similar to: strange behaviour of is.factor()

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, 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, 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, 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 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, 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, 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) >

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, 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 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 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 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, 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 Mailing-List, I think this is a newbie question. However, i would like to integrate a loop in the function below. So that the script calculates for each variable within the dataframe df1 the connecting data in df2. Actually it takes only the first row. I hope that's clear. My goal is to apply the function for each data in df1. Many thanks in advance. An example is as follows: df1