similar to: IWSM 2013: LAST call for papers

dear all, apologizes for this off topic. I would like to inform you that registration and paper submission for the 28th International Workshop on Statistical Modelling (IWSM) to be held in Palermo (Italy) 8-12 July 2013 is now open at http://iwsm2013.unipa.it Register at http://iwsm2013.unipa.it/?cmd=registration and then submit your abstract. Deadlines for Abstract submission is February 4,

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, 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 --

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, 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 R users, I am pleased to announce that segmented 2.1-0 is now available on CRAN. segmented focuses on estimation of breakpoints/changepoints of segmented, i.e. piecewise linear, relationships in (generalized) linear models. Starting with version 2.0-0, it is also possible to model stepmented, i.e. piecewise constant, effects. In the last release both models may be fitted via a formula

dear R users, I am pleased to announce that segmented 2.1-0 is now available on CRAN. segmented focuses on estimation of breakpoints/changepoints of segmented, i.e. piecewise linear, relationships in (generalized) linear models. Starting with version 2.0-0, it is also possible to model stepmented, i.e. piecewise constant, effects. In the last release both models may be fitted via a formula

dear all, I do not if it is a nonsense question.. Is it possible in the R session to get the name of the current .Rdata file that I ran? I mean: suppose I double click the file myfile.Rdata. ls() returns the names of the objects in the current workspace (that is saved in myfile.Rdata). In the current R session, I would like to obtain "myfile.Rdata". Is it possible? Thanks in

dear all, It appears that glmnet(), when "selecting" the covariates entering the model, skips from K covariates, say, to K+2 or K+3. Thus 2 or 3 variables are "added" at the same time and it is not possible to obtain a ranking of the covariates according to their importance in the model. On the other hand lars() "adds" the covariates one at a time. My question

dear all, Is the following intentional? Am I missing anything in documentation? d<-data.frame(y=rnorm(10,5,.5),exp=rnorm(10), age=rnorm(10)) formula(lm(exp(y)~exp+age, data=d)) #--> exp(y) ~ exp + age formula(lm(exp(y)~., data=d)) #--> exp(y) ~ age variable 'exp' (maybe indicating "experience") is not included in the model. The same happens with 'log' (and

Dear list, I would like to know: 1. After I have used the R code (http://pscl.stanford.edu/zeroinfl.r) to fit a zero-inflated negative binomial model, what criteria I should follow to compare and select the best model (models with different predictors)? 2. How can I compare the model I get from question 1 (zero-inflated negative binomial) to other models like glm family models or a logistic

Dear all, This is off-topic, however I hope someone can give me useful suggestion.. Given the regression model y = b0 + b1*x + e I am interested in testing for positive coeffs, namely H0: b0>0 AND b1>0 H1: b0,b1 unconstrained It is simple to estimate the model under H0 and H1 (there are several suggestions on the Rlist about estimation but nothing about testing..) perform a likelihood

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, I am stuck on the following problem. Give a string like ss1<- "z:f(5, a=3, b=4, c='1:4', d=2)" or ss2<- "f(5, a=3, b=4, c=\"1:4\", d=2)*z" I would like to remove all entries within parentheses.. Namely, I aim to obtain respectively "z:f()" or "f()*z" I played with sub() and gsub() but without success.. Thank you very

Dear all, I have tried to consider a correlation structure in lme (package lme4), but without success. I have used something like: > risul<-lme(y~x+ z , data=mydata, random=~ x | g, correlation = corAR1()) but the result is the same as: > risul<-lme(y~x+ z , data=mydata, random=~ x | g). Can anybody help me? Antonella ************************************************** Prof.

dear all, I apologize for my delay in replying you. Here my contribution, maybe just for completeness: Similar to "earth", "segmented" also fits piecewise linear relationships with the number of breakpoints being selected by the AIC or BIC (recommended). #code (example and code from Martin Maechler previous email) library(segmented) o<-selgmented(y, ~x, Kmax=20,

Dear list, I do not know if this is a bug. Let's suppose mat is a matrix derived from this code: > x<-rnorm(10) > y<-rnorm(10) > names(x)<-LETTERS[1:10] > names(y)<-LETTERS[1:10] > mat<-cbind(x,y) Now in R 2.0.1 I have: > names(dist(mat)) "A" "B" "C" "D" "E" "F" "G" "H"

Hi everybody, I have the following problem. I have a vector containing character elements, such as: list = c("aa","bb","cc","dd","ee") I want to create an index which identifies the elements that are different from, e.g. "aa" and "bb". When I do the following: jj = list!="aa" & list!="bb" > jj

Dear all, there are R packages able to simulate or estimate bilinear model for time series? I know it is an open problem, but do exist something for very simplified bilinear models? Alternatively, what kinfd of non linear time series models are performed in R? If R is not able, could someone suggest me for some commercial softwares to deal with bilinear models? i'm afraid of a negative

Hello, I have the following problem. I am running simulations on possible states of a set of agents (1=employed, 0=unemployed). I store these simulated time series in a matrix like the following, where rows indicates time periods, columns the number of agents (4 agents and 8 periods in this case): Atr=[ 1 1 1 1 1 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0