[R-sig-Geo] Meaning of specialization eigenvalues in ENFA
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
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,
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
> For example:
> ev_mar<-sum(enfa1$pr)/length(enfa1$pr) * sum(enfa1$li$Mar^2)/sum((niche1$Mar -
> and then to calculate the proportion for the example of the first
> eigenvalue of 1.565:
>  0.2045417
> *Gesendet:* Donnerstag, 01. März 2012 um 16:18 Uhr
> *Von:* "Mathieu Basille" <basille at ase-research.org>
> *An:* r-sig-geo at r-project.org
> *Betreff:* Re: [R-sig-Geo] Meaning of specialization eigenvalues in ENFA
> Dear Lisa,
> Please use a sensible subject in the future (I corrected it for better
> archiving and to help you get more answers).
> The specialization can not be interpreted in terms of explained variance.
> The specialization on any axis is the ratio of the available variance over
> the used variance, which do not sum up to 1 on every axis. The
> specialization values are given by the eigenvalues of an enfa object. Let's
> take the example in ?enfa:
> map <- lynxjura$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)
> 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$Mar -
> Hope this helps,
> Le 01/03/2012 05:42, Lisa Freudenberger a écrit :
> > Dear all,
> > I hope this is the right place to post my question and that someone might
> > be able to help me. I am performing an ENFA with species data and
> > environmental variables unsing the adehabitat package. What I know how to
> > get are the eigenvalues of the specialization axes and the coefficients for
> > all environmental variables.
> > However I was wondering how you can calculate the percentage of variance
> > (specialization) explained by each axis from these data. As I know the
> > marginality axis always explains the whole marginality as well as part of
> > the specialization. My initial thought was to sum up all eigenvalues and
> > calculate the proportional values for each eigenvalue, meaning
> > eigenvalue/sum of eigenvalues. But as I do not get an eigenvalue for the
> > marginality axis I can´t build the sum of all of them.
> > Does anyone of you have an idea how to solve this problem?
> > Any help is appreciated!
> > All the best
> > Lisa
> > Ihr WEB.DE Postfach immer dabei: die kostenlose WEB.DE Mail App für iPhone
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> ~$ whoami
> Mathieu Basille, Post-Doc
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> Laboratoire d'Écologie Comportementale et de Conservation de la Faune
> + Centre d'Étude de la Forêt
> Département de Biologie
> Université Laval, Québec
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> ``If you can't win by reason, go for volume.''
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Mathieu Basille, Post-Doc
Laboratoire d'Écologie Comportementale et de Conservation de la Faune
+ Centre d'Étude de la Forêt
Département de Biologie
Université Laval, Québec
``If you can't win by reason, go for volume.''
Calvin, by Bill Watterson.
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