Hello,
I've got some problems. I hope someone can help me.
First question:
I am trying to get grid on scatterplot3d (from scatterplot3d package). It
seems that scatterplot3d draw on grip on X and Z side. Is it possible to get
Grid on the whole Box?
Second question:
Is it possible to use the
Unfortunately your proposed change changes the type of the output:
simplification is intended in many applications of by().
Before:
str(by(mytimes$date[1], mytimes$set[1], function(x)x))
by [, 1] 1.21e+09
- attr(*, dimnames)=List of 1
..$ mytimes$set[1]: chr 1
- attr(*, call)=
Greetings all,
I am trying to use Deal to learn a Bayesian Network for discrete data. I
came across the following problem with jointprior function.
library(deal)
a - read.csv(prepared.cluster1.csv)
fit - network(a)
fit.prior - jointprior(fit)
Error in array(1, Dim) : 'dim' specifies too large
On Thu, 17 Apr 2008, Phil taylor wrote:
Hi folks,
when trying to update packages in version 2.6.1 I recuieve the following
error messages:
Actually, they are *warnings*.
Warning: unable to access index for repository
http://www.sourcekeg.co.uk/cran/bin/windows/contrib/2.6
Warning: unable
On Wed, 16 Apr 2008 18:51:30 +0200 Agustin Lobo wrote:
AL 2. I need the figure displayed on the screen, I'm using
AL Impress (the equivalent of ppt). Thus EPS is not an option,
1. So what please is wrong with png? I am using it with Word (If for
some reasons I cannot use LaTeX -- some coauthors
Hello,
I've found the function isPeak from the Bioconductor package
PROCess very useful for peak finding. My data was similar to yours
(spectroscopy, i'm a physicist) but not time dependent. It may be a
starting point to work on one spectrum, retrieve a few relevant
informations (peak
Dear Bryan,
For work in that general direction, look at the R News that was on 'R in
chemistry' (Volume 6/3, August 2006,
http://www.r-project.org/doc/Rnews/Rnews_2006-3.pdf) and the special vol.
of the Journal of Statistical Software on 'Spectroscopy and Chemometrics
in R' (vol 18,
Edwin Sendjaja wrote:
Hello,
I've got some problems. I hope someone can help me.
First question:
I am trying to get grid on scatterplot3d (from scatterplot3d package). It
seems that scatterplot3d draw on grip on X and Z side. Is it possible to get
Grid on the whole Box?
At least
Stefan,
Stefan Grosse escribió:
On Wed, 16 Apr 2008 18:51:30 +0200 Agustin Lobo wrote:
AL 2. I need the figure displayed on the screen, I'm using
AL Impress (the equivalent of ppt). Thus EPS is not an option,
1. So what please is wrong with png? I am using it with Word (If for
some reasons
Thanks, actually I had thought on Lynx as an option, so
your opinion encourages me to try.
But wanted to solve the problem within the OO environment
as well.
Agus
Marc Schwartz escribió:
See comments inline:
Agustin Lobo wrote:
Thanks for your answers, let me summarize the situation:
1.
Dear list:
Is there any way of getting the equivalent to what you get with bty=l
for the right and bottom axes? and the equivalent to
bty=7 for the upper and left axes?
Thanks!
Agus
--
Dr. Agustin Lobo
Institut de Ciencies de la Terra Jaume Almera (CSIC)
LLuis Sole Sabaris s/n
08028 Barcelona
Hello everyone,
I was trying to calculate linear contrasts with coxph via the contrasts
function, but I'm not sure if it is correct.
First, all the statistics change, if I state 2 contrasts instead of 3.
Second, stipulating common linear contrasts does not even nearly produce an
expected result
The newsletter of the sections Statistical Computing and Graphics of
the ASA (American Statistical Association) is currently looking for
contributions. The newsletter is widely read by statisticians all over
the world (thousands of downloads).
The forum is ideal for ongoing research and also for
Agustin Lobo wrote:
Dear list:
Is there any way of getting the equivalent to what you get with bty=l
for the right and bottom axes? and the equivalent to
bty=7 for the upper and left axes?
Thanks!
Agus
Not using bty, but you can get the coordinates of the boundary via
par(usr) and draw
Rolf Turner wrote:
On 17/04/2008, at 9:33 AM, Charles C. Berry wrote:
snip
I'll lay odds that Matthew's 'matrix' is actually a data.frame, and
I'll not be surprised if the columns are factors.
snip
I suspect that you're right.
***Why*** can't people distinguish
Not sure who to send this to...
On the books page http://www.r-project.org/doc/bib/R-books.html the entries
from [54] down have three forward slashes in the hyperlinks to publisher
info and so they don't work
Robin
[[alternative HTML version deleted]]
Adding a simplify argument to by would suit me fine.
In my (limited) experience in using R, the automatic simplification
that R does in various situations is one of it's most troublesome
features. It means that I cannot expect a program to work even if I
give it data of the same types as I
On Thu, 17 Apr 2008, Alex Brown wrote:
Adding a simplify argument to by would suit me fine.
In my (limited) experience in using R, the automatic simplification that R
does in various situations is one of it's most troublesome features. It
means that I cannot expect a program to work even
There is a Defaults package on CRAN that allows one
to set default arguments for any function.
On Thu, Apr 17, 2008 at 6:49 AM, Alex Brown [EMAIL PROTECTED] wrote:
Adding a simplify argument to by would suit me fine.
In my (limited) experience in using R, the automatic simplification
that R
Also, here is a similar solution, even more compact,
not involving zoo:
plot(y ~ x, na.omit(data.frame(x, y)), lty = 3, type = l)
lines(y ~ x)
On Wed, Apr 16, 2008 at 11:18 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Try this which interpolates the NAs using na.approx from zoo
drawing the
On Wed, Apr 16, 2008 at 03:56:26PM -0600, Matthew Keller wrote:
Yes Chuck, you're right.
just a comment:
Thanks for the help. It was a data.frame not a matrix (I had called
as.matrix() in my script much earlier but that line of code didn't run
because I misnamed the object!). My bad.
Hi,
I am working in different formula substitutions using R i have applied
linear[lm(y~x)], Quadratic[lm(y~x+I(x^2))] and Cubic[lm(y~x+I(x^2)+I(x^3))]
How to use/apply One Term Exponential, Two Term Exponential, Three Term
Exponential formulas in R Stat.
Regards,
Guru
That works for me. Very awesome. Most appreciated.
Please follow the attached hyperlink to an important disclosure:
http://www.credit-suisse.com/legal/marketcommentary
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Wednesday, April 16, 2008 9:52 AM
To:
Dear R-users,
I have noticed small discrepencies in the reported estimate of the
variance of the frailty by the print method for survreg() and the
'theta' component included in the object fit:
# Examples in R-2.6.2 for Windows
library(survival) # version 2.34-1 (2008-03-31)
# discrepancy
Gad Abraham wrote:
Hi,
Design isn't strictly an R base package, but maybe someone can explain
the following.
When lrm is called within a function, it can't find the dataset dd:
library(Design)
age - rnorm(30, 50, 10)
cholesterol - rnorm(30, 200, 25)
ch - cut2(cholesterol,
Dear R users,
I was wondering from where I could get the C source code to compute
pnbinom() and qnorm() ?
(I would use R in batch mode but I find the startup time prohibitive, unless
there is a way to speed it up)
I searched the Web and it clearly is part of the R distribution, I just
don't know
On 4/17/2008 9:16 AM, Markus Loecher wrote:
Dear R users,
I was wondering from where I could get the C source code to compute
pnbinom() and qnorm() ?
(I would use R in batch mode but I find the startup time prohibitive, unless
there is a way to speed it up)
I searched the Web and it clearly
Hi,
I had a matrix with NULL values, which I wanted to replace with NA. Is there an
efficient way to do this?
Small sample input matrix:
A B C D E
1 5222.18 6355.10 4392.68 2607.41 4524.09
2NULL 257.33NULL 161.51 119.44
3NULL 274.80 305.28
Tim Smith wrote:
Hi,
I had a matrix with NULL values, which I wanted to replace with NA. Is there
an efficient way to do this?
Small sample input matrix:
A B C D E
1 5222.18 6355.10 4392.68 2607.41 4524.09
2NULL 257.33NULL 161.51 119.44
3
Hi everybody. I'm looking for someone who want to do a little script in R for
money. The work is about kmeans algorithms and it will not take too much
time to do it. If you are interested just send me a pm and we will discuss
about everything.
Thank you, regards
Jimmy Nox
--
View this message
On Thu, 17 Apr 2008, Dimitris Rizopoulos wrote:
Dear R-users,
I have noticed small discrepencies in the reported estimate of the
variance of the frailty by the print method for survreg() and the
'theta' component included in the object fit:
This is the print method for class survreg.penal.
Hi Uwe,
Thanks for your answer.
What is the different between rgl and scatterplot3d? I dont need a graphik
like vulcano. I just need 3D-dot-plot.
Am Donnerstag, 17. April 2008 09:03:17 schrieb Uwe Ligges:
Edwin Sendjaja wrote:
Hello,
I've got some problems. I hope someone can help me.
Tim Smith wrote:
Hi,
I had a matrix with NULL values, which I wanted to replace with NA. Is there
an efficient way to do this?
Small sample input matrix:
A B C D E
1 5222.18 6355.10 4392.68 2607.41 4524.09
2NULL 257.33NULL 161.51 119.44
3
On Thu, 17 Apr 2008, Uwe Ligges wrote:
Tim Smith wrote:
Hi,
I had a matrix with NULL values, which I wanted to replace with NA. Is there
an efficient way to do this?
Small sample input matrix:
A B C D E
1 5222.18 6355.10 4392.68 2607.41 4524.09
2NULL
Edwin Sendjaja wrote:
Hi Uwe,
Thanks for your answer.
What is the different between rgl and scatterplot3d? I dont need a graphik
like vulcano. I just need 3D-dot-plot.
Sure, rgl can do it as well.
Difference is that scatterplot3d is based on R's standard devices while
rgl is based on
if your matrix is stored in a text file then
xx - read.table(x, na.strings=NULL)
where x is the name of the text file.
On Thu, 17 Apr 2008, Tim Smith wrote:
Hi,
I had a matrix with NULL values, which I wanted to replace with NA. Is there
an efficient way to do this?
Small sample input
Hi to all,
I recently created a small package which contains some simple R
functions in order to help students in my research unit to use R. The
package is pure R, it passes R CMD check, but it is really designed
for internal use and, moreover, it is entirely documented in french.
Is it possible
Thank you very much, Dr. Ripley. The solution ifelse() you provided
is exactly what I want. I am so happy this morning for that I recieved
your email. Yesterday night I was trying to write a loop to substitute
NA. But now I learn that ifelse() does a much more efficient work.
Really appreciate
ML == Markus Loecher [EMAIL PROTECTED]
on Thu, 17 Apr 2008 09:16:50 -0400 writes:
ML Dear R users,
ML I was wondering from where I could get the C source code to compute
ML pnbinom() and qnorm() ?
ML (I would use R in batch mode but I find the startup time prohibitive,
df1 - data.frame(a = LETTERS[1:2], b = LETTERS[3:4], c = 1:2)
I am looking for an idiom that swaps the elements of df$a and df$b
when (e.g.) df$c == 2, resulting in
df2 - data.frame(a = LETTERS[c(1, 4)], b = LETTERS[c(3, 2)], c = 1:2)
_
Professor Michael Kubovy
Suppose that we have o 2-D contingency table where the row variable is nominal
and the column one is ordinal. In SAS it is possible to compute the statistic
named as row mean scores differ. How can we programmed it in R?
(See also Aggresti (2002), Categorical Data Analysis, p. 302)
With regards
temp-df1$a
df$a-ifelse(df1$c == 2, df$b, df$a)
df$b-ifelse(df1$c == 2, temp, df$b)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Michael Kubovy
Sent: Thursday, April 17, 2008 10:53 AM
To: [EMAIL PROTECTED]
Subject: [R] Conditionally swap items in a
Dear Ben,
Thank you very much for your reply to my R2WinBUGS query. As requested, I
am attaching the following files to this e-mail:
1) The .odc file containing the WinBUGS code I am trying to run from R. The
nodes to be monitored for posterior inference are: T, best, d, lor,
Thanks for the suggestion, Somon! I did try glht from multcomp
package, but the problem is that for the hypothesis
H0: TypeT1 =0 and TypeT2 = 0
it gives results for two separate hypotheses H01: TypeT1 =0 and H02:
TypeT2 = 0, not exactly one statistic for the original hypothesis H0.
So my
Dear R developpers,
I am a user of Splus since many years and I have developped lots of
functions to plot graph of data and model results of irregular or
regular times series.
In Splus regular times series are created using the rts function and
irregular time series using the its functions.
Suppose that we have o 2-D contingency table where the row variable is nominal
and the column one is ordinal. In SAS it is possible to compute the statistic
named as row mean scores differ. How can we programmed it in R?
(See also Aggresti (2002), Categorical Data Analysis, p. 302)
There is a
Thanks both Harold Doran and Prof. Ripley for the suggestion.
Time*Group - 1 or Time*(Group-1) does seem better. However as Prof.
Ripley pointed out, it is a little complicated with the interactions.
For example,
==
set.seed(1)
group - as.factor (sample (c(M,F), 12, T))
y - rnorm(12)
On Thu, Apr 17, 2008 at 9:03 AM, Stefan Grosse [EMAIL PROTECTED] wrote:
1. So what please is wrong with png? I am using it with Word (If for
some reasons I cannot use LaTeX -- some coauthors are unfortunately
quite resistant there...)
.. for the reticent, there is always LyX.. [1] To a
The place to look is the CRAN Task View 'ExperimentalDesign'.
There are several packages there related to design and analysis of
experiments.
The package 'AlgDesign' appears to have a function for generating mixture
designs, and there might be others in other packages.
Good luck!
-Christos
Try also:
df1 - data.frame(a = LETTERS[1:2], b = LETTERS[3:4], c = 1:2)
df1$a - as.character(df1$a) #if is factor
df1$b - as.character(df1$b) #if is factor
df1[df1$c == 2, ] - c(subset(df1, c == 2, select = c(b, a)), 2)
On Thu, Apr 17, 2008 at 11:52 AM, Michael Kubovy [EMAIL PROTECTED]
wrote:
Hello,
I am having trouble with creating a simple bar chart using R. My file just
consists of 1 column called reason. Within this column, there are 2
responses(Room for Improvement and Info entered is all relevant). My code
for creating the charts is as follows:
d
Julien Barnier wrote:
Hi to all,
I recently created a small package which contains some simple R
functions in order to help students in my research unit to use R. The
package is pure R, it passes R CMD check, but it is really designed
for internal use and, moreover, it is entirely documented in
Try:
pie(table(d$Reason),
main=Results)
On Thu, Apr 17, 2008 at 1:20 PM, hoogeebear [EMAIL PROTECTED] wrote:
Hello,
I am having trouble with creating a simple bar chart using R. My file just
consists of 1 column called reason. Within this column, there are 2
responses(Room for
Have a look at the zoo package and read its three vignettes.
library(zoo)
z - zoo(1:10, Sys.Date() + 1:10)
zz - zoo(c(11, 13, 15), Sys.Date() + c(1, 3, 5))
plot(cbind(z, zz), screen = 1, type = p, col = 1:2)
# or omit screen = 1 for a multi-panel plot
vignette(package = zoo)
On Thu, Apr 17,
Hi Pedro,
There's no particularly easy way to do this. However, you can create
the scales, train them manually, and then draw them in any way you
choose:
colour - scale_colour_hue(My scale)
colour$train(factor(c(a,b)))
grid.newpage()
grid.draw(gglegend(colour$legend_desc(), list(colour=point)))
Dear All,
Using odfWeave is working fine for me using code chunks ...= ... @
But the inline \Sexpr{date()} is not evaluated and returns again
\Sexpr{date()} instead of Thu Apr 17 18:15:47 2008.
What might be the reason for that, what do I have to look for?
I'm using Windows XP, R 2.6.1,
I have a spatialpixelsdataframe containing bathymetry data. Is it
possible to save this as something similar to the meuse.grid so it can
be loaded without having to run the spatial interpolation each time the
script is called?
I don't see a straightforward method in the sp documentation.
Dear netters, suppose I have a matrix X [1,] 'c1' 'r6' '150'[2,] 'c1' 'r4'
'70'[3,] 'c1' 'r2' '20'[4,] 'c1' 'r5' '90'[5,] 'c2' 'r2' '20'[6,] 'c3' 'r1'
'10'I want to apply some funciton to groups of rows by the first column.If the
function is just to calculate the average X[,3], it will
Does something like this work for you? It is using 'lapply' with the
indices of the rows:
x - matrix(c( 'c1' , 'r6', '150', 'c1' , 'r4' ,'70' ,'c1' , 'r2' ,'20',
+ 'c1' , 'r5' ,'90', 'c2' ,'r2' ,'20', 'c3' , 'r1' ,'10'),
byrow=TRUE, ncol=3)
# use lapply
result - lapply(split(seq(nrow(x)),
Hi all.
The problem is:
I have, let´s say a number of 1000 subjects within a cross-sectional study.
These subjects (=animals) are housed in 50 Farms. I want to model the
influence of farm-specific factors on a binary health outcome (incidence).
How can I implement the farm effects which are
Dear R-listers,
I am trying to compute interaction effects in a probit model, and
conduct hypothesis tests on these effects correctly.
Specifically, I have a model of the form y = a + b1 m + b2 x + b3 m*x,
where both y and m are 0-1 dummies, x is continuous, and I am
interested in the sign and
Daniel,
thank you!
I want to perfrom the simplest way of matching:
a one-to-one exact match (by age and school):
for every case in treat find ONE case (if there is one) in control .
The cases in control that could be matched, should be tagged as
not available or taken away (deleted) from the
On 18/04/2008, at 2:19 AM, Julien Barnier wrote:
Hi to all,
I recently created a small package which contains some simple R
functions in order to help students in my research unit to use R. The
package is pure R, it passes R CMD check, but it is really designed
for internal use and,
Hi -
I'm having a really hard time w/understanding R's get function, and would
appreciate any help with this.
Specifically, I'm using a for loop to call a function. I'd like the
function to have access to the variable being incremented in the for-loop,
i.e.
t.fn - function() return( get( i ) )
Dear All,
After doing more try and error ...
This problem occurred as I was loading R2HTML before the odfWeave function.
This issue is also described in the Sweave FAQ.
From the FAQ
A.16 After loading package R2HTML Sweave doesn’t work properly!
Package R2HTML registers an Sweave driver for
Hello,
Does anyone know of a way of finding all the nodes that are between a pair
of specified nodes in the excellent graph package (R vers 2.5.0).
I have a class(graphAM) object and need to determine all possible pathways
in this object.
Here's an example: In the simple case of a--b--c
Hi,
I have count data that have been meddled with enough to make them non
integers. Using glm(poisson) returns a non integer error but
glm(quasipoisson) does not. Just wondering if anyone knows if I am
violating the assumptions of a quasipoisson error structure by using
these non-integer
Have a look at the RBGL package
(but you probably want to update your R etc to something more recent)
[EMAIL PROTECTED] wrote:
Hello,
Does anyone know of a way of finding all the nodes that are between a pair
of specified nodes in the excellent graph package (R vers 2.5.0).
I have a
Dear R People:
I was looking to see if there are any functions for Vector ARMA modeling.
I found Vector AR(p) but no Vector ARMAs.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: [EMAIL PROTECTED]
Dear All,
I have a list of models(1000) which have variable scores from 20 different
method. I would like to rank models using consensus approach based on high
scores from different methods.Is there any function available in R for this
purpose? I will appreciate any pointers in this
This should be very easy, but alas, I'm very new to R. My end goal is to
calculate p-values from arima().
Let's say I just ran this:
MyModel - arima(y[1:58], order=c(1,0,0), xreg=MyData[1:58,7:14],
method=ML)
MyModel
And I see:
arima(x = y[1:58], order = c(1, 0, 0), xreg = MyData[1:58,
Hi Uwe,
I decided to use scatterplot3d, because it looks better.
I have some questions:
Is it possible to get 1 axis( for example: z-axis) not as numeric, but as
character.
Because I have date set like this:
x=relative Time: 0,3 ms; 0,5ms, etc
y=Delay:10 ms, 20 ms, etc
z= Host: cnn.com,
Hi,
I am trying to generate a group of graphics with an iteration. Some
thing like this...
x=1
y=1
max=10
myfiles - paste(foo, x:max, .png, sep=)
while (x = max)
{
png(file=myfiles, pointsize = 20, width = 600, height = 600,
units = px, bg=#eaedd5)
plot(x,y)
Frank E Harrell Jr wrote:
Gad Abraham wrote:
Hi,
Design isn't strictly an R base package, but maybe someone can explain
the following.
When lrm is called within a function, it can't find the dataset dd:
library(Design)
age - rnorm(30, 50, 10)
cholesterol - rnorm(30, 200, 25)
This has been solved, and I'm thankful for the help. The solution is:
z = MyModel$coef/diag(MyModel$var.coef)
...and from there I will use a for loop and pnorm to get the p-values.
Thanks again!
Byron
zerfetzen wrote:
This should be very easy, but alas, I'm very new to R. My end goal
Dear R users,
I require your expertise to evaluate an integral in the attached file.
I would very much appreciate it as I have a very limited knowledge about
R software.
Thank you very much.
Regards,
Salen
__
R-help@r-project.org mailing list
On Fri, 18 Apr 2008, Gad Abraham wrote:
Frank E Harrell Jr wrote:
Gad Abraham wrote:
Hi,
Design isn't strictly an R base package, but maybe someone can explain
the following.
When lrm is called within a function, it can't find the dataset dd:
library(Design)
age - rnorm(30, 50, 10)
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