similar to: new package: segmented

I appreciate to Roger Koenker, Achim Zeileis and Vito Muggeo for their informative answers. Listed below is unedited replies I got followed by the question I posted. Kyong 1. Roger Koenker: You might try rqss() in the quantreg package. It gives piecewise linear fits for a nonparametric form of median regression using total variation of the derivative of the fitted function as a penalty

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

> Subject: [R] LRT in arima models > Date: Mon, 17 Feb 2003 11:53:04 +0100 > From: "vito muggeo" <vito.muggeo at giustizia.it> > To: <r-help at stat.math.ethz.ch> > > Dear all, > > For some reason I'm evaluating the size of the LRT testing for the effect of > some explanatory variable in arima models. > I performed three different simulations

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,

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

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'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

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 all, I'm looking for someone that help me to write an R function to simulate survival data under complex situations, namely time-varying hazard ratio, marginal distribution of survival times and covariates. The algorithm is described in the reference below and it should be not very difficult to implement it. However I tried but without success....;-( Below there the code that I used; it

Below. -- Bert On Tue, Apr 3, 2012 at 1:47 PM, Danielle Duncan <dlduncan2 at alaska.edu> wrote: > Thanks for the response, I should have clarified that the NOEL is the > smallest dose above which there is a statistically significant effect. > This is not a scientifically meaningful nor defensible definition as it is stochastic, depends on the test used, design, level chosen, etc.

Dear R-users, I am trying to understand how the 'segmented'-package works to determine breakpoints and slopes of regression lines in broken-line regression models. However, I am not able to repeat the example on the "plant"-dataset, which was reported in the accompanying paper of the package. (V.M.R Muggeo, "Segmented: an R package to fit regression models 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

--- On Mon, 3/12/12, aleksandr shfets <a_shfets at mail.ru> wrote: > From: aleksandr shfets <a_shfets at mail.ru> > Subject: Fwd: Re[2]: [R] B-spline/smooth.basis derivative matrices > To: "Vassily Shvets" <shv736 at yahoo.com> > Received: Monday, March 12, 2012, 5:15 PM > > > > -------- ???????????? ????????? > -------- > ?? ????:

Dear all, It appears that MASS::polr() and Design::lrm() return the same point estimates but different st.errs when fitting proportional odds models, grade<-c(4,4,2,4,3,2,3,1,3,3,2,2,3,3,2,4,2,4,5,2,1,4,1,2,5,3,4,2,2,1) score<-c(525,533,545,582,581,576,572,609,559,543,576,525,574,582,574,471,595, 557,557,584,599,517,649,584,463,591,488,563,553,549) library(MASS) library(Design)

Functions are first class objects, so some kind of collision is bound to happen if you do this... so don't. -- Sent from my phone. Please excuse my brevity. On January 30, 2018 3:11:56 AM PST, "Vito M. R. Muggeo" <vito.muggeo at unipa.it> wrote: >dear all, >Is the following intentional? Am I missing anything in documentation? >

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, 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

On 15 Jun 2004 at 13:52, Vito Muggeo wrote: > Dear Petr, > Probably I don't understand exactly what you are looking for. > > However your "plot(x,c(y,z))" suggests a broken-line model for the > response "c(y,x)" versus the variables x. Therefore you could estimate > a segmented model to obtain (different) slope (and breakpoint) > estimates. See

Well... ?terms.formula says: "data: a data frame from which the meaning of the special symbol . can be inferred. It is unused if there is no . in the formula." So this seems to me to be an obscure bug, as I have found no warning against this admittedly confusing but still, I think, legal syntax. Note: > d <- data.frame(log = runif(10), x = 1:10) > y <- rnorm(10,5) >

dear all, This is a probably a silly question. If I type > grep("x",c("a.x" ,"b.x","a.xx"),value=TRUE) [1] "a.x" "b.x" "a.xx" Instead, I would like to obtain only "a.x" "b.x" How is it possible to get this result with grep()? many thanks for your attention, best, vito --