# R for Ecologists: Simulating Species-Area Curves (linear vs. nonlinear regression)

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This post is about basic model simulation so we can get a feel for how curves are supposed to look given certain processes assumed by the model. One of the most prevalent patterns in ecology is the species-area (SAR) curve, which plots the number of species (species richness) against the area sampled. Unfortunately, SAR curves can arise from a number of competing mechanisms (niche partitioning, neutral processes, pure randomness). The curve alone cannot distinguish between these processes, but we can use models to determine how curves will look given underlying, known processes.

I will simulate a SAR curve using pure randomness as the underlying process to show what this curve may look like. Let’s assume we have 20 one meter square plots. Ignoring spatial distribution of the plots, we can choose plots randomly to sample areas of different sizes. For example, I can randomly choose one plot to sample a 1 sq m. area, or I can randomly choose two plots to sample a 2 sq m. area, and so on all the way up to 20 sq m. **So the first assumption is a lack of spatial patterns (i.e. neutral/dispersion processes)**. We aren’t taking into account any spatial arrangement of plots. Also, we will assume a **purely random** distribution of species, wherein species occur in plots according to their overall abundance (*i.e.* species ‘A’ has an overall relative abundance of 0.8 across all plots, then it is 80% likely to occur in any given plot). This assumptions means **a lack of niche partitioning, no biotic/abiotic interactions**.

OK, so here’s how to simulate this in R. As stated above, we have 20 plots. We’ll assume for simplicity that we sample five individuals within each plot (equal sampling effort across plots). So set up an empty 5 x 20 array:

plots <- array(dim=c(5,20))

Now let’s assume that we have an overall species richness of 20 (we’ll just call them species a, b, c, etc…)

species <- c(letters[1:20])

We need to assign a probability of occurrence to each species, making sure the probabilities sum to one. So we simulate 20 probabilities:

probs <- numeric(20) probs[1:8] <- runif(8, 0, 0.1) for(i in 9:20){ probs[i] <- runif(1, 0, 1-sum(probs[1:i-1])) }

The first line sets up an empty container vector. The second line randomly assigns the first 8 species a probability between 0 and 0.1 (this is to prevent any species from having a probability of 0.7 or higher and dominating all the plots because that would be boring). The for( ) loop makes sure each remaining species has a probability between 0 and one minus the sum of all other probabilities (constraining them to equal one). You can check the constraint:

sum(probs)

which should equal 1 (more or less).

Now for each plot (column) in our plot array, we want to sample five individuals randomly using the probabilities we just created (because the probability of picking any one species is equal to its overall probability).

for(i in 1:20){ plots[,i] <- sample(species, size=5, replace=T, prob=probs) }

We sample with replacement to account for the possibility that species can occur multiple times within a plot. Essentially, this says the probability of an individual in a plot belonging to a given species is equal to the relative abundance or each species.

Now we work on generating our random SAR curve. Let’s assume that we will randomly sample plots 20 times, and that we will increase the number of plots sampled (*i.e.* we will sample 1 plot 20 times, then 2 plots 20 times, then 3 plots 20 times, etc..). For each sampling event, we will calculate species richness.

We set up an empty 20 x 20 container (20 sampling events for 20 different plot numbers):

SAR.mat <- array(dim=c(20,20))

We use a nested for( ) loop to simulate the sampling:

for(j in 1:20){ for(i in 1:20){ plot.index <- sample(1:20, j, replace=F) SAR.plot <- c(plots[,plot.index]) SAR.mat[i,j] <- length(unique(SAR.plot)) } }

The first loop tells the program to sample j plots going from 1 to 20. This is our ‘plot area’ going from 1 sq m to 20 sq m. The second loop is the sampling event, going from 1-20. plot.index is the index of sampled plots (i.e. randomly pull j plots from the 20 that we have). The length(unique( )) command just calculates the number of unique species *i.e.* species richness.

Now we have a 20 x 20 array containing 20 sampling events (rows) for each possible area (columns). Set up a vector relating the columns to areas, calculate the mean species richness of each column (area), calculate the 95% confidence interval, and then plot:

areas <- 1:20 means <- apply(SAR.mat, MARGIN=2, mean) lower.95 <- apply(SAR.mat, MARGIN=2, function(x) quantile(x, 0.025)) upper.95 <- apply(SAR.mat, MARGIN=2, function(x) quantile(x, 0.975)) par(mar=c(4,4,1,1)+0.2) plot(areas, means, type='n', xlab=expression('Area '*(m^-2)), ylab='Species Richness', cex.lab=1.2, ylim=c(0,12)) polygon(x=c(areas, rev(areas)), y=c(lower.95, rev(upper.95)), col='grey90') lines(areas, means, lwd=2)

Should look nice. If it doesn’t, rerun the above code in its entirety and you’ll get a different plot.

Now we want to fit a model to this. The common SAR model is

We can log transform each side to get the following model to fit with a linear regression:

We can fit the model and then plot the curve

SAR.mod <- lm(log(means) ~ log(areas)) summary(SAR.mod) curve(exp(coef(SAR.mod)[1])*x^coef(SAR.mod)[2], add=T, from=0, to=20, col='red', lwd=2)

Unfortunately, the log-transformed model *actually *fits a model assuming multiplicative errors:

The model we really want to fit is:

See Xiao et al (2011) in *Ecology* for a good description of the differences between the two models above and how they have been misused in allometric studies. We can fit the second model using a non-linear regression, using the log-linear model parameters as reasonable starting estimates. We then plot the nls( ) curve and tack on a legend.

SAR.nls <- nls(means ~ a*areas^b, start=list('a'=exp(coef(SAR.mod)[1]), 'b'=coef(SAR.mod)[2])) curve(coef(SAR.nls)[1]*x^coef(SAR.nls)[2], add=T, from=0, to=20, col='blue', lwd=2) legend('topleft', lty=1, col=c('black', 'red', 'blue'), legend=c('Median Species Richness', 'Linear Model Fit', 'Nonlinear Model Fit'), cex=0.8, bty='n')

And there you have it. In my version, the log-linear model and the nls( ) model both fit pretty well, but the nls( ) model fit *ever so* *slightly* better than the log-linear model (although the difference between the two is probably trivial). Also, the scaling exponent *z* from the nls( ) model was estimated as ~0.44. which is within the range of values predicted by island biogeography. As always, email me or comment if I screwed something up or my code could be improved.

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