[Rd] lm() gives different results to lm.ridge() and SPSS

Viechtbauer Wolfgang (SP) wolfgang.viechtbauer at maastrichtuniversity.nl
Fri May 5 14:43:57 CEST 2017

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)

The interaction term is the product of ZMEAN_PA and ZDIVERSITY_PA. This is not a standardized variable itself and not the same as "ZINTER_PA_C" in the png you showed, which is not a variable in the dataset, but can be created with:

dat$ZINTER_PA_C <- with(dat, scale(ZMEAN_PA * ZDIVERSITY_PA))

If you want the same results as in SPSS, then you need to fit:


This yields:

              Estimate Std. Error t value Pr(>|t|)    
(Intercept)    6.41041    0.01722  372.21   <2e-16 ***
ZMEAN_PA      -1.62726    0.04200  -38.74   <2e-16 ***
ZDIVERSITY_PA -1.50082    0.07447  -20.15   <2e-16 ***
ZINTER_PA_C   -0.58955    0.05288  -11.15   <2e-16 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Exactly the same as in the png.

Peter already mentioned this as a possible reason for the discrepancy: https://stat.ethz.ch/pipermail/r-devel/2017-May/074191.html ("Is it perhaps the case that x1 and x2 have already been scaled to have standard deviation 1? In that case, x1*x2 won't be.")


-----Original Message-----
From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Nick Brown
Sent: Friday, May 05, 2017 10:40
To: peter dalgaard
Cc: r-devel at r-project.org
Subject: Re: [Rd] lm() gives different results to lm.ridge() and SPSS


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             -0.1425772      ## This is what I thought was the problem originally. :-) 

> coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)) (Intercept)               ZMEAN_PA          ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA 
            0.07342198            -0.39650356            -0.36569488            -0.09435788 > coefficients(lm.ridge(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)) ZMEAN_PA          ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA 
            0.07342198            -0.39650356            -0.36569488            -0.09435788 The equivalent from SPSS is attached. The unstandardized coefficients in SPSS look nothing like those in R. The standardized coefficients in SPSS match the lm.ridge()$coef numbers very closely indeed, suggesting that the same algorithm may be in use. 

I have put the dataset file, which is the untouched original I received from the authors, in this Dropbox folder: https://www.dropbox.com/sh/xsebjy55ius1ysb/AADwYUyV1bl6-iAw7ACuF1_La?dl=0. You can read it into R with this code (one variable needs to be standardized and centered; everything else is already in the file): 

s1 <- read.csv("Emodiversity_Study1.csv", stringsAsFactors=FALSE) s1$ZDEPRESSION <- scale(s1$DEPRESSION) 
Hey, maybe R is fine and I've stumbled on a bug in SPSS? If so, I'm sure IBM will want to fix it quickly (ha ha ha). 


----- Original Message -----

From: "peter dalgaard" <pdalgd at gmail.com> 
To: "Nick Brown" <nick.brown at free.fr> 
Cc: "Simon Bonner" <sbonner6 at uwo.ca>, r-devel at r-project.org 
Sent: Friday, 5 May, 2017 10:02:10 AM 
Subject: Re: [Rd] 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". 


> 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 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" <sbonner6 at uwo.ca> 
> To: "Nick Brown" <nick.brown at free.fr>, r-devel at 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$coef),coefficients(fit2)) 
> [,1] [,2] [,3] 
> (Intercept) 0.99276001 NA 0.99276001 
> X1 -0.03980772 -0.04282391 -0.03980772 
> X2 1.11167179 1.06200476 1.11167179 
> -- 
> Simon Bonner 
> Assistant Professor of Environmetrics/ Director MMASc 
> Department of Statistical and Actuarial Sciences/Department of Biology 
> University of Western Ontario 
> Office: Western Science Centre rm 276 
> Email: sbonner6 at uwo.ca | Telephone: 519-661-2111 x88205 | Fax: 519-661-3813 
> Twitter: @bonnerstatslab | Website: http://simon.bonners.ca/bonner-lab/wpblog/ 
>> -----Original Message----- 
>> From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Nick 
>> Brown 
>> Sent: May 4, 2017 10:29 AM 
>> To: r-devel at r-project.org 
>> Subject: [Rd] lm() gives different results to lm.ridge() and SPSS 
>> Hallo, 
>> I hope I am posting to the right place. I was advised to try this list by Ben Bolker 
>> (https://twitter.com/bolkerb/status/859909918446497795). I also posted this 
>> question to StackOverflow 
>> (http://stackoverflow.com/questions/43771269/lm-gives-different-results- 
>> from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first 
>> program in 1975 and have been paid to program in about 15 different 
>> languages, so I have some general background knowledge. 
>> I have a regression from which I extract the coefficients like this: 
>> lm(y ~ x1 * x2, data=ds)$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 

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