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

[Mathieu Basille]

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
>
>
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-- 

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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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http://ase-research.org/basille

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``If you can't win by reason, go for volume.''
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