similar to: Breakpoint in logistic GLM with 'segmented' package - error: replacement length zero

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

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

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,

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

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:

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 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

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

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

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

Thank you for the suggestion, Bill. The result is not quite what I would like. Here's sample code for you or anyone else who may be interested: Al1 = c('A','C','C','C') Al2 = c('G','G','G','T') Freq1 = c(0.0078,0.0567,0.9434,0.9908) MAF = c(0.0078,0.0567,0.0566,0.0092) m1 = data.frame(Al1=Al1,

Hello all, I have 3 time series (tt) that I've fitted segmented regression models to, with 3 breakpoints that are common to all, using code below (requires segmented package). However I wish to specifiy a zero coefficient, a priori, for the last segment of the KW series (green) only. Is this possible to do with segmented? If not, could someone point in a direction? The final goal is to

Hi everyone. I'm using segmented package to find break point in a bi-linear relationship. In a particular case, I find 1 pointcut (so 2 slopes). I would like to know if it is possible to retrieve information in the segmented object that could let me to plot 1 particular segment with a different color. For example, in that folowing example, I would like to plot the second segment in red.

Data and code as promised: #Creating a test dataset Year<- c(2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014) Lake1<- c(2, 4, 5, 2, 1, 1, 2, 3, 4, 5, 6, 2, 3, 1, 2) Lake2<- c(1, 3, -1, 4, -2, 1, 2, 3, 4, 5, 6, 2, 3, 1, 2) Lake3<- c(1, 2, 5, -3, 1, 1, 2, 3, 4, 5, 6, 2, 3, 1, 2) Lake4<- c(1, 1, 1, 1, 1, 1, 1, 250, 240, 240, 240, 240, 240, 239,

Hey, all. I'm working on a data set with a broken stick linear regression where I know one of the two slopes. It is a negative linear function until the line intersects with the x-axis, at which point it becomes 0. It is not a nonlinear asymptotic function, and, indeed, using negative exponential or logistic types of fits as an approximation has tended to lead to an under or

No. If your ouput is a numeric "matrix", it cannot include alpha. Columns in a data frame can be of different classes, but each column must be single class. and finally, of course, see ?cat -- I think you are misusing it. If you simply want to return "somestuff", your function should be: function(w) {"somestuff"} not function(w) {cat("somestuff")} As

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

I am new in R and i am having trouble here. I?ve already searched in the list but hasn?t helped When i run this script above i get the message "Error in gen[j, i] : incorrect number of dimensions". However gen is 1000x200 (ind x loc) and so is g could anybody help me for (i in 1 : loc) { #loc=200 for (j in 1 : ind) { #ind=1000 g1 <= function ( gen ) matrix ( if (gen[j, i]

Hello, I have produced some segmented regressions with the segmented package by Viggo Mutteo. I have some example data and code below. I want to annotate the individual segments with the slope parameter (actually it would be nicer to annotate with 1000*slope and add some small amount of text as well). How can I do it? Reading the docs for segmented I can access all of the slope parameters via a

I am encountering an error when I attempt to reference a glm model within a function. The function uses the segmented.glm command (package = segmented). Within the segmented.glm command one specifies an object, in this case a logistic regression model, and specifies a starting threshold term (psi). I believe this is an environment problem, but I do not have a solution. Any assistance