On Wed, 9 Feb 2005 [EMAIL PROTECTED] wrote:
I recently encountered a similar problem to
http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg31960.html
One conclusion from that thread seems to be
that if XP users need to update the index,
after installing a package,
they have to install R into
x - log(1:10)
y - log(1:10)
z - 1:100
contour(matrix(z,ncol=10))
did not do what I expected it to do. I expected the contour lines to
become compressed gradually.
I believe you want to do:
contour(x,y,matrix(z,ncol=10))
This will show the desired contour-plot.
Kind regards, Darius
Hi Darius
On 8 Feb 2005 at 17:43, [EMAIL PROTECTED] wrote:
I understand that I need to have a (in this case) square matrix with
all the data. But the question now is;
- can the contourplot not interpolate the missing values
or alternatively
- I have fit a model to the z data (z = 100
I am not sure if this is what you want, but try:
library(lattice)
xyplot(conc~uptake|Treatment+Type, data=CO2,
scales=list(x=list(log=TRUE)))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
On Wed, 9 Feb 2005, michael watson (IAH-C) wrote:
Hi Guys
Thanks for this! As I am a beginner, I bet I'm running into some really
basic problems. Using the example from the Map2poly function in
maptools:
try - read.shape(euadmll.shp)
mappolys - Map2poly(try)
# this produces 14
Dear useRs
I am looking for a way to randomise the values within a matrix:
the conditions are that the sums of the rows and the sums of the columns should
remain the same as in the original matrix.
Any help would be appreciated
Cheers
Yann
__
Yann Clough [EMAIL PROTECTED] writes:
Dear useRs
I am looking for a way to randomise the values within a matrix:
the conditions are that the sums of the rows and the sums of the columns
should
remain the same as in the original matrix.
Any help would be appreciated
Er, explain
Dear Yann,
Whether you can do this depends upon what you mean. If you mean randomly
permute the elements of the matrix preserving row and column sums, then this
generally won't be possible. Consider, e.g., matrix(1:4, 2, 2).
If you mean generate a matrix with specified row and column sums but
I am working on an ecological problem and dealing with a matrix where
rows correspond to samples, and columns correspond to species.
The values in the matrix are recorded abundances of the organisms.
I want to create a series of randomised datasets where total
This is not trivial. There is a really nice algorithm (should be simple to
code up) due to Persi Diaconis, and described in example 1.22 (page 1-14) of
Joseph Chang's notes on Markov chains available at
http://www.stat.yale.edu/~jtc5/251/mc.pdf
-Original Message-
From: [EMAIL
Dear all,
I am trying to extract rows from a data.frame based on the
rowSums != 0. I want to preserve rownames in the first column in the subset.
Does anyone know how to extract all species that don't have rowSums equal
to zero? Here it is:
# dataset
x - data.frame(
then try this:
library(lattice)
xyplot(conc~uptake|Treatment+Type, data=CO2,
scales=list(x=list(log=TRUE, at=c(10^1, 10^1.2, 10^1.4, 10^1.6),
labels=letters[1:4])))
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
On Wed, 2005-02-09 at 14:27 +0100, Yann Clough wrote:
I am working on an ecological problem and dealing with a matrix where
rows correspond to samples, and columns correspond to species.
The values in the matrix are recorded abundances of the organisms.
I want to create a
Yann Clough y.clough at NS1.uaoe.gwdg.de writes:
:
: Dear useRs
:
: I am looking for a way to randomise the values within a matrix:
: the conditions are that the sums of the rows and the sums of the columns
should
: remain the same as in the original matrix.
:
?r2dtable
Selon Yann Clough [EMAIL PROTECTED]:
Dear useRs
I am looking for a way to randomise the values within a matrix:
the conditions are that the sums of the rows and the sums of the columns
should
remain the same as in the original matrix.
Any help would be appreciated
Cheers
Yann
subset(x, rowSums(x[,-1], na.rm=TRUE) != 0)
Rogerio Rosa da Silva wrote:
Dear all,
I am trying to extract rows from a data.frame based on the
rowSums != 0. I want to preserve rownames in the first column in the subset.
Does anyone know how to extract all species that don't have rowSums equal
to
you have answer it yourself:
x[rowSums(x[,2:4])!=0,]
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/336899
Fax: +32/16/337015
Web:
Something like:
x[rowSums(x[,-1]) 0,]
species site1 site2 site3 site4
1sp.1 2 0 0 0
2sp.2 3 0 1 0
4sp.4 0 0 6 0
Andy
From: Rogerio Rosa da Silva
Dear all,
I am trying to extract rows from a data.frame based on the
rowSums
Dear Rogerio,
x[rowSums(x[,2:5]) != 0,] should do what you want.
I hope this helps,
John
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
Is there any way to control the spacing between bars in a histogram, or
change the border width (I'm assuming the hist() function, though
alternatives are welcome)? I'm interested in changing the visual spacing
between columns in a plotted histogram.
The general effect I'm looking for can be
Hi
I'm trying to debug one of my functions, which stopped with the
following message:
Error in is.vector(x.star) : subscript out of bounds
So I put a print(x.star) statement in just before the is.vector bit
and got
Error in print(x.star) : subscript out of bounds
So I put dput(x.star) in and
On Wednesday 09 February 2005 04:07, Christoph Scherber wrote:
Hi,
I´ll try to rephrase my question using the following example:
xyplot(conc~uptake|Treatment+Type,data=CO2)
Would it be possible to make each of the x axes log-scaled
Yes, with
xyplot(conc ~ uptake | Treatment + Type, data
Hi,
I'm still trying to get RMySQL installed on my Mac. When I try and
install from R (version 2.0.1) I'm told that the library libz can't be
located. It suggests checking in /usr/lib to see if I have the
necessary libraries. However, if I check /usr/lib I find I have the
following:
Thanks to Gabor, Francois, Peter D., Jari, Peter R.and John
r2dtable is what I was looking for, it implements the Patefield algorithm.
Patefield, W. M. (1981) Algorithm AS159. An efficient method of generating r x c
tables with given row and column totals. Applied Statistics 30, 91 97.
This
Petr,
It works perfectly! But I still have a question;
I have fit the following data;
x,y,z
1,10,11
2,11,15
3,12,21
4,13,29
5,14,39
6,15,51
7,16,65
8,17,81
9,18,99
10,19,119
dat.lm - lm(z~I(x^2)+y, data=dat)
dat.lm
Call:
lm(formula = z ~ I(x^2) + y, data = dat)
Coefficients:
(Intercept)
Hi all,
I'm a little surprised at
NULL == c(a, b)
logical(0)
all(NULL == c(a, b))
[1] TRUE
Reading the documentation for all() this was not clear for me to be expected.
Originally the question came up when using
match.arg(NULL, c(a, b))
[1] a
where I had thought an error would occur.
So could
Last Friday, Gregory Chaitin (http://www.umcs.maine.edu/~chaitin/lm.html)
mentioned that there can be no proof that a given code is the shortest for
a problem, even within a language. Still, the script below, a replacement
of the TDT, one of the most frequently used tests in genetics
How can I get lattice plots (with xyplot, etc.) to be produced in black
and white, rather than in color? Im using R 2.0.1 under windows XP.
(I'm not subscribed to the list, so would appreciate direct replies.)
Thanks for any help.
David Parkhurst
__
A histogram is a density estimate (at least as defined in the Encyclopedia
of Statistics Sciences, if not in many US Universities). It is an area,
not a series of unrelated bars, so it makes no sense to have spaces
between the subareas.
Unfortunately, hist() will also produce barplots of
Tom,
I dealt with this once.
barplot.data- c(2,3,4,5,6)
x- barplot(barplot.data, ylim = c(0,10), space= .9,.9,.9,.9)
use the space= to define the spacing between each of your barplot values.
If you have groups of bars that you want together, with spaces between the
groups,
you have to put
Robin Hankin [EMAIL PROTECTED] writes:
Hi
I'm trying to debug one of my functions, which stopped with the
following message:
Error in is.vector(x.star) : subscript out of bounds
So I put a print(x.star) statement in just before the is.vector bit
and got
Error in
Dear R-help..
I am rather new in R so i would appreciate your help in my problem..
I cant seem to be able to write a function that has arguments being objects and
column names of these ojbects...
A simple example code that doesnt work is the following..
auto_plot - function
Matthias Burger [EMAIL PROTECTED] writes:
Hi all,
I'm a little surprised at
NULL == c(a, b)
logical(0)
all(NULL == c(a, b))
[1] TRUE
Reading the documentation for all() this was not clear for me to be expected.
all() over a vector of length zero is TRUE for much the same reason
Insightful is searching for a candidate to fill
a technology position whose primary function is
to develop commercial software for mixed-effects
models and methods for analyzing longitudinal data,
survival data, and missing data.
Utilize your background in statistical algorithms,
data analysis,
Dear R-help..
I am rather new in R so i would appreciate your help in my problem..
I cant seem to be able to write a function that has arguments being objects
and
column names of these ojbects...
A simple example code that doesnt work is the following..
auto_plot - function
On Wed, 9 Feb 2005, Matthias Burger wrote:
I'm a little surprised at
NULL == c(a, b)
logical(0)
NULL is of length 0.
== coerces arguments and recycles as required, so you have done
as.character(NULL) == character(0)
all(NULL == c(a, b))
[1] TRUE
Reading the documentation for all() this was not
On Wednesday 09 February 2005 09:50, David Parkhurst wrote:
How can I get lattice plots (with xyplot, etc.) to be produced in
black and white, rather than in color? Im using R 2.0.1 under
windows XP.
This happens automatically for postscript output. Otherwise, start your
device with
On Wed, 9 Feb 2005, Matthias Burger wrote:
Hi all,
I'm a little surprised at
NULL == c(a, b)
logical(0)
all(NULL == c(a, b))
[1] TRUE
Reading the documentation for all() this was not clear for me to be expected.
This is related to the question about sum(numeric(0)) that came up a few
days ago.
How about data.lm$model ?
On Wed, 2005-02-09 at 16:32 +0100, [EMAIL PROTECTED] wrote:
Petr,
It works perfectly! But I still have a question;
I have fit the following data;
x,y,z
1,10,11
2,11,15
3,12,21
4,13,29
5,14,39
6,15,51
7,16,65
8,17,81
9,18,99
10,19,119
dat.lm -
On Wed, 9 Feb 2005, Knut M. Wittkowski wrote:
Last Friday, Gregory Chaitin (http://www.umcs.maine.edu/~chaitin/lm.html)
mentioned that there can be no proof that a given code is the shortest for a
problem, even within a language.
I presume he said something more like `for some problems there is
thanks to Peter Daalgard, Brian Ripley, and Thomas Lumley for their quick,
accurate and very helpful replies.
My understanding of NULL was incorrect, thanks for making this clear.
As for the question on match.arg I will use pmatch instead.
Best,
Matthias
Thomas Lumley wrote:
On Wed, 9 Feb
Hi all,
I created a graph by plotting 4 time series on top of each other but don't
know how to add extra axes on the right hand side of the plot (or leftside).
For the first two time series it axis are plotted but the last two time
series don't have and axis. How can these be added and where?
[EMAIL PROTECTED] wrote:
Dear R-help..
I am rather new in R so i would appreciate your help in my problem..
I cant seem to be able to write a function that has arguments being objects and
column names of these ojbects...
A simple example code that doesnt work is the following..
auto_plot -
Hello,
I will really appreciate some help with the following question.
let's say cor.m is a correlation coefficient matrix.
tree-hclust(as.dist(1-cor.m));
plot(tree);
my question is: is it possible to add the correlation coefficient information
to the plot?
thanks,
S. Ma
I am trying to use the function gl (generate levels), and would like to make
levels with different number of replications. Does anyone know how to
generate different number of replications in each level?
Something like:
[1] Site1 Site1 Site1 Site1 Site1 Site1 Site1
[8] Site2 Site2 Site2 Site2
hello all...
i am trying to install r v2.0.1 on my suse 9.2 pro box... when i run
configure, i get the following error:
checking how to get verbose linking output from g77... configure:
WARNING: compilation failed
checking for Fortran libraries of g77...
checking for dummy main to
Can't do it. gl is based on rep(), which easily does what you want:
factor(rep(c('Site1','Site2'),c(7,5)))
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
An R program needs to accumulate a list (length unknown until done) of
objects. How efficiently and repeatedly can one append new objects to the
list?
I have been using something like
mylist[[length(mylist)+1]] = newobject
but I wonder if there is something better.
--
Michael Prager
NOAA
I understand the problem from a statistical perspective, and you make
an excellent point (as I have come to expect, reading this list).
However, I'm thinking about it from a visual/aesthetic perspective.
Let me try this. Plot two histograms side-by-side:
x - rnorm(10)
par(mfcol=c(1,2))
I see where you're going...yes, I could create my own histogram
distribution from the data and plot it using barplot (in fact, I've
already achieved the visual effect I want with barplot, so I could just
expand on that). I'm hoping that I can find a way to do it with hist()
to avoid
Jan Verbesselt wrote:
Hi all,
I created a graph by plotting 4 time series on top of each other but don't
know how to add extra axes on the right hand side of the plot (or leftside).
For the first two time series it axis are plotted but the last two time
series don't have and axis. How can these be
[EMAIL PROTECTED] wrote:
Hello,
I will really appreciate some help with the following question.
let's say cor.m is a correlation coefficient matrix.
tree-hclust(as.dist(1-cor.m));
plot(tree);
my question is: is it possible to add the correlation coefficient information
to the plot?
Which
A hint: hist() does no plotting. That is done by the histogram method
of plot(), which you can copy and modify to your own needs/desires.
Alternatively you can create myplot() and call
myplot(hist(..., plot=TRUE), ...)
On Wed, 9 Feb 2005, Thomas Hopper wrote:
I understand the problem from a
Rogrio Rosa da Silva rrsilva at ib.usp.br writes:
:
: I am trying to use the function gl (generate levels), and would like to make
: levels with different number of replications.Does anyone know how to
: generate different number of replications in each level?
:
: Something like:
:
: [1]
Mike Prager Mike.Prager at noaa.gov writes:
:
: An R program needs to accumulate a list (length unknown until done) of
: objects. How efficiently and repeatedly can one append new objects to the
: list?
:
: I have been using something like
:
: mylist[[length(mylist)+1]] = newobject
:
: but
Dear useRs,
How come the first attempt to sort a POSIXt vector fails (Error:
non-atomic type in greater), while the second succeeds? (Code inserted
below.) The documentation says that POSIXt is used to allow operations
such as subtraction, so I'd expect sorting to work. Is this perhaps an
OS
bogdan romocea br44114 at yahoo.com writes:
:
: Dear useRs,
:
: How come the first attempt to sort a POSIXt vector fails (Error:
: non-atomic type in greater), while the second succeeds? (Code inserted
: below.) The documentation says that POSIXt is used to allow operations
: such as
Hi-
I've been trying to install the hier.part package on
my mac (OSX 10.3.7) and it is not working for some
reason. I am downloading the package source called :
hier.part_1.0.tar.gz. When I try to auto install from
the cran site, I get this message:
* Installing *source* package 'hier.part' ...
Hi All:
I am interested in tools available in R that will enable me to run analysis of
case crossover studies. I searched the web (http://www.google.com), and found
that a tool called funfit was available for Splus. Searching the R archives,
I also found mention of a funfit library.
However,
Hm. Everyone else got it (the reasons for the change).
The reason is that beginners often fail to get it (package vs.
library), and one of the contributing factors is the naming approach,
which isn't clear. Certain people are rather sensitive to mistakes of
this nature, so we end up with 2-3
I am not sure if I understad you correctly. Are you inrested in getting a
list of units that belond to each group? Then do:
branches-cutree(hc,k=2)
group1-which(branches==1)
group2-which(branches==2)
Hope this helps!
Ales Ziberna
- Original Message -
From: T. Murlidharan Nair [EMAIL
On Wed, 9 Feb 2005, Spencer Graves wrote:
The reasons to 'introduce package() and deprecate library()' may be
OBVIOUS to you, but they completely escape me. Could you please clarify why
that's obvious? I've seen many admonitions on this list that the term is
package NOT library, but I
You can order (which is not the same as sort) POSIXct vectors but not
POSIXlt ones.
class() on your objects (or str(), but that only shows one class) would
have been revealing.
It is somewhat fortuitous that you can order and sort POSIXct vectors, as
you are ordering the underlying numeric
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