similar to: glm inside one self-defined function

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

Dear experts, I have a few quick questions related to GLMs: 1) Suppose my response is of the type Yes/No, How can I control which response I'm modelling? 2) How can I perform Type III tests? Is it with -> drop1(mymodel, test="Chisq") ? 3) I have a numerical variable which I converted to an ordered factor as shown below, but in the summary results of the logistic regression I

I am using nlsLM {minpack.lm} to find the values of parameters a and b of function myfun which give the best fit for the data set, mydata. mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) myfun=function(a,b,r,t){ prd=a*b*(1-exp(-b*r*t)) return(prd)} and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05), lower = c(1000,0),

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

Hello, 1) What is the difference between a "data frame" (J H Maindonald, Using R, p. 12) and a "vector"? In Using R, the author asks the reader to enter the following data in a data frame, which I will call "mydata": year snow.cover 1970 6.5 1971 12.0 1972 14.9 1973 10.0 1974 10.7 1975 7.9 ... mydata=data.frame(year=c(1970,...),snow.cover=c(6.5,...)) 2) How to

Hi everyone, Can someone please help me in these questions?: 1)if I use crossvalidation with svm, do I have to use this equation to calculate RMSE?: mymodel <- svm(myformula,data=mydata,cross=10) sqrt(mean(mymodel$MSE)) But if I don’t use crossvalidation, I have to use the following to calculate RMSE: mymodel <- svm(myformula,data=mydata) mytest

Hello, I'm trying to adjust a non linear model in which the biological response variable (ratio of germinated fungus spores) is dependent on 2 covariates (temperature and time). The response to temperature is modeled by a kind of beta function with 2 parameters (optimal and maximum temperatures) and the time function is a 2-parameter Weibull. Adjustments with nls or gnls work, but I need to

Dear All, I am trying to reproduce the example that I found online here http://bit.ly/11VG4ha However, when I run my script (pasted at the end of the email), I notice that there is a factor 2 between the values for the coefficients for the categorical variable female calculated by my script and in the online example. Any idea about where this difference comes from? Besides, how can I

Hello, I'd like to create a model with a factor-type response variable. This is an example: > mydata <- data.frame(factor_var = as.factor(c(rep('one', 100), rep('two', 100), rep('three', 100))), real_var = c(rnorm(150), rnorm(150) + 5)) > summary(mydata) factor_var real_var one :100 Min. :-2.742877 three:100 1st Qu.:-0.009493 two

Hi I wish to read a file from my local directory from inside a function. I am passing the filename as the argument but this does not work. Say for example function(dat) { dat1=read.csv("D:\\dat.csv",header=TRUE) } If I call funtion(dat) I get the following error. 'Intuitively' i understand this is a mistake but how do I overcome this and how can I read a file name passed as an

Dear R community members and R experts I am stuck at a point and I tried with my colleagues and did not get it out. Sorry, I need your help. Here my data (just created to show the example): # generating a dataset just to show how my dataset look like, here I have x variables # x1 .........to X1000 plus ind and y ind <- c(1:100) y <- rnorm(100, 10,2) set.seed(201) P <-

I was unable to build the whole Mumble system on CentOS but it's available for Fedora Development. I just wanted the server part on my headless server, and a static build is available from the Mumble project on Sourceforge. So I grabbed the Mumble SRPM from Fedora Development, the static build from Sourceforge, and stripped the spec file down to the minimum needed to just install the

Dear users, I'll appreciate your help with this (hopefully) simple problem. I have a model object which was fitted to inputs X1, X2, X3. Now, I'd like to use this object to make predictions on a new data set where only X1 and X2 are available (just use the estimated coefficients for these variables in making predictions and ignoring the coefficient on X3). Here's my attempt but, of

This should be very easy, but alas, I'm very new to R. My end goal is to calculate p-values from arima(). Let's say I just ran this: > MyModel <- arima(y[1:58], order=c(1,0,0), xreg=MyData[1:58,7:14], > method="ML") > MyModel And I see: arima(x = y[1:58], order = c(1, 0, 0), xreg = MyData[1:58, 7:14], method = "ML") Coefficients: ar1

Hi everybody, I want to ask your help to explain what is going on with my following code: > mydata <- data.frame(y=rbinom(100, 1, 0.5), x1=rnorm(100), x2=rnorm(100)) > glm.fit.method <- function(model,data,...){glm(formula=model,data=data,family="binomial",. ..)} > fit1 <- glm(y ~ x1 + x2, data=mydata, family=binomial()) > update(fit1, .~1) Call: glm(formula =

Hello, I am new to R. I would like others to explain to me how to add absolute values inside the individual stacked bars in a consistent way using the basic R plotting function (R base). I tried to plot a stacked bar graph using R base but the values appear in an inconsistent/illogical way in such a way that its supposed to be 100% for each village but they don't sum up to 100%. Here is the

Dear list, after fitting an lrm with the Design package (stored as "mymodel") I try running a summary, but I get the following error: dim(mydata) [1] 235 9 names(mydata) [1] "id" "VAR1" "VAR2" "VAR3" "VAR4" "VAR5" "VAR6" "VAR7" "VAR8" summary(mymodel) Error in `contrasts<-`(`*tmp*`,