[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$ZINTER_PA_C <- with(dat, scale(ZMEAN_PA * ZDIVERSITY_PA))
If you want the same results as in SPSS, then you need to fit:
res <- lm(DEPRESSION ~ ZMEAN_PA + ZDIVERSITY_PA + ZINTER_PA_C, data=dat)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.41041 0.01722 372.21 <2e-16 ***
ZMEAN_PA -1.62726 0.04200 -38.74 <2e-16 ***
ZDIVERSITY_PA -1.50082 0.07447 -20.15 <2e-16 ***
ZINTER_PA_C -0.58955 0.05288 -11.15 <2e-16 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Exactly the same as in the png.
Peter already mentioned this as a possible reason for the discrepancy: https://stat.ethz.ch/pipermail/r-devel/2017-May/074191.html ("Is it perhaps the case that x1 and x2 have already been scaled to have standard deviation 1? In that case, x1*x2 won't be.")
From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Nick Brown
Sent: Friday, May 05, 2017 10:40
To: peter dalgaard
Cc: r-devel at r-project.org
Subject: Re: [Rd] lm() gives different results to lm.ridge() and SPSS
Here is (I hope) all the relevant output from R.
> mean(s1$ZDEPRESSION, na.rm=T)  -1.041546e-16 > mean(s1$ZDIVERSITY_PA, na.rm=T)  -9.660583e-16 > mean(s1$ZMEAN_PA, na.rm=T)  -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,
> ----- 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 $coef or via the coefficients() function. The latter does produce results that match the output from lm.
> I hope that helps.
> ## Load packages
> ## Set seed
> ## Set parameters
> n <- 100
> beta <- c(1,0,1)
> sigma <- .5
> rho <- .75
> ## Simulate correlated covariates
> Sigma <- matrix(c(1,rho,rho,1),ncol=2)
> X <- mvrnorm(n,c(0,0),Sigma=Sigma)
> ## Simulate data
> mu <- beta + X %*% beta[-1]
> y <- rnorm(n,mu,sigma)
> ## Fit model with lm()
> fit1 <- lm(y ~ X)
> ## Fit model with lm.ridge()
> fit2 <- lm.ridge(y ~ X)
> ## Compare coefficients
> [,1] [,2] [,3]
> (Intercept) 0.99276001 NA 0.99276001
> X1 -0.03980772 -0.04282391 -0.03980772
> X2 1.11167179 1.06200476 1.11167179
> Simon Bonner
> Assistant Professor of Environmetrics/ Director MMASc
> Department of Statistical and Actuarial Sciences/Department of Biology
> University of Western Ontario
> Office: Western Science Centre rm 276
> Email: sbonner6 at uwo.ca | Telephone: 519-661-2111 x88205 | Fax: 519-661-3813
> Twitter: @bonnerstatslab | Website: http://simon.bonners.ca/bonner-lab/wpblog/
>> -----Original Message-----
>> From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Nick
>> Sent: May 4, 2017 10:29 AM
>> To: r-devel at r-project.org
>> Subject: [Rd] lm() gives different results to lm.ridge() and SPSS
>> I hope I am posting to the right place. I was advised to try this list by Ben Bolker
>> (https://twitter.com/bolkerb/status/859909918446497795). I also posted this
>> question to StackOverflow
>> from-lm-ridgelambda-0). I am a relative newcomer to R, but I wrote my first
>> program in 1975 and have been paid to program in about 15 different
>> languages, so I have some general background knowledge.
>> I have a regression from which I extract the coefficients like this:
>> lm(y ~ x1 * x2, data=ds)$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
>> 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,
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