Yes, it is possible, and always has been when I have checked (which is not a
proof). You can check this by seeing that it has no negative eigenvalues in
principal coordinates analysis (apart from occasional negative almost-zero).
Legendre & Legendre book discuss this.
Cheers, Jari Oksanen
>
Hi all,
I have a very simple question: is it possible that the square-root of
Bray-Curtis values is Euclidean? if not, is there a way to transform
bray-curtis which is semi-quantitative in Euclidean?
thanks a lot!
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