similar to: a question on plotting nonlinear regression

The cautions people have given about starting values are worth heeding. That nlxb() does well in many cases is useful, but not foolproof. And John Fox has shown that the problem can be tackled very simply too. Best, JN On 2023-08-19 18:42, Paul Bernal wrote: > Thank you so much Dr. Nash, I truly appreciate your kind and valuable contribution. > > Cheers, > Paul > > El El

Thank you so much Dr. Nash, I truly appreciate your kind and valuable contribution. Cheers, Paul El El s?b, 19 de ago. de 2023 a la(s) 3:35 p. m., J C Nash < profjcnash at gmail.com> escribi?: > Why bother. nlsr can find a solution from very crude start. > > Mixture <- c(17, 14, 5, 1, 11, 2, 16, 7, 19, 23, 20, 6, 13, 21, 3, 18, 15, > 26, 8, 22) > x1 <- c(69.98, 72.5,

The PDF file on the web, which is an appendix on nonlinear regression associated with the CAR book, is very nice. When I ran through the code presented there, I found something odd. The code does a certain model in 3 ways: Vanilla NLS (using numerical differentation), Analytical derivatives (where the user supplies the derivatives) and analytical derivatives (using automatic differentiation). The

Dear friends, Hope you are all doing well and having a great weekend. I have data that was collected on specific gravity and spectrophotometer analysis for 26 mixtures of NG (nitroglycerine), TA (triacetin), and 2 NDPA (2 - nitrodiphenylamine). In the dataset, x1 = %NG, x2 = %TA, and x3 = %2 NDPA. The response variable is the specific gravity, and the rest of the variables are the predictors.

Hello, I would like to use the parfm package: https://cran.r-project.org/web/packages/parfm/parfm.pdfhttps://cran.r-project.org/web/packages/parfm/parfm.pdf in my work. This package fits parametric frailty models to survival data. To ensure I was using it properly, I started by running some small simulations to generate some survival data (without any random effects), and analyse the data using

Hello to all, I have been using an optim with the following call: optim(param_ini,fun_errores2,Precio_mercado=Precio,anos_pagosE2=anos_pagos,control=list(maxit=10000,reltol=1e-16)) depending on the intial values I'm getting the same solution but once I get the convergence message=10 (no convergence) and for the others I get convergence message = 0 Solution1: $par beta1

Hello everybody, i already searched the archieves, but i still don't know what is wrong in my implementation, mybe anybody coud give me some advice ll1<-function(rho,theta,beta1,beta2,beta3,beta4,t,Szenariosw5,Testfaellew5,X1,X2) { n<-length(t) t<-cumsum(t) tn<-t[length(t)] Szenn<-Szenariosw5[length(Szenariosw5)]

What's up fellows... I am a begginer in R and i am trying to find the parameters of one likelihood function, but when i otimize it, always appers a error or advertisement and the solve does not occur. The problem seems like that: "lMix<-function(pars,y){ beta1<-pars[1] beta2<-pars[2] beta3<-pars[3] beta4<-pars[4] beta5<-pars[5] alfa1<-pars[6]

Dear all I used R2WinBUGS package's bugs() function to generate MCMC results. Then I tried to save the simulation draws in R, using read.bugs() function. Here is a simple test: ###################### library(coda) library(R2WinBUGS) #fake some data to test beta0=1 beta1=1.5 beta2=-1 beta3=2 N=200 x1=rnorm(N, mean=0,sd=1) x2=rnorm(N, mean=0,sd=1) x3=rnorm(N, mean=0,sd=1) lambda2= exp(beta0+

Hi, I am trying to covert a Winbugs code into R code. Here is the winbugs code model{# model’s likelihoodfor (i in 1:n){time[i] ~ dnorm( mu[i], tau ) # stochastic componenent# link and linear predictormu[i] <- beta0 + beta1 * cases[i] + beta2 * distance[i]}# prior distributionstau ~ dgamma( 0.01, 0.01 )beta0 ~ dnorm( 0.0, 1.0E-4)beta1 ~ dnorm( 0.0, 1.0E-4)beta2 ~ dnorm( 0.0, 1.0E-4)#

Dear all, I considered an ordinary ridge regression problem. I followed three different ways: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-t(as.matrix(cbind(2,3,4,5))) n<-nrow(X) p<-ncol(X) #Without

Dear all, For an ordinary ridge regression problem, I followed three different approaches: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-as.matrix(c(2,3,4,5)) n<-nrow(X) p<-ncol(X) #Without standardization

Here is my code, which plots three lines with errorbar. How could I add an extra line without errorbar to the plot? Thank you very much. beta.data <- data.frame ( method = rep(c("Wrong", "Correct", "Full Bayes"), each = T_obs), mean.beta = c(mean.beta1, mean.beta2, mean.beta3), t = rep(points, 3), std.beta = c(std.beta1, std.beta2, std.beta3) ) limits =

Is there a way to do hierarchical (bayesian) logistic regression in R, the way we do it in BUGS? For example in BUGS we can have this model: model {for(i in 1:N) { y[i] ~ dbin(p[i],n[i]) logit(p[i]) <- beta0+beta1*x1[i]+beta2*x2[i]+beta3*x3[i] } sd ~ dunif(0,10) tau <- pow(sd, -2) beta0 ~ dnorm(0,0.1) beta1 ~ dnorm(0,tau) beta2 ~ dnorm(0,tau) beta3 ~

Hi there, Suppose mu is constant, and error is normally distributed with mean 0 and fixed variance s. I need to find a statistics that: Y_i = mu + beta1* I1_i beta2*I2_i + beta3*I1_i*I2_i + +error, where I_i is 1 Y_i is from group A, and 0 if Y_i is from group B. It is large when beta1=beta2=0 It is small when beta1 and/or beta2 is not equal to 0 How can I find it by R? Thank you very much

I need to make nonlinear regression with the posterior script, but how is the problem? I have error in library (nls), package 'nls' has been merged into 'stats'. I need help? What other forms I have to make nonlinear regression? and how I find to calculate statistics y residuals, scatterplot. thanks SCRIPT ros<-read.table("Dataset.csv",header=T,sep=",")

Hi, Reading http://ssc.wisc.edu/~ahanna/20_newsgroups.csv after downloading it using data <- read.csv("20_newsgroups.csv",header=TRUE) throws this. Warning message: In scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : EOF within quoted string So, for example, the first line in the file is this. This column contains only such text. Is there a way read it ?

Dear all,I am a new user of WinBUGS and need your help. After running the following code, I got parameters of beta0 through beta4 (stats, density), but I don't know how to get the prediction of the last value of h, the variable I set to NA and want to model it using the following code.Does anyone can given me a hint? Any advice would be greatly appreciated.Best

You might want to try some of the suggestions mentioned in this post: https://stackoverflow.com/q/17414776/2140956 Jean On Thu, Aug 10, 2017 at 7:59 AM, <Mohan.Radhakrishnan at cognizant.com> wrote: > Hi, > > Reading http://ssc.wisc.edu/~ahanna/20_newsgroups.csv after downloading > it using > > data <- read.csv("20_newsgroups.csv",header=TRUE) > >

Yes. I tried that already. Not straightforward. data <- read.csv("20_newsgroups.csv",fill=TRUE,as.is=T,header=F, quote="", sep=",", encoding="UTF-8") This line does read it haphazardly. The emails in the column are split into multiple columns and there are several columns with just ?NA?. Totally 202 columns. And then I removed columns with NA?s and