[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$s+ev_mar)
> > [1] 0.2045417
> >
> > Greetings
> > Lisa
> >
> > *Gesendet:* Donnerstag, 01. März 2012 um 16:18 Uhr
> > *Von:* "Mathieu Basille" <basille at ase-research.org>
> > *An:* r-sig-geo at r-project.org
> > *Betreff:* Re: [R-sig-Geo] Meaning of specialization eigenvalues in ENFA
> > Dear Lisa,
> >
> > Please use a sensible subject in the future (I corrected it for better
> > archiving and to help you get more answers).
> >
> > The specialization can not be interpreted in terms of explained variance.
> > The specialization on any axis is the ratio of the available variance over
> > the used variance, which do not sum up to 1 on every axis. The
> > specialization values are given by the eigenvalues of an enfa object. Let's
> > take the example in ?enfa:
> >
> > library(adehabitatHS)
> >
> > data(lynxjura)
> > map <- lynxjura$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$Mar))^2)
> >
> > Hope this helps,
> > Mathieu.
> >
> >
> > Le 01/03/2012 05:42, Lisa Freudenberger a écrit :
> > > Dear all,
> > >
> > > I hope this is the right place to post my question and that someone might
> > > be able to help me. I am performing an ENFA with species data and
> > > environmental variables unsing the adehabitat package. What I know how to
> > > get are the eigenvalues of the specialization axes and the
> coefficients for
> > > all environmental variables.
> > >
> > > However I was wondering how you can calculate the percentage of variance
> > > (specialization) explained by each axis from these data. As I know the
> > > marginality axis always explains the whole marginality as well as part of
> > > the specialization. My initial thought was to sum up all eigenvalues and
> > > calculate the proportional values for each eigenvalue, meaning
> > > eigenvalue/sum of eigenvalues. But as I do not get an eigenvalue for the
> > > marginality axis I can´t build the sum of all of them.
> > >
> > > Does anyone of you have an idea how to solve this problem?
> > >
> > > Any help is appreciated!
> > >
> > > All the best
> > > Lisa
> > >
> > >
> > > Ihr WEB.DE Postfach immer dabei: die kostenlose WEB.DE Mail App für
> iPhone
> > > und Android.
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> > >
> > >
> > >
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> >
> > --
> >
> > ~$ 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.
> >
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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.
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