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

[Lisa Freudenberger]

Ok thanks again for your help. I will talk to the other coauthors what 
to do and if to include the percentage values. At least I know now how 
to calculate them. It has been a pleasure discussing this issue with you 
:-)
Lisa

On 02.03.2012 18:47, Mathieu Basille wrote:
> 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$Mar))^2)
>> >
>> > Hope this helps,
>> > Mathieu.
>> >
>> >
>> > Le 01/03/2012 05:42, Lisa Freudenberger a écrit :
>> > > Dear all,
>> > >
>> > > I hope this is the right place to post my question and that 
>> someone might
>> > > be able to help me. I am performing an ENFA with species data and
>> > > environmental variables unsing the adehabitat package. What I 
>> know how to
>> > > get are the eigenvalues of the specialization axes and the
>> coefficients for
>> > > all environmental variables.
>> > >
>> > > However I was wondering how you can calculate the percentage of 
>> variance
>> > > (specialization) explained by each axis from these data. As I 
>> know the
>> > > marginality axis always explains the whole marginality as well as 
>> part of
>> > > the specialization. My initial thought was to sum up all 
>> eigenvalues and
>> > > calculate the proportional values for each eigenvalue, meaning
>> > > eigenvalue/sum of eigenvalues. But as I do not get an eigenvalue 
>> for the
>> > > marginality axis I can´t build the sum of all of them.
>> > >
>> > > Does anyone of you have an idea how to solve this problem?
>> > >
>> > > Any help is appreciated!
>> > >
>> > > All the best
>> > > Lisa
>> > >
>> > >
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>> > >
>> > >
>> > > _______________________________________________
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>> >
>> > --
>> >
>> > ~$ 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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