[Rd] complex tests failure

[Kasper Daniel Hansen]

For a while I have been getting that the complex tests fails on RHEL 6.
The specific issue has to do with tanh (see below for full output from
complex.Rout.fail).

This is both with the stock compiler (GCC 4.4.7) and a compiler supplied
through the conda project (GCC 4.8.5).  The compiler supplied through conda
ends up linking R to certain system files, so the binary is not completely
independent (although most dynamically linked libraries are coming from the
conda installation).

A search on R-devel reveals a discussion in April on an issue reported on
Windows with a bug in tanh in old versions of the GNU C standard library;
this seems relevant.  The discussion by Martin Maechler suggest "using R's
internal substitute".  So how do I enable this?  Or does this requires
updating the C standard library?

** From complex.Rout.fail

> stopifnot(identical(tanh(356+0i), 1+0i))
Error: identical(tanh(356 + (0+0i)), 1 + (0+0i)) is not TRUE
In addition: Warning message:
In tanh(356 + (0+0i)) : NaNs produced in function "tanh"
Execution halted

Best,
Kasper

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Re: [Rd] complex tests failure

[Tomas Kalibera]

As a quick fix, you can undefine HAVE_CTANH in complex.c, somewhere 
after including config.h

An internal substitute, which is implemented inside complex.c, will be used.

Best
Tomas



On 05/04/2017 02:57 PM, Kasper Daniel Hansen wrote:

For a while I have been getting that the complex tests fails on RHEL 6.
The specific issue has to do with tanh (see below for full output from
complex.Rout.fail).

This is both with the stock compiler (GCC 4.4.7) and a compiler supplied
through the conda project (GCC 4.8.5).  The compiler supplied through conda
ends up linking R to certain system files, so the binary is not completely
independent (although most dynamically linked libraries are coming from the
conda installation).

A search on R-devel reveals a discussion in April on an issue reported on
Windows with a bug in tanh in old versions of the GNU C standard library;
this seems relevant.  The discussion by Martin Maechler suggest "using R's
internal substitute".  So how do I enable this?  Or does this requires
updating the C standard library?

** From complex.Rout.fail


stopifnot(identical(tanh(356+0i), 1+0i))

Error: identical(tanh(356 + (0+0i)), 1 + (0+0i)) is not TRUE
In addition: Warning message:
In tanh(356 + (0+0i)) : NaNs produced in function "tanh"
Execution halted

Best,
Kasper

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Re: [Rd] complex tests failure

[Kasper Daniel Hansen]

Thanks.

I assume there is no way to control this via. environment variables or
configure settings?  Obviously that would be great for something like this
which affects tests and seems to be a known problem for older C standard
libraries.

Best,
Kasper

On Thu, May 4, 2017 at 9:12 AM, Tomas Kalibera 
wrote:

>
> As a quick fix, you can undefine HAVE_CTANH in complex.c, somewhere after
> including config.h
> An internal substitute, which is implemented inside complex.c, will be
> used.
>
> Best
> Tomas
>
>
>
>
> On 05/04/2017 02:57 PM, Kasper Daniel Hansen wrote:
>
>> For a while I have been getting that the complex tests fails on RHEL 6.
>> The specific issue has to do with tanh (see below for full output from
>> complex.Rout.fail).
>>
>> This is both with the stock compiler (GCC 4.4.7) and a compiler supplied
>> through the conda project (GCC 4.8.5).  The compiler supplied through
>> conda
>> ends up linking R to certain system files, so the binary is not completely
>> independent (although most dynamically linked libraries are coming from
>> the
>> conda installation).
>>
>> A search on R-devel reveals a discussion in April on an issue reported on
>> Windows with a bug in tanh in old versions of the GNU C standard library;
>> this seems relevant.  The discussion by Martin Maechler suggest "using R's
>> internal substitute".  So how do I enable this?  Or does this requires
>> updating the C standard library?
>>
>> ** From complex.Rout.fail
>>
>> stopifnot(identical(tanh(356+0i), 1+0i))
>>>
>> Error: identical(tanh(356 + (0+0i)), 1 + (0+0i)) is not TRUE
>> In addition: Warning message:
>> In tanh(356 + (0+0i)) : NaNs produced in function "tanh"
>> Execution halted
>>
>> Best,
>> Kasper
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-devel@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>
>
>

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Re: [Rd] complex tests failure

[Tomas Kalibera]

There is no way to control this at runtime.
We will probably have to add a configure test.

Best,
Tomas

On 05/04/2017 03:23 PM, Kasper Daniel Hansen wrote:
> Thanks.
>
> I assume there is no way to control this via. environment variables or 
> configure settings?  Obviously that would be great for something like 
> this which affects tests and seems to be a known problem for older C 
> standard libraries.
>
> Best,
> Kasper
>
> On Thu, May 4, 2017 at 9:12 AM, Tomas Kalibera 
> mailto:tomas.kalib...@gmail.com>> wrote:
>
>
> As a quick fix, you can undefine HAVE_CTANH in complex.c,
> somewhere after including config.h
> An internal substitute, which is implemented inside complex.c,
> will be used.
>
> Best
> Tomas
>
>
>
>
> On 05/04/2017 02:57 PM, Kasper Daniel Hansen wrote:
>
> For a while I have been getting that the complex tests fails
> on RHEL 6.
> The specific issue has to do with tanh (see below for full
> output from
> complex.Rout.fail).
>
> This is both with the stock compiler (GCC 4.4.7) and a
> compiler supplied
> through the conda project (GCC 4.8.5).  The compiler supplied
> through conda
> ends up linking R to certain system files, so the binary is
> not completely
> independent (although most dynamically linked libraries are
> coming from the
> conda installation).
>
> A search on R-devel reveals a discussion in April on an issue
> reported on
> Windows with a bug in tanh in old versions of the GNU C
> standard library;
> this seems relevant.  The discussion by Martin Maechler
> suggest "using R's
> internal substitute".  So how do I enable this?  Or does this
> requires
> updating the C standard library?
>
> ** From complex.Rout.fail
>
> stopifnot(identical(tanh(356+0i), 1+0i))
>
> Error: identical(tanh(356 + (0+0i)), 1 + (0+0i)) is not TRUE
> In addition: Warning message:
> In tanh(356 + (0+0i)) : NaNs produced in function "tanh"
> Execution halted
>
> Best,
> Kasper
>
> [[alternative HTML version deleted]]
>
> __
> R-devel@r-project.org  mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
> 
>
>
>
>


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

[Nick Brown]

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

[Duncan Murdoch]

On 04/05/2017 10:28 AM, Nick Brown wrote:

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.


I don't know about this instance, but a common cause of this sort of 
difference is a different parametrization.  If that's the case, then 
predictions in the two systems would match, even if coefficients don't.


Duncan Murdoch




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

[peter dalgaard]

Um, the link to StackOverflow does not seem to contain the same question. It 
does contain a stern warning not to use the $coef component of lm.ridge...

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.

Also notice that SPSS tends to use "Beta" for standardized regression 
coefficients, and (AFAIR) "b" for the regular ones.

-pd

> On 4 May 2017, at 16:28 , Nick Brown  wrote:
> 
> 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

-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd@cbs.dk  Priv: pda...@gmail.com

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

[Simon Bonner]

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()

## 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

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

[Nick Brown]

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"  
To: "Nick Brown" , 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() 

## 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 
> 
> 
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