similar to: Linear Regression Problem

Dear all, It appears that MASS::polr() and Design::lrm() return the same point estimates but different st.errs when fitting proportional odds models, grade<-c(4,4,2,4,3,2,3,1,3,3,2,2,3,3,2,4,2,4,5,2,1,4,1,2,5,3,4,2,2,1) score<-c(525,533,545,582,581,576,572,609,559,543,576,525,574,582,574,471,595, 557,557,584,599,517,649,584,463,591,488,563,553,549) library(MASS) library(Design)

dear all, This is a probably a silly question. If I type > grep("x",c("a.x" ,"b.x","a.xx"),value=TRUE) [1] "a.x" "b.x" "a.xx" Instead, I would like to obtain only "a.x" "b.x" How is it possible to get this result with grep()? many thanks for your attention, best, vito --

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,

dear all, Could anyone explain me the behaviour of median() within by()? (I am running R.2.7.0) thanks, vito > H<-cbind(rep(0:1,l=20),matrix(rnorm(20*2),20,2)) > by(H[,-1],H[,1],mean) INDICES: 0 V1 V2 -0.2101069 0.2954377 --------------------------------------------------------------------------------------------------------------------- INDICES: 1 V1

Dear all, I am dealing with the following (apparently simple problem): For some reasons I am interested in passing variables from a dataframe to a specific environment, and in fitting a standard glm: dati<-data.frame(y=rnorm(10),x1=runif(10),x2=runif(10)) KK<-new.env() for(i in 1:ncol(dati)) assign(names(dati[i]),dati[[i]],envir=KK) #Now the following two lines work correctly:

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'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.

Dear R Users, I have a CO2 time series. I want to fit this series seasonal cycle and trend with fourth harmonic function, and then compute residuals. I am doing something like: file<-read.csv("co2data.csv") names(file) attach(file) fit<-lm(co2~1+time+I(time^2)+sin(2*pi*time)+cos(2*pi*time)+sin(4*pi*time)+cos(4*pi*time)+

dear all, When I use all.vars(), I am interest in extracting only the variable names.. Here a simple example all.vars(as.formula(y~poly(x,k)+z)) returns [1] "y" "x" "k" "z" and I would like to obtain "y" "x" "z" Where is the trick? many thanks vito -- ==================================== Vito M.R. Muggeo Dip.to Sc

dear all, I would like to get the lme call without fitting the relevant model. library(nlme) data(Orthodont) fm1 <- lme(distance ~ age, random=list(Subject=~age),data = Orthodont) To get fm1$call without fitting the model I use call(): my.cc<-call("lme.formula", fixed= distance ~ age, random = list(Subject = ~age)) However the two calls are not the same (apart from the data

Dear all, I have the following (rather) strange problem.. For some reasons, I finally work with a variable whose name includes an R function, "a.log(z)", say. And that is a problem when I call it in a formula, for instance: > myname<-"a.log(z)" > dd<-data.frame("a.log(z)"=1:10,y=rnorm(10)) > o<-lm(y~1,data=dd) >

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

In SAS, for a two-way (or 3-way, stratified) table, the CMH option in SAS PROC FREQ gives 3 tests that take ordinality of the factors into account, for both variables, just the column variable or neither. Is there an equivalent in R? The mantelhaen.test in stats gives something quite different (a test of conditional independence for *nominal* factors in a 3-way table). e.g. I'd like to

Hello, I would like to test some data to see whether it has the shape of a step function (i.e. y1 up until x_th and then y2 where x_th is the threshold). The threshold x_th is unknown and the x values can only take discrete values (0,1,2,3,4). An example would be: data<- data.frame(x=1:20,y=c(rnorm(10),rnorm(10,10))) I was thinking along the lines of fitting some sort of piiecewise linear

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

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,

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?m trying to run a logistic joinpoint regression utilising the ljr package. I?ve been using the forward selection technique to get the number of knots for the analysis, but I?m uncertain as to my results and the interpretation. The documentation is rather brief ( in the package and the stats in medicine article is quite technical) and without any good examples. At the moment I?m thinking 1)find