Daniel,
I can help somewhat I think. PC-ORD also allows data input in what
it calls database format, where each row is
sample, taxon, abundance
There as many rows/sample as there are non-zero species, and only three
columns. To get your taxon data.frame (currently samples as rows,
Those are pretty interesting approaches Ted. An alternative is to
establish a maximum length for any segment and delete all segments
longer than that. Then, find the shortest connected path such that no
segments are longer than your threshold. In many cases this would
result in the same
Sotiris,
R is generally fairly graceful about FORTRAN in linux; Windows is
another matter. For example, R/linux will allow you to write to the R
console as a file device without using the special I/O routines often
needed in R. There are many very complicated FORTRAN routines currently
In addition to the references from Professor Ripley, you might be
interested in the R packages and pages maintained by ecologists for such
work (even if you're doing movies). Packages labdsv, vegan, and ade4
both have a broad variety of distance/dissimilarity indices and numerous
alternative
Colby,
Function surf() in package labdsv uses the gam() function from mgcv
to do this in conjunction with akima. You might want to look at that
routine for an idea. Currently it fits the gam as z - gam(s(x) + s(y),
but it's possible in the mgcv version of gam to fit z - gam(s(x,y)) as
You can just use data.frame(). If (using your example) your dataframes
are called first and second, your could
new - dataframe(first$A,second$Z,first$B,second$Y,first$C,second$X...)
followed by
names(new) - c('A','Z','B','Y','C','X')
If you have an enormous number of columns that's a pain,
You might prefer boxplot(insolation~veg_type) as a graphic. That will
give you quantiles. To get the actual numeric values you could
for (i in levels(veg_type)) {
print(i)
quantile(insolation[veg_type==i])
}
see ?quantile for more help.
Dylan Beaudette wrote:
Greetings,
I have a
Christian,
One of the arguments to prcomp() is retx, with a default value of
TRUE. As explained in the help file, if retx is TRUE the prcomp object
reurned by the function contains the projection of the original data
along the principal components (which many of us call scores). Thus
Norman,
You're missing a step. You need to convert the data file into a
'dist' object, which is either a distance or dissimilarity matrix. This
is typically done by function dist(), but may also be done by other
functions which produce dist objects, like daisy() in package cluster,
This would be extraordinarily helpful, but I have not thought of a
graceful way to do it. Everything in R now has a class attribute, but a
timestamp for such simple things as vectors seems like overkill. On the
other hand, those of us writing packages could implement this pretty
easily for
Well, this has been an interesting thread. I guess my own perspective
is warped, having never been a C programmer. My native languages are
FORTRAN, python, and R, all of which accept (or demand) a linefeed as a
terminator, rather than a semicolon, and two of which are very
particular about
Graham,
It's relatively easily done, especially the first one.
Let's suppose your veg data frame is called veg
dom1 - apply(veg,1,which.max)
returns a vector with the column number of the species with the highest
abundance for each sample (if there are ties, it returns the first one).
If
Mike,
It's not clear whaty way you are doing it now, but this works
x - matrix(c(0,2,0,0,0,4,3,0,0),nrow=3)
x
[,1] [,2] [,3]
[1,]003
[2,]200
[3,]040
y - as.vector(t(x))
z - y[y!=0]
z
[1] 3 2 4
Dave
Mike Jones wrote:
hi all,
i'm
Eszter,
I suspect the problem is different than you think. It's possible that
when you read in the data it assumed that the first column was data, not
row names, and so when you transpose the first column becomes the first
row. Since it is alpha, all the columns become factors rather then
cmdscale calculates an eigenanalysis of the dissimilarity matrix, and
does not employ stress per se. Rather, it attempts to maximize
variability along axes.
If you call the the cmdscale() function with eig=TRUE it returns a
list object with the coordinates called points and the eigenvalues
There are several bootstrap packages available at CRAN that probably
provide an elegant solution, but simply permuting the matrix is pretty easy
data - matrix(1:100,nrow=5) # matrix of 5 rows and 20 columns
x - data[sample(1:5),] # permute the rows
y - x[,sample(1:20)] # permute the
AT XX TT CC NA NA TT
8 TT XX TT AC AG AG TT
9 AT XX TT CC AG NA TT
10 TT XX TT CC GG GG TT
Notice that the instances of 'CC' in tmp$V7 did not change.
HTH, Dave Roberts
Federico Calboli wrote:
Hi All,
I have the following problem, that's driving me mad.
I have a dataframe
neck a few times. It does not work with acroread (the
linux Acobat Reader program) however.
Dave Roberts
Thomas Schönhoff wrote:
Hi,
2005/10/21, Thomas Schönhoff [EMAIL PROTECTED]:
Hello again,
2005/10/21, Thomas Schönhoff [EMAIL PROTECTED]:
2005/10/21, Ted Harding [EMAIL PROTECTED
If I understand the request, he wants to take a large number of matrices
of identical size and stack them into a three dimentional array, and
then calculate statistics on the the third dimension. If the multiple
arrays have object names they can be combined into a 3-d array
a -
Martin,
If the data are actually coded 0/1, the tree function would
probably intepret them as integers and try a regression instead of a
classification. If the dependent variable is called var, try
x - tree(factor(var)~species)
0.0 1.0 0.0
0.0 ) *
I'll try agin with a larger dataset and see if it's a memory limitation.
Dave Roberts
Martin Wegmann wrote:
On Friday 23 September 2005 17:08, Dave Roberts wrote:
Martin
locations, but I think it
should work.
Good luck, Dave Roberts
Martin Wegmann wrote:
On Friday 23 September 2005 17:08, Dave Roberts wrote:
Martin,
If the data are actually coded 0/1, the tree function would
probably intepret them as integers and try a regression instead
? Perhaps something altogether different will work better.
Thanks, Dave Roberts
On Mon, 2005-09-19 at 09:41 +0200, [EMAIL PROTECTED] wrote:
Hi,
I'm trying to find out what threshold of indicator value in labadsv should be
used to accept a specie as an indicator one? So far I assumed that indval
.
Thanks, Dave Roberts
On Mon, 2005-09-19 at 09:41 +0200, [EMAIL PROTECTED] wrote:
Hi,
I'm trying to find out what threshold of indicator value in labadsv should
be
used to accept a specie as an indicator one? So far I assumed that
indval=0.5
is high enough to avoid any mistakes
24 matches
Mail list logo
