similar to: Looking for constrained optimisation code

_______________________________________________________________________________________ Hi, I'm writing up the online help for a package I'm developing (in-house only, sorry), and I've come across an odd glitch when trying to nest a list inside the "arguments" section of the .Rd file. I was just wondering if anyone could provide some insights. I'm using R 2.2.0 on

Thomas, As F Harrel in a preceding message told me that you are the maintainer of Hmisc package, I write directly to you, with copy to the Rhelp list in case someone encountered the same problem and find some benefit in the response. I used quite often the function sas.get in Hmisc library which I is very, very useful. But, as trying to reuse it today it seems not to function anymore. I tried

# Hello R Users, # # I would like to fit a glm model with quasi family and # logistical link function, but this does not seam to work # with binary data. # # Please don't suggest to use the quasibinomial family. This # works out, but when applied to the true data, the # variance function does not seams to be # appropriate. # # I couldn't see in the # theory why this does not work. # Is

i want to estimate parameters with maximum likelihood method with contraints (contant numbers). for example sum(Ai)=0 and sum(Bi)=0 i have done it without the constraints but i realised that i have to use the contraints. Without constraints(just a part-not complete): skellamreg_LL=function(parameters,z,design) { n=length(z); mu=parameters[1]; H=parameters[2]; Apar=parameters[3:10];

_______________________________________________________________________________________ Note: This e-mail is subject to the disclaimer contained at the bottom of this message. _______________________________________________________________________________________ Hi, I came across some rather unexpected behaviour the other day with assigning names and lists. Here's an example. > z

Hi list, could anyone please educate me on the following: lst<-seq(47, 239, by=12) for(n in lst) { lower=n; upper=lower+10 for(i in lower+2 : upper) { print(paste(n, " i: ", i, " lower: ",lower, " upper :", upper)) } } does not stop when i = upper A while loop fixes this but, I still don't understand why the for loop doesn't stop

Hello, I'm a grad student in the Intelligent Transportation Systems lab at Portland State Univ. in Portland, OR, USA. I'm trying to learn the basics of R to run a negative binomial in the near future, and so I ran a test regression on roadway crash data obtained from "Statistical and Econometric Methods for Transportation Data Analysis" by Washington et al (p. 250). I ran the

Dear all, I tried to use "step" function to do model selection, but I got an error massage. What I don't understand is that data as data.frame worked well for my other programs, how come I cannot make it run this time. Could you please tell me how I can fix it? ***************************************************************************************************

I am just starting with R and have the following problem: given a matrix of ones and zeroes, say mdim=4 m<-matrix(round(runif(mdim^mdim)),mdim,mdim) how to construct the matrix of those rows of m, whose elements sum to 2. Contrary to the random matrix above, the actual matrix always has at least one such row. Serguei Kaniovski

Dear List: We are generating data such that students are clustered in schools for some item response data for a simulation study. One component of our simulation is to generate measurement error from a logistic distribution with a mean of 0 and standard deviation of 1.7 to match the logistic curve of the Rasch model. We are generating an error term for each of the 40 hypothetical test items a

dear all, how can i get a vector that shows the number of the column of matrix that contains the maximum of the row ?? can?t believe in need a loop for this... i have a 100 x 3 matrix and want to get a 100 x 1 vector with values 1,2,3 . there must be a simple solution. i just cannot find it. i think am searching on the wrong end. thx for help in advance. m.

Dear All, I would like to minimise a nonlinear function subject to linear inequality constraints as part of an R program. I have been using the constrOptim function. I have tried all of the methods that come with Optim, but nothing finds the correct solution. If I use the correct solution as the vector of starting values, though, my program does output the correct solution and optimum - the

Hi there, I have a piece of Matlab code I use to optimise a trding strategy. If there are any Matlab/R specialists out there, I would appreciate your help in doing the exact same optimisation in R. I suspect I would use nlm() in R but am not sure where to define my constraints. I have attached my Matlab code below for reference. Many thanks. Constraints function [c,ceq]=TriskellConstraints(X)

Dear Friends. I found a puzzling phenomenon in R when you use 'if' within a function: # defining a function aaa aaa=function(a) {if (a==1) {aaa=1}; if (a!=1) {aaa=2} } # using the function: > b=20 > bbb=aaa(b) > bbb [1] 2 > typeof(bbb) [1] "double" > > > c=1 > ccc=aaa(c) > ccc NULL > typeof(ccc) [1] "NULL" It seems that only the last

_______________________________________________________________________________________ Hello, Not sure if this counts as a bug or not, but I just noticed in R 2.4.0 that : and interaction are not quite equivalent. For example: > x <- factor(letters[1:4]) > y <- factor(letters[11:14]) > x:y [1] a:k b:l c:m d:n Levels: a:k a:l a:m a:n b:k b:l b:m b:n c:k c:l c:m c:n d:k d:l d:m

What is a simple way to convert a factor into a matrix of dummy variables? fm<-lm(y~f) where f is a factor takes care of this in the estimation. I'd like to save the result of expanding f into a matrix for later use. Thanks. Charles -- Charles H. Franklin Professor, Political Science University of Wisconsin, Madison franklin at polisci.wisc.edu chfrankl at wisc.edu

Hello, I've got to compute a minimization equation under an equality constraint (Min g(x1,x2,x3) with x1+x2=const). The Constroptim function does not authorize an equality condition but only inequality conditions. Which function can I use instead? Thank you very much for your help. Gael Robert - +33 1 42 14 27 96 ****************************************************************** This

_______________________________________________________________________________________ Hi, When I create multiple graphs subsequent graphs overwrite previous graphs. How do I keep all my graphs in the current workspace (but not all on the same page) so I can scroll through them? thanks, Rohini Rohini Mulford Senior Research Analyst Technical Research Research &

Dear R helpers I have a question regarding the constrainOptim. I'm coding the nested logit and would like to set a bound of rho to (0,1] as an extreme value distribution where rho = exp(lambda)/1+exp(lambda) I wonder if I can do that directly in optim (say rho > 0 & <= 1) or need to use constrainOptim I read the help but still don't know how to set ui and ci Thanks, June

Dear R wizards: I am playing (and struggling) with the example in the constrOptim function. simple example. let's say I want to constrain my variables to be within -1 and 1. I believe I want a whole lot of constraints where ci is -1 and ui is either -1 or 1. That is, I have 2*N constraints. Should the following work? N=10 x= rep(1:N) ci= rep(-1, 2*N) ui= c(rep(1, N), rep(-1, N))