similar to: Piecewise Regression with a known slope

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

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,

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 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 the segmented regression. Each row in the matrix would be the results

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 All, I have been trying to use a for loop to run segmented regressions (from?R package segmented)?on many columns of data in a data frame with the end goal of writing a new file with the following columns: column title, breakpoint year, slope, and difference in slope. Unfortunately, when one of the columns doesn't have a breakpoint the code stops and provides?an error or warning. I

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

Others may have greater insight, but my response is: Exactly what did or didn't happen that makes you say the code didn't work? That is, what did or didn't you get when you ran it compared to your expectations? 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

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

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 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

Dear all, I'm attempting to use a piecewise regression to model the trajectory of reproductive traits with age in a longitudinal data set using a mixed model framework. The aim is to find three slopes and two points- the slope from low performance in early age to a point of high performance in middle age, the slope (may be 0) of the plateau from the start of high performance to the

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

Hi - I'd like to construct and plot the percents by year in a small data set (d) that has values between 1988 and 2007. I'd like to have a breakpoint (buy no discontinuity) at 1996. Is there a better way to do this than in the code below? > d year percent se 1 1988 30.6 0.32 2 1989 31.5 0.31 3 1990 30.9 0.30 4 1991 30.6 0.28 5 1992 29.3 0.25 6 1994 30.3

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:

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 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

R-experts, I have fitted many different lines. The fast-tau estimator (yellow line) seems strange to me?because this yellow line is not at all in agreement with the other lines (reverse slope, I mean the yellow line has a positive slope and the other ones have negative slope). Is there something wrong in my R code ? Is it because the Y variable is 1 vector and should be a matrix ? Here is the

Dear R-Users, I have a question concerning the determination of breakpoints and comparison of slopes from broken-line regression models. Although this is rather a standard problem in data analysis, all information I gathered so far, did not answer my questions. I added a subsetted example of my data. Basically it is a timeseries of recorded phenotypes in three different groups of plants. You