similar to: OFF topic testing for positive coeffs

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, 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, apologizes for this OT =========================== 28th International Workshop on Statistical Modelling (IWSM), Palermo (Italy) 8-12 July 2013. http://iwsm2013.unipa.it Dear friend, For your information, I would like to bring to your attention that deadline for submission of abstracts is FEBRUARY 18 If you are still interested to visit Palermo (and taste its specialities :-))

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 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, 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 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, 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, Is there *any* reason explaining what I describe below? I have the following line myfun(x) If I type them directly in R (or copy/past), it works.. However if I type in R 2.6.1 > source("code.R") ##code.R includes the above line Error in inherits(x, "data.frame") : object "d" not found namely myfun() does not work correctly. In particular the

Dear all, It appears that the function is.factor() returns different results when used inside the apply() function: that is, is.factor() fails to recognize a factor.. Where is the trick? many thanks, vito > df1<-data.frame(y=1:10,x=rnorm(10),g=factor(c(rep("A",6),rep("B",4)))) > is.factor(df1[,1]) [1] FALSE > is.factor(df1[,2]) [1] FALSE >

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:

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 am stuck on the following problem with integrate(). I have been out of luck using RSiteSearch().. My function is g2<-function(b,theta,xi,yi,sigma2){ xi<-cbind(1,xi) eta<-drop(xi%*%theta) num<-exp((eta + rep(b,length(eta)))*yi) den<- 1 + exp(eta + rep(b,length(eta))) result=(num/den)*exp((-b^2)/sigma2)/sqrt(2*pi*sigma2)

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 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) >

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, 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

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"