gmane.comp.lang.r.robust

Re: Are PcaHubert and PCAproj randomized algorithms?

Kaveh Vakili — 2012-04-25T19:35:52

yes, the solutions will not be the same but they should be fairly similar.

Re: In robust PCA methods, how to get variance explained?

Valentin Todorov — 2012-04-24T19:07:21

library(rrcov)
data(bus)
p <- ncol(bus)
rpca <- PcaHubert(bus, k=p, kmax=p)
summary(rpca)

Hope this helps.
Best regards,
Valentin


On Tue, Apr 24, 2012 at 6:03 PM, Michael <comtech.usa-Re5JQEeQqe8AvxtiuMwx3w< at >public.gmane.org> wrote:


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Are PcaHubert and PCAproj randomized algorithms?

Michael — 2012-04-24T16:05:27

I am reading about these robust PCA functions in R...

I read words like "random directions"...

It seems that these algos are random algorithms, i.e their results will be
different each time we run them?

Thank you!

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In robust PCA methods, how to get variance explained?

Michael — 2012-04-24T16:03:47

In robust PCA methods, how to get variance explained?

For example, PcaHubert,

how to get the variance explained which are similar to those concepts in
traditional PCA?

In traditional PCA, you have a bunch of eigenvalue lambdas...

and you sort the lambdas from the biggest to the smallest,

the lambda_i / (sum of all lambdas) is the variance explained by that
principal component...

how to obtain the equivalent concepts in PcaHubert?

Thanks a lot!

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library cox robust : weights?

Elisabetta Mattiolo — 2012-02-07T21:12:34

I'm a italian student and I use R for my final degree thesis ... I want
study a Robustly Proportional Hazards (Bednarski's teory) for a dataset
of medical result. 
I have exstimate with library coxrobust a coxr model (exponential
weights)


site , cyto5.6 and zeb1 are my esplicative variables so the model is:

0.95, f.w= "exp" , singular.ok = TRUE, model = FALSE)

with plot(exp) I can see 5 graphs 
- the firs shows the standardizzed survival differences : one with Cox
model, and one green with Kaplan -Meir stimator 
-other four show the same differences for four strata, defined by the
quartiles of the estimated linear predictor. 

But my problem is the I want a graphic the robust exponential weight
(log trasformed) versus case number for the dataset..
If I ask R about weight of the model exp: 
NULL
If I write 
Error in plot.window(...) : 'xlim' devono essere finiti
Inoltre: Warning messages:
1: In min(x) : no non-missing arguments to min; returning Inf
2: In max(x) : no non-missing arguments to max; returni

Re: [R] nlrob problem

Martin Maechler — 2011-12-23T16:21:22

    > Dear all, I am not sure if this mail is for R-help or
    > should be sent to R-devel instead, and therefore post to
    > both.

    > While using nlrob from package 'robustbase', I ran into
    > the following problem:

    > For psi-functions that can become zero
    > (e.g. psi.bisquare), weights in the internal call to nls
    > can become zero. Example:






You are right.
The next version of robustbase (0.8-1) will have this fixed.

Note however, that for me, in your example and in other more
interesting ones, as soon as I use a redescending  psi() function,
the IRLS iterations do *not* converge...
but rather strangely ``cycle'' around the true solution.
As a redescending psi() has a non-convex rho(), and non-linear
problems depend quite a bit on "feasible"/"good" starting
values, I currently think I would discourage from using such psi().


As others, some possibly more expert in robust non-linear fitting,
may be interested, and have interesting feedback,
I'm crossposting this reply to the R-

get outlier list without plot

Ben qant — 2011-09-28T23:12:19

Hello,

First time posting on the robust list. New to R.

How do I get the outliers list generated via uni.plot without opening a
graphics device?

Regards,

Ben

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Call for Papers ICPRAM 2012

Pedro Latorre Carmona — 2011-07-27T10:17:01

Dear all,

My name is Pedro Latorre Carmona, program co-chair of the "2012  
International Conference on Pattern Recognition Applications and  
Methods (ICPRAM2012)". I send you the Call for Papers for this  
conference and would like to draw your attention in particular to the  
six special sessions that are also organised. I hope you will find it  
interesting and may contribute to ICPRAM 2012.


*********************************************************************
        2012 International Conference on Pattern Recognition
                  Applications and Methods (ICPRAM2012)

                            February 6-8, 2012
                      Vilamoura, Algarve, Portugal
                         http://www.icpram.org
*********************************************************************

ICPRAM (1st International Conference on Pattern Recognition Applications and
Methods - http://www.icpram.org/) has an open call for papers, whose
deadline is set for July 26, 2011. We hope you can participate in this

Re: minimum sample size for the robust counterpart of thet-test #3

Richard Friedman — 2011-06-21T22:56:12

Dear Manuel (and list).

Thank you for doing so much  the work to answer my question.
Much of the theory is beyond me at present although I plan on learning
more about robust methods with time along with many other fields
I have to learn. I would like to rephrase the question- your response
deals mainly with power. What I need to know about more is Tyoe I
error. Do robust methods ever increase type I -error lead to a greater
number of false positives, for n=5 than does the classical t-test?
In particular does the Huber with k=1.345? I have a particular
reason for returning to this method even though it is apparently
no longer the robust method of choice. A collaborator used it to find
significance where the classical t-test did not, and I am wondering
which test to believe.

Thanks and best wishes,
Rich



On Jun 17, 2011, at 11:55 AM, Manuel Koller wrote:

Re: minimum sample size for the robust counterpart of the t-test #2

Manuel Koller — 2011-06-17T15:55:57

Dear Richard,

Since we did not quite cover your specific case, I ran another small
simulation. See the attached file. It is basically the simulation
study of our paper, but for models having only an intercept. I hope I
did not overlook anything when I did this. There was a lot going on
today... I apologize for the overloaded plots. I guess Figures 4, 7
and 8 are the most interesting figures.

As Rand already stated, the asymmetric error distributions are a
problem: all the methods perform quite badly. Otherwise, the levels of
the tests are pretty much ok (even for OLS, i.e., t-test). But of
course, the power will be pretty bad. In numbers, for n = 5 you will
have approximately the correct level (+/- 2%), but a power of about
40% only for an effect size of 1 (10% for an effect size of 0.4). And
this does not really depend on which method you are using.

To conclude, I would recommend to use lmrob from robustbase with the
argument setting="KS2011".

I hope this helps,

Manuel

On Thu, Jun 16, 2011 at 8:19 PM,

Re: minimum sample size for the robust counterpart of thet-test #2

Richard Friedman — 2011-06-16T18:19:56

Rand,

Thanks, I know very little about robust methods. I am interested in  
whether rlm can be used in its default
state or if I have to tearn much more to do use the methods correctly.

Best wishes,
Rich

On Jun 16, 2011, at 2:14 PM, Rand Wilcox wrote:

Re: minimum sample size for the robust counterpart of the t-test #2

Rand Wilcox — 2011-06-16T18:14:39

When dealing with M-estimators and the goal is to compute confidence intervals, one thing you have to be careful about is skewed distributions. Have not encountered any non-bootstrap method that performs well in simulations where the confidence interval is based on an estimate of the standard error. Just how symmetric the distribution must be seems unclear. What works better is a percentile bootstrap method, even with fairly small sample sizes. This is why the methods in my book focus on bootstrap techniques when dealing with M-estimators.


However, have not yet seen the Koller and Stahel paper. Maybe this problem has been addressed.

Rand

Rand Wilcox
Professor
Dept of Psychology
USC
Los Angeles, CA 90089-1061

FAX: 213-746-9082
For information about statistics books and software, see http://www-rcf.usc.edu/~rwilcox/
as well as
http://college.usc.edu/labs/rwilcox/home

----- Original Message -----
From: Richard Friedman <friedman-VS9ntHdz6ztaNOFNA11YEsysmGwsrwg7h13vi7wywA4< at >public.gmane.org>
Date: Thursday,

Re: minimum sample size for the robust counterpart of thet-test

Martin Maechler — 2011-06-16T16:43:46

    > Dear List, I am a beginner in the use of robust methods. Is
    > there a minimum sample size for which the robust analog of a two
    > sample t-test using rlm with default parameters and categorical
    > explanatory variables may be trusted to yield reliable p-values?
    > Is so, can you please point me at a reference which treats this
    > problem.

It's a bit more complicated, because "the" robust analog does
not exist:  There are an infinite number of possible robust
analogues to the t-test,
and my two colleagues have been actively researching on this,
not just for the two-sample case, but the general  lm() case,
*with* an emphasis on small-sample performance:

Originally, (I think) they started answering the question (+/-): 

  How do you have to estimate  sj^2 := \Var(\hat{\beta_j}) such that
    \hat{\beta_j}  +/- 1.96 * sj
  has the correct coverage probability of 95%, also for small
  samples (and of course generalizing to other probs. alpha).
 
Here's their main publication :     

 Koll

Re: minimum sample size for the robust counterpart of thet-test #2

Richard Friedman — 2011-06-16T16:02:04

Dear Rand (and List),

I read the relevant sections of your book and while informative it  
did not answer my question
directly as best I can see. I will restate the question more explicitly:

A robust analog of the two sample  t-test is performed with the rlm  
function with the default parameters of
the Huber method with K=1.345. Is there a minimum sample size for  
which it should be trusted?
are 5 samples enough? 10 samples?

If this question does not have a simple answer please let me know.

Thanks and best wishes,
Rich


On Jun 15, 2011, at 3:19 PM, Rand Wilcox wrote:

minimum sample size for the robust counterpart of the t-test

Richard Friedman — 2011-06-15T19:10:03

Dear List,

I am a beginner in the use of robust methods. Is there a minimum  
sample size
for which the robust analog of a two sample t-test using rlm with  
default parameters and categorical
explanatory variables may be trusted to yield reliable p-values?
Is so, can you please point me at a reference which treats this problem.

Thanks and best wishes,
Rich
------------------------------------------------------------
Richard A. Friedman, PhD
Associate Research Scientist,
Biomedical Informatics Shared Resource
Herbert Irving Comprehensive Cancer Center (HICCC)
Lecturer,
Department of Biomedical Informatics (DBMI)
Educational Coordinator,
Center for Computational Biology and Bioinformatics (C2B2)/
National Center for Multiscale Analysis of Genomic Networks (MAGNet)
Room 824
Irving Cancer Research Center
Columbia University
1130 St. Nicholas Ave
New York, NY 10032
(212)851-4765 (voice)
friedman-VS9ntHdz6ztaNOFNA11YEsysmGwsrwg7h13vi7wywA4< at >public.gmane.org
http://cancercenter.columbia.edu/~friedman/

I am a Ba

Re: VIF for robust regression?

Olivier Renaud — 2011-04-06T09:33:36





I don't know if there is work on a robust VIF. Here are my two cents:
- What type of variable do you have ? If some of them are discrete with 
very few different values, this may cause problem to robust covariance 
matrices estimation.
- Your approach to apply the "partialling out"  of the robust matrix, as 
in the OLS case,  might or might not be correct, I don't know.
- If you believe that 1/(1-R^2_i) is a good measure, then you might want 
to compute its direct robust equivalent.  The output of lmrob does not 
provide a R^2, but the output of lmRob does. We have recently published 
a paper however that shows that the robust R-squared provided by lmRob 
is biased, sometimes to a large extent. We provide a consistent and 
robust estimator of R-squared and a version adjusted for the sample 
size.  See my previous post for an example and the code at
https://stat.ethz.ch/pipermail/r-sig-robust/2010/000290.html

Olivier

ref: Renaud, O. & Victoria-Feser, M.-P. (2010). A robust coefficient of 
determination

Estimating robust distances in R (MVE vs. MCD)

James Shaw — 2011-04-01T21:47:16

I have been trying to estimate robust Mahalanobis distances in R for a
set of three regressors that includes one dummy variable.  Initially,
I tried generating robust MCD estimates and their associated VCE using
cob.rob.  However, when I did so I received the following error
message:  "Error in solve.default(cov, ...) :  Lapack routine dgesv:
system is exactly singular".  I believe that the MCD estimator
involves subsampling and that the parameter for the discrete variable
could not be identified in one of the subsamples due to insufficient
variance.  When using the minimum volume ellipsoid (MVE) estimator, I
did not experience any problems.  My code is given below.


x<-cbind(c0[,3], c0[,7], c0[,8])
rest<-cov.rob(x, method = "mve", nsamp = "exact", cor=FALSE)
xrd<-mahalanobis(x, rest$center, rest$cov, inverted=FALSE)
xrd<-xrd^.5
d0<-ifelse(xrd> 3.0575159,1,0)


Can anyone explain to me why the MVE estimator is able to accommodate
discrete variables, whereas the MCD estimator cannot do so?  I would
like to b

VIF for robust regression?

Nicholas Lewin-Koh — 2011-03-31T15:35:10

Hi,
I was looking at the vif function in the car package
and it it is trivial to modify to make a version for robust
regression. However, after trying it out I noticed  that what 
were reasonable values under ols, jumped way up. 
So my thought is that either,
I made a coding error, and the weights attribute needs to be used
to modify the variance covariance matrix of the coefficients
Or, the reduced variance from the robust regression, causes peripheral
points
(outside the mve) to have much more influence in the r^2's for each
predictor.
So that the standard vif measure, 1/(1-R^2_i) is not relevant in this
context.
Am I off base here? 

Thanks
Nicholas

Re: hubers m-estimator in R / SPSS

Rudolf Dutter — 2011-02-23T20:53:41

Martin,

I agree fully with you. It reminds me that some 30 years ago I puzzled
around why SPSS produces negative variances. It was not too difficult to
find out the reason by trial and error. Conclusion: We didn't recommend
the usage SPSS at a scientific level any more, and some years ago, we
cancelled all licenses from our institute. I am not surprised that
Hubers algorithm remains mysterious in that package.
Sorry to say that.

Rudi

Martin Maechler wrote:

Re: hubers m-estimator in R / SPSS

Martin Maechler — 2011-02-23T20:25:29

    MH> thanks. of course, that was my idea too, but i couldn't find out which 
    MH> scale estimator spss uses. no manual, book, paper or whatever contains 
    MH> information about that. i tried other programs (like R or the ACM-tool), 
    MH> which use MAD, SMAD or an iterated standard deviation, but i always get 
    MH> different results with spss...

so why on earth are you using SPSS at all?
...
actually it's hard to believe that they don't document what
their M-estimate does ...
but that would really belong to an SPSS help list rather than an
R one, right?

Martin

    MH> manfred


    MH> Am 22.02.2011 00:42, schrieb Matias Salibian-Barrera:
    >> Hello Manfred,
    >> 
    >> I'm not familiar with SPSS, but based on my experience, I'll go out on a
    >> limb and say that the difference may be on the scale estimator that it's
    >> used (either a preliminary estimator like the MAD) or a simultaneously
    >> computer M-scale. Hopefully the SPSS manual will have some details.
    >> 
    >>

hubers m-estimator in R / SPSS

Manfred Hammerl — 2011-02-19T02:30:13

hello, i'm new to robust statistics but found out very quick that R 
(used huber, hubers and huberM) and SPSS (huber m-estimator) calculate 
different location estimates, given the same tuning constant k. since 
the differences a really not very small, i wanted to get some detailed 
information about this but i couldn't find out which algorithm SPSS uses 
to calculate hubers estimator so far...

does anyone know something about SPSS's huber function?

greetings,

manfred