# [R-sig-Geo] Meaning of specialization eigenvalues in ENFA

Mathieu Basille basille at ase-research.org
Fri Mar 2 14:40:44 CET 2012

```Well, of course you can, technically, no-one's stopping you :)

However, the sum of specialization values (in your case: 7.65) does not
represent anything at all. Each specialization value is a variance ratio,
so that I don't know what would be a sum of variance ratio (maybe a mean
specialization axes are only orthogonal to the marginality axis, and not
necessarily to each other (while the scores on each of them are
uncorrelated). I don't quite know if the sum, or even the mean, of such
not-orthogonal axes would make any sense, in a geometric way...

Finally, if you're interested to test the significance of the ENFA, you can
have a look at 'randtest.enfa', which tests the first specialization
eigenvalue (it follows that, if the first is significant, every eigenvalue
that you selected with a sensible method is significant too). I would not
especially recommend it, since the ENFA is essentially an exploratory
analysis, but it's better to know that this function exists.

Hope this helps,
Mathieu.

Le 02/03/2012 03:16, Lisa Freudenberger a écrit :
> Dear Mathieu,
>
> thank you for your quick response and the clarification. Sorry I did not
> fill in a subject. I have actually never been active in a mailing list before.
>
> You are saying that the eigenvalues are the specialization values and that
> the specialization accounted for by the marginality can also be calculated
> as described by your script. Can´t I then use the value of the
> specialization accounted for by the marginality axis to sum it up with the
> eigenvalues of the specialization axes to calculate a proportion of
> specialization explained by the marginality axis and by each specialization
> axis?
>
> For example:
>
> ev_mar<-sum(enfa1\$pr)/length(enfa1\$pr) * sum(enfa1\$li\$Mar^2)/sum((niche1\$Mar -
> mean(niche1\$Mar))^2)
>
> and then to calculate the proportion for the example of the first
> eigenvalue of 1.565:
>
> 1.565/sum(enfa1\$s+ev_mar)
> [1] 0.2045417
>
> Greetings
> Lisa
>
> *Gesendet:* Donnerstag, 01. März 2012 um 16:18 Uhr
> *Von:* "Mathieu Basille" <basille at ase-research.org>
> *An:* r-sig-geo at r-project.org
> *Betreff:* Re: [R-sig-Geo] Meaning of specialization eigenvalues in ENFA
> Dear Lisa,
>
> Please use a sensible subject in the future (I corrected it for better
>
> The specialization can not be interpreted in terms of explained variance.
> The specialization on any axis is the ratio of the available variance over
> the used variance, which do not sum up to 1 on every axis. The
> specialization values are given by the eigenvalues of an enfa object. Let's
> take the example in ?enfa:
>
>
> data(lynxjura)
> map <- lynxjura\$map
> locs <- lynxjura\$locs
> locs <- locs[slot(locs, "data")[,2]!="D",]
> slot(map,"data")[,4] <- sqrt(slot(map,"data")[,4])
> tab <- slot(map, "data")
> pr <- slot(count.points(locs, map), "data")[,1]
> pc <- dudi.pca(tab, scannf = FALSE)
> (enfa1 <- enfa(pc, pr, scannf = FALSE, nf = 3))
>
> Here, the variance available on the first axis of specialization is 1.57
> times higher than the variance used on this axis. We can actually compute
> these values using the ratio of variances:
>
> niche1 <- enfa1\$li[rep(1:length(enfa1\$pr), enfa1\$pr), ]
> sum(enfa1\$pr)/length(enfa1\$pr) * sum(enfa1\$li\$Spe1^2)/sum(niche1\$Spe1^2)
> sum(enfa1\$pr)/length(enfa1\$pr) * sum(enfa1\$li\$Spe2^2)/sum(niche1\$Spe2^2)
> enfa1\$s
>
> Note, however, that the marginality axis also accounts for some
> marginality. To compute the specialization on the marginality axis, it is
> exactly the same approach:
>
> sum(enfa1\$pr)/length(enfa1\$pr) * sum(enfa1\$li\$Mar^2)/sum((niche1\$Mar -
> mean(niche1\$Mar))^2)
>
> Hope this helps,
> Mathieu.
>
>
> Le 01/03/2012 05:42, Lisa Freudenberger a écrit :
>  > Dear all,
>  >
>  > I hope this is the right place to post my question and that someone might
>  > be able to help me. I am performing an ENFA with species data and
>  > environmental variables unsing the adehabitat package. What I know how to
>  > get are the eigenvalues of the specialization axes and the coefficients for
>  > all environmental variables.
>  >
>  > However I was wondering how you can calculate the percentage of variance
>  > (specialization) explained by each axis from these data. As I know the
>  > marginality axis always explains the whole marginality as well as part of
>  > the specialization. My initial thought was to sum up all eigenvalues and
>  > calculate the proportional values for each eigenvalue, meaning
>  > eigenvalue/sum of eigenvalues. But as I do not get an eigenvalue for the
>  > marginality axis I can´t build the sum of all of them.
>  >
>  > Does anyone of you have an idea how to solve this problem?
>  >
>  > Any help is appreciated!
>  >
>  > All the best
>  > Lisa
>  >
>  >
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> --
>
> ~\$ whoami
> Mathieu Basille, Post-Doc
>
> ~\$ locate
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> + Centre d'Étude de la Forêt
> Département de Biologie
> Université Laval, Québec
>
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--

~\$ whoami
Mathieu Basille, Post-Doc

~\$ locate
Laboratoire d'Écologie Comportementale et de Conservation de la Faune
+ Centre d'Étude de la Forêt
Département de Biologie
Université Laval, Québec

~\$ info
http://ase-research.org/basille

~\$ fortune
``If you can't win by reason, go for volume.''
Calvin, by Bill Watterson.

```