similar to: glm output format

Suppose we have some glm object such as: myglm <- glm( y ~ x, data=DAT) Is there an elegant way--or the "right way" within the R way of thinking--to obtain the names of the response variable, the predictor variables, and the dataset, as character strings? For instance, suppose the "right way" was to use the (currently fictitious) functions theresponse(), thepredictors(),

Hello, Is there a way to obtain the variance-covariance matrix of the estimated parameters from GLM? my.glm<-glm(mat ~X,family = binomial, data =myDATA) out1<-predict(my.glm,se.fit = TRUE) std<-out1$se.fit se.fit is for getting the standard errors of the estimated parameters (\betas). Is there a way to get the variance-covariance matrix of the estimated parameters? Many thanks,

Hi there I want to pass arguments (i.e. the response variable and the subset argument) in a self-made function to glm. Here is one way I can do this: f.myglm <- function(y,subfact,subval) { glm(d.mydata[,y]~d.mydata[,'x1'],family=binomial,subset=d.mydata[,subfact]==subval) } > str(d.mydata) `data.frame': 15806 obs. of 3 variables: $ y : Factor w/ 2 levels

In summary.glm I'm trying to get a better feel for the z output. The following lines can be found in the function 1 if (p > 0) { 2 p1 <- 1:p 3 Qr <- object$qr 4 coef.p <- object$coefficients[Qr$pivot[p1]] 5 covmat.unscaled <- chol2inv(Qr$qr[p1, p1, drop = FALSE]) 6 dimnames(covmat.unscaled) <- list(names(coef.p), names(coef.p))

Dear all, I have a NIS server which shares a database of users between some computers (nodes exactly) and I would like that, on the first login, the user changes its password. So, on the NIS server I have made: chage -d 0 USER Then: # cd /var/yp # make On the NIS server I have: chage -l USER Last password change : password must be changed Password expires

Is it possible with "chage" to configure a password caducity for, at most, 2 hours? I think "chage" only allows caducity for, at least, one day.

Hi I have a little R problem. I have created a GLM model in R and now I want to predict some values outside the values I have in the model (extrapolate). I have this code: fitted.model4 <- glm(Yval ~ time, family=gaussian, data=Fuel) The question is - How do I predict a value of Yval ie with a value of time = 340 and also get confidence/prediction intervals for Yvar? I have tried the

Greetings: We are evaluating the performance of R matrix algebra es as we port a MATLAB R14 script into R. The MATLAB code basically evaluates the AX=B system on sparse matrices that result in output matrices of 100 to 1,000,000 rows/columns. Our R prototype script uses spase base matrices and the methods qr() and qr.coeff(). The following statements are called inside a doubly-nested loop: G

In the iterative weighted least squares loop of the glm.fit function, a QR decomposition of the matrix X*sqrt(w) is obtained at each iteration, using fitted values of the mean response mu from the previous interation. Here "X" is the design matrix, w = wt*(d(mu)/d(eta))/variance(mu), "wt" is the vector of weights, and "d(mu)/d(eta)" is the derivative of mu wrt the

Version 1.34.6 (on Debian Stretch) Using virt-builder I tried to create a user and inject my ssh key: virt-builder -v -x centos-7.4 --output testimage --format qcow2 --selinux-relabel --hostname testimage --firstboot-command 'useradd -m -p "" dummy' --firstboot-command 'chage -d 0 dummy' --ssh-inject dummy ... [ 16.1] Installing firstboot command: useradd -m -p

G'day all, I had a look at the GLM code of R (1.4.1) and I believe that there are problems with the function "glm.fit" that may bite in rare circumstances. Note, I have no data set with which I ran into trouble. This report is solely based on having a look at the code. Below I append a listing of the glm.fit function as produced by my system. I have added line numbers so that I

We have Centos 5.8 on LInux server.I setup /etc/login.defs following: PASS_MAX_DAYS?? 3 PASS_MIN_DAYS?? 0 PASS_MIN_LEN??? 8 PASS_WARN_AGE?? 1 after that I chack user password policy and it show:# chage -l user1 Last password change??????????????????????????????????? : Jun 29, 2015 Password expires??????????????????????????????????????? : never Password

With this example set.seed(123) A <- matrix(runif(40), nrow = 8) y <- 1:nrow(A) A.laqr <- qr(A, LAPACK=TRUE) both qr.qy(A.laqr,y) and qr.qty(A.laqr,y) give the respective error messages Error in qr.qy(A.laqr, y) : 'b' must be a numeric matrix Error in qr.qty(A.laqr, y) : 'b' must be a numeric matrix However when Lapack is not used as in A.liqr <- qr(A,

On Thu, 19 Apr 2007, brech at delphioutpost.com wrote: > Full_Name: Christian Brechbuehler > Version: 2.4.1 Patched (2007-03-25 r40917) > OS: Linux 2.6.15-27-adm64-xeon; Ubuntu 6.06.1 LTS > Submission from: (NULL) (24.61.47.236) > > > Splus and R have different ideas about what qr.coef(qr()) should return, > which is fine... but I believe that R has a bug in that it is not

Hello, I'm using dual boot system and have windows partitions on my disk. The problem is that I can't write to it as user (copy files etc). I'm also unable to chage permissions or owner. So wine can' access them. I looked up fstab man and related issues but still didn't find solution. Thanks in advance Serge Yastreilov

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-

Dear R people, I do not have much knowledge about linear algebra but currently I need to understand what the function qr.qty is actually doing. The documentation states that it calculates t(Q) %*% y via a previously performed QR matrix decomposition. In order to do that, I tried following basic example: m<-matrix(c(1,0,0,0,1,0,0,0,1,0,0,1),ncol=3) # 4x3 matrix

The following snippet suggests that there is either a bug in qr(,LAPACK=T), or some bug in my understanding. Note that the detected rank is correct (= 2) using the default LINPACK qr, but incorrect (=3) using LAPACK. This is running on Linux Redhat 9.0, using the lapack library that comes with the Redhat distribution. I'm running R 1.7.1 compiled from the source. If the bug is in my

Hi, I'm having some trouble setting methods for the qr family of functions. I believe I have distilled my misunderstanding in the code snippet below: foo <- function(x, ...) UseMethod("foo") foo.default <- function(x) { } # foo setGeneric(name = "foo", useAsDefault = foo) setMethod("foo", signature(x="bar"), function(x, ...) { } )

Might it be possible to tighten the definition of "is.qr". I noticed that after I mistakenly typed example(lm) # make lm object named lm.D9 qr.Q(lm.D9) which exhausted the heap memory and produced two warning messages. As an object of class "lm" has a "qr" component, "is.qr" failed to detect that "lm.D9" was not a "qr" object. The