similar to: using names with functions..

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, 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, 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, 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, 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, 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 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)

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,

With version 8 of acrobat reader, it is now possible to have 3D in PDf documents. Does it exist already an R package who manage to produce 3D plots which can be saved as interactive 3D graphs in a PDF file? Best Regards Bruno Cavestro ------------------------------------------------------ Leggi GRATIS le tue mail con il telefonino i-mode? di Wind http://i-mode.wind.it/

I have estimated a multiple nonparametric regression using the loess command in R. I have also estimated an additive version of the model using the gam function. Is there a way of using the output of these two models to test the restrictions imposed by the additive model?

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 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, 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

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

Hi all, I am using the package ?Segmented? to estimate logistic regression models with unknown breakpoints (see Muggeo 2003 Statistics in Medicine 22:3055-3071). In the documentation it suggests that it might be possible to include several variables with breakpoints in the same model: ?Z = a vector or a matrix meaning the (continuous) explanatory variable(s) having segmented relationships with

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,

Hi all, I'm having sometrouble with managing the seach path, in a function , I need to attach some data set at the begining and detach them at the end, say, myfunction<- function() { attach(mylist); ............. detach(mylist) } , the problem is, since I am still debugging this code, sometimes it got error and ended before reaching the end, thus the data is left in the