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 ----- From: "Simon Bonner" <sbonn...@uwo.ca> To: "Nick Brown" <nick.br...@free.fr>, r-devel@r-project.org Sent: Thursday, 4 May, 2017 7:07:33 PM Subject: RE: [Rd] 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 lm.ridge differs depending on whether the coefficients are accessed via $coef or via the coefficients() function. The latter does produce results that match the output from lm. I hope that helps. Cheers, Simon ## Load packages library(MASS) ## Set seed set.seed(8888) ## Set parameters n <- 100 beta <- c(1,0,1) sigma <- .5 rho <- .75 ## Simulate correlated covariates Sigma <- matrix(c(1,rho,rho,1),ncol=2) X <- mvrnorm(n,c(0,0),Sigma=Sigma) ## Simulate data mu <- beta[1] + X %*% beta[-1] y <- rnorm(n,mu,sigma) ## Fit model with lm() fit1 <- lm(y ~ X) ## Fit model with lm.ridge() fit2 <- lm.ridge(y ~ X) ## Compare coefficients cbind(fit1$coefficients,c(NA,fit2$coef),coefficients(fit2)) [,1] [,2] [,3] (Intercept) 0.99276001 NA 0.99276001 X1 -0.03980772 -0.04282391 -0.03980772 X2 1.11167179 1.06200476 1.11167179 -- Simon Bonner Assistant Professor of Environmetrics/ Director MMASc Department of Statistical and Actuarial Sciences/Department of Biology University of Western Ontario Office: Western Science Centre rm 276 Email: sbonn...@uwo.ca | Telephone: 519-661-2111 x88205 | Fax: 519-661-3813 Twitter: @bonnerstatslab | Website: http://simon.bonners.ca/bonner-lab/wpblog/ > -----Original Message----- > From: R-devel [mailto:r-devel-boun...@r-project.org] On Behalf Of Nick > Brown > Sent: May 4, 2017 10:29 AM > To: r-devel@r-project.org > Subject: [Rd] lm() gives different results to lm.ridge() and SPSS > > Hallo, > > I hope I am posting to the right place. I was advised to try this list by Ben > Bolker > (https://twitter.com/bolkerb/status/859909918446497795). I also posted this > question to StackOverflow > (http://stackoverflow.com/questions/43771269/lm-gives-different-results- > from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first > program in 1975 and have been paid to program in about 15 different > languages, so I have some general background knowledge. > > > I have a regression from which I extract the coefficients like this: > lm(y ~ x1 * x2, data=ds)$coef > That gives: x1=0.40, x2=0.37, x1*x2=0.09 > > > > When I do the same regression in SPSS, I get: > beta(x1)=0.40, beta(x2)=0.37, beta(x1*x2)=0.14. > So the main effects are in agreement, but there is quite a difference in the > coefficient for the interaction. > > > X1 and X2 are correlated about .75 (yes, yes, I know - this model wasn't my > idea, but it got published), so there is quite possibly something going on > with > collinearity. So I thought I'd try lm.ridge() to see if I can get an idea of > where > the problems are occurring. > > > The starting point is to run lm.ridge() with lambda=0 (i.e., no ridge > penalty) and > check we get the same results as with lm(): > lm.ridge(y ~ x1 * x2, lambda=0, data=ds)$coef > x1=0.40, x2=0.37, x1*x2=0.14 > So lm.ridge() agrees with SPSS, but not with lm(). (Of course, lambda=0 is > the > default, so it can be omitted; I can alternate 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 all bets are off in > terms > of what the betas mean, but I would really expect lm() and lm.ridge() to give > the same results. (I would be happy to ignore SPSS, but for the moment it's > part of the majority!) > > > > Thanks for reading, > Nick > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-devel@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-devel [[alternative HTML version deleted]] ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel