search for: vmuggeo

...that results from lrm() are right.. Am I wrong? Thanks, vito -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 6626240 fax: 091 485726/485612 http://dssm.unipa.it/vmuggeo

...sult with grep()? many thanks for your attention, best, vito -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 6626240 fax: 091 485726/485612 http://dssm.unipa.it/vmuggeo

...ault(data[x, , drop = FALSE], ...) : need numeric data > -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 6626240 fax: 091 485726/485612 http://dssm.unipa.it/vmuggeo

..." "z" Where is the trick? many thanks vito -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 23895240 fax: 091 485726/485612 http://dssm.unipa.it/vmuggeo

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 <-update(fit.glm,.~.-x) fit.seg1 <-update(fit.seg) If I got well, the two last lines are supposed to constrain the slope, but it did not change a...

...the one from fm1$call) by means of call()? thanks, vito -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 23895240 fax: 091 485726/485612 http://dssm.unipa.it/vmuggeo

Hi all, Could anybody recommend some easy-to-understand and example based notes/tutorials on how to use cubic splines to do filtering on multi-dimension data? I am confused by the 1-dimensional case, and more confused by multi-dimensional case. I found all the books suddenly become very abstract when it comes to this subject. They don't provide examples in R or Splus at all. Specifically,

One of the assumptions, on which the (General) Linear Modelling is based is that the response variable is measured with some uncertainties (or weighted), but the explanatory variables are fixed. Is it possible to extend the model by assigning the weights to the explanatory variables as well? Is there a package for doing such a model fit? Thanks

Hello, I'm trying to find an algorithm to estimate a switching regression model based on the 1990 Economics Letters paper by Ohtani/Kakimoto/Abe or the earlier version from 1985 (Ohtani/Katayama, Economic Studies Quarterly; assuming as a transition path a polynomial of order 1). I found an idea for using nls here: http://www.biostat.wustl.edu/archives/html/s-news/2000-04/msg00223.html.

Dear All, I have a matrix say, X ( 100 X 40,000) and a vector say, y (100 X 1) . I want to perform linear regression. I have scaled X matrix by using scale () to get mean zero and s.d 1 . But still I get very high values of regression coefficients. If I scale X matrix, then the regression coefficients will bahave as a correlation coefficient and they should not be more than 1.

Dear R, Is there an efficient way to make a list that each element is from the corresponding column of a matrix. For example, if I have a matrix "a" > a <- matrix(1:10, 5, 2) > a [,1] [,2] [1,] 1 6 [2,] 2 7 [3,] 3 8 [4,] 4 9 [5,] 5 10 I would like to have a list "b" like this > b <- list(a[, 1], a[, 2]) > b [[1]] [1] 1 2 3

Hi I was wondering whether anyone can help me with this problem....it's been driving me nuts, I've been trying to figure it out for months and months without success!! Basically I have a group of participants who attended 2 experimental sessions a few months apart. I took measures of the way they approach two tasks at Time 1 and the same two tasks at Time 2. I have categorical data (a

...me any advices or suggest me references? Many thanks, vito -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 6626240 fax: 091 485726/485612 http://dssm.unipa.it/vmuggeo

Dear All, I'm trying to do the estimation in a changepoint regression problem via R, but never found any suitable function which might help me to do this. Could someone give me a hand?on this matter? Thank you.

Hello list. I was trying to see some of the code for plot.glmnet in package glmnet (this function name is in the documentation). After loading the library, I tried the obvious typing in the name, but I received a message telling me it could not be found. So I fiddled around a little, and noticed that R does not recognize ''plot'' as a generic function, and as such,

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

Dear all, can somebody give me some pointer how I can fit a "linear-by-linear association model" (i.e. loglinear model for the ordinal variables) in R? A brief description can be found here 'https://onlinecourses.science.psu.edu/stat504/node/141'. Thanks for your help

Hello, I am currently working with the betareg package, which allows the fitting of a variable dispersion beta regression model (Simas et al. 2010, Computational Statistics & Data Analysis). I was wondering whether there is any package in R that allows me to fit variable dispersion parameters in the standard logistic regression model, that is to make the dispersion parameter contingent upon

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

...quot;myfile.Rdata". Is it possible? Thanks in advance, vito -- ==================================== Vito M.R. Muggeo Dip.to Sc Statist e Matem `Vianelli' Universit? di Palermo viale delle Scienze, edificio 13 90128 Palermo - ITALY tel: 091 23895240 fax: 091 485726 http://dssm.unipa.it/vmuggeo