I also tried to compile things differently,
f95 -fno-second-underscore -c -fPIC pattern_dist.f90
gcc -std=gnu99 -shared -L/usr -o pattern_dist.so pattern_dist.o
now, I get the following error:
Error in dyn.load(~/myfortran/pattern_dist.so) :
unable to load shared library
Dear R gurus, particularly those of generous M$ tolerance and diverse
gifts and knowledge!
I have an interesting challenge that I will end up crunching in R
involving service usage by patients. Maybe I can do all of it in R but
I can't see how yet.
My situation is that our IT Department
Hi all
I have a question regarding the Fligner-Killeen test. I am using
- a PC with Windows XP (Build 20600.xpsp080413-2111 (Service Pack 3);
- the following R version:
sessionInfo()
R version 2.7.0 (2008-04-22)
i386-pc-mingw32
I have a vector LENGTH and a factor RELATION that are distributed
Dear all;
First of all, this is probably a more conceptual question than a
R-related one, but still want to give it a try. When working with the
interpolation function interp from the package akima and the
triangulation function tri.mesh from package tripack I've got NA's
for the interpolation
Have you tried to merge the two datasets? See ?merge for the details.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie
Hallo all,
I am looking for an R implementation of the four parameter kappa
distribution (cdf, pdf, quantile, and ransom numbers), or at a minimum,
the generalized logistic distribution.
Any suggestions?
Thank you very much,
Serguei
__
Hi,
How can I create a construct argument passing in R script.
So that later I can call it like this:
$ R --vanilla myscript.R ARGUMENT1 ARGUMENT2
--
Gundala Viswanath
__
R-help@r-project.org mailing list
Gundala Viswanath gundalav at gmail.com writes:
How can I create a construct argument passing in R script.
So that later I can call it like this:
$ R --vanilla myscript.R ARGUMENT1 ARGUMENT2
The example
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/121806.html
is for Windows, but
You have repeated (u,v) values in your data frame. It is not possible to
interpolate such data, and I get an error from 'interp' (not NAs).
*** SDTRAN Error 2: The first three data points are collinear.
There is no problem with interpolating from a regular grid, but be aware
that the
Is there a way to do it?
For example I tried this:
args-commandArgs()
fname - args[6].-.args[9]
# then I would like to create a file
# using the above created name
postscript(filename=fname)
dev.off()
But that concatenation doesn't seem to do it.
--
Gundala Viswanath
Hello R-user community!
I am running R 2.7.0 on a Power Book (Tiger). (I am still R and
statistics beginner)
I did the following :
pdf(InLnegthMaxHomogeneity.pdf)
boxplot(inflorescence_length_Max~Sex, main=Bartletts Homogeneity for
inflorescence length,data=FemMal_Sex)
On 29 May 2008, at 10:39, Gundala Viswanath wrote:
Is there a way to do it?
For example I tried this:
args-commandArgs()
fname - args[6].-.args[9]
This would work under Perl :)
Look for details: ?paste
Try this:
fname - paste(args[6], ., args[9], sep=)
--Hans
Hello,
I'm not a guru but we have to do reporting with data in Excel files. We use
xlsReadWrite package.
Below an exemple, with an Excel.R file which import MySheet sheet from a
MyFile.xls workbook. First column are numeric, Second are integer. Third
character and so on.
library(xlsReadWrite)
I think the issue here is that the Excel spreadsheets as described do not
contain simple tables of the form that the various .xls readers described
in 'R Data Import/Export' handle. Your example is not applicable (and if
it were there are several alternatives in that manual).
I would look
1. read.xls in the gdata package has a pattern= argument that
will skip everything before the first cell containing that pattern.
It calls xls2csv from the same package which in turn uses
perl so it works on all platforms and does not require a copy
of Excel.
2. If that is not good enough you
Hi,
How can I use a substring match as a condition in a subset command?
Sth like this:
subset(input, field1==blah1 field2==blah2) # but now with substring
match in field2
subset(input, field1==blah1 field3 *substringmatch* blah3)
I've tried with gsub, but it won't work:
subset(input,
Hi Kevin,
thanks a lot for the advices. The first suggestion worked, overall, I
have to find a way to scale the names/numbers on the interval
considered. But it was the good way to look at the problem.
The second suggestion didn't really work. It is not a problem of
custom fitting, it is
Something like this may be close to what you want:
subset(input, field1==blah1 !is.na(charmatch(blah3,input$field3)))
---
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Albert Vilella
Sent: 29 May 2008 11:38
To: r-help@r-project.org
Subject: [R]
Dear R-helpers,
I'm having a problem in using plot.design in Design Library. Tho
following example code produce the error:
n - 1000# define sample size
set.seed(17) # so can reproduce the results
age- rnorm(n, 50, 10)
blood.pressure - rnorm(n, 120, 15)
cholesterol-
1. See ?locator
2. Try this:
plot(1:2, pch = c(\u2640, \u2642))
On Thu, May 29, 2008 at 4:40 AM, Birgit Lemcke
[EMAIL PROTECTED] wrote:
Hello R-user community!
I am running R 2.7.0 on a Power Book (Tiger). (I am still R and statistics
beginner)
I did the following :
Hi I am using ubuntu 8.04 and faced some problems when installing rJava. What
I did first was install.packages(rJava) but I could not proceed since:
checking Java support in R... configure: error: absent
R was configured without Java support. Please run
R CMD javareconf
as root to add Java
I am looking for an R implementation of the four parameter kappa
distribution (cdf, pdf, quantile, and ransom numbers), or at a minimum,
the generalized logistic distribution.
I searched for kappa distribution on Rseek (http://www.rseek.org) and
the functions you want appear to be in the
eesteves at ualg.pt writes:
Dear All,
I'me having (much) trouble understanding why it happened and answering
a referee's comment to part of a submitted manuscript. I've tried to
google for help but... I'm really confident that although this is a
R-Help list someone can help me!
I
Ben Bolker wrote:
eesteves at ualg.pt writes:
Dear All,
I'me having (much) trouble understanding why it happened and answering
a referee's comment to part of a submitted manuscript. I've tried to
google for help but... I'm really confident that although this is a
R-Help list someone
Wittner, Ben, Ph.D.:
Is there a way to display an image (such as is done with the function
image()) in a grid package viewport?
Yes. Use levelplot() from the lattice package.
--
Karl Ove Hufthammer
__
R-help@r-project.org mailing list
dear Harrell,
thank you for quick reply and suggestions. I still have the problem:
library(Design)
x = rnorm(100)
y = runif(100)(exp(x)/(1+exp(x)))
y = 0*y+1*y
d = datadist(x,y)
options(datadist=d)
fit = lrm(y~x)
# works fine, but
plot(fit) #produce the error message
Error in
Frank, I believe, is correct. Using the AIC/BIC for data-driven model
selection does NOT solve the stepwise problem. This is because the
distribution of the sample AIC is changed from its original distribution to an
extreme-value distribution, e.g.., min (AIC1, AIC2, ..., AICn). Thus,
Hello,
I have quite a simple problem that I believe can be solved quite easily. I
have a dataframe as such:
Symbol Date Time Exchange TickType ReferenceNumber Price Size
1 3:YMZ7.EC 12/03/2007 08:30:00 ECB83916044 133879
2 3:YMZ7.EC 12/03/2007 08:30:00 EC
The low R2 says the model does not explain much of the variance.
But the high significance arises from the very large number of degrees
of freedom.
This is not an 'incompatibility'; just what happens with large
dispersion, small effects and a very large number of observations.
But you clearly
I need to repeat an experiment 1000 times. Each experiment involves randomly
selecting one ball each from two separate bags. Each bag contains 10 balls,
numbered 1, 2, 3, ... , 10. So the probability of selecting any one pair of
balls is equal to all others.
For each experiment, what I need to
Hi all,
I am using RPART for my genetic study under ANOVA method. I wanted to know
if it is possible to see r-squared or the amount of the variance in the data
explained by a model (or a tree in this case from the RPART package. I am
guessing that there has to be one since I am using ANOVA to
yoo wrote:
Hi,
I'm able to create a library with R CMD INSTALL cmd, etc... I'm just
You probably mean you are able to *install* a *package*, I guess.
wondering.. is it possible that when the user says library(boo), it runs
some initialization code?
Yes: See the manual
What you have sent is almost unreadable.
Please do not send HTML messages (as the posting guide suggests)!
Wen-Ching Lin wrote:
Hi,
I am reading the source code of rpart. I have problems understand the following
code and would appreciate for any helps. In rpart.s, there is a line:
rpfit
[EMAIL PROTECTED] wrote:
*Thanks* all those who took the time to help me (even if the
question was not related to - the use of - R).
Now I think I can soundly make my point w/ the referee (can I use your
replies? If so I intend to properly cite its use?!?).
In general, I think it is best not
Perhaps you want to read the files only once, and save those 85 or
whatever versions you want.
For SearchReplace you might want to take a look at ?gsub and friends.
Uwe Ligges
Romain wrote:
Hi there,
I would like to know if it is possible to modify a text file with a R function.
In fact I
On Thu, 29 May 2008, Albert Vilella wrote:
Hi,
How can I use a substring match as a condition in a subset command?
Perhaps
subset(input, field1==blah1 regexpr(blah2,field3) != -1 )
??
Study in
example(gsub)
the regexpr example
and in
?gsub
the 'Value'
If you mean a package ... 'Writing R Extensions' says
Ideally, the @R{} code files should only directly assign @R{} objects
and definitely should not call functions with side effects such as
@code{require} and @code{options}.
Note that you are expecting the side-effect of 'print', which is
Dear all,
I am fairly new to R and this list (this is my first post), so I am wondering
whether
there is a possibility to view posts on this list conveniently on a website
besides
reading my email.
I have in mind something like this :
http://www.ruby-forum.com/forum/4
for the Ruby
Hi. I am plotting graphs for values ranging between -1 and 10, for example:
(1,2,1,1,6,7,-1,-1,5,-1)
I am trying to plot the graphs so that the points with value of -1 will be
in one specific color, and the rest of the points will be in one different
specific color. I would be grateful for any
Message: 24
Date: Wed, 28 May 2008 05:53:26 -0700 (PDT)
From: Philip Twumasi-Ankrah [EMAIL PROTECTED]
Subject: [R] rbinom not using probability of success right
To: r-help@r-project.org
Message-ID: [EMAIL PROTECTED]
Content-Type: text/plain
I am trying to simulate a series of ones and zeros (1
On Thu, 29 May 2008, Axel Etzold wrote:
Dear all,
I am fairly new to R and this list (this is my first post), so I am wondering
whether
there is a possibility to view posts on this list conveniently on a website
besides
reading my email.
See
Yes, starting from
http://www.r-project.org/
you can go to the mailing lists, choose the appropriate one, go to the
corresponding archives and read all of it sorted by the most common
criteria.
Best wishes,
uwe
Axel Etzold wrote:
Dear all,
I am fairly new to R and this list (this is my
Hi Axel,
Perhaps http://www.nabble.com/R-f13819.html can be useful.
Thanks,
Jorge
On Thu, May 29, 2008 at 12:03 PM, Axel Etzold [EMAIL PROTECTED] wrote:
Dear all,
I am fairly new to R and this list (this is my first post), so I am
wondering whether
there is a possibility to view posts
I'm attempting to plot a cubic relationship between two variables
controlling for the effects of a third variable. In this short
example, I'm trying to use AGE to predict CORTEX while controlling for
the effects of TIV (total intracranial volume):
cortex =
I am fairly new to R and this list (this is my first post), so I am
wondering whether
there is a possibility to view posts on this list conveniently on a
website besides
reading my email.
You can see them on Nabble, but there is a delay of a couple of hours.
Rnabble, search on the CRAN site, and some others
On Thu, May 29, 2008 at 12:03 PM, Axel Etzold [EMAIL PROTECTED] wrote:
Dear all,
I am fairly new to R and this list (this is my first post), so I am
wondering whether
there is a possibility to view posts on this list conveniently on a
Amarjit Singh Sethi set_alt at yahoo.co.in writes:
Dear all
I wish to carry out sigma- and beta-convergence analysis in respect of
panel data [wherein current value of
one of the variables needs be regressed upon suitably transformed lagged
values of another
variable(s)]. I am quite new
Hello R-Users,
I am new to R and trying my best however I need help with this simple task.
I have a dataset, YM1207.
X.Symbol Date Time Exchange TickType
ReferenceNumber Price Size
12491 3:YMZ7.EC 12/03/2007 08:32:50 ECB
8598577013379
In the past few weeks I have had to give myself a crash course in R, in order
to accomplish some necessary tasks for my job. During that time, I've found
this forum to be helpful time and time again - usually I find the answer to
my problem by searching the archives; once or twice I've posted
My stat textbook tells me that using Shapiro-Wilk test for each variable
one by one is not equal to a test for multivariate normality as a whole.
Does R have a function of testing for multivariate normality? Thanks.
Hongsheng (Hank) Liao, Ph.D.
Lab Manager
Center for Quantitative Fisheries
Hi Neil,
as i am not an advanced user,
i find reference cards very handy
(google: reference card R)
hth a bit,
Wim
Message: 70
Date: Wed, 28 May 2008 15:25:36 -0500
From: Neil Gupta [EMAIL PROTECTED]
Subject: [R] R reference Books
To: R-help@r-project.org
Message-ID:
[EMAIL PROTECTED]
Hi,
I would like to use the sem package to perform a path analysis (no
latent variables) with a mixture of 2 nominal exogenous, 1 continuous
exogenous, and 4 continuous endogenous variables. I seek advice as to
how to calculate the appropriate covariance matrix for use with the sem
package.
Patrizio Frederic wrote:
dear Harrell,
thank you for quick reply and suggestions. I still have the problem:
library(Design)
x = rnorm(100)
y = runif(100)(exp(x)/(1+exp(x)))
y = 0*y+1*y
d = datadist(x,y)
options(datadist=d)
fit = lrm(y~x)
# works fine, but
plot(fit) #produce the error
Yes. You could install mvnormtest Package and perform the
multivariate normality test. By using mshapiro.test
I wish this is helpful!
Chunhao Tu
Quoting HongSheng Liao [EMAIL PROTECTED]:
My stat textbook tells me that using Shapiro-Wilk test for each variable
one by one is not equal to
thanks.
one more thing w/r this particular data set:
tri.mesh doesn't work in R but triangulation (S+SpatialStats) does work
interp.old (diff ncp) doesn't work in R but it does in Splus for the
same ncp values???
so,R or SPlus? who's right?
On Thu, May 29, 2008 at 4:34 AM, Prof Brian
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in prices and
positions?
if (nAssetPositions != nAssetPrices) {
cat(Different number of assets! \n\n)
Hi
Bill Cunliffe wrote:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in prices and
positions?
if (nAssetPositions != nAssetPrices) {
cat(Different number of assets! \n\n)
See
?return
HTH,
Chuck
On Thu, 29 May 2008, Bill Cunliffe wrote:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in prices and
positions?
if (nAssetPositions != nAssetPrices) {
Tubin wrote:
In the past few weeks I have had to give myself a crash course in R, in order
to accomplish some necessary tasks for my job. During that time, I've found
this forum to be helpful time and time again - usually I find the answer to
my problem by searching the archives; once or twice
stop('Different number of assets! \n\n')
X
Bill Cunliffe 写道:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in prices and
positions?
if (nAssetPositions != nAssetPrices) {
Bill Cunliffe wrote:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in prices and
positions?
if (nAssetPositions != nAssetPrices) {
cat(Different number of assets! \n\n)
It looks like you want to stop the function execution on detecting an
error condition, in which case the appropriate function to call is
stop(), as in
if (nAssetPositions != nAssetPrices)
stop(Different number of assets!)
An alternative, if you don't want to write the error messages for each
Hi, thanks for your replay the previous post of mine. Let me ask you one more
question similar to the previous one, based on a naive example:
f1-function(theta, theta1)
{theta[1]+theta[2]+theta1[1]}
f2-function(theta, x, theta1)
{f1(theta, theta1)*exp(x)*theta1[2]}
function to be optimized
You might try to use on.exit or stop?
# on.exit
if (nAssetPositions != nAssetPrices) {
on.exit(cat(Different number of assets! \n))
}
# stop
if (nAssetPositions != nAssetPrices) {
stop(Different number of assets!)
}
You could find these in S programming
Try using stop:
if (nAssetPositions != nAssetPrices) {
stop(Different number of assets! \n\n)
On Thu, May 29, 2008 at 3:23 PM, Bill Cunliffe [EMAIL PROTECTED]
wrote:
For example, based on a certain condition, I may want to exit my code
early:
# Are
Try:
x - c(1,2,1,1,6,7,-1,-1,5,-1)
plot(x, col = ifelse(x == -1, red, black), pch = 16)
On Thu, May 29, 2008 at 1:23 PM, uv [EMAIL PROTECTED] wrote:
Hi. I am plotting graphs for values ranging between -1 and 10, for example:
(1,2,1,1,6,7,-1,-1,5,-1)
I am trying to plot the graphs so that
I totally agree both of you. This is a super place to mature the R.
I learn a lot from this R heaven!
Chunhao
Quoting Esmail Bonakdarian [EMAIL PROTECTED]:
Tubin wrote:
In the past few weeks I have had to give myself a crash course in
R, in order
to accomplish some necessary tasks for my
I´m trying to find datasets that will give me residuals, after applying
the lm function, with no normality, non linearity, and heteroscedacity
so I can try to exemplify
those cases in the linear regression model. Can you give any advice on
what datasets would be appropiate? I can´t use the ones
Hi,
I think Ray has answered this question in the previous e-mail.
Because optim can only use one single parameter thus you can not have
the parameters: theta, theta1 and x at the same time.
such as:
f1-function(theta)
{theta[1]+theta[2]}
f2-function(theta)
{f1(theta)*3}
f3-function(theta)
Hi Ashish,
I am rather more concerned about whether what you outlined is legitimate
(your question 1 below). If you are looking at children, higher AGE will
be associated with higher TIV, so both variables would essentially
measure the same thing (see Miller Chapman, Misunderstanding
Neil Gupta wrote:
Hello R-Users,
I am new to R and trying my best however I need help with this simple task.
I have a dataset, YM1207.
X.Symbol Date Time Exchange TickType
ReferenceNumber Price Size
12491 3:YMZ7.EC 12/03/2007 08:32:50 ECB
85985770
Dear Gus,
If the nominal variables are exogenous then you need not use polychoric and
polyserial correlations. (Indeed, if they are polytomous and unordered, then
it would be inappropriate to do so.) Simply use dummy exogenous variables,
as you would in a dummy regression.
You could use a
Did you copy-paste that error message?
I get:
optim(par=c(1,1), f3)
Error in f1(theta, theta1) : object theta not found
You seem to be confusing your formal and actual parameters.
When you invoke f3 (within optim()), theta is not defined within f3().
Ray
On Fri, 30 May 2008, threshold
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use of:
mymodel-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same length
the result is a=NA.so i want to know where is the problem.
--
View this message in context:
Dear Hanen,
You don't need 1 in your R code. Try this:
# Model
mymodel-lm(y~x1+x2)
# Coefficients
summary(mymodel)
See also ?lm.
Thanks,
Jorge
On Thu, May 29, 2008 at 5:15 PM, hanen [EMAIL PROTECTED] wrote:
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use
hanen wrote:
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use of:
mymodel-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal to one?
run the command: round( sum(
On 30/05/2008, at 9:57 AM, Jorge Ivan Velez wrote:
Dear Hanen,
You don't need 1 in your R code. Try this:
# Model
mymodel-lm(y~x1+x2)
# Coefficients
summary(mymodel)
See also ?lm.
That's not the problem. The ``1'' is redundant but does no harm.
There is presumably something ``wrong''
Carlos López wrote:
I´m trying to find datasets that will give me residuals, after applying
the lm function, with no normality, non linearity, and heteroscedacity
so I can try to exemplify
those cases in the linear regression model. Can you give any advice on
what datasets would be appropiate?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
This is not really addressing your problem, but I thought you might want to
know that the rtracklayer package in Bioconductor already supports parsing
BED files, as well as GFF and WIG. It's main purpose is to load the tracks
into genome browsers, like UCSC.
Michael
On Wed, May 28, 2008 at 1:11
Your problem can be easily solved analytically
yielding that E(A/B=b) = (110 - b*b)/(21 - 2*b) - no
programming needed!
If you insist on writing a program, you could do
something like:
x - sample(1:10,2000,replace=TRUE)
M - matrix(x,nrow=2) #each column of M represents
two balls
A -
Hi,
Cleber Nogueira Borges wrote:
hanen wrote:
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use of:
mymodel-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]0) log(x[i]/(number of consecutive zeros preceding it +1))
x-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use
of:
mymodel-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same
length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal to one?
run the command: round( sum( c(x1,x2) ),12)==1
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
2. if (x[i]==0) NA
x-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA,
Hi Folks,
I need help with a query about R on Windows, specifically
about graphics devices.
I'm advising someone remotely (so it's all by email) who
is running R on Windows, while I am not (Linux only).
Things have reached the stage where saving graphics plots
as Windows metafiles is looming.
Hi Mike,
if you can decompose the bimodal distribution into (eg two) known
forms, then you could try a stepwise approach, eg:
If uniform 0.5 then double it and use it to draw from the inverse
cdf of A,
else, double (uniform - 0.5) and use it to draw from the inverse cdf of B.
You can change
I need to estimate maximum tree crown radius and am looking for a package to
prepare stochastic frontier models in R. I have not found any package
references on Nabble R help, google, or R help. Any tips on a package for
this?
With regards,
Aaron Trowbridge
Researcher
BV Research Centre
On 5/29/2008 7:48 PM, Ted Harding wrote:
Hi Folks,
I need help with a query about R on Windows, specifically
about graphics devices.
I'm advising someone remotely (so it's all by email) who
is running R on Windows, while I am not (Linux only).
Things have reached the stage where saving
I have been attempting to do some work using hclust, and have run
into a (possibly subtle) problem.
The background is that I constructed a dissimilarity matrix ``d1''
(it involved something called the ``Jaccard similarity coefficient'';
I won't go
into the details unless requested). I then
All:
I am new to R and would like your help in identifying the appropriate
process to follow in order to modify the output from an existing
package. I've had difficulty finding an answer online, perhaps because
I am using incorrect terminology.
A package that I am using (mmlcr) invokes
Hi Jenny,
A simple solution is to add your line to the function, re-load/source
the modified function to the console (e.g. by copy and paste). Then the
new function with the same name as the original one will be called next
time. If you don't want to use modified function any longer, just use
Yes, here are 3 functions that can do it
grid::grid.rect
RGraphics::grid.imageFun
lattice::panel.levelplot
On Thu, May 29, 2008 at 10:48 PM, Wittner, Ben, Ph.D.
[EMAIL PROTECTED] wrote:
Is there a way to display an image (such as is done with the function image())
in a grid package viewport?
Yes I know the problem can be solved analytically -- the point was to see how
well R could simulate the theory.
In any case, your assistance is greatly appreciated and it worked well.
Thankyou. R does a good job of simulating the experiment!
--
View this message in context:
On May 29, 2008, at 9:56 PM, lek2k wrote:
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
I (and certainly many others) have been using multiple points calls
for a while now with no problems at
On May 29, 2008, at 11:02 PM, [EMAIL PROTECTED] wrote:
Suppose I have a plot
plot(1:10, pch = )
And I want some text to indicate a Normal distrubition. I could do
this:
text(5, 6, substitute(XN(mu, sigma^2)), adj = 0)
text(5.35, 6, ~, adj = 0)
But that's clumsy, and depending on your
I'm stumbling my way through manipulating data in multiply imputed datasets,
and have run into a problem translating code I used to run on my pre-imputed
dataset to multiple datasets. The imputation runs just fine, as does the
reading of the mi data sets into an imputationList. I run into
Hello,
I have been trying to make a graph that have error bars and text at
specific position.
I used the following code from the help file of xYplot(Hmisc) as an
example except I add a myPanel function, which is just supposed to add
letters from the alphabet at the position aligned at y
Dear R Gurus,
I am having a little difficulty with nlm. I've searched the archives and
found nothing that tells me why this is occuring -- though there are
some slightly similar issues.
A simple example:
lev2-function(aaa,bbb,ccc,ddd,eee){
res-aaa+bbb+ccc+ddd+eee
res
}
Thanks, Dieter,
but as far as I understand, 'glht' does not support objects of class
'aovlist' either. I mean, I know there is a TukeyHSD function out
there, but that's the problem: repeated measures ANOVA yields an
aovlist object, and TukeyHSD calls for an aov object.
And I don't know if
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