similar to: package segmented problem

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 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

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

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

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

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:

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

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

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

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

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

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 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 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

On 5/22/2018 11:32 AM, Bailey Hewitt wrote: > Hi Bert, > > Thank you for the quick response! > > In its current?state?the code?prints three lines that say "warning". What I was expecting is that I would?get?a matrix with 4 columns, 1. column names (from the original data, ex. Lake1) 2. breakpoint year 3. slope 4. slope difference from the first to the?second segment of