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Consider again Petrus’s hunting spider data of Exercise 15.2. He established (e.g. in Code Box 15.7) that different species respond differently to environmental gradients, especially so for soil dryness, herb layer and reflectance. He would like to know what the main types of environmental response were, across species.
What approach should he use to answer this question?
He wants to understand how taxa vary in environmental response, so he could use a Species Archetype Model to classify species according to type of environmental response.
Run the below lines to install ecomix:
Code to fit ecomix will only be run if it is installed
(the if statement is included because CRAN policies require vignettes to
be able to be run only using packages that can be installed from
CRAN)
is_ecomix_installed = requireNamespace("ecomix", quietly = TRUE)
library(mvabund)
data(spider)
SpiderDF=spider$x
SpiderDF$abund=as.matrix(spider$abund)
spiderFormula = abund ~ soil.dry + bare.sand + fallen.leaves + moss +
herb.layer + reflection
if(is_ecomix_installed)
{
ft_Mix = ecomix::species_mix(spiderFormula, data=SpiderDF, family="negative.binomial", nArchetypes=2, control=list(init_method='kmeans',ecm_refit=5, ecm_steps=2) )
coef(ft_Mix)$beta
}
#> SAM modelling
#> There are 2 archetypes to group the species into
#> There are 28 site observations for 12 species
#> The model for the archetype (grouping) is ~soil.dry + bare.sand + fallen.leaves + moss + herb.layer + reflection
#> The model for the species is ~1
#> You are implementing a negative.binomial Species Archetype Model.
#> Using ECM algorithm to find starting values; using 5 refits
#> ECM restart 1 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -1041.731 | Ratio loglik 0
#> Iteration: 2 | New loglik -834.997 | Ratio loglik 0.801548
#> ECM restart 2 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -1041.731 | Ratio loglik 0
#> Iteration: 2 | New loglik -834.997 | Ratio loglik 0.801548
#> ECM restart 3 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -1041.731 | Ratio loglik 0
#> Iteration: 2 | New loglik -834.997 | Ratio loglik 0.801548
#> ECM restart 4 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -1041.731 | Ratio loglik 0
#> Iteration: 2 | New loglik -834.997 | Ratio loglik 0.801548
#> ECM restart 5 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -1041.731 | Ratio loglik 0
#> Iteration: 2 | New loglik -834.997 | Ratio loglik 0.801548
#> initial value 834.997322
#> iter 10 value 822.933991
#> iter 20 value 818.054221
#> iter 30 value 816.602567
#> iter 40 value 812.844421
#> iter 50 value 793.214165
#> iter 60 value 791.057454
#> iter 70 value 786.683339
#> iter 80 value 778.413169
#> iter 90 value 777.956005
#> final value 777.933344
#> converged
#> soil.dry bare.sand fallen.leaves moss herb.layer reflection
#> Archetype1 1.5629184 -0.05470042 -0.2784744 -0.1449658 0.7081518 -0.4730736
#> Archetype2 -0.1069941 0.22420728 -0.2855582 0.4186874 0.5894873 0.4759734Which predictors vary the most across archetypes (hence across taxa)? Is this what you saw in Code Box 15.7?
soil.dry had the biggest change, jumping from
2.15 to 0.10. The next biggest changes were in
herb.layer and reflection. This is pretty much
what was happening in Code Box 15.7 as well.
if(is_ecomix_installed)
plot(ft_Mix, type="link")
#> Infinite residuals removed from residual plots: 82 in total.
Do model assumptions seem reasonable here?
This generally looks good, no fan-shape in residual vs fits plot, points close(ish) to a straight line on normal quanitle plot.
Consider again Anthony’s revegetation study. We would like to better characterise how different invertebrate orders respond to the revegetation treatment. Fit a Species Archetype Model with two archetypes.
library(ecostats)
data(reveg)
revegDF=data.frame(reveg)
revegDF$abundMat=as.matrix(reveg$abund)
if(is_ecomix_installed)
{
ft_revegMix = ecomix::species_mix(abundMat~treatment, data=revegDF, family="negative.binomial", nArchetypes=2, control=list(init_method='kmeans',ecm_refit=5, ecm_steps=2) )
coef(ft_revegMix)$beta
}
#> SAM modelling
#> There are 2 archetypes to group the species into
#> There are 10 site observations for 24 species
#> The model for the archetype (grouping) is ~treatment
#> The model for the species is ~1
#> You are implementing a negative.binomial Species Archetype Model.
#> Using ECM algorithm to find starting values; using 5 refits
#> ECM restart 1 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -932.547 | Ratio loglik 0
#> Iteration: 2 | New loglik -745.016 | Ratio loglik 0.798905
#> ECM restart 2 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -932.547 | Ratio loglik 0
#> Iteration: 2 | New loglik -745.016 | Ratio loglik 0.798905
#> ECM restart 3 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -932.547 | Ratio loglik 0
#> Iteration: 2 | New loglik -745.016 | Ratio loglik 0.798905
#> ECM restart 4 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -932.547 | Ratio loglik 0
#> Iteration: 2 | New loglik -745.016 | Ratio loglik 0.798905
#> ECM restart 5 of 5
#> Initialising starting values
#> Initial groups parameter estimates by K-means clustering
#> Iteration: 1 | New loglik -932.547 | Ratio loglik 0
#> Iteration: 2 | New loglik -745.016 | Ratio loglik 0.798905
#> initial value 745.015688
#> final value 745.015688
#> converged
#> treatmentImpact
#> Archetype1 5.350399
#> Archetype2 1.441362Explain what the main two types of response to revegetation were, across invertebrate taxa.
Oh that’s odd – the estimated trend is almost identical for these two groups.
Look at the posterior probabilities for each Order, identify a few taxa that characterise each response type, and plot the raw data for these taxa. Does the Species Archetype Model seem to capture the main trends in response to revegetation treatment?
This doesn’t seem worthwhile when there are no differences in response across these archetypes!!
This code chunk has not been run, as it takes a minute or so to complete
if(is_ecomix_installed)
{
nClust=rep(2:6,3)
bics = rep(NA, length(nClust))
for(iClust in 1:length(nClust))
{
fti_Mix = ecomix::species_mix(spiderFormula, data=SpiderDF,
family="negative.binomial", nArchetypes=nClust[iClust],
control=list(init_method='kmeans',ecm_refit=5, ecm_steps=2))
bics[iClust] = BIC(fti_Mix)
}
plot(bics~nClust, ylab="BIC", xlab="# archetypes")
}How many archetypes would you use?
I guess I’d go for four archetypes, that looks like the winner here.
Note that repeat runs didn’t always give the same BIC. Why did this happen?
As in Maths Box 16.1, mixture model likelihoods can be messy with multiple maxima. Hence it’s a good idea to run multiple times and keep the best solution.
Consider again a Species Archetype Model for Anthony’s revegetation data. In Exercise 16.2 we assumed there were only two archetypes. How many archetypes should be used for Anthony’s data? Answer this question using the model selection technique of your choice.
OK I’m going for BIC as above.
This code chunk has not been run, as it takes a minute or so to complete
if(is_ecomix_installed)
{
nClust=rep(1:5,3)
bics = rep(NA, length(nClust))
for(iClust in 1:length(nClust))
{
fti_Mix = ecomix::species_mix(abundMat~treatment, data=revegDF,
family="negative.binomial", nArchetypes=nClust[iClust])
bics[iClust] = BIC(fti_Mix)
}
plot(bics~nClust, ylab="BIC", xlab="# archetypes")
}Well it looks like there should be two archetypes (presumably, one of these is increasing with revegetation, the other not).
Consider again Petrus’s hunting spider data of Exercise 15.2… He wants to know the extent to which species traits explain interspecific variation in environmental response. What approach should he use to answer this question?
He wants to know why species vary in their environmental response, so a good idea would be to include predictors on species (species traits) and to use a fourth corner model to understand the extent to which interspecific variation in environmental response can be explained by these traits.
library(gllvm)
#>
#> Attaching package: 'gllvm'
#> The following objects are masked from 'package:VGAM':
#>
#> AICc, nobs, predict, vcov
#> The following objects are masked from 'package:stats4':
#>
#> nobs, vcov
#> The following object is masked from 'package:vegan':
#>
#> ordiplot
#> The following object is masked from 'package:mvabund':
#>
#> coefplot
#> The following objects are masked from 'package:stats':
#>
#> nobs, predict, simulate, vcov
data(spider)
X = spider$x[,c("soil.dry","herb.layer")]
ft_trait = gllvm(spider$abund, X, spider$trait, randomX=~soil.dry+herb.layer, family="negative.binomial")
logLik(ft_trait)
#> 'log Lik.' -711.2906 (df=60)
library(lattice)
a = max( abs(ft_trait$fourth.corner) )
colort = colorRampPalette(c("blue","white","red"))
plot.4th = levelplot(ft_trait$fourth.corner, col.regions=colort(100),
at=seq(-a, a, length=100), scales = list( x=list(rot = 45)) )
print(plot.4th)
Heloise collected counts of 41 species of ants at 30 sites across
south-eastern Australia, stored as antTraits in the
mvabund package. She also recorded five environmental
variables at each site, and five functional traits for each species. She
would like to know if these traits explain why ant species differ in
environmental response. Fit a fourth corner model to this
dataset.
(The below code chunk takes about a minute to run.)
ft_anttrait = gllvm(antTraits$abund, envMat, antTraits$traits, randomX=~envMat, family="negative.binomial", starting.val="zero")
logLik(ft_anttrait)
#> 'log Lik.' -1919.352 (df=223)
library(lattice)
a = max( abs(ft_anttrait$fourth.corner) )
colort = colorRampPalette(c("blue","white","red"))
plot_4th = levelplot(t(as.matrix(ft_anttrait$fourth.corner)),
col.regions=colort(100), at=seq(-a, a, length=100),
scales = list( x= list(rot = 45)) )
print(plot_4th)
This doesn’t always produce a coefplot because of
infinite values – this is symptomatic of non-convergence. Make sure you
end up with a logLik value of about -2300 or
better, if this doesn’t happen then re-run. I ended up using
starting.val="zero" which usually gave better fits (it
doesn’t always but did this time around).
What are the key traits that capture why (and how) ants differ in environment response?
A big one seems to be Canopy.cover:polymorphism.
Canopy.cover:Pilosity also seemed to be very negative in
most runs, but with a fairly large standard error.
Don’t forget to check your P’s and Q’s!
The below code chunk has not been evaluated because it is a function of the previous one
This looks good to me. The points on the normal quantile plot don’t all stay in the envelope (which is pointwise not global, so is tighter than it should be), but the violations we have are not of the alarming kind – residuals are smaller in magnitude than expected, rather than too large. This is often symptomatic of overfitting rather than lack-of-fit – the model does have a lot of parameters in it!
nVars = dim(spider$abund)[2]
newTraits = spider$trait
# set factors not of interest here to be a constant value
newTraits$length= mean(spider$trait$length) #set length to its mean
newTraits$colour=factor(rep(levels(spider$trait$colour)[1],nVars),
levels=levels(spider$trait$colour)) #set to first level of factor
# set starting rows of ’marks’ to take all possible values
nMarks = nlevels(spider$trait$marks)
newTraits$marks[1:nMarks]=levels(spider$trait$marks)
# create a new env dataset where the only thing that varies is soil:
newEnv = spider$x[1:2,c("soil.dry","herb.layer")]
newEnv[,"soil.dry"]=range(scale(spider$x[,"soil.dry"]))
newEnv[,"herb.layer"]=0
#make predictions and plot:
newPreds = predict(ft_trait,newX=newEnv,newTR=newTraits,type="response", level=0)
matplot(newEnv[,1], newPreds[,1:nMarks],type="l", log="y")
legend("topright",levels(newTraits$marks),lty=1:nMarks,col=1:nMarks)
Recall that fourth corner model you fitted to Heloise’s data in Exercise 16.5. For an interaction of your choice, construct a plot to visualise how the environment-abundance association changes for different values of the functional trait.
I’m going with Canopy.cover:Polymorphism, as the one
with the biggest (and only significant) interaction term:
The below code chunk has not been evaluated because it is a function of the previous one
nVars = dim(antTraits$abund)[2]
newTraits = antTraits$traits
#set length vars to their mean
newTraits$Femur.length= mean(antTraits$traits$Femur.length)
newTraits$Webers.length= mean(antTraits$traits$Webers.length)
# set factors not of interest here to be a constant value
newTraits$No.spines=rep(0,nVars)
newTraits$Pilosity = factor(rep(levels(antTraits$traits$Pilosity)[1],nVars),
# set first few levels of Polymorphism to all possible values
levels=levels(antTraits$traits$Pilosity)) #set to first level of factor
nPoly = nlevels(antTraits$traits$Polymorphism)
newTraits$Polymorphism[1:nPoly]=levels(antTraits$traits$Polymorphism)
# create a new env dataset where the only thing that varies is Canopy.cover:
newEnv = antTraits$env
newEnv$Canopy.cover=range(scale(newEnv$Canopy.cover))
newEnv$Bare.ground = mean(newEnv$Bare.ground)
newEnv$Shrub.cover = mean(newEnv$Shrub.cover)
newEnv$Volume.lying.CWD = mean(newEnv$Volume.lying.CWD)
newEnv$Feral.mammal.dung = mean(newEnv$Feral.mammal.dung)
newEnv=as.matrix(newEnv)
#make predictions and plot:
newPreds = predict(ft_anttrait,newX=newEnv,newTR=newTraits,type="response", level=0)
matplot(newEnv[,"Canopy.cover"], newPreds[,1:nPoly],type="l", ylab="Mean abundance", xlab="Canopy cover", log="y")
legend("topright",levels(newTraits$Polymorphism),lty=1:nPoly,col=1:nPoly)We can see there that when there is no polymorphism, abundance is relatively high with low canopy cover.
ft_spp = gllvm(spider$abund, X, family="negative.binomial")
ft_trait = gllvm(spider$abund, X, spider$trait, family="negative.binomial")
ft_main = gllvm(spider$abund, X, spider$trait, family="negative.binomial", formula=~soil.dry+herb.layer)
an_spider4th = anova(ft_main, ft_trait, ft_spp)
#> Model 1 : ~ soil.dry + herb.layer
#> Model 2 : y ~ soil.dry + herb.layer + (soil.dry + herb.layer):(length + colour + marks)
#> Model 3 : y ~ X
an_spider4th
#> Resid.Df D Df.diff P.value
#> 1 287 0.00000 0
#> 2 279 40.12996 8 3.02999e-06
#> 3 265 76.84430 14 1.08408e-10
an_spider4th$D[2]/sum(an_spider4th$D)
#> [1] 0.3430666Consider again Heloise’s ant data. What proportion of the variation in environmental response is explained by the measured species traits?
ft_antspp = gllvm(antTraits$abund, envMat, family="negative.binomial")
ft_anttrait = gllvm(antTraits$abund, envMat, antTraits$traits, family="negative.binomial")
mainFormula=as.formula(paste("~", paste(colnames(envMat),collapse="+")))
ft_antmain = gllvm(antTraits$abund, envMat, antTraits$traits, family="negative.binomial", formula=mainFormula)
an_ant4th = anova(ft_antmain, ft_anttrait, ft_antspp)
#> Warning in anova.gllvm(ft_antmain, ft_anttrait, ft_antspp): This test was not designed for tests with a df.diff larger than 20 so the P-value should be treated as approximate.
#> Model 1 : ~ Bare.ground + Canopy.cover + Shrub.cover + Volume.lying.CWD + Feral.mammal.dung
#> Model 2 : y ~ Bare.ground + Canopy.cover + Shrub.cover + Volume.lying.CWD + Feral.mammal.dung + (Bare.ground + Canopy.cover + Shrub.cover + Volume.lying.CWD + Feral.mammal.dung):(Femur.length + No.spines + Pilosity + Polymorphism + Webers.length)
#> Model 3 : y ~ X
an_ant4th
#> Resid.Df D Df.diff P.value
#> 1 1062 0.00000 0
#> 2 1022 75.73794 40 0.000548758
#> 3 862 271.85048 160 8.23701e-08
an_ant4th$D[2]/sum(an_ant4th$D)
#> [1] 0.2178955So it looks like about a quarter of interspecific variation can be explained by traits, which is not super, especially considering that the trait model uses a quarter as many parameters. This suggests we are not really onto a winner here…
#remove this chunk once gllvm has been updated on CRAN:
These binaries (installable software) and packages are in development.
They may not be fully stable and should be used with caution. We make no claims about them.