sum" of x[1], x[2], etc. And then the sum of x[1:2], x[2:3], etc. And then...so on. The result would be: 1 2 3 4 2 5 7 6 9 10 I can

dear all, consider a simple function like func = function(){ return(runif()) } and say that you would like to create a vector of size N, each entry being a call to this function "func". What is the equivalent of the python code vect = [ func() for i in range(N)] Indeed, you cannot just call vect<-rep( func(), N) - I guess that there also exists a one-liner to do that