similar to: variable dispersion in glm models

My data: I have raw data points that form a logit style curve as if they were a time series. Which is to say they form 3 distinct lines with 3 distinct slopes in backwards z pattern. A certain class of my data looks essentially flat to the eye with marginal oscillation. What is important to me is the x value at which the state change is occurring, in other words, the break point Use of

Hello all, I've been having trouble with assessing a breakpoint in a logistic GLM with two explanatory variables. For this analysis I've been using the 'segmented' package version 0.2-9.1. But I keep getting an error and I don't see where I would be going awry. The situation is the following: Two explanatory variables: bedekking - a variable with possible values between 0 and

Hello, I have a dataset with proportions that vary around a fixed mean, is it possible to use betareg to look at variance in the dispersion parameter while keeping the mean fixed? I am very new to R but have tried the following: svec<-c(qlogis(mean(data1$scaled)),0,0,0) f<-betareg(scaled~-1 | expt_label + grouped_hpi, data=data1, link.phi="log",

dear all, When I use all.vars(), I am interest in extracting only the variable names.. Here a simple example all.vars(as.formula(y~poly(x,k)+z)) returns [1] "y" "x" "k" "z" and I would like to obtain "y" "x" "z" Where is the trick? many thanks vito -- ==================================== Vito M.R. Muggeo Dip.to Sc

dear all, I would like to get the lme call without fitting the relevant model. library(nlme) data(Orthodont) fm1 <- lme(distance ~ age, random=list(Subject=~age),data = Orthodont) To get fm1$call without fitting the model I use call(): my.cc<-call("lme.formula", fixed= distance ~ age, random = list(Subject = ~age)) However the two calls are not the same (apart from the data

I am running a binomial glm with response variable the no of mites of two species y->cbind(mitea,miteb) against two continuous variables (temperature and predatory mites) - see below. My model shows overdispersion as the residual deviance is 48.81 on 5 degrees of freedom. If I use quasibinomial to account for overdispersion the dispersion parameter estimate is 2501139, which seems

Dear R, Is there an efficient way to make a list that each element is from the corresponding column of a matrix. For example, if I have a matrix "a" > a <- matrix(1:10, 5, 2) > a [,1] [,2] [1,] 1 6 [2,] 2 7 [3,] 3 8 [4,] 4 9 [5,] 5 10 I would like to have a list "b" like this > b <- list(a[, 1], a[, 2]) > b [[1]] [1] 1 2 3

Hi I was wondering whether anyone can help me with this problem....it's been driving me nuts, I've been trying to figure it out for months and months without success!! Basically I have a group of participants who attended 2 experimental sessions a few months apart. I took measures of the way they approach two tasks at Time 1 and the same two tasks at Time 2. I have categorical data (a

Hello list. I was trying to see some of the code for plot.glmnet in package glmnet (this function name is in the documentation). After loading the library, I tried the obvious typing in the name, but I received a message telling me it could not be found. So I fiddled around a little, and noticed that R does not recognize ''plot'' as a generic function, and as such,

Hi everyone. I'm trying to use the segmented function with the following data: For instance, I use segmented package as follow: myreg2 = lm(xy$y ~ xy$x) mysegmented = segmented(myreg2, seg.Z=~x, psi=c(245000), control = seg.control(display=FALSE)) Which get me to the following error : As a break point, a starting guess of 245000 seems fair. Anyone has an idea why I'm getting such

Dear all, can somebody give me some pointer how I can fit a "linear-by-linear association model" (i.e. loglinear model for the ordinal variables) in R? A brief description can be found here 'https://onlinecourses.science.psu.edu/stat504/node/141'. Thanks for your help

Hi everyone, I have encountered this problem while using 'segmented' plugin for R i386 2.15.0 (for Windows 32bit OS) and I just cannot find neither explanation nor solution for it. I am trying to run this data gpp temp 1.661 5 5.028 10 9.772 15 8.692 20 5.693 25 6.293 30 7.757 5 4.604 10 8.763 15 8.134 20 4.616 25 8.417 30 3.483 5 5.046 10 8.306 15 9.142 20 4.686 25 7.301 30 and with

Hi All, I am trying to fit a piecewise lasso regression, but package Segmented does not work with Lars objects. Does any know of any package or implementation of piecewise lasso regression? Thanks, Lucas

Hi I want to compare the survival curves in two groups. Because the hazards are not proportional (the curves cross) the log rank test or Cox proportional hazard test are not suitable. How should such curves be compared? Comands are welcome.... Thanks in advance

Hello, I'm attempting to find multiple breakpoints in an association of my response variable (R.AUC) with two explanatory variables ('s.size' and 'bedekking'). The association between 's.size' and 'R.AUC' shows a plateau, but the value when this plateau is reached is differs for different values of 'bedekking'. Initially I thought these different

I?m trying to run a logistic joinpoint regression utilising the ljr package. I?ve been using the forward selection technique to get the number of knots for the analysis, but I?m uncertain as to my results and the interpretation. The documentation is rather brief ( in the package and the stats in medicine article is quite technical) and without any good examples. At the moment I?m thinking 1)find

Hi, I would like to obtain the dispersion parameter for a quasipoisson model for later use in calculating QAIC values for model comparison.Can anyone suggest a method of how to go about doing this? The idea I have now is that I could use the residual deviance divided by the residual degrees of freedom to obtain the dispersion parameter. The residual deviance is available in the summary

I am trying to fit a very simple broken stick model using the package "segmented" but I have hit a roadblock. > str(data) 'data.frame': 18 obs. of 2 variables: $ Bin : num 0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 ... $ LnFREQ: num 5.06 4.23 3.50 3.47 2.83 ... I fit the lm easily: > fit.lm<-lm(LnFREQ~Bin, data=id07) But I keep getting an error

Hello I'm using function gam from package mgcv to fit splines. ?When I try to make a prediction slightly beyond the original 'x' range, I get this error: > A = runif(50,1,149) > B = sqrt(A) + rnorm(50) > range(A) [1] 3.289136 145.342961 > > > fit1 = gam(B ~ s(A, bs="ps"), outer.ok=TRUE) > predict(fit1, newdata=data.frame(A=149.9), outer.ok=TRUE) Error

Hi everyone, while trying to use 'segmented' (R i386 2.15.0 for Windows 32bit OS) to determine the breakpoint I got stuck with an error message and I can't find solution. It is connected with psi value, and the error says: Error in seg.glm.fit(y, XREG, Z, PSI, weights, offs, opz) : (Some) estimated psi out of its range This is the code I am using: