Hi,
I'm probably being dense, but could somebody tell me how I can get tick marks on the
axes of my wireframe plot to show up?
This gets me a beautiful looking plot, but no ticks:
wireframe(z ~ x * y, data, drape=T, colorkey=F)
Thanks in advance,
-James
[[alternative HTML
I would like to do some fairly basic stem-and-leaf displays in R.
I am aware (I might even say painfully aware) of stem(base) and
have tried it. That's why I'm hoping someone has a usable stem-
and-leaf display for R so that I don't have to write my own.
r-project.org Search R Site Search
try
wireframe(z ~ x * y, data, drape=T, colorkey=F, scales=list(arrows=FALSE))
see the help for scales under ?xyplot for other goodies.
Cheers,
Simon.
Simon Blomberg, PhD
Depression Anxiety Consumer Research Unit
Centre for Mental Health Research
Australian National University
?optim
optim(par, fn, gr = NULL,
method = c(Nelder-Mead, BFGS, CG, L-BFGS-B, SANN),
lower = -Inf, upper = Inf,
control = list(), hessian = FALSE, ...)
.
fn: A function to be minimized (or maximized), with first
argument the vector of parameters
Roger == Roger D Peng [EMAIL PROTECTED]
on Tue, 15 Jul 2003 10:19:38 -0700 writes:
Roger I think at version 1.7.0 R started using LAPACK for
Roger its eigen/svd routines. I think using `eigen(x,
Roger EISPACK = TRUE)' uses the previous version.
Yes (2 x).
But it the eigen
At 18:08 16/07/03 +1200, Richard A. O'Keefe wrote:
I would like to do some fairly basic stem-and-leaf displays in R.
I am aware (I might even say painfully aware) of stem(base) and
have tried it. That's why I'm hoping someone has a usable stem-
and-leaf display for R so that I don't have to write
Is there an option for running SUR and 2SLS regressions with weighting
(I am analysing mortality in towns, hence want to weight by population size)
Many thanks
Jon Anson
--
Yonathan (Jon) Anson
Department of Social Work
Ben Gurion University of the Negev
84105 Be'er Sheva, Israel.
Tel: +972 8
I'd appreciate any advice people can give me on the following problem :-
I've recently started using R, as support for WinBugs software, on my PC
which runs Windows XP.
Whatever I've done has resulted in the loss of the GUI display, although
other functionality appears to be okay. This occurred
Dear Andrew,
Assuming your variables are called V1 to V4, and are vectors,
I'd use something like
plot(V1,V2,xlim=range(V1,V3),ylim=range(V2,V4),type=n)
segments(V1,V2,V3,V4)
HTH
Thomas
---
Thomas Hotz
Research Associate in Medical Statistics
University of Leicester
United Kingdom
As the list of packages on CRAN becomes longer almost on a weekly
basis, we need a formal mechanism to handle the case when somebody
wants to resign from maintaining a package.
Possible reasons for orphanizing a package:
1) The current maintainer actively wants to orphanize the package,
[EMAIL PROTECTED] wrote:
I'd appreciate any advice people can give me on the following problem :-
I've recently started using R, as support for WinBugs software, on my PC
which runs Windows XP.
Whatever I've done has resulted in the loss of the GUI display, although
other functionality appears
Sorry
I'm student in biomedical engineer and i have to solve this formula
for immuno-assay. I need to design a calibration curve
But i don't understand How can i write this formula in R language:
y = a + (c - a) /(1+ e[-b(x-m])
where
x = ln(analyte dose + 1)
y = the optical absorbance data
a =
This group impresses me, so far I have been helped with all my questions
within 24 hours. Thanks.
Therefore another one.
I am used to programs (such as STATA) where observations with missing values
that are included in a model are simply ignored in the analysis. So far I
have not been able to
There are many packages that handles missing values, aka imputations.
From memory Hmisc is one of them. But you can certainly search through
CRAN.
On Wed, 16 Jul 2003, Tor A Strand wrote:
Date: Wed, 16 Jul 2003 12:11:32 +0200
From: Tor A Strand [EMAIL PROTECTED]
To: R-list [EMAIL
Dear R users,
I am running a maximum likelihood model with optim. I chose the
simulated annealing method (method=SANN).
SANN is not performing bad, but I guess it would be much more effecive
if I could set the `parscale' parameter.
The help sais:
`parscale' A vector of scaling values for the
Shi, Tao [EMAIL PROTECTED] writes:
Hi, Ted:
I guess this problem is platform-dependent. I just tied it on a R
1.6.1 runing on Win2K, it gave me two different p values. But when I
tried it on R1.7.0 on a Linux Server, I got the similar result as
you did. I have filed a bug-report as Peter
This really depends on what you want to do. I will try to give some
example below.
1. Coding the missing values
But you definitely do not need to delete observations BEFORE loading
them into R.
By default any empty cells or NA is treated as NA, when you load the
data using read.delim(). You
Andrea Calandra [EMAIL PROTECTED] writes:
I'm student in biomedical engineer and i have to solve this formula
for immuno-assay. I need to design a calibration curve
But i don't understand How can i write this formula in R language:
y = a + (c - a) /(1+ e[-b(x-m])
where
x = ln(analyte
On 7/11, I replied to one of your earlier posts on this problem wiht the
following:
chemYield -
function(a, x)(a[1]+(a[3]-a[2])/(1+exp(-a[2]*(x-a[4]))
This can be used in optim but not nls. Mimicking an example in the
documentation for nls (package nls), I suspect the following should
work:
The phrase:
f - 10*(Wt-Wtmod)^2)/Wt) + (((Hgt-Hgtmod)^2)/Hgt))2) ; f
is an immediate computation, not a function. If you want a function,
try something like the following:
f - function(x){
Wt - x[1]
Wtmod - x[2]
Hgt - x[3]
Hgtmod -
Hi there R-Helpers,
Does anyone know if it is possible to sort a dataframe?
I.e. Sort alphabetically column 1 ( which has some reocurring elements) then
sort alphabetically column2 but keeping the order of column 1 constant;
much the same way that the sort function works in Excel.
Regards,
Hi.
Can anyone help me?
I want to create a vector from specific matrix-elements( e.g.[1,1]) these matrices are
elements of a list.
Is there any possibility to work without a loop? (e.g. with lapply?)
hope, you can help me
thank you
Michael
[[alternative HTML version deleted]]
Having read previous correspondance on this topic, am I right in using a
gaussian distribution for a tobit model, one article suggests a normal distribution?
Also, I want to censure at the upper bound, so, using the survival5 package I use:
survreg(Surv(y,yc,type=right)~x) for a censored
?order
hope this helps. spencer graves
Wayne Jones wrote:
Hi there R-Helpers,
Does anyone know if it is possible to sort a dataframe?
I.e. Sort alphabetically column 1 ( which has some reocurring elements) then
sort alphabetically column2 but keeping the order of column 1 constant;
much the
On Wed, 16 Jul 2003 14:42:09 +0100, Wayne Jones [EMAIL PROTECTED]
wrote :
Hi there R-Helpers,
Does anyone know if it is possible to sort a dataframe?
I.e. Sort alphabetically column 1 ( which has some reocurring elements) then
sort alphabetically column2 but keeping the order of column 1
Spencer Graves [EMAIL PROTECTED] writes:
On 7/11, I replied to one of your earlier posts on this problem wiht
the following:
chemYield -
function(a, x)(a[1]+(a[3]-a[2])/(1+exp(-a[2]*(x-a[4]))
This can be used in optim but not nls. Mimicking an example in
the documentation for nls
On Wed, 2003-07-16 at 08:42, Wayne Jones wrote:
Hi there R-Helpers,
Does anyone know if it is possible to sort a dataframe?
I.e. Sort alphabetically column 1 ( which has some reocurring elements) then
sort alphabetically column2 but keeping the order of column 1 constant;
much the same
On Tue, 15 Jul 2003, Roger D. Peng wrote:
I think at version 1.7.0 R started using LAPACK for its eigen/svd
routines. I think using `eigen(x, EISPACK = TRUE)' uses the previous
version.
Yes, but as the sign of eigenvectors is not well-defined it may be better
to fix the scripts so that they
Gaussian and normal are two different words for the same thing.
The example at
http://www.biostat.wustl.edu/archives/html/s-news/1999-06/msg00125.html
should give you plenty of clues.
David
On Wednesday, Jul 16, 2003, at 16:26 Europe/London, Andrew Barnes wrote:
Having read previous
Michael Kirschbaum [EMAIL PROTECTED] writes:
I want to create a vector from specific matrix-elements( e.g.[1,1])
these matrices are elements of a list.
Is there any possibility to work without a loop? (e.g. with lapply?)
Sounds like you want
lapply(mlist, [, i = 1, j = 1)
where mlist is
Dear Doug:
Thanks for your reply and clarifications. I greatly appreciate them.
I preach the Gospel of Bates Watts every chance I get, but life has
given me so many other problems to solve that I don't know your works as
well as I would like.
Best Wishes,
Spencer Graves
Douglas Bates
Have you considere unlist? Also, note that a matrix is a vector with
a dim attribute, and a data.frame is actually a list with some other
attributes.
hope this helps.
spencer graves
Michael Kirschbaum wrote:
Hi.
Can anyone help me?
I want to create a vector from specific matrix-elements(
Something like this?
myvec - sapply(list.of.mat, function(x) x[i, j])
If all the matrices are of the same dimension, it might be easier to make
them into an array, and just slice through the array.
Andy
-Original Message-
From: Michael Kirschbaum [mailto:[EMAIL PROTECTED]
Sent:
Michael Kirschbaum wrote:
Hi.
Can anyone help me?
I want to create a vector from specific matrix-elements( e.g.[1,1]) these matrices are
elements of a list.
Is there any possibility to work without a loop? (e.g. with lapply?)
hope, you can help me
thank you
Michael
[[alternative HTML version
hi Emkiba, try this:
x - matrix(1:12, 4, 3)
y - matrix(x, 12, 1)
yy - y[,1]
regards,
Huan
Internet
[EMAIL PROTECTED]@stat.math.ethz.ch - 07/16/2003 03:24 PM
Sent by:[EMAIL PROTECTED]
To:r-help
cc:
Subject:[R] turning list-elements into a vector
Hi.
Can anyone help me?
I
Dear all
I would like to ask you about possible bug in using bwplot (from lattice) together
with Hmisc library attached. I found it in my actual data, but here is a toy
example. It appears only when some levels are missing.
library(lattice)
library(Hmisc)
# preparing data
x1-rnorm(10,5,1)
Hi, Spencer
I know I submitted a beastly ammount of code, but I'm not sure how to simplify
it much further, and still sucessfully address the problem that i am having.
The reason being is that the funciton begins
f- function (q)
At the top of the iterative loop. This is what takes q and
I was wondering if is there any way of plotting line segments on a plot.
Actually I need a graph that shows the advantage of using a different method
for estimate the parameter of my databse, so I've plotted the N estimatives
for M areas in my database and then I've plotted the new estimation
Standard advice: plot to set xlim and ylim, then a loop with a
separate call to lines in a loop for each line desired. Someone else
may have something better for your current needs, but this works for me.
hope this helps. spencer graves
Carlos Rios wrote:
I was wondering if is there any way
Segments should do what you want. Try ?segments for help.
Jim
James W. MacDonald
Affymetrix and cDNA Microarray Core
University of Michigan Cancer Center
1500 E. Medical Center Drive
7410 CCGC
Ann Arbor MI 48109
734-647-5623
Carlos Rios [EMAIL PROTECTED] 07/16/03 12:26PM
I was wondering if
The following snippet suggests that there is either a bug in qr(,LAPACK=T), or some
bug in my understanding. Note that the detected rank is correct (= 2) using the
default LINPACK qr, but incorrect (=3) using LAPACK. This is running on Linux Redhat
9.0, using the lapack library that comes
As the help page for qr says, LAPACK does not attempt to detect
linear dependencies or rank deficiencies, so you should not use the
value of rank obtained with argument, LAPACK = TRUE. Computing the
rank of a matrix using finite precision is difficult, as the example on
the help page for qr
Dear r-helpers,
I have been trying to invoke R from Java in a Windows 2000 computer
(unfortunately). All my environment variables seem to be properly set,
everything seems to be in order, but I obtaining a
Fatal error: unable to open the base package
error window.
Also, the output of the
Several people have kindly (and gently) pointed out that the ?qr documentation states
that rank detection does not work for the LAPACK case. Its my fault for assuming that
rank detection did work. --Mike
On Wed, 16 Jul 2003 09:54:39 -0700
Mike Meyer [EMAIL PROTECTED] wrote:
The following
Perhaps you forget the entry in your /rw1071/.Renviron like:
JAVA_HOME=c:/yourPath/Java/j2re1.4.1_02 ?
library(SJava)
using JAVA_HOME = c:/Programme/Java/j2re1.4.1_02
regards,christian
- Original Message -
From: Vicky Albornoz [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday,
Hello,
I apologize in advance if what I'm about to ask is trivial or has been
answered before. In the latter case I would appreciate a pointer to the
right list/location
I'm trying to model the following experimental design with groupedData
and lme in R:
Subjects were measured on two tasks
Hi, Reid
Do the values of W and Hg over time for a given q agree between R and Excel?
Not the optimal value of q, just the trajectories for fixed q (trying
several values for q).
If I take the iterative loop out of the function, and ask for values of
Hgmod, Hgtmod, and f, then I get EXACTLY what
I'm confused:
I've done this type of thing by programming the same objective
function in R (or S-Plus) and Excel. After the answers from my
objective function in R match the answers in Excel, then I pass that
objective function to something like optim, which then finds the same
answers
R Help,
How does one do a least squares regression with an AR component (for any
order) in R?
Thanks,
Steve Chriss
Steve W. Chriss
Oregon Public Utility Commission
P: 503.378.3778
F: 503.373.7752
E: [EMAIL PROTECTED]
Dear Steve,
At 11:45 AM 7/16/2003 -0700, CHRISS Steve wrote:
R Help,
How does one do a least squares regression with an AR component (for any
order) in R?
Thanks,
Steve Chriss
Do you mean a model with AR errors? If so, you can use the gls (generalized
least squares) function in the nlme package.
Dear Ross,
I think I can not answer your question, but I have three ideas that
might be useful:
1. Cluster randomized trials often ask similar questions. Having a look
at CRT methods may suggest you a solution.
2. Similar questions may arise in time series analysis. I wouldn't be
surprised if
Dear R helpers
I am trying to convert a list into a data frame but when I try, I get a
stack overflow error (Error: protect(): stack overflow). My list contains
about 17000 rows and looks like shown at the bottom. The reason that I
want to convert it in to a data frame is that I want to export it
Clearly there is much I don't understand about your problem. I
suspect you are doing something that isn't correct in R, and the
complexity of the problem makes it difficult for you or anyone else to
isolate the likely error.
Have you tried discarding Hg (or W) and simplifying the
Hi, Reid and Spencer-
I think I've figured something out pretty critical to the problem.
Loking at my 'solver' options, I have a condition added that 'Hgtmod = Hgt'.
Without this conditional statement, I have to run solver 3-4 times before I get
a final solution. MEANING- solver and R,
I am looping over many data files and reading in the data with F -
read.table(filename)to read in a 22000 by 15 matrix. Works fine on the
first matrix F, but I get the following error when the second file is read
into F:
Error in row.names-.data.frame(*tmp*, value = row.names) :
Jesper Runge Madsen [EMAIL PROTECTED] writes:
Dear R helpers
I am trying to convert a list into a data frame but when I try, I get a
stack overflow error (Error: protect(): stack overflow). My list contains
about 17000 rows and looks like shown at the bottom. The reason that I
want to
I don't expect you to have a complete solution from the simplifications.
I expect you to learn something from the toy problems that can help
you solve the real problems.
Michael Rennie wrote:
Hmmm.
I tried entering 'Hgtmod = Hgt' at the end of my 'optim' function, but that
didn't help me
The row.names argument (if present) defines which column is to be used
as the row names. The default in read.table() is missing row.names and
hence row.names is not detected. The row names (if present) need to be
unique.
Giving us your code would have been helpful, but I am guessing you have
set
I'm having a little difficulty understanding this thread. If Excel can
do the job correctly and suits your needs, why not just use Excel?
As far as I know, 'optim' cannot optimize a function subject to
arbitrary equality constraints. The 'constrOptim' function allows for
linear inequality
Jesper Runge Madsen wrote:
Dear R helpers
I am trying to convert a list into a data frame but when I try, I get a
stack overflow error (Error: protect(): stack overflow). My list contains
about 17000 rows and looks like shown at the bottom. The reason that I
want to convert it in to a data frame
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