From: Achim Zeileis <Achim.Zeileis_at_wu-wien.ac.at>

Date: Sat 09 Jul 2005 - 05:29:25 EST

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 Received on Sat Jul 09 05:34:45 2005

Date: Sat 09 Jul 2005 - 05:29:25 EST

On Fri, 8 Jul 2005, K. Steinmann wrote:

> Dear R-helpers,

*>
**> Using the package Lattice, I performed a PCA.
*

In my R installation, the function princomp() is contained in the stats package, not lattice (sic!).

> For example

*> pca.summary <- summary(pc.cr <- princomp(USArrests, cor = TRUE))
**>
**> The Output of "pca.summary" looks as follows:
**>
**> Importance of components:
**> Comp.1 Comp.2 Comp.3 Comp.4
**> Standard deviation 1.5748783 0.9948694 0.5971291 0.41644938
**> Proportion of Variance 0.6200604 0.2474413 0.0891408 0.04335752
**> Cumulative Proportion 0.6200604 0.8675017 0.9566425 1.00000000
**>
**> How can I extract the proportion of variance?
*

The standard deviations are in

pc.cr$sdev

(see also ?princomp), thus the variances are the squared values and the
proportions can be computed as

pc.cr$sdev^2/sum(pc.cr$sdev^2)

and by taking the cumsum() the cumulative proportion can be computed. This
is exactly what the the print method for "summary.princomp" objects does
(see stats:::print.summary.princomp).

Best,

Z

> Since names(pca.summary) or str(pca.summary) do not contain the proportion of

*> variance,
**> it seems I can not use something similar like pca.summary[[2]]$Comp.1[3].
**> I can't see how the values are stored.
**>
**>
**> Thanks in advance.
**>
**> K. St.
**>
**> ______________________________________________
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**>
*

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 Received on Sat Jul 09 05:34:45 2005

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