similar to: using variables in obj$model

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:

I've redirected this reply from r-help to the bugs list. On 11/3/2006 8:25 AM, vito muggeo wrote: > 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)) >

Dear all, I'm looking for someone that help me to write an R function to simulate survival data under complex situations, namely time-varying hazard ratio, marginal distribution of survival times and covariates. The algorithm is described in the reference below and it should be not very difficult to implement it. However I tried but without success....;-( Below there the code that I used; it

Hello I'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

Hi, How can I specify a Cox proportional hazards model with a covariate which i believe its strength on survival changes/diminishes with time? The value of the covariate was only recorded once at the beginning of the study for each individual (e.g. at the diagnosis of the disease), so I do not have the time course data of the covariate for any given individual. For example, I want to state at the

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 R-users, I've built these functions usefell for me to import/export data from/to Excel: importa.da.excel<-function(){read.delim2("clipboard", dec=",") ## questa funzione consente di importare dati da Excel in R ## selezionare in Excel le celle che contengono i dati, ## compresi in nomi delle colonne ## Autore: Vito Ricci email:vito_ricci at yahoo.com ## Data di

Dear all, I would like to write a function like: myfun<-function(x,fn) {xx<-exp(x); x*fn(xx)} where fn is a symbolic description of any function with its argument to be specified. Therefore myfun(5,"2+0.3*y^2") should return 5*(2+0.3*exp(5)^2), myfun(5,"log(y)") should return 5*log(exp(5)) and so on. I tried with "expression" and others, but without success.

DeaR R-useRs, I'm trying to fit a logist model with these data: > dati y x 1 1 37 2 1 35 3 1 33 4 1 40 5 1 45 6 1 41 7 1 42 8 0 20 9 0 21 10 0 25 11 0 27 12 0 29 13 0 18 I use glm(), having this output: > g<-glm(y~x,family=binomial,data=dati) Warning messages: 1: Algorithm did not converge in: glm.fit(x = X, y = Y, weights = weights, start = start, etastart =

dear all, Given a matrix A, say, I would like to apply a bivariate function to each combination of its colums. That is if myfun<-function(x,y)cor(x,y) #computes simple correlation of two vectors x and y then the results should be something similar to cor(A). I tried with mapply, outer,...but without success Can anybody help me? many thanks in advance, vito

Dear all, This is a pure statistical question, not necessarly related to R. I could not find it in literature. Suppose I'm intersted in a parameter rho, say, equal to: r=beta1/beta2, where beta1 and beta2 come from a linear model y=beta0+beta1X1+beta2X2+.... Fitting the model I can get the (biased) estimate of r=b1/b2, where b1 and b2 are the estimates in the regression model; I can get the

Dear all, The help file for the generic function vcov states "Classes with methods for this function include: 'lm', 'glm', 'nls', 'lme', 'gls', 'coxph' and 'survreg' (the last two in package 'survival')." Since, I am not able to use vcov.coxph(), I am wondering whether I am missing something (as I suspect..) regards, vito

Hi, you can address to a single ts in a multivariate ts object by namets[,index]. See this example: > dati X Y 1 100 200 2 150 210 3 180 220 4 200 230 5 220 250 > serie<-ts(dati,start=1999) > serie Time Series: Start = 1999 End = 2003 Frequency = 1 X Y 1999 100 200 2000 150 210 2001 180 220 2002 200 230 2003 220 250 > serie[,1] ## first ts Time Series: Start =

Hi. I have to forecast a time series of a Internet network traffic bitrate. The data are in file http://www.forumaltavilla.it/joomla/datitesi/dati.datand the sampling time is every 0.05 seconds. Now, i want to use HoltWinters forecasting. This is my script. dt=1.58443823e-9 #0.05 seconds in years dati.ts=ts(scan("dati.dat"),start=0,deltat=dt) model=HoltWinters(dati.ts)

Dear All, I'm happily extracting data of temperature from an oracle db under R via RODBC. After manipulating the extracted data I put them into a data.frame 'dati' which is as follows: > dati DATA tm. UDINE/RIVOLTO tm.TORINO/CASELLE 1 2005-07-01 22.35 23.80 2 2005-07-02 22.70 22.85 3 2005-07-03 23.80

I'm learning how to use tapply. Now I'm having a go at the following code in which dati contains almost 600 lines, Pot - numeric - are the capacities of power plants and SGruppo - text - the corresponding six technologies ("CCC", "CIC","TGC", "CSC","CPC", "TE"). .....................................................

Have you seen : ( centos/redhat ) https://outsideit.net/realmd-sssd-ad-authentication/ ( debian/ubuntu ) http://www.alandmoore.com/blog/2015/05/06/joining-debian-8-to-active-directory/ but i must say, i havent tested/tried these, i dont use sssd. But i think these are usefull for you to read at least. If you use the debian variant, you may need to install also : One or more of these :

Hi all, I'd like to fit model where the terms both are in the data.frame, mydata say, and are vectors *not in the data.frame*. >obj<-glm(y~x, data=mydata) #works >Z<-pmax(mydata$x-20,0) >(length(Z)==length(obj$y)) >[1] TRUE >update(obj,.~.+Z) #works However for some subset it doesn't works: >obj<-glm(y~x, data=mydata, subset=f==1) #works

On a Centos 7 minimal fresh install and samba 4.4.4 I have follow this howto: http://www.hexblot.com/blog/centos-7-active-directory-and-samba and I have Joining to an Active Directory server and login to it with domain user without problem. My problem occur when I try from windows to modify some new rights (ACL's) to new folder on samba share. The folder is created correctly but if I add

dear all, Is it possible to extract the response vector from a fitted arima object? For instance in glm it is allowed, by: obj.glm<-glm(y~x) obj.glm$y #the response vector In arima I can't find it: obj.arima<-arima(y, order=c(1,0,1)) #say names(obj.arima) doesn't seem to include the response. Am I wrong? Many thanks for your help, best, vito