Hi everyone,
I think Lou Josts numbers equivalents should help solve this. He
suggested that the Shannon diversity index is very helpful, but it needs to
be translated back to the effective number of species (=e^H, assuming youve
calculated it using the natural log).
This way, you can interpret diversity in terms of the effective number of
species, after accounting for relative abundance.
It also means that one species will always have a diversity of one, as it
should be. Zero species = zero diversity.
Consider this example with perfect evenness, the effective richness is the
same as species richness (R):
library(vegan)
> ex1
[,1] [,2] [,3] [,4]
[1,] 0 0 0 1
[2,] 0 0 1 1
[3,] 0 1 1 1
[4,] 1 1 1 1
> H = diversity(ex1)
> H
[1] 0.0000000 0.6931472 1.0986123 1.3862944
> exp(H)
[1] 1 2 3 4
If you use the diversity function in R, youll probably have to fix the
cases where richness = 0 because it returns H = 0 for zero species:
richness = specnumber(ex1)
H = diversity(ex1)
eff.rich = exp(H)
for (i in 1:length(richness)) {
if (richness[i] == 0) {eff.rich[i] = 0} else
{eff.rich[i] = eff.rich[i]}
}
eff.rich
See:
Jost, L. 2006. Entropy and diversity. Oikos 113: 363375.
http://www.loujost.com/Statistics%20and%20Physics/Diversity%20and%20Similarity/PopGen.htm
For evenness, use (e^H)/S or Wilsons Evar.
I think you should be very careful about interpreting evenness in
one-species assemblages. I concur that evenness doesn't really make sense
with zero species.
Sincerely,
Evan Weiher
Professor, Plant & Community Ecology
University of Wisconsin - Eau Claire
