I am interested in getting evenness based on Inverse Simpson index. Would appreciate clarification on the simpsoneven command- is it a measure of evenness based on Simpson index or Inverse Simpson index? Thanks.
The Simpson eveness is just a 0 - 1 value obtained by dividing the Simpson index by the theoretical maximum diversity you could achieve in the sample data. The calculation is actually trivial (once you have your SImpson index), it basically comes down to:
Simpson’s E = Simpsons D / [number of OTUs]
Although the equation can be written a few different ways (see here). The logic is that in a perfectly even community (all OTUs are present in equal ratios), the Simpsons D would just be 1 / [number of OTUs] so any deviation in your observed Simpsons D from this value is the result of uneveness. It would be easy to convert your inverse Simpsons D back to the basic Simpsons D and do the eveness calculation yourself - I usually just work these out in Excel when needed.
If you’re interested, this is also how the Shannon E is calculated as well, but dividing the observed Shannon index by the natural log of the total species.
Thanks for the reply
Thanks for the reply.
Using Mothur’s calculators, I got the following values for my sample:
Observed OTUs: 478
Inverse Simpson (1/D): 6.3944
Simpson evenness: 0.013378
If I tried to manually calculate Simpson evenness using the above values:
Simpson evenness = [1/(1/D)]/ [number of OTUs] = (1/6.3944)/478 = 0.000327
The value “0.000327” does not match what I got from Mothur (i.e., 0.013378)
But if I manually calculate Inverse Simpson evenness using the above values:
Inverse Simpson evenness = (1/D)/[number of OTUs] = (6.3944)/478 = 0.01337
This value now coincides with what I got from Mothur
Based on this observation, does it mean that the simpsoneven command actually calculates Inverse Simpson evenness?