similar to: treatment of factors and errors in ridge() function with coxph

Hi, I have a character vector that contains the names of several objects that I would like to pass to a function (specifically, the ridge function in the survival package, but cbind is a similar example). I've been struggling with how to do this so that the object values get interpreted by the function, rather than the object names. For example, x1 <- 1:4 x2 <- 2:5 x3 <-

Hi, I am using coxph from the survival package to fit a large model (100,000 observations, ~35 covariates) using both ridge regression (on binary covariates) and penalized splines (for continuous covariates). In fitting, I get a strange error: Error in if (abs((y[nx] - target)/(y[nx - 1] - target)) > 0.6) doing.well <- FALSE else doing.well <- TRUE : missing value where

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$loglik[2] is unpenalized log-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected.

Dear all, I need to calculate likelihood ratio test for ridge regression. In February I have reported a bug where coxph returns unpenalized log-likelihood for final beta estimates for ridge coxph regression. In high-dimensional settings ridge regression models usually fail for lower values of lambda. As the result of it, in such settings the ridge regressions have higher values of lambda (e.g.

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

I need to add a belated acknowledgement to my prior comments. Thomas Lumley has also been a significant contributer to the survival code. Until I moved my development from Splus to R he was the primary maintainer of the R code, he tightened up a lot of the C code to make sure it would work on multiple architectures, and added several functions. Merging our two code bases back together was a

For an application of ridge regression, I need to get the covariance matrices of the estimated regression coefficients in addition to the coefficients for all values of the ridge contstant, lambda. I've studied the code in MASS:::lm.ridge, but don't see how to do this because the code is vectorized using one svd calculation. The relevant lines from lm.ridge, using X, Y are:

Hi, I'm a beginner in the field, I have to perform the ridge regression with lm.ridge for many datasets, and I wanted to do it in an automatic way. In which way I can automatically choose lambda ? As said, right now I'm using lm.ridge MASS function, which I found quite simple and fast, and I've seen that among the returned values there are HKB estimate of the ridge constant and L-W

Dear all, I want to get the likelihood (or AIC or BIC) of a ridge regression model using lm.ridge from the MASS library. Yet, I can't really find it. As lm.ridge does not return a standard fit object, it doesn't work with functions like e.g. BIC (nlme package). Is there a way around it? I would calculate it myself, but I'm not sure how to do that for a ridge regression. Thank you in

Hi Nick, I think that the problem here is your use of $coef to extract the coefficients of the ridge regression. The help for lm.ridge states that coef is a "matrix of coefficients, one row for each value of lambda. Note that these are not on the original scale and are for use by the coef method." I ran a small test with simulated data, code is copied below, and indeed the output from

I asked you before, but in case you missed it: Are you looking at the right place in SPSS output? The UNstandardized coefficients should be comparable to R, i.e. the "B" column, not "Beta". -pd > On 5 May 2017, at 01:58 , Nick Brown <nick.brown at free.fr> wrote: > > Hi Simon, > > Yes, if I uses coefficients() I get the same results for lm() and

Hello, I have posted this mail a few days ago but I did it wrong, I hope is right now: I have the following doubts related with lm.ridge, from MASS package. To show the problem using the Longley example, I have the following doubts: First: I think coefficients from lm(Employed~.,data=longley) should be equal coefficients from lm.ridge(Employed~.,data=longley, lambda=0) why it does not happen?

Hi, i have now a retriever of Rock Ridge names from ISO directory records and their eventual Continuation Areas. Further i have a detector for SUSP and Rock Ridge signatures. Both have been tested in libisofs by comparing their results with the Rock Ridge info as perceived by the library. 50 ISO images tested. Some bugs repaired. Now they are in sync. (The macro case

Dear all I'm familiarising myself with Ridge Regressions in R and the following is bugging me: How does one get p-values for the coefficients obtained from MASS::lm.ridge() output (for a given lambda)? Consider the example below (adapted from PRA [1]): > require(MASS) > data(longley) > gr <- lm.ridge(Employed ~ .,longley,lambda = seq(0,0.1,0.001)) > plot(gr) > select(gr)

Dear all, I considered an ordinary ridge regression problem. I followed three different ways: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-t(as.matrix(cbind(2,3,4,5))) n<-nrow(X) p<-ncol(X) #Without

I had no problems running regression models in SPSS and R that yielded the same results for these data. The difference you are observing is from fitting different models. In R, you fitted: res <- lm(DEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=dat) summary(res) The interaction term is the product of ZMEAN_PA and ZDIVERSITY_PA. This is not a standardized variable itself and not the same as

I am trying to understand the code for lm.ridge from the MASS package. Here is the part I am having trouble understanding: if(Inter <- attr(Terms, "intercept")) { Xm <- colMeans(X[, -Inter]) Ym <- mean(Y) p <- p - 1 X <- X[, -Inter] - rep(Xm, rep(n, p)) Y <- Y - Ym } else Ym <- Xm <- NA Xscale <- drop(rep(1/n, n) %*% X^2)^0.5 X <- X/rep(Xscale, rep.int(n,

Dear all, For an ordinary ridge regression problem, I followed three different approaches: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-as.matrix(c(2,3,4,5)) n<-nrow(X) p<-ncol(X) #Without standardization

Dear R users. I have a doubt about the use of the sequence option on Ridge regression. I'm trying to understand the use of this option when variables are highly linear correlated. I'm running a model where the variables HtShoes and Ht have high VIF values. My program is written below, but I'm not sure about the correct way of using the sequence option: library (faraway) data (seatpos)