search for: ridges

Displaying 20 results from an estimated 509 matches for "ridges".

Did you mean: bridges
2009 Aug 01
2
Cox ridge regression
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
2017 May 04
4
lm() gives different results to lm.ridge() and SPSS
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
2017 May 04
2
lm() gives different results to lm.ridge() and SPSS
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 -----
2005 Aug 24
1
lm.ridge
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?
2009 Mar 17
1
Likelihood of a ridge regression (lm.ridge)?
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
2011 Nov 24
3
How to deal with package conflicts
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
2011 Aug 06
0
ridge regression - covariance matrices of ridge coefficients
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:
2013 Mar 31
1
Rock Ridge for core/fs/iso9660
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
2017 May 05
6
lm() gives different results to lm.ridge() and SPSS
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
2017 May 05
1
lm() gives different results to lm.ridge() and SPSS
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
2007 Apr 12
1
Question on ridge regression with R
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
2017 May 05
0
lm() gives different results to lm.ridge() and SPSS
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
2010 Dec 09
1
survival: ridge log-likelihood workaround
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.
2009 Jun 04
0
help needed with ridge regression and choice of lambda with lm.ridge!!!
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
2017 May 05
0
lm() gives different results to lm.ridge() and SPSS
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
2017 May 04
0
lm() gives different results to lm.ridge() and SPSS
Hi Nick, I think that the problem here is your use of $coef to extract the coefficients of the ridge regression. The help for lm.ridge states that coef is a "matrix of coefficients, one row for each value of lambda. Note that these are not on the original scale and are for use by the coef method." I ran a small test with simulated data, code is copied below, and indeed the output from
2010 Dec 02
0
survival - summary and score test for ridge coxph()
It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version
2017 May 05
1
lm() gives different results to lm.ridge() and SPSS
Hi John, Thanks for the comment... but that appears to mean that SPSS has a big problem. I have always been told that to include an interaction term in a regression, the only way is to do the multiplication by hand. But then it seems to be impossible to stop SPSS from re-standardizing the variable that corresponds to the interaction term. Am I missing something? Is there a way to perform the
2010 Feb 16
1
survival - ratio likelihood for ridge coxph()
It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$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.
2017 May 05
0
lm() gives different results to lm.ridge() and SPSS
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