R help - Aug 2007 - Comparing long species lists via Sorenson’s dissimilarity

I have 4 very large species lists and I would like to compare them.
   
  I have the following results from running Sorenson’s dissimilarity tests:
   
  Norfolk Fens compared to Suffolk Coastal Fens:               
QS=0.583961142689298
  Norfolk Fens compared to Breckland Edge Fens:               
QS=0.714896020281379
  Norfolk Fens compared to Other Fens:                            
QS=0.78572348898302
   
  Suffolk Coastal Fens compared to Breckland Edge Fens:    QS=0.78572348898302
  Suffolk Coastal Fens compared to Other Fens:                 
QS=0.855011155835705
   
  Breckland Edge Fens compared to Other Fens:                 
QS=0.175091076893185
   
  Does anyone know how to interpret the above results? I have read a paper that
states that when comparing two samples if QS is less than 50% than the samples
are considered dissimilar, what is 100%? I mean 50% of what?
   
  There are a number of other tests that I can run using R. These are squared
euclidan, manhattan, bray-curtis, jaccard, ruzicka, (dis)similarity ratio,
ochiai, cosine compliment, and Raup-crick. Would it be advantageous to run these
now that I have my sorensons result? It is especially easy for me to run the
Bray-Curtis all I would need to do is change the terms from ‘binary’ to
‘minimum’ and re-run. Would Bray-Curtis be a Percentage Similarity unlike the
others?
   
  Here is an example of what happens when I run sorensons:
   
  #Compare Breckland with Other
  > a=table(breckland)
  > J=sum(breckland*other)
  > A=sum(breckland^2)
  > B=sum(other^2)
   
  > brecklandOther <- designdist(a,method=(A+B-2*J)/(A+B), terms =
c("binary"))
  [1] 0.1750911
  attr(,"call")
  designdist(x = a, method = (A + B - 2 * J)/(A + B), terms =
c("binary"))
  attr(,"method")
  [1] "0.175091076893185 binary"
   
  I do not know why the following lines are included (perhaps R is trying to
calculate the header row? But it doesn’t seem to stop the result getting out.
   
  attr(,"call")
  designdist(x = a, method = (A + B - 2 * J)/(A + B), terms =
c("binary"))
  attr(,"method")
   
  I’ve also managed to get diversity indices from R but someone told me that
comparing diversity indices is not good because of the way that they are
calculated.
   
  >norfolkShannon
  3.653768 
  > suffolkShannon
   Suffolk 
  3.646138 
  > brecklandShannon
  Breckland 
   4.051742 
  > otherShannon
     Other 
  3.587403
   
  For the above statistical methods is there a way of laying out the data
perhaps via a graph that will show how similar/dissimilar the sites are?
   
  Many thanks for yesterdays help and in advance for any help given today.
   
  Richard Price
  University of Birmingham.
   
   
   

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