[R] lm.ridge fit with some dummy variables (w/o intercept)
I have columns which sum to one. They are membership dummies, fractions are
allowed - I made an example:
x - c( 9.899898,6.9555431,-1.251,0.5200,0.480,0.000,-2.2384737,
16.791361,6.8924369,-3.286,0.78846154,0.2115385,0.000,-0.4720061,
6.115735,-5.8381799,-1.176,1.,0.000,0.000,-0.6312019,
10.325595,5.4950276,-2.634,1.,0.000,0.000,1.4729420,
3.800141,4.1287662,-2.243,0.8300,0.170,0.000,,0.9314859,
2.159567,-2.3952889,-4.645,0.5300,0.000,0.470,0.7252069,
21.536111,3.3844964,-4.352,1.,0.000,0.000,-0.9931833,
7.526573,-1.1675684,-5.023,1.,0.000,0.000,0.2397390,
28.684897,-0.4594389,-3.233,0.8900,0.070,0.040,0.6017004,
0.894931,-0.9059129,-5.023,0.04347826,0.000,0.9565217,0.6081505)
x = matrix(x,ncol=10)
x = t(x)
colnames(x) = c('a','b','c','d','e','f','y')
x
a b c d e f y
[1,] 9.899898 6.9555431 -1.251 0.5200 0.480 0.000 -2.2384737
[2,] 16.791361 6.8924369 -3.286 0.78846154 0.2115385 0.000 -0.4720061
[3,] 6.115735 -5.8381799 -1.176 1. 0.000 0.000 -0.6312019
[4,] 10.325595 5.4950276 -2.634 1. 0.000 0.000 1.4729420
[5,] 3.800141 4.1287662 -2.243 0.8300 0.170 0.000 0.9314859
[6,] 2.159567 -2.3952889 -4.645 0.5300 0.000 0.470 0.7252069
[7,] 21.536111 3.3844964 -4.352 1. 0.000 0.000 -0.9931833
[8,] 7.526573 -1.1675684 -5.023 1. 0.000 0.000 0.2397390
[9,] 28.684897 -0.4594389 -3.233 0.8900 0.070 0.040 0.6017004
[10,] 0.894931 -0.9059129 -5.023 0.04347826 0.000 0.9565217 0.6081505
apply(x[,4:6],1,sum)
[1] 1 1 1 1 1 1 1 1 1 1
I am trying to use lm.ridge and got some problems on how to extract
parameter estimates. E.g., for lambda = 0 case (I cut and pasted at the
bottom), how to backout the coef estimate to match them with lm fit? In
general, for any given lambda, how to back out the original scale coef
estimates?
lm.fit = lm(y~.-a-1,data=data.frame(x),weights=a)
ridge.fit = lm.ridge(y~.-a-1,data=data.frame(x),weights=a,lambda=0)
ridge.fit
b c d e f
0.1125886 0.1748883 0.9122774 -5.9140208 1.8784332
lm.fit
Call:
lm(formula = y ~ . - a - 1, data = data.frame(x), weights = a)
Coefficients:
b c d e f
0.04232 0.32343 1.36039 -4.67399 3.29727
Thanks so much in advance!
Young
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Re: [R] lm.ridge
On Sat, 8 Oct 2005, Erin Hodgess wrote: Dear R People: I have a question about the lm.ridge function, please. In the example, there is one set of output values in the select function but another in the comment section. Am I missing something please? The values in the examples were computed in S-PLUS. Apparently the dataset in R is not the same as that in S-PLUS. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] lm.ridge
Dear R People: I have a question about the lm.ridge function, please. In the example, there is one set of output values in the select function but another in the comment section. Am I missing something please? R Version 2.1.1 Windows Thanks, Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] lm.ridge
G'day Daniel,
DR == daniel [EMAIL PROTECTED] writes:
DR First: I think coefficients from lm(Employed~.,data=longley)
DR should be equal coefficients from
DR lm.ridge(Employed~.,data=longley, lambda=0) why it does not
DR happen?
Which version of R and which version of MASS are you using?
lm(Employed~.,data=longley)
Call:
lm(formula = Employed ~ ., data = longley)
Coefficients:
(Intercept) GNP.deflator GNPUnemployed Armed.Forces
-3.482e+03 1.506e-02-3.582e-02-2.020e-02-1.033e-02
Population Year
-5.110e-02 1.829e+00
lm.ridge(Employed~.,data=longley, lambda=0)
GNP.deflator GNPUnemployed Armed.Forces
-3.482259e+03 1.506187e-02 -3.581918e-02 -2.020230e-02 -1.033227e-02
Population Year
-5.110411e-02 1.829151e+00
These coefficients look pretty identical to me, except that they are
printed to different numbers of significant digits.
In fact, the following shows that they are identical (upto numerical
precision):
fm1 - lm(Employed~.,data=longley)
fm2 - lm.ridge(Employed~.,data=longley, lambda=0)
coef2 - print(fm2)
GNP.deflator GNPUnemployed Armed.Forces
-3.482259e+03 1.506187e-02 -3.581918e-02 -2.020230e-02 -1.033227e-02
Population Year
-5.110411e-02 1.829151e+00
max(abs(coef(fm1)-coef2))
[1] 7.275958e-12
DR Second: if I have for example Ridge-lm.ridge(Employed~.,
DR data=longley, lambda = seq(0,0.1,0.001)), I suppose intercept
DR coefficient is defined implicit,
Yes.
DR why it does not appear in Ridge$coef?
If you look at the code of lm.ridge, you will see that, if an
intercept is included in the model, all non-constant regressors are
centered (i.e. made orthogonal to the intercept term) and scaled to
have the same variance. Further more, the intercept term is typically
*not* penalised. The components in Ridge$coef are the coefficients on
this transformed scale. No need of including the intercept here,
since it is the same for all values of lambda. If you print the
model, then the ridge coefficients on the original scale are
calculated, see:
getAnywhere(print.ridgelm)
A single object matching 'print.ridgelm' was found
It was found in the following places
registered S3 method for print from namespace MASS
namespace:MASS
with value
function (x, ...)
{
scaledcoef - t(as.matrix(x$coef/x$scales))
if (x$Inter) {
inter - x$ym - scaledcoef %*% x$xm
scaledcoef - cbind(Intercept = inter, scaledcoef)
}
print(drop(scaledcoef), ...)
}
environment: namespace:MASS
DR Third: I suppose that if I define
DR 1) y-longley$Employed
DR 2) X-as.matrix(cbind(1,Longley[,1:6])
DR 3) I = identity matrix the
DR following should be true: Coef=(X'X+kI)^(-1) X'y
No, as noted above, the intercept term is usually not penalised.
DR and if a take k=Ridge$kHKV, Coef should be approx equal to
DR Ridge$Coef[near value of kHKV]
No, as noted above the estimates in the coef component of an object
returned by lm.ridge are the coefficients on a different scale.
DR and it does not seem to happen, why?
Because the intercept is not penalised by lm.ridge and the
non-constant columns of the design matrix are rescaled; hence the
returned coefficients are on another scale.
DR Any help, suggestion or orientation?
HTH.
Cheers,
Berwin
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[R] lm.ridge
Hello, I have posted this mail a few days ago but I did it wrong, I hope is right now: I have the following doubts related with lm.ridge, from MASS package. To show the problem using the Longley example, I have the following doubts: First: I think coefficients from lm(Employed~.,data=longley) should be equal coefficients from lm.ridge(Employed~.,data=longley, lambda=0) why it does not happen? Second: if I have for example Ridge-lm.ridge(Employed~., data=longley, lambda = seq(0,0.1,0.001)), I suppose intercept coefficient is defined implicit, why it does not appear in Ridge$coef? Third: I suppose that if I define 1) y-longley$Employed 2) X-as.matrix(cbind(1,Longley[,1:6]) 3) I = identity matrix the following should be true: Coef=(X'X+kI)^(-1) X'y and if a take k=Ridge$kHKV, Coef should be approx equal to Ridge$Coef[near value of kHKV] and it does not seem to happen, why? Values: Ridge$kHKB [1] 0.004275357 Using the calculation above (third question, third point): Coef= [,1] 1-0.095492310 GNP.deflator -0.052759002 GNP 0.070993540 Unemployed -0.004244391 Armed.Forces -0.005725582 Population -0.413341544 Year 0.048420107 And if I take from Ridgecoef: Ridge$coef[0.004] GNP.deflator -0.03098507 GNP -1.32553151 Unemployed -1.53237769 Armed.Forces -0.63334911 Population -0.88690241 Year 6.82105049 Any help, suggestion or orientation? Thanks in advance Daniel Rozengardt __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] lm.ridge again
Hello, I have posted this mail a few days ago without any answer: I have the following doubts related with lm.ridge, from MASS package. To show the problem using the Longley example, I have the following doubts: First: I think coefficients from lm(Employed~.,data=longley) should be equal coefficients from lm.ridge(Employed~.,data=longley, lambda=0) why it does not happen? Second: if I have for example Ridge-lm.ridge(Employed~., data=longley, lambda = seq(0,0.1,0.001)), I suppose intercept coefficient is defined implicit, why it does not appear in Ridge$coef? Third: I suppose that if I define 1) y-longley$Employed 2) X-as.matrix(cbind(1,Longley[,1:6]) 3) I as the identity the following should be true: Coef=(X'X+kI)^(-1) X'y and if a take k=Ridge$kHKV, the coefficients should be approx equal to Ridge$Coef[near value of kHKV] and it does not seem to happen, why? Values: Ridge$kHKB [1] 0.004275357 Using the calculation above (third question, third point): Coef= [,1] 1-0.095492310 GNP.deflator -0.052759002 GNP 0.070993540 Unemployed -0.004244391 Armed.Forces -0.005725582 Population -0.413341544 Year 0.048420107 And if I take from Ridgecoef: Ridge$coef[0.004] GNP.deflator -0.03098507 GNP -1.32553151 Unemployed -1.53237769 Armed.Forces -0.63334911 Population -0.88690241 Year 6.82105049 Any help, suggestion or orientation? Thanks in advance Daniel Rozengardt __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
