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

Mathieu Basille basille at ase-research.org
Fri Mar 2 18:47:51 CET 2012

```Well, that is a tricky question! :)

Alexandre Hirzel et al. did exactly what you wanted to achieve, i.e. they
computed the proportion of each eigenvalue to the sum of eigenvalues. But
to me, it is just a ad-hoc measurement, with no real meaning (in this case,
it is quite different from, e.g. PCA eigenvalues). Instead, I would rely on
the value itself, i.e. the ratio of variances on each axis (including
marginality), which conveys an immediate meaning easy to interpret.

Mathieu.

Le 02/03/2012 12:25, lisa_web_de a écrit :
> Thank you Mathieu,
> sorry for pushing this point. I understand why you think it does not make
> any sense to compute the proportion. I am just wondering what Hirzel for
> example did in his paper
> Hirzel et al 2002 Ecological Niche Factor Analysis: How to compute habitat
> suitability maps without absence data? Ecology 83(7) 2027-2036.
>
> Greetings
> Lisa
>
>
> *Gesendet:* Freitag, 02. März 2012 um 14:40 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
> 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
> would make more sense, but it does not help you here). In addition,
> 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
>  > archiving and to help you get more answers).
>  >
>  > 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
>  > >
>  > >
>  > > Ihr WEB.DE Postfach immer dabei: die kostenlose WEB.DE Mail App für
> iPhone
>  > > und Android.
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>  > >
>  > >
>  > >
>  > > _______________________________________________
>  > > R-sig-Geo mailing list
>  > > R-sig-Geo at r-project.org
>  > > https://stat.ethz.ch/mailman/listinfo/r-sig-geo
>  >
>  > --
>  >
>  > ~\$ 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.
>  >
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>  >
>  >
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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.

```