Dear all,
Just to expand the reply by Zoltan.
NMDS1 and NMDS2 in the envfit() output are the positions of heads of
arrows of unit length.
They show the direction of the "best" linear fit into the n-dimensional
solution of the ordination.
The r2 then says how good the fit is. I.e. it refers to
(Sorry for double-posting, the original message does not seem to appear
in the forum.)
"Everything" seems to be possible with the dendextend package.
Check its extensive vignette at CRAN.
Vit
Dne 2017-10-10 14:54, Tania Bird napsal:
I am trying to control the order and colour of a
"Everything" seems to be possible with the dendextend package.
Check its extensive vignette at CRAN.
Vit
Dne 2017-10-10 16:01, Clifford Beall napsal:
Try the reorder function, the dendrogram specific help is at
?reorder.dendrogram. I’m not sure about the symbols on the labels
though.
On Oct
Hi Bjorn,
A common practice in simplifying ordinatin plots is to keep species that
show the best fit into the ordination space.
The position of such species in the ordination diagram really (may) tell
something.
In linear methods like PCA a linear fit is suitable. This can be
computed using
Thank you very much for your explanations. So if I understood it
correctly, a significant outcome from adonis() (or capscale()) should
indicate me that my species community changes along the gradient?
Yes, you understand it correctly.
Nice that you also mention
betadisper, because I'd like
Hi Valérie,
adonis is analogous to RDA or CCA, as it directly estimates the variance
in the distance matrix attributable to an independent variables(s), with
the advantage that one may use any distance measure. It parallels r2 in a
linear model.
Mantel test simply calculates the correlation
Dear Caroline,
It seems that you response (community-like) data are Depth and Velocity.
Than a dataframe with these two variables should be on the left hand side
of the formula, while SFT on the right hand side.
Something like
adonis(LB[,2:3] ~ SFT, data=LB, method=canberra)
should work.
Cheers,
Hi Roman,
Thank you.
My question should be rather if something has changed since then.
The web page of the book by Alain F. Zuur (chapter 17 and 33) is exactly
where I was looking for the R code.
Cheers,
Vit
This is from the highstat website on DFA (Analyzing ecological
Hi Mango,
If I understand it well, the following loop should do the work. I include
a small example.
a- c(1,0,0,0,'A',0,0,0,2,0,0,0,1,0,0,0,0,2,0,0,0,1)
for(i in (1:length(a)-1)){
if (a[i+1]==0) a[i+1]-a[i]}
a
[1] 1 1 1 1 A A A A 2 2 2 2 1 1 1 1 1
2 2 2 2 1
Cheers,
Vit
--
Vit Syrovatka
Dear listers,
I intended to send a separate question but it fell under comments on
Eduard Szöcs's question. Sorry for that.
The problem:
I collected aquatic insect larvae from 17 sites three times (spring,
summer, autumn), giving a total of 51 samples. During each sampling
campaign several
My apologies for sending my reply to Jari Oksanen only.
See it below, please.
Vit
Dear Jari Oksanen,
Thank you for your suggestions!
You really have to define what is a good fit. Having a large change in
probability is different from explaining a large part of the variation of
observations.
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