similar to: Problem with segmented function

I am having trouble predicting new data with a model created from package mboost: > mb1<-glmboost(as.formula(formula1),data=data_train,control=boost_control(mstop=400,nu=.1)) > f.predict<-predict(mb1,newdata=data_train) Error in scale.default(X, center = cm, scale = FALSE) : length of 'center' must equal the number of columns of 'x' Ultimately I want to predict

Hi all I am trying to use R for some data editing using the "array" function to write binary data to a text file. I realise that R was not designed for this purpose but I am no C programmer and would prefer to use R (as I know how to do it and hate c). Basically I get the following error. Error: cannot allocate vector of size 1634069 kb It seems that R is not keen on allocating

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

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 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, How can a function in R handle different types of input? I have written a function, which should calculate the slope from several 3-time-point measurements by linear regression 4 three-time-point-measurements: x<-cbind(c(1,2,3,4),c(2,3,4,5),c(3,4,5,6)) time points: time<-c(1,3,9) function for calculating the slope by linear regression: fit<-function(xx,t){slope <-

Hi, I have a list of names (alphanumeric characters) and each row by itself contains a different number of names seperated by a semicolon (";"). There are also rows with no entry. I would like to have the name before the first semicolon in each row, if there is any entry. Does anyone have an idea how to solve this in R? Thank you! Stella -- View this message in context:

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 wrote a function with three inputs fun(x,y,z). x is a matrix of three vectors combined with cbind. e.g. x1<-(1,2,3,4) x2<-(2,3,4,5) x3<-(3,4,5,6) x<-cbind(x1,x2,x3) y is a vector e.g y<-c(7,8,9) z is a real number e.g. z<-2.5 If a give the function an input like this, I get 'NA' in return. If I give the function a vector e.g c(1,2,3) instead of 'x'

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

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 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

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 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

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 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.

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 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

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, I would like to calculate the area under segmented regression lines (single breakpoints). I had thought that I could do this using the predict() function to ascertain some key x y values in order to determine the dimensions for two trapezoids (under each segment of my regression). However, I've not had any success with this and have just read online that the predict function does not work