R help - Jan 2003 - factor analysis (pca): how to get the 'communalities'?

Dear expe-R-ts,

I try some test data for a factorAnalysis (resp. pca) in the sense of Prof. 
Ripley's MASS ? 11.1, p. 330 ff., just to prepare myself for an analysis of
my
own empirical data using R (instead of SPSS).

1. the data.

## The test data is (from the book of Backhaus et al.: Multivariate ## 
Analysemethoden. Springer 2000 [9th ed.], p. 300 ff):

a<-c(4.5,5.167,5.059,3.8,3.444,3.5,5.25,5.857,5.083,5.273,4.5)
b<-c(4.0,4.25,3.824,5.4,5.056,3.5,3.417,4.429,4.083,3.6,4.0)
c<-c(4.375,3.833,4.765,3.8,3.778,3.875,4.583,4.929,4.667,3.909,4.2)
d<-c(3.875,3.833,3.438,2.4,3.765,4.0,3.917,3.857,4.0,4.091,3.9)
e<-c(3.25,2.167,4.235,5.0,3.944,4.625,4.333,4.071,4.0,4.091,3.7)
f<-c(3.75,3.75,4.471,5.0,5.389,5.250,4.417,5.071,4.25,4.091,3.9)
g<-c(4.0,3.273,3.765,5.0,5.056,5.5,4.667,2.929,3.818,4.545,3.6)
h<-c(2.0,1.857,1.923,4.0,5.615,6.0,3.25,2.091,1.545,1.6,1.5)
i<-c(4.625,3.75,3.529,4.0,4.222,4.75,4.5,4.571,3.75,3.909,3.5)
j<-c(4.125,3.417,3.529,4.6,5.278,5.375,3.583,3.786,4.167,3.818,3.7)

m<-data.frame(a,b,c,d,e,f,g,h,i,j)

2. My try of a pca with R.

## My R input was:

m
cor(m)
library(mva)
m.pca<-princomp(m,cor=T)
m.pca
summary(m.pca)
loadings(m.pca)
m.pca$scores
m.FA <- factanal(factors = 3, covmat=cov(m))
m.FA

3. Here are my questions. 

Q1. 
The cor(m)-Matrix is the same as reported by using SPSS (or OpenStats2). 
But in R I get other eigenvalues compared with the following SPSS output:

Original matrix trace =  10,00
Roots (Eigenvalues) Extracted:
   1  5,052
   2  1,771
   3  1,427
   4  0,819
   5  0,430
   6  0,247
   7  0,159
   8  0,062
   9  0,029
  10  0,003

- What is going behind the scene? 
- Or what I am doing wrong in my use of R?
- If I am doing the pca correct, can I use the R results as equally aceptable 
  without further discussion? 
  Maybe a different 'hidden' algorithm is the reason for different
results?

Q2. How to get the so called 'Communality Estimates' with R?

Here the values reported by SPSS for the above test data.frame m:
Communality Estimates as percentages:
  1 88,619
  2 76,855
  3 89,167
  4 85,324
  5 76,043
  6 84,012
  7 80,223
  8 92,668
  9 63,297
 10 88,786

Any help, suggestions or hints are very welcome.
Best regards and happy new year for you and R

Wolfgang

--
Wolfgang Lindner                           Lindner at math.uni-duisburg.de
   Gerhard-Mercator-Universitaet Duisburg  Tel: +49 0203 379-1326
   Fakultaet 4 - Naturwissenschaften       Fax: +49 0203 379-2528
   Institut fuer Mathematik,  LE 424
   Lotharstr. 65
   D 47048  Duisburg (Germany)

On Fri, 3 Jan 2003, Wolfgang Lindner wrote:
> I try some test data for a factorAnalysis (resp. pca) in the sense of Prof.
Well, factor analysis and pca are different things, and only one
is appropriate in a given problem.
> Ripley's MASS § 11.1, p. 330 ff.,
Eh?  Would that be *Venables & Ripley's* MASS, and if so which edition
(it
is not the current one).  Those editions which cover factor analysis do
explain the difference.
>just to prepare myself for an analysis of my
> own empirical data using R (instead of SPSS).
>
> 1. the data.
>
> ## The test data is (from the book of Backhaus et al.: Multivariate ##
> Analysemethoden. Springer 2000 [9th ed.], p. 300 ff):
>
> a<-c(4.5,5.167,5.059,3.8,3.444,3.5,5.25,5.857,5.083,5.273,4.5)
> b<-c(4.0,4.25,3.824,5.4,5.056,3.5,3.417,4.429,4.083,3.6,4.0)
> c<-c(4.375,3.833,4.765,3.8,3.778,3.875,4.583,4.929,4.667,3.909,4.2)
> d<-c(3.875,3.833,3.438,2.4,3.765,4.0,3.917,3.857,4.0,4.091,3.9)
> e<-c(3.25,2.167,4.235,5.0,3.944,4.625,4.333,4.071,4.0,4.091,3.7)
> f<-c(3.75,3.75,4.471,5.0,5.389,5.250,4.417,5.071,4.25,4.091,3.9)
> g<-c(4.0,3.273,3.765,5.0,5.056,5.5,4.667,2.929,3.818,4.545,3.6)
> h<-c(2.0,1.857,1.923,4.0,5.615,6.0,3.25,2.091,1.545,1.6,1.5)
> i<-c(4.625,3.75,3.529,4.0,4.222,4.75,4.5,4.571,3.75,3.909,3.5)
> j<-c(4.125,3.417,3.529,4.6,5.278,5.375,3.583,3.786,4.167,3.818,3.7)
>
> m<-data.frame(a,b,c,d,e,f,g,h,i,j)
>
> 2. My try of a pca with R.
>
> ## My R input was:
>
> m
> cor(m)
> library(mva)
> m.pca<-princomp(m,cor=T)
> m.pca
> summary(m.pca)
> loadings(m.pca)
> m.pca$scores
> m.FA <- factanal(factors = 3, covmat=cov(m))
> m.FA
>
> 3. Here are my questions.
>
> Q1.
> The cor(m)-Matrix is the same as reported by using SPSS (or OpenStats2).
> But in R I get other eigenvalues compared with the following SPSS output:
You don't get eigenvalues at all in R.  You do get `Proportion of
Variance' which are these numbers divided by their total.
> Original matrix trace =  10,00
> Roots (Eigenvalues) Extracted:
>    1  5,052
>    2  1,771
>    3  1,427
>    4  0,819
>    5  0,430
>    6  0,247
>    7  0,159
>    8  0,062
>    9  0,029
>   10  0,003
>
> - What is going behind the scene?
Why don't you ask the SPSS people that?  R at least gives you sensible
labels on the output.
> - Or what I am doing wrong in my use of R?
> - If I am doing the pca correct, can I use the R results as equally
aceptable
>   without further discussion?
No, as more acceptable: at least they have meaningful labels.
>   Maybe a different 'hidden' algorithm is the reason for different
results?
Ask SPSS that.  R's code is open, and nothing is hidden.  You have not
demonstrated that the results are different, anyway!
> Q2. How to get the so called 'Communality Estimates' with R?
First, use the data as in
> (m.FA <- factanal(m, factors=3))
and where did the number of factors come from?

100*(1 - m.FA$uniquenesses) gives the communalities.  They are different
from SPSS, because (1) R uses maximum likelihood FA and (2) tries a lot
harder to find a maximum and there are many local maxima in most FA
problems.

In this case you have fitted too many factors, and just one suffices.
> Here the values reported by SPSS for the above test data.frame m:
> Communality Estimates as percentages:
>   1 88,619
>   2 76,855
>   3 89,167
>   4 85,324
>   5 76,043
>   6 84,012
>   7 80,223
>   8 92,668
>   9 63,297
>  10 88,786
>
> Any help, suggestions or hints are very welcome.
1) Be a lot more accurate.
2) Read the help pages to find out what the output means.  In the case of
R the information is there, but you may well have to post on an SPSS help
list to find out why SPSS gives different output from R.
3) Don't believe SPSS knows what it is doing.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
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, UK                Fax:  +44 1865 272595

Dear Prof. Ripley,

many thanks for your prompt answer and the valuable hints and pointers, I will 
study them and try it again.

1.
Sorry, for not giving the full quote; my book is:

Venables & Ripley: MASS. Springer 1999, 3rd Ed. (Corr. 3rd printing 2001); 
           ? 11.1, p. 330 ff.                   (?11 written by B.D.R.)
> Those editions which cover factor analysis do explain the difference.
So I will look for the latest edition ..

2.
> 3) Don't believe SPSS knows what it is doing.
o.k. I see ;-) 

Best regards
Wolfgang

--
Wolfgang Lindner                           Lindner at math.uni-duisburg.de
   Gerhard-Mercator-Universitaet Duisburg  Tel: +49 0203 379-1326
   Fakultaet 4 - Naturwissenschaften       Fax: +49 0203 379-2528
   Institut fuer Mathematik,  LE 424
   Lotharstr. 65
   D 47048  Duisburg (Germany)

Scot,

thank you very much for your wonderful clear and short fix of my first problem: 
seeing your solution as one-liner in the impressive insightful syntax of R is 
really an aesthetic experience for me:

|  I ran your example and found that you can get the eigenvalues SPSS by [..]
|    m.pca$sdev^2
|  So squaring the standard deviations (sdev) of the components gives you the
|  eigenvalues SPSS reports.

I am a little sorrow of not having seen it for myself ;-) - but I think
that's
live in becoming a friend of R and making the first steps with pca, fa, ca &
co.
R is indeed a first choice tool in doing understandable statistics and Prof 
Ripley's indication to R's open code points definitive in the same
direction for
me. Now the two worlds become reconciled and the fog gets thinner for me.
Thank you both.

Wolfgang
--
Wolfgang Lindner                           Lindner at math.uni-duisburg.de
   Gerhard-Mercator-Universitaet Duisburg  Tel: +49 0203 379-1326

If interested, on my web site I have code to do factor analysis by PC.  Does
exactly as below, but a nice wrapper to print methods, rotations, sorting, and
other conveniences.

  home.earthlink.net/~bmagill/MyMisc.html 

The relevant code snipets are "prinfact", "plot.pfa", and
"print.pfa", along
with the other required functions as indiciated on the web site.


On Fri, 3 Jan 2003 21:04:21 +0100 Wolfgang Lindner <LindnerW at
t-online.de>
wrote:
> Scot,
> 
> thank you very much for your wonderful clear
> and short fix of my first problem: 
> seeing your solution as one-liner in the
> impressive insightful syntax of R is 
> really an aesthetic experience for me:
> 
> |  I ran your example and found that you can
> get the eigenvalues SPSS by [..]
> |    m.pca$sdev^2
> |  So squaring the standard deviations (sdev)
> of the components gives you the
> |  eigenvalues SPSS reports.
> 
> I am a little sorrow of not having seen it for
> myself ;-) - but I think that's 
> live in becoming a friend of R and making the
> first steps with pca, fa, ca & co. 
> R is indeed a first choice tool in doing
> understandable statistics and Prof 
> Ripley's indication to R's open code points
> definitive in the same direction for 
> me. Now the two worlds become reconciled and
> the fog gets thinner for me.
> Thank you both.
> 
> Wolfgang
> --
> Wolfgang Lindner                          
> Lindner at math.uni-duisburg.de
>    Gerhard-Mercator-Universitaet Duisburg  Tel:
> +49 0203 379-1326
> 
> ______________________________________________
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>