similar to: A problem about outer()

Dear R-help, I am trying to write a function to simulate datasets of size n which contain two time-to-event outcome variables with associated 'Event'/'Censored' indicator variables (flag1 and flag2 respectively). One of these indicator variables needs to be dependent on the other, so I am creating the first and trying to use this to create the second using an if/else statement.

I have the following setup: DS3 -> Cisco AS5400 -> H.323 (ooh323) -> Asterisk Inbound calls work great but outbound calls fail. So to check and make sure we have outbound calling ability on the DS3 we setup a Cisco Call Manager Express and it can make outbound calls both local and long distance with no problems. The failure code is Cause i = 0x8381 - Unallocated/unassigned number. We

http://cgit.freedesktop.org/~darktama/nouveau/commit/?id=4d60422 also broken with this commit

Dear Wiz[R]ds, I am deeply grateful for the help from Duncan Murdoch, Gray Calhoun, and others. We are almost there. For whatever reason, I can't change the symbol from a circle to a triangle in the upright posture plots. Any ideas? I have included the problem in full. # tritiated (3H)-Norepinephrine(NE) disappearance from plasma # concentrations supine and upright # supine datasu <-

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 =

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

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

Hi there, I have a numeric vector let say: Vect <- c(12.234, 234.5675, 1.5) Now I want a string vector like: changedVec <- c("012.234", "234.568", "001.500") Would be grateful if somebody help me how can I do that. Thanks and regards, [[alternative HTML version deleted]]

Hello I'm a population ecologist and use R for all my stats and modelling. Recently I have been using R to numerically solve integral projection models. This involves constructing several large matrices. The current code by Easterling (Size-specific sensitivity: Applying a new structured population model. Ecology, 2000, 81, 694-708) uses nested loops to construct the matrices. To speed up the

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 All, I want to draw samples (n=4) from one of 2 triangular distributions for each value in a matrix. I am using an ifelse statement to try to define which distribution to draw from. >From the output, I can see that the ifelse statement is choosing the correct distribution, however, my n=4 simulations aren't occurring. Is there a way to adjust the ifelse statement to fix this, or must

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 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 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

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, You can use the projectLinear argument in BB::spg to optimize with linear equality/inequality constraints. Here is how you implement the constraint that all parameters sum to 1. require(BB) spg(par=p0, fn=myFn, project="projectLinear", projectArgs=list(A=matrix(1, 1, length(p0)), b=1, meq=1)) Hope this is helpful, Ravi [[alternative HTML version deleted]]

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