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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

...ate between including or deleting ".ridge" in the function call, and watch the coefficient for the interaction change.) What seems slightly strange to me here is that I assumed that lm.ridge() just piggybacks on lm() anyway, so in the specific case where lambda=0 and there is no "ridging" to do, I'd expect exactly the same results. Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will not be easy to make, but I can share the data via Dropbox or something if that would help. I appreciate that when there is strong collinearity then...

...t;.ridge" in the function call, and watch the coefficient for the interaction > change.) > > > > What seems slightly strange to me here is that I assumed that lm.ridge() just > piggybacks on lm() anyway, so in the specific case where lambda=0 and there > is no "ridging" to do, I'd expect exactly the same results. > > > Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will > not be easy to make, but I can share the data via Dropbox or something if that > would help. > > > > I apprecia...

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?

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

In my genridge package, I define a function ridge() for ridge regression, creating objects of class 'ridge' that I intend to enhance. In a documentation example, I want to use some functions from the car package. However, that package requires survival, which also includes a ridge() function, for coxph models. So, once I require(car) my ridge() function is masked, which means I have to

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 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

...on call, and watch the coefficient for the interaction >> change.) >> >> >> >> What seems slightly strange to me here is that I assumed that lm.ridge() just >> piggybacks on lm() anyway, so in the specific case where lambda=0 and there >> is no "ridging" to do, I'd expect exactly the same results. >> >> >> Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will >> not be easy to make, but I can share the data via Dropbox or something if that >> would help. >> &gt...

...on call, and watch the coefficient for the interaction >>> change.) >>> >>> What seems slightly strange to me here is that I assumed that lm.ridge() just >>> piggybacks on lm() anyway, so in the specific case where lambda=0 and there >>> is no "ridging" to do, I'd expect exactly the same results. >>> >>> Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will >>> not be easy to make, but I can share the data via Dropbox or something if that >>> would help. >&gt...

Hi, I am working on a project about hospital efficiency. Due to the high multicolinearlity of the data, I want to fit the model using ridge regression. However, I believe that the data from large hospital(indicated by the number of patients they treat a year) is more accurate than from small hosptials, and I want to put more weight on them. How do I do this with lm.ridge? I know I just need

...on call, and watch the coefficient for the interaction >> change.) >> >> >> >> What seems slightly strange to me here is that I assumed that lm.ridge() just >> piggybacks on lm() anyway, so in the specific case where lambda=0 and there >> is no "ridging" to do, I'd expect exactly the same results. >> >> >> Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will >> not be easy to make, but I can share the data via Dropbox or something if that >> would help. >> &gt...

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.

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

..." in the function call, and watch the coefficient for the interaction >> change.) >> >> What seems slightly strange to me here is that I assumed that lm.ridge() just >> piggybacks on lm() anyway, so in the specific case where lambda=0 and there >> is no "ridging" to do, I'd expect exactly the same results. >> >> Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will >> not be easy to make, but I can share the data via Dropbox or something if that >> would help. >> >> I ap...

...".ridge" in the function call, and watch the coefficient for the interaction > change.) > > > > What seems slightly strange to me here is that I assumed that lm.ridge() just > piggybacks on lm() anyway, so in the specific case where lambda=0 and there > is no "ridging" to do, I'd expect exactly the same results. > > > Unfortunately there are 34,000 cases in the dataset, so a "minimal" reprex will > not be easy to make, but I can share the data via Dropbox or something if that > would help. > > > > I appreciate t...

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

...>>>> change.) >>>> >>>> What seems slightly strange to me here is that I assumed that >>>>lm.ridge() just >>>> piggybacks on lm() anyway, so in the specific case where lambda=0 and >>>>there >>>> is no "ridging" to do, I'd expect exactly the same results. >>>> >>>> Unfortunately there are 34,000 cases in the dataset, so a "minimal" >>>>reprex will >>>> not be easy to make, but I can share the data via Dropbox or >>>>someth...

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.

...ion >>>> change.) >>>> >>>> What seems slightly strange to me here is that I assumed that >>>>lm.ridge() just >>>> piggybacks on lm() anyway, so in the specific case where lambda=0 and >>>>there >>>> is no "ridging" to do, I'd expect exactly the same results. >>>> >>>> Unfortunately there are 34,000 cases in the dataset, so a "minimal" >>>>reprex will >>>> not be easy to make, but I can share the data via Dropbox or >>>>something...