similar to: Replies of the question about robustness of segmented regression

Hello, R users, I applied segmented regression method contributed by Muggeo and got different slope estimates depending on the initial break points. The results are listed below and I'd like to know what is a reasonable approach handling this kinds of problem. I think applying various initial break points is certainly not a efficient approach. Is there any other methods to deal with segmented

Dear List: First, many thanks to those who offered assistance while I constructed code for the simulation. I think I now have code that resolves most of the issues I encountered with memory. While the code works perfectly for smallish datasets with small sample sizes, it arouses a windows-based error with samples of 5,000 and 250 datasets. The error is a dialogue box with the following: "R

Hello, I have some trouble using step() and stepAIC() functions. I'm predicting recruitment against several factors for different plant species using a negative binomial glm. Sometimes, summary(step(model)) or summary(stepAIC(model) does not select the best model (lowest AIC) but just stops before. For some species, step() works and stepAIC don't and in others, it's the opposite.

Hello R users, I've used the following help two compare two regression line slopes. Wanted to test if they differ significantly: Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope

Dear R-users, I am facing problem with integrating in R a likelihood function which is a function of four parameters. It's giving me the result at the end but taking more than half an hour to run. I'm wondering is there any other efficient way deal with. The following is my code. I am ready to provide any other description of my function if you need to move forward.

Hi, for testing if the slope of experimental data differs from a given slope I'm using the function "test_regression_against_slope" (see below). I am now confronted with the problem that I have data which requires a modelII regression (also called reduced major axes regression (RMA) or geometric mean regression). For this I use the function "modelII" (see below). What

This isn't a question about R, but I'm hoping someone will be willing to help. I've been looking at calibration plots in multiple regression (plotting observed response Y on the vertical axis versus predicted response [Y hat] on the horizontal axis). According to Frank Harrell's "Regression Modeling Strategies" book (pp. 61-63), when making such a plot on new data

Hello, I would like to write a loop to 1) run 100 linear regressions, and 2) compile the slopes of all regression into one vector. Sample input data are: y1<-rnorm(100, mean=0.01, sd=0.001) y2<-rnorm(100, mean=0.1, sd=0.01) x<-(c(10,400)) #I have gotten this far with the loop for (i in 1:100) { #create the linear model for each data set model1<-lm(c(y1[i],y2[i])~x)

Dear, i am a student and I need help in comparing between different slopes and finding whther there is a significant difference between them? Thanks a lot [[alternative HTML version deleted]]

dear R users, I am pleased to announce that segmented 2.1-0 is now available on CRAN. segmented focuses on estimation of breakpoints/changepoints of segmented, i.e. piecewise linear, relationships in (generalized) linear models. Starting with version 2.0-0, it is also possible to model stepmented, i.e. piecewise constant, effects. In the last release both models may be fitted via a formula

Hi all, I am trying to implement the following matlab code with Ryacas : syms U x x0 C d1=diff(U/(1+exp(-(x-x0)/C)),x); pretty(d1) d2=diff(U/(1+exp(-(x-x0)/C)),x,2); pretty(d2) solx2 = solve(d2 == 0, x, 'Real', true) pretty(solx2) slope2=subs(d1,solx2) I have tried the following : library(Ryacas) x <- Sym("x");U <- Sym("U");x0 <-

[message bounced because of "octet-stream" attachment which are not allowed in our mailing lists; manually fixed and approved, MM] Dear R-list, R: 1.5.1 OS: Windows NT additional packages needed: tseries for those of you who are interested, pls. find attached a function for estimating VECM's by employing the method of Johansen (see for example: Hamilton,

Have you studied the "Introduction to Ryacas" vignette that come with the package? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Sep 19, 2017 at 2:37 AM, Vivek Sutradhara <viveksutra at gmail.com> wrote:

You also need to reply-all so the mailing list stays in the loop. -- Sent from my phone. Please excuse my brevity. On December 19, 2017 4:00:29 PM PST, Timothy Axberg <axbergtimothy at gmail.com> wrote: >Sorry about that. Here is the code typed directly on the email. > >qe = (Qmax * Kl * ce) / (1 + Kl * ce) > >##The data >ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)

Hi, I appreciate your help with the segmented function. I am relatively new to R. I followed the introduction of the 'segmented'-package by Vito Muggeo, but still it does not work. Here are the lines I wrote: data_test<-data.frame(x=c(1:10),y=c(1,1,1,1,1,2,3,4,5,6)) lr_test<-lm(y~x,data_test) seg_test<-segmented(lr_test,seg.Z~x,psi=1) /error in segmented.lm(lr_test, seg.Z ~ x,

Thanks for the response. Yes, I did study the vignette but did not understand it fully. Anyway, I have tried once again now. I am happy to say that I have got what I wanted. library(Ryacas) x <- Sym("x");U <- Sym("U");x0 <- Sym("x0");C <- Sym("C") my_func <- function(x,U,x0,C) { return (U/(1+exp(-(x-x0)/C)))} FirstDeriv <-

Hi, while using package "segmented" (version 0.2-4) by Vita Muggeo to investigate a possible change point (around time = 222) in admissions for a specific medical condition I had the following error message: fit2.seg<-segmented(fit2, seg.Z=~time,psi=222) Error in segmented.lm(fit2, seg.Z = ~time, psi = 222) : (Some) estimated psi out of its range "fit2" is a simple

Hello, I'm currently using the segmented package of M.R. Muggeo to fit a two-slope segmented regression. I would like to constrain a null-left-slope, but I cannot make it. I followed the explanations of the package (http://dssm.unipa.it/vmuggeo/segmentedRnews.pdf) to write the following code : fit.glm <- glm(y~x) fit.seg <- segmented(fit.glm, seg.Z=~x,psi=0.3) fit.glm

Hi all, I am using the package ?Segmented? to estimate logistic regression models with unknown breakpoints (see Muggeo 2003 Statistics in Medicine 22:3055-3071). In the documentation it suggests that it might be possible to include several variables with breakpoints in the same model: ?Z = a vector or a matrix meaning the (continuous) explanatory variable(s) having segmented relationships with

A few days ago I uploaded to CRAN a new package called segmented. The package contains functions to fit (generalized) linear models with `segmented' (or `broken-line' or `piecewise linear') relationships between the response and one or more explanatory variables according to methodology described in Muggeo VMR (2003), Estimating regression models with unknown break-points, Statistics