similar to: ridge regression

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.

Hallo, I hope I am posting to the right place. I was advised to try this list by Ben Bolker (https://twitter.com/bolkerb/status/859909918446497795). I also posted this question to StackOverflow (http://stackoverflow.com/questions/43771269/lm-gives-different-results-from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first program in 1975 and have been paid to program in about

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.

Hi Simon, Yes, if I uses coefficients() I get the same results for lm() and lm.ridge(). So that's consistent, at least. Interestingly, the "wrong" number I get from lm.ridge()$coef agrees with the value from SPSS to 5dp, which is an interesting coincidence if these numbers have no particular external meaning in lm.ridge(). Kind regards, Nick ----- Original Message -----

I am using GLM to calculate logit models based on cross-sectional data. I am now down to the hard work of making the results intelligible to very average readers. Is there any way to calculate a psuedo analoque to the R^2 in standard linear regression for use as a purely descriptive statistic of goodness of fit? Most of the readers of my report will be vaguely familiar and more comfortable with

Thanks, I was getting to try this, but got side tracked by actual work... Your analysis reproduces the SPSS unscaled estimates. It still remains to figure out how Nick got > coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)) (Intercept) ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA 0.07342198 -0.39650356

Hi John, Thanks for the comment... but that appears to mean that SPSS has a big problem. I have always been told that to include an interaction term in a regression, the only way is to do the multiplication by hand. But then it seems to be impossible to stop SPSS from re-standardizing the variable that corresponds to the interaction term. Am I missing something? Is there a way to perform the

Hi, Here is (I hope) all the relevant output from R. > mean(s1$ZDEPRESSION, na.rm=T) [1] -1.041546e-16 > mean(s1$ZDIVERSITY_PA, na.rm=T) [1] -9.660583e-16 > mean(s1$ZMEAN_PA, na.rm=T) [1] -5.430282e-15 > lm.ridge(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)$coef ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA -0.3962254 -0.3636026

Christoph Lehman had problems with seperated data in two-class logistic regression. One useful little trick is to penalize the logistic regression using a quadratic penalty on the coefficients. I am sure there are functions in the R contributed libraries to do this; otherwise it is easy to achieve via IRLS using ridge regressions. Then even though the data are separated, the penalized

Hello. I have an XYY series that I would like to graph with matplot() or some other single function that will do the trick. The X in question is a vector of POSIXt values obtained from strptime(). Is it possible to tell matplot() how to handle POSIXt x values? I have examined the examples at http://lark.cc.ukans.edu/~pauljohn/R/statsRus.html#5.22 , but would prefer not have to overlay the

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)

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

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

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 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

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

Apologies if this is this too obscure for R-help. In package splines, ns(x,,knots,intercept=TRUE) produces an n by K+2 matrix N, the values of K+2 basis functions for the natural splines with K (internal) knots, evaluated at x. It does this by first generating an n by K+4 matrix B of unconstrained splines, then postmultiplying B by H, a K+4 by K+2 representation of the nullspace of C (2 by K+4),

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

Dear Nick, On 2017-05-05, 9:40 AM, "R-devel on behalf of Nick Brown" <r-devel-bounces at r-project.org on behalf of nick.brown at free.fr> wrote: >>I conjecture that something in the vicinity of >> res <- lm(DEPRESSION ~ scale(ZMEAN_PA) + scale(ZDIVERSITY_PA) + >>scale(ZMEAN_PA * ZDIVERSITY_PA), data=dat) >>summary(res) >> would reproduce the

Dear all, I am a newcomer to R. I intend to using R to do stepwise regression and PLS with a data set (a 55x20 matrix, with one dependent and 19 independent variable). Based on the same data set, I have done the same work using SPSS and SAS. However, there is much difference between the results obtained by R and SPSS or SAS. In the case of stepwise, SPSS gave out a model with 4 independent