From jfox at mcmaster.ca  Fri May  5 20:22:53 2017
From: jfox at mcmaster.ca (Fox, John)
Date: Fri, 5 May 2017 18:22:53 +0000
Subject: [Rd] lm() gives different results to lm.ridge() and SPSS
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Message-ID: <D5323A89.71B2%jfox@mcmaster.ca>

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 SPSS Beta values.
>
>Yes, that works. Thanks!

That you have to work hard in R to match the SPSS results isn?t such a bad
thing when you factor in the observation that standardizing the
interaction regressor, ZMEAN_PA * ZDIVERSITY_PA, separately from each of
its components, ZMEAN_PA and ZDIVERSITY_PA, is nonsense.

Best,
 John

-------------------------------------
John Fox, Professor
McMaster University
Hamilton, Ontario, Canada
Web: http://socserv.mcmaster.ca/jfox/


> 
>
>----- Original Message -----
>
>From: "peter dalgaard" <pdalgd at gmail.com>
>To: "Viechtbauer Wolfgang (SP)"
><wolfgang.viechtbauer at maastrichtuniversity.nl>, "Nick Brown"
><nick.brown at free.fr>
>Cc: r-devel at r-project.org
>Sent: Friday, 5 May, 2017 3:33:29 PM
>Subject: Re: [Rd] 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 -0.36569488 -0.09435788
>
>
>which does not match your output. I suspect that ZMEAN_PA and
>ZDIVERSITY_PA were scaled for this analysis (but the interaction term
>still obviously is not). 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 SPSS Beta values.
>
>
>> On 5 May 2017, at 14:43 , Viechtbauer Wolfgang (SP)
>><wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
>> 
>> 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 "ZINTER_PA_C"
>>in the png you showed, which is not a variable in the dataset, but can
>>be created with: 
>> 
>> dat$ZINTER_PA_C <- with(dat, scale(ZMEAN_PA * ZDIVERSITY_PA))
>> 
>> If you want the same results as in SPSS, then you need to fit:
>> 
>> res <- lm(DEPRESSION ~ ZMEAN_PA + ZDIVERSITY_PA + ZINTER_PA_C,
>>data=dat) 
>> summary(res) 
>> 
>> This yields: 
>> 
>> Coefficients: 
>> Estimate Std. Error t value Pr(>|t|)
>> (Intercept) 6.41041 0.01722 372.21 <2e-16 ***
>> ZMEAN_PA -1.62726 0.04200 -38.74 <2e-16 ***
>> ZDIVERSITY_PA -1.50082 0.07447 -20.15 <2e-16 ***
>> ZINTER_PA_C -0.58955 0.05288 -11.15 <2e-16 ***
>> --- 
>> Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1
>> 
>> Exactly the same as in the png.
>> 
>> Peter already mentioned this as a possible reason for the discrepancy:
>>https://stat.ethz.ch/pipermail/r-devel/2017-May/074191.html ("Is it
>>perhaps the case that x1 and x2 have already been scaled to have
>>standard deviation 1? In that case, x1*x2 won't be.")
>> 
>> Best, 
>> Wolfgang 
>> 
>> -----Original Message-----
>> From: R-devel [mailto:r-devel-bounces at r-project.org] On Behalf Of Nick
>>Brown 
>> Sent: Friday, May 05, 2017 10:40
>> To: peter dalgaard
>> Cc: r-devel at r-project.org
>> Subject: Re: [Rd] lm() gives different results to lm.ridge() and SPSS
>> 
>> Hi, 
>> 
>> Here is (I hope) all the relevant output from R.
>> 
>>> mean(s1$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)$coef ZMEAN_PA ZDIVERSITY_PA
>>>ZMEAN_PA:ZDIVERSITY_PA
>> -0.3962254 -0.3636026 -0.1425772 ## This is what I thought was the
>>problem originally. :-)
>> 
>> 
>>> coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1))
>>>(Intercept) ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA
>> 0.07342198 -0.39650356 -0.36569488 -0.09435788 >
>>coefficients(lm.ridge(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1))
>>ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA
>> 0.07342198 -0.39650356 -0.36569488 -0.09435788 The equivalent from SPSS
>>is attached. The unstandardized coefficients in SPSS look nothing like
>>those in R. The standardized coefficients in SPSS match the
>>lm.ridge()$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).
>> 
>> Nick 
>> 
>> ----- 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".
>> 
>> -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
>>>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 
>> ______________________________________________
>> R-devel at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>
>-- 
>Peter Dalgaard, Professor,
>Center for Statistics, Copenhagen Business School
>Solbjerg Plads 3, 2000 Frederiksberg, Denmark
>Phone: (+45)38153501
>Office: A 4.23 
>Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
>
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[Rd] lm() gives different results to lm.ridge() and SPSS
Fox, John