metaMDS: unrealistic good stress

Hello,

i'm using metaMDS to ordinate trait mean data of plants in different 
ecosystems.
I used different calibrations but the stress differs only  between 0,001 
an 0,1e-19.
I have never seen such a good stress, so I wonder about possible 
mistakes, but i cannot find one.

Data looks like that:

plots      trait1      trait2      traits3
1            2,4           0,2         20
2            0,5           0,3         21 
.                .               .            .
300          .               .            .

My standard calibration is:
metaMDS(test
,distance = "euclidean"   # because metric data
, k = 3                                 # because I have 3 traits, but I 
couldn't plot a scree plot
, trymax = 3-5                     # not more, because too large dataset
, autotransform =TRUE
, noshare =1    # Ordination may be very difficult if a large proportion 
of sites have no shared species. In this case, the results may be 
improved with stepacross dissimilarities, or flexible shortest paths 
among all sites. No manipilation: noshare=1.0
,trace = TRUE         # will monitor the final stresses

Partitioning variance with VarPart - permutation test on difference between fractions

Hi there!

I am analyzing a dataset on hoverflies along an altitudinal gradient and
I managed to partition the variance explained among four predictor
variable sets (CLIMATE, RESOURCES, STRUCTURE, and SPACE) in R with
varpart and tested the significance of the (testable) fractions by
single (partial) RDAs.

However, how can I test whether two fractions differ significantly from
each other in R with the permutation test proposed in the Ecology paper
by Peres-Neto et al. 2006?

I know there is an executable program in the appendix of that paper, but
I would like to continue my analysis in R. Does anybody know of an
available R command or R code to test whether two fractions differ?

Best wishes from windy Berlin (+2 degrees, no snow),

Frank

--

-- 

Prof. Dr. Frank Dziock
Animal Ecologist and Head of Department (Juniorprofessor)

Department of Biodiversity Dynamics
Technische Universitaet Berlin
Sekr. AB 1
Rothenburgstr. 12

Re: metaMDS: unrealistic good stress

Daniel <daniel_ho@...> writes:

> i'm using metaMDS to ordinate trait mean data of plants in different
> ecosystems. I used different calibrations but the stress differs only
> between 0,001 an 0,1e-19. I have never seen such a good stress, so I
> wonder about possible mistakes, but i cannot find one.

You've got three traits, and you're using MDS to project them into three
dimensions (k = 3). There's no stress, because you've got as many
dimensions as variables. You only get stress when you try and squeeze
the variation from three variables into less than three dimensions. Try
again with k = 2 and you should get some stress.

Put another way, why are you trying to project your data into three
dimensions? If you want to plot it on paper, you need to project it into
two dimensions. If you actually are using the three ordination axes in
some subsequent analysis, you could just use your variables (possibly
transformed) instead.

HTH,

Tyler

>
> Data looks like that:
>
> plots      trait1      trait2      traits3
> 1            2,4           0,2         20
> 2            0,5           0,3         21 .                .               .            .
> 300          .               .            .

colorkey for image plots

Hi
I'm developing a model of leaf photosynthesis based on leaf nitrogen (N) 
and phosphorus (P). I've created a figure with a large range of possible 
combinations of N and P (x and y axes) with the outputs of my model as 
colors from the image function.
I couldn't include on the figure a color key that would give the range 
of the model outputs. I tried other functions (filled.contour, 
levelplot, wireframe) but couldn't get around the need for ascending 
order of values.

My code follows below. I'd appreciate any comments. Also, ideas on how 
to include a third leaf trait on this.

Cheers,
Tomas

###################################
# figure with the variation of Vcmax with N and P

rm(list=ls(all=TRUE))
detach()

##the libraries
library(akima)
library(stats)

## the data
data1 <- read.csv(file.name <- choose.files(),na.strings = "NA" , header 
= TRUE)
attach(data1)

Re: colorkey for image plots

If you can't figure out how to resolve the issue with optional arguments 
to a plotting function, you can copy and modify the plotting function 
itself.

The component of code that creates legends for filled.contour() is a 
useful template and so is the snippet of my code, below, which is 
executable and creates a shaded bar showing progress of data collection. 
The 3 examples show that you can specify any colors or ordering you wish.

foo <- function(colors){
#For purposes of this example:
N <- 100
n <- 75

#Display percent of plots completed in a
  #bar at the bottom of figure
  plot.new()
  par(fig=c(0,1,0,0.5))
  plot.window(c(0,1),c(0,1))

  #N <- sum(progress$n.scheduled)
  #n <- sum(progress$n.completed)
  progress <- n/N
  if(n>0){
    breaks <- seq(0,progress,length=n+1)
    xleft <- breaks[-(n+1)]
    xright <- breaks[-1]
    ybottom <- rep(0.75,n)
    ytop <- rep(1,n)
    col <- colors

Re: colorkey for image plots

Maybe  image.plot in package fields could help you

Cheers,

Marcelino

At 17:45 04/02/2009, Glen A Sargeant wrote:
>If you can't figure out how to resolve the issue with optional arguments
>to a plotting function, you can copy and modify the plotting function
>itself.
>
>The component of code that creates legends for filled.contour() is a
>useful template and so is the snippet of my code, below, which is
>executable and creates a shaded bar showing progress of data collection.
>The 3 examples show that you can specify any colors or ordering you wish.
>
>foo <- function(colors){
>#For purposes of this example:
>N <- 100
>n <- 75
>
>#Display percent of plots completed in a
>   #bar at the bottom of figure
>   plot.new()
>   par(fig=c(0,1,0,0.5))
>   plot.window(c(0,1),c(0,1))
>
>
>   #N <- sum(progress$n.scheduled)

About biplot analysis in ADE4GUI

I am new to R, but learned to use ADE4GUI, and run PCA analysis. I donot
know how to create a BIPLOT using Classes.
I used soil fauna count data -abundance- the details are as follows
number of rows 60 (from 15 habitats x 4 replicates)
number of columns (17) ie, 17 soil faunal group
the scatter file is attached for reference.

*in the example file explained in Interactive Multivariate Data Analysis in R
with the ade4 and ade4TkGUI Packages by usinf file meaudret, they created
the biplot. I tried the procedure, but unsuccessful.*
**
*Awaiting for your suggestions*

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R-sig-ecology mailing list
R-sig-ecology@...
https://stat.ethz.ch/mailman/listinfo/r-sig-ecology

About biplot analysis in ADE4GUI

---------- Forwarded message ----------
From: mujeeb rahman <mujeebrahmanp@...>
Date: Fri, Feb 6, 2009 at 12:32 PM
Subject: About biplot analysis in ADE4GUI
To: r-sig-ecology@...

I am new to R, but learned to use ADE4GUI, and run PCA analysis. I donot
know how to create a BIPLOT using Classes.
I used soil fauna count data -abundance- the details are as follows
number of rows 60 (from 15 habitats x 4 replicates)
number of columns (17) ie, 17 soil faunal group
the scatter file is attached for reference.

*in the example file explained in Interactive Multivariate Data Analysis in R
with the ade4 and ade4TkGUI Packages by usinf file meaudret, they created
the biplot. I tried the procedure, but unsuccessful.*
**
*Awaiting for your suggestions*
**
Mujeeb Rahman P
KFRI, Peechi, Thrissur, Kerala, India

_______________________________________________
R-sig-ecology mailing list
R-sig-ecology@...
https://stat.ethz.ch/mailman/listinfo/r-sig-ecology

significance of penalized regression coefficients from ridge regression

Hi all,

I've been struggling with multicollinearity in a time series data set.
My 2 predictors are collinear because there is a biennial pattern in the
data (masting shrubs), and one of the predictors is just the response
variable (berry production) at t-1, and the other (a measure of soil
moisture) is functionally related to the response at time t.  I'm
tempted to proceed with a linear regression model, present the partial
regression coefficients, and argue that the effect of collinearity would
be to reduce power to identify significant effects of either predictor,
but have also been working with ridge regression. I think I've managed
to estimate lambda using boot::cv.glm, and then estimate the penalized
regression coefficients using lm.ridge, but am uncertain how to get
effective degrees of freedom, t-, and p-values for those coefficients. 

Any advice would be appreciated.

Thanks,

Eric   

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Review of R books

Greetings!
Looking for books on R software for statistics and modeling?

If so, check out our review of 12 of the most recent and popular books on R
at:
http://ecotope.org/blogs/page/R-Book-Review.aspx

In our review, we review each book and select what we consider to be:
* The Best Introductory book on R software
* The Best R Reference book
* Best Introductory Statistics book based on R

There is also a new blog post on R:
http://ecotope.org/blogs/post/R-is-for-statistics-Revolution.aspx

Onward with the R*evolution!
-Erle
-----
 Erle C. Ellis
 Associate Professor
 Dept. of Geography & Environmental Systems
 211 Sondheim Hall
 University of Maryland, Baltimore County
 1000 Hilltop Circle, Baltimore, MD 21250 USA
 Tel: 410 455-3078 | Fax: 410 455-1056
->http://www.umbc.edu/ges/people/ellis
-->http://ecotope.org/people/ellis