similar to: None-linear equality constrained optimisation problems

Dear List, I have the code below, where I am using the constrained optimisation package, 'constrOptim.nl' to find the values of two values, b0 and b1. I have no problems when I enter further variable information DIRECTLY into the functions, fn, and heq. In this instance I require fn to have -0.0075 appended to it, and in the case of heq, h[1] has -0.2. library(alabama)

Dear List, I am using constrOptim to solve the following fr1 <- function(x) { b0 <- x[1] b1 <- x[2] ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 } As you can see, my objective function is ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 and I would like to solve for both b0 and b1. If I were to use optim then I would derive the gradient of the

Good afternoon list, I have the following character strings; one with spaces between the maths operators and variable names, and one without said spaces. form<-c('~ Sentence + LEGAL + Intro + Intro / Intro1 + Intro * LEGAL + benefit + benefit / benefit1 + product + action * mean + CTA + help + mean * product')

Good morning R list, My apologies if this has *already* answered elsewhere, but I have not found the answer that I am looking for. I have a character string, i.e. form<-c('~ A + B + C + C / D + E + E / F + G + H + I + J + K + L * M') Now, my aim is to find the position of all those instances of '*' and to remove said '*'. However, I would also like to remove the

Hi, I have a matrix of say 1024x1024 and I want to look at it in chunks. That is I'd like to divide into a series of submatrices of order 2x2. | 1 2 3 4 5 6 7 8 ... | | 1 2 3 4 5 6 7 8 ... | | 1 2 3 4 5 6 7 8 ... | | 1 2 3 4 5 6 7 8 ... | ... So the first submatrix would be | 1 2 | | 1 2 | the second one would be | 3 4 | | 3 4 | and so on. That is I want the matrix to be evenly divided

Hi all A couple of questions about string processing from someone who has only scratched the surface so far. 1) I am wanting to send some strings into a function to allow flexibility inside. My first idea has been e.g. auto_io <- function( var_string, factors ) { # e.g. var_string sent as "test_file.txt" factors sent as "x1 + x2 + x3" # input data_name <- get(

Hi R experts I am just wondering if something is already available (or easily adaptable) to do the following. I am planning to build linear models for all possible combinations of terms, so for example if the terms are sent into a function as this string " X1 + X2 + X3 + X4 + X1:X2" I would want to build models for all possible combinations of these 5 terms, e.g. m1 <- lm( y ~

Hi R people I am using regsubsets() to fit large numbers of models and collect summary statistics in order to perform a Bayesian analysis of multi-way ANOVA with specific prior information. In general the variables have differing numbers of levels >=2. This works well but with variable of more than 2 levels there are naturally some arbitrary decisions about which treatment contrasts to

Dear R useRs, I have a problem with nls.lm function of minpackl.lm package. I need to fit the Van Genuchten Model to a set of data of Theta and hydraulic conductivity with nls.lm function of minpack.lm package. For the first fit, the parameter estimates keep changing even after 1000 iterations (Th) and I have a following error message for fit of hydraulic conductivity (k); Reason for

Hi there, I'm a PhD student investigating growth patterns in fish. I've been using the minpack.lm package to fit extended von Bertalanffy growth models that include explanatory covariates (temperature and density). I found the nls.lm comand a powerful tool to fit models with a lot of parameters. However, in order to select the best model over the possible candidates (without covariates,

Hi, Is this method broken in R? I am using it to find roots of the following function: f(x) = 2.5*exp(-0.5*(2*0.045 - x)) + 2.5*exp(-0.045) + 2.5*exp(-1.5*x) - 100 It is giving an answer of -38.4762403 which is not even close (f(x) = 2.903809e+25 for x=-38.4762403). The answer should be around 0.01-0.1. This function should converge.. Even for a simple function like f(x) = exp(-x) * x, it gives

> > > > The approach I was thinking would look something like this: > > * instead of Features, which is really a namespace implemented as a > class, we separate out the calculation of the different features > into distinct subclasses of Feature, whose only job is to calculate > a single feature. Currently the FeatureManager calls these (via >

Deal all: i want to do the probit with sample selection estimation, the following is my code: probit with sample selection can be done by stata :heckprob The heckprobll is the likelihood function shown in W.H. Greene 5th p714 ¡´ The question is the convergence is very slow compare with Stata using likellihood only. ¡´ Second i did the similar way in matlab using fminsearch , the estimated

Buenos días: Por favor, ¿alguien podría crear un script de r con este código que envío e intentar paquetizarlo?. Yo siempre lo conseguía, pero con la versión actual de R (3.1.0), una vez generado el zip del paquete y cargado desde la RGui, no me detecta la existencia del environment que creé (.Ch.env). Es como si ahora el pkt sólo pudiera estar integrado (a efectos prácticos) por funciones, sin

Queridos, Me apunté a esta lista de correo con el objetivo de avanzar en mi proceso de aprendizaje con R y R-Commander. Mis limitados conocimientos me hacen ver que este no es el foro que necesito, su nivel es demasiado avanzado para mí. Si me podeis orientar para estar en el foro adecuado que sería tener oportunidad de comentar problemas de los primeros pasos con R y R-Commander os lo

Look back to the SO link [1] you posted, I assume you register your pass by writing something like below already, INITIALIZE_PASS_BEGIN ... INITIALIZE_PASS_END ModulePass *llvm::createYourPass() { return new YourPass(); } The only suggestion I can give is looking at other existing pass to see what you might miss. For example, take a look on X86OptimizeLEAs.cpp. It works like

Hola Eva. Me alegro de que hayas resuelto el problema! De todas formas, no tendrá algo que ver con los problemas de "globalización" de las variables que ya has tenido en .ChrL.env (hace un par de años, creo recordar)?. Un Saludo, Miguel. -----Mensaje original----- De: r-help-es-bounces en r-project.org [mailto:r-help-es-bounces en r-project.org] En nombre de Eva Prieto Castro

Does ActiveRecord.connection.execute(string) perform any modifications on the given string? If so, what quoting do I need to add? Here''s why: I''m trying to use execute(string) to load a stored procedure. I can hand-load the procedure without error (i.e. via dbconsole), but when I call the function below, it throws the error: ==== ActiveRecord::StatementInvalid: Mysql::Error:

Hi, I have an issue with *samba-3.6.6*. It has been installed with : ./configure --prefix="/usr/local/samba-3.6.6" --without-winbind --with-fhs --sysconfdir=/etc --localstatedir=/var --with-piddir=/var/run/samba smb.conf contains : private dir = /var/lib/samba/private lock directory = /var/run/samba state directory = /var/lib/samba cache directory = /var/cache/samba So

Hi everybody I have a very big problem: With R 3.0.2 I could construct the package for this code: if (exists('.ChrL.env') == TRUE) {   rm(.ChrL.env) } .ChrL.env <- new.env() .ChrL.env$lGlo <- list() .ChrL.env$bStarted <- FALSE CheckGloCreated <- function() {   if (.ChrL.env$bStarted == TRUE) {     stop("Data structures were already initialized.", call.=FALSE)