[R-sig-Geo] randtest.enfa {adehabitat}

Mathieu Basille basille at ase-research.org
Tue Jun 8 20:09:53 CEST 2010 

Consuelo,

Here are some elements that you might find useful.

> I had already read the paper. I understood the idea of the scatter plot,
> that the arrows were a projection of the marginality and specialization but
> I didn't understand where the length of arrows came from. I was somehow lost
> thinking how their length represent the marginality and specialization. Now
> you tell me that they "correspond to the scores of the environmental
> variables on the axes of the ENFA". Hmmm.... Are these scores given
> somewhere in the output of enfa?

Yes they are. Look at the $co element of the enfa object. Following the 
example in ?enfa:

 > enfa1$co
               Mar        Spe1
forets  0.7178829  0.29304902
hydro   0.0977121 -0.94018385
routes -0.5394778  0.15001749
artif   0.4290224 -0.08758622

It gives you the coordinates of the arrow. The starting point of the 
vector is the average availability which is centred on zero.

> You see, the main issue for me is that if I put that graph in a manuscript,
> I'd like to be able to tell, the arrow length represent XX times the
> marginality and XX times the specialization of each variable or something
> like that. In order to give an idea of the real value of marginality and
> specialization. Does it make any sense? Maybe it doesn't make any sense and
> it's not necessary, what do you think? I hope you understand me this time...
> otherwise, I'll give up!!!

You should interpret the scatter plot qualitatively. E.g. in the example 
of ?enfa, the marginality is very strong on "forets" (forest) with a 
shift towards high values, on "routes" (roads) with a shift on low 
values, and to a lesser extent on "artif" (artificial areas) with a 
shift on high values. On the other hand, the specialization is very 
pronounced on "hydro" only, with a restriction around the mean (you have 
to look at the specific distributions on this particular variable to say 
more about it).

> Regarding the global specialization issue. Not sure about something again.
> You told me that there's not global specialization, but Hirzel et al (2002),
> described it. Why you say "it cannot be measured globally over the
> ecological space"? Is it wrong if I use the global specialization formula of
> Hirzel et al (square root of the sum of the eigeinvalues divided by the
> number of variables) to estimate this global specialization (sensu Hirzel et
> al)? Can I do that?

You can indeed compute such an index. But now, how would you interpret 
it? The only case it could be useful is the comparison of two niches in 
the same environment, i.e. same availability. Two study sites cannot be 
compared on such basis. So that it would give you an idea of a global 
specialization that is totally context dependent. This is why it is not, 
from my point of view, very useful.

> And regarding the tolerance, just to be sure we're talking about the same,
> is it the inverse of the specialization, right? if I calculate it, could I
> use it to estimate the specialization? Actually, I have tried to calculate
> the tolerance, but I think I have used the wrong "wei". Can you please
> confirm I'm using the correct terms?

It is conceptually the inverse of the specialization: the specialization 
is a ratio of variances (available over used) while the tolerance is the 
sum of used variances over all variables (which is similar to divide it 
by an available variance of 1). You might have a look at ?niche.test for 
a global measure of tolerance.

> You told me to use this formula: sum(dudi.pca(tab, row.w=wei,
> scan=FALSE)$eig); in which "tab" is the dataframe containing the values of
> environmental variables and "wei" is vector describing the utilization
> weight of each pixel. Using the manual example for enfa:
> 
> data(lynxjura)
> map <- lynxjura$map
> tmp <- lynxjura$locs[,4]!="D"
> locs <- lynxjura$locs[tmp, c("X","Y")]
> dataenfa1 <- data2enfa(map, locs)
> enfa1 <- enfa(pc, dataenfa1$pr,+ scannf = FALSE)
> 
> For me, in this case, "tab" would be kasc2df(map)  and "wei" would be
> dataenfa1$pr
> so, the tolerance would be: sum(dudi.pca(kasc2df(map), row.w=dataenfa1$pr,
> scan=FALSE)$eig)
> Is this correct? I got this huge value (2355). It's kind of high, isn't it?

Maybe try:

sum(dudi.pca(kasc2df(map), row.w = dataenfa1$pr/sum(dataenfa1$pr), 
scan=FALSE)$eig)

which gives you 5 only. The weights should sum to 1 (i.e. they are 
proportions). But then, how would you interpret this? This is the same 
as for the "global specialization".

> And about the specialization axis you keep in enfa (or madifa or gnesfa or
> any other one of these analysis), is there a way to get the % variation
> explained by each eigenvalue? I mean, a referee can ask that, right? I have
> tried to figure it out, but without any luck... I have chosen only one
> eigenvalue of specialization, but I wasn't able to calculate how much
> variation it accounted for.

There is no way to give a percentage of variation for a specialization 
axis, by nature: the eigenvalue of each axis gives you a ratio of 
variance. E.g. an eigenvalue of 12 tells you that the available variance 
on this axis is 12 times higher than the used variance. It is up to you 
to interpret this (is it a strong specialization or not).

> I hope you don't mind all these question... I need to be sure of what I'm
> doing if I want to publish these results...

No problem about it. The ENFA, however, should be seen as a very elegant 
way of exploring the data, at the manner of a PCA (and the 
interpretation is in many ways similar, when you take into account the 
specificity of the analysis).

Hope this helps a bit,
Mathieu.


> Thanks!!!
> 
> Consuelo
> 
> -------------
> Consuelo Hermosilla
> PhD student
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> 
> 2010/6/7 Clément Calenge <clement.calenge at gmail.com>
> 
>>  On 06/04/2010 09:40 AM, Consuelo Hermosilla wrote:
>>
>> Hummm, true, I got confused! Sorry!! I meant scatter.enfa.
>> What I don't understand is the length of the arrows. The grid is d=2,
>> different from the grid set with the biplot of marginality and
>> specialization (d=0.5). In that case, the length make senses to me, but I
>> don't understand it in the scatter.enfa. They don't have the same lenght
>> /value as they have in the biplot. Do you understand me?
>>
>>
>> I do not understand you... The function scatter.enfa draws the biplot
>> associated with the results of the ENFA. The following paper explains this
>> graph in detail:
>>
>> Basille, M., Calenge, C., Marboutin, E., Andersen, R. and Gaillard, J.M.
>> 2008. Assessing habitat selection using multivariate statistics: some
>> refinements of the ecological niche factor analysis. Ecol. model. 211:
>> 233--240.
>>
>>
>>
> 
>> I am not sure what you mean by "biplot of marginality and specialization".
>> There is only one way to draw a biplot with the ENFA: it is provided by
>> scatter.enfa (see the paper cited above). So I do not understand how the
>> result of scatter.enfa could be inconsistent with the biplot, since the
>> result of scatter.enfa *is* the biplot.
>> Best,
>>
>>
>> Clément Calenge
>>
>> --
>> Clément CALENGE
>> Cellule d'appui à l'analyse de données
>> Office national de la chasse et de la faune sauvage
>> Saint Benoist - 78610 Auffargis
>> tel. (33) 01.30.46.54.14
>>
>>
>>
>>
>>
>>
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+ Centre d'Étude de la Forêt
Département de Biologie
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