Gareth Davies a écrit :
Hi everyone.
The function image() seems not to be correctly plotting some
matrices that I give it. (I’m using R 2.1.1)
..where the first column in x (vector ccc) is depicted
horizontally along the bottom of the image.
Hello,
here's a way to have an image
Hello,
Can anybody tell me, please, how to get a matrix of SE of differences (or
any SE) from a GLM object? Both model.tables and se.contrast work only for
ANOVA objects. I remember there was a disp s directive in GLIM package.
I would need something like that.
Many thanks.
Wishes,
Stano Pekar
Dear all,
I'd like to get a linear regression of some data, and impose that the line
goes through a given point P. I've tried to use the lm() method in the
package stats, but I wasn't able to specify the coordinates of the point
P. Maybe I should use another method?
I also have another question:
Hi !
We study the effect of several variables on fruit set for 44 individuals
(plants). For each individual, we have the number of fruits, the number
of flowers and a value for each variable.
Here is our first model in R :
y - cbind(indnbfruits,indnbflowers);
model1
Thanks a lot, really, Philippe, for this explanation... and sorry for
taking some of your time.
Of course your guess is right : I turned to French (and forgot to
tell...). So I'll use your second suggestion (turn to english in MDI).
Once again this list works amazingly well !
Olivier
Philippe
On Mon, 10 Oct 2005, Bernd Weiss wrote:
I successfully set up a local MySQL-database. Connecting via RODBC is
not problem, the same in fetching 3 of 4 tables. But trying to
connect to table 4 fails.
author-sqlFetch(test,author)
Error in fromchar(unclass(x)) : character string is not in a
I guess what you see is the limited resolution of your screen rather
than a bug in R.
Try to produce a reliably image, e.g. some pdf file:
pdf(test.pdf)
cc=runif(n=1500,min=0.1,max=1.2)
ccc=ceiling(cc-1)
dd=runif(n=1500,min=0.1,max=1.2)
ddd=ceiling(dd-1)
Bing Ho wrote:
Hello,
I noticed that the README found in /bin/windows/contrib/ATLAS indicates that
the ATLAS version is 3.4.1. According to the ATLAS sourceforge site, 3.6.0
the latest stable version.
Does anybody know if the ATLAS Rblas.dll are 3.4.1 or 3.6.0, and if they are
not the
hi all
an integration problem. i would like an exact or good approximation for
the following, but i do not want to use a computer. any suggestions:
integral of exp(b*x)/sqrt(1-x^2)
where b is a constant greater than or equal to 0
and
the integral runs from 0 to 1
any help would be
Domenico Cozzetto a écrit :
Dear all,
I'd like to get a linear regression of some data, and impose that the line
goes through a given point P. I've tried to use the lm() method in the
package stats, but I wasn't able to specify the coordinates of the point
P. Maybe I should use another
We study the effect of several variables on fruit set for 44
individuals (plants). For each individual, we have the number
of fruits, the number of flowers and a value for each variable.
...
- Glm does not take account of the correlation between the
flowers of a unique individual. So we
[EMAIL PROTECTED] wrote:
Domenico Cozzetto a écrit :
Dear all,
I'd like to get a linear regression of some data, and impose that the line
goes through a given point P. I've tried to use the lm() method in the
package stats, but I wasn't able to specify the coordinates of the point
P. Maybe I
Hi
I have a loop which is doing time consuming calculations and I would
like to be able to have some feedback on where it is in it's
calculations. I tried to simply show the counter variable in the loop,
but id doesn't work as all display seems to be delayed until the loop is
completed. Is
Rainer M. Krug [EMAIL PROTECTED] writes:
Hi
I have a loop which is doing time consuming calculations and I would
like to be able to have some feedback on where it is in it's
calculations. I tried to simply show the counter variable in the loop,
but id doesn't work as all display seems
Leite,Walter wrote:
Dear R users
I am trying to write an R function to solve for a,b,c in the following
system of equations, given any value of x1, x2 and x3:
b^2 + 6*b*a + 2*c^2 + 15*a^2 = x1
2*c*(b^2 + 24*b*a + 105*a^2 + 2) = x2
24*(b*a + c^2*(1 + b^2 + 28*b*a) + a*(12 + 48 *b*a +
Rainer M. Krug wrote:
Hi
I have a loop which is doing time consuming calculations and I would
like to be able to have some feedback on where it is in it's
calculations. I tried to simply show the counter variable in the loop,
but id doesn't work as all display seems to be delayed until
I put print(i) in the loop instead of i, but it still only prints (in
the Windows R GUI) i after it finished the calculations.
I guess it might be due to the output buffering you mentioned - but how
do I unset it?
Rainer
Peter Dalgaard wrote:
Rainer M. Krug [EMAIL PROTECTED] writes:
Hi
I
Clark Allan wrote:
hi all
an integration problem. i would like an exact or good approximation for
the following, but i do not want to use a computer. any suggestions:
integral of exp(b*x)/sqrt(1-x^2)
Sounds like the problem of integrating the Gaussian density...
Uwe Ligges
where
Thanks - flush.console() did the trick.
As you might guess, I am quite new to R. I like the idea of vectorizing
the calculation, but I guess it is not possible in this case - I will
ask in a new thread.
Thanks,
Rainer
Uwe Ligges wrote:
Rainer M. Krug wrote:
Hi
I have a loop which is
Hi
I have the following loop and would like to vectorize it. Any ideas if
it is possible?
Thanks,
Rainer
Tha Loop:
for (i in 2:Result$NoSims)
{
ppp - runifpoint(Result$NoPlants)
K - Kest(ppp)
Result$LSim[i,] - sqrt(K$iso / pi) - K$r
CM - (Result$LSim[i,] *
Hello,
You must explicitly use print(), show() on an object -here, use
print(i)- in a loop or alternatively, use cat() to display string like:
cat(loop, i, \n)
With RGui under Windows, there is another subtility: if you have turn on
'Misc - Buffered output' (it is ON by default), all output
Rainer M. Krug [EMAIL PROTECTED] writes:
I put print(i) in the loop instead of i, but it still only prints (in
the Windows R GUI) i after it finished the calculations.
I guess it might be due to the output buffering you mentioned - but how
do I unset it?
Using the user friendly
Rainer M. Krug a écrit :
Hi
I have a loop which is doing time consuming calculations and I would
like to be able to have some feedback on where it is in it's
calculations. I tried to simply show the counter variable in the loop,
but id doesn't work as all display seems to be delayed until
Rainer M. Krug wrote:
Hi
I have a loop which is doing time consuming calculations and I would
like to be able to have some feedback on where it is in it's
calculations. I tried to simply show the counter variable in the loop,
but id doesn't work as all display seems to be delayed until the
On Mon, 10 Oct 2005, Philippe Grosjean wrote:
Hello,
You must explicitly use print(), show() on an object -here, use
print(i)- in a loop or alternatively, use cat() to display string like:
cat(loop, i, \n)
With RGui under Windows, there is another subtility: if you have turn on
'Misc -
Hi all.
I have defined a plot thus:
par(mar=c(5,5,4,5),las=1, xpd=NA)
plot(Day, Ym1Imp, ylim=c(0,100), type=b, bty=l, main=Ym1
Expression, cex=1.3, xaxt=n, yaxt=n) #plot implant data
axis(side=1, at=c(0,1,3,5,7,10,14,21), labels=c(0,1,3,5,7,10,14,21)) #
label x axis
mtext(Day, side =1, at=10,
Dear list,
I would like to plot points with two types of labels, one at the data
point (the name of the point) and another offset a bit with another
factor which is either of the two greek characters alpha or beta. I have
tried to get the routine to plot a greek character with expression() or
Iain Gallagher wrote:
Hi all.
I have defined a plot thus:
par(mar=c(5,5,4,5),las=1, xpd=NA)
plot(Day, Ym1Imp, ylim=c(0,100), type=b, bty=l, main=Ym1
Expression, cex=1.3, xaxt=n, yaxt=n) #plot implant data
axis(side=1, at=c(0,1,3,5,7,10,14,21), labels=c(0,1,3,5,7,10,14,21)) #
label x
There a re a few ways to do it without loop. Here is one:
dat = matrix(runif(100), 50,2)
dat[,1] = dat[,1] = dat[,2]
Jarek
\
Jarek Tuszynski, PhD. o / \
Science Applications International Corporation
Hi. Sorry (esp to Uwe for the repeated messages!)
Here is the data and my code in full. Thanks for the
help.
Data.
Day Ym1Imp Ym1sham Semimp Semsham
0 5.785.781.221.36
1 44.36 42.116.26 18.83
3 38.39 14.66 18.02 2.86
5 57.76 1.03
pekar at sci.muni.cz writes:
Can anybody tell me, please, how to get a matrix of SE of differences (or
any SE) from a GLM object? Both model.tables and se.contrast work only for
ANOVA objects. I remember there was a disp s directive in GLIM package.
I would need something like that.
Not unless we know what runifpoint() and Kest() are. AFAIK these are not
part of base R. If you use functions from add-on packages, please state
them so as not to leave others guessing. (This is in the Posting Guide,
which you were asked to read.)
Andy
From: Rainer M. Krug
Hi
I have
Sorry
runifpoint() and Kest are from the package spatstat
Rainer
Liaw, Andy wrote:
Not unless we know what runifpoint() and Kest() are. AFAIK these are not
part of base R. If you use functions from add-on packages, please state
them so as not to leave others guessing. (This is in the
On Mon, 2005-10-10 at 07:59 -0400, Roy Little wrote:
Dear list,
I would like to plot points with two types of labels, one at the data
point (the name of the point) and another offset a bit with another
factor which is either of the two greek characters alpha or beta. I have
tried to get
Uwe Ligges a écrit :
[EMAIL PROTECTED] wrote:
Domenico Cozzetto a écrit :
Dear all,
I'd like to get a linear regression of some data, and impose that the
line
goes through a given point P. I've tried to use the lm() method in the
package stats, but I wasn't able to specify the
To whom it may concern,
I have a question referring to the calculation of variance estimation of the
survey package
I need to estimate the variance for different Domains but for a stratified
sampling desing in several stages. Särndal et al (1992), CAP 10, makes
reference to this
Hello,
has someone written by chance a function to extract the
variance-covariance matrix from a lmer-object? I've noticed the VarCorr
function, but it gives unhandy output.
Regards,
Roel de Jong
__
R-help@stat.math.ethz.ch mailing list
[EMAIL PROTECTED] wrote:
Uwe Ligges a écrit :
[EMAIL PROTECTED] wrote:
Domenico Cozzetto a écrit :
Dear all,
I'd like to get a linear regression of some data, and impose that
the line
goes through a given point P. I've tried to use the lm() method in the
package stats, but I wasn't
Hi all,
This is a question for any of you who use R.app (OS X). Is there any
way to resize the quartz plot window from within R? I know that you
can resize the window by dragging the corner of the window, and fro
the preferences panel. But is there a way to specify the window size
from
Dear useRs,
I'm wondering why the for() loop below runs slower as it progresses.
On a Win XP box, the iterations at the beginning run much faster than
those at the end:
1%, iteration 2000, 10:10:16
2%, iteration 4000, 10:10:17
3%, iteration 6000, 10:10:17
98%, iteration 196000, 10:24:04
99%,
All,
Is there are a wildcard in R for varible names as in unix? For example,
rm(results*)
to remove all variable or function names that begin w/ results?
cheers,
Dave
ps - please respond directly to [EMAIL PROTECTED]
Nevermind, I found the fix. Declaring the length for out eliminates
the performance decrease,
out - vector(mode=numeric,length=length(test))
On 10/10/05, bogdan romocea [EMAIL PROTECTED] wrote:
Dear useRs,
I'm wondering why the for() loop below runs slower as it progresses.
On a Win XP
On Mon, 2005-10-10 at 10:37 -0400, Afshartous, David wrote:
All,
Is there are a wildcard in R for varible names as in unix? For example,
rm(results*)
to remove all variable or function names that begin w/ results?
cheers,
Dave
ps - please
Am 10 Oct 2005 um 7:51 hat christian schulz geschrieben:
Hi,
there is a problem with the type of attributes - is it varchar!? IMHO
you should play a bit with different type's in mysql and the
consequence in R.
I recognize similar problems with RMySQL, if you have variables
with type
rm() can take a list of object names as an argument and ls(pattern='^results')
gives such a list. So
rm(ls(pattern='^results'))
would remove all objects that are matched by the regular expression, that is all
that begin with 'results'
HTH, Johan
2005/10/10, Marc Schwartz (via MN) [EMAIL
On 10/10/05, Marc Schwartz (via MN) [EMAIL PROTECTED] wrote:
On Mon, 2005-10-10 at 10:37 -0400, Afshartous, David wrote:
All,
Is there are a wildcard in R for varible names as in unix? For
example,
rm(results*)
to remove all variable or function names
On Mon, 10 Oct 2005, Marc Schwartz (via MN) wrote:
On Mon, 2005-10-10 at 10:37 -0400, Afshartous, David wrote:
All,
Is there are a wildcard in R for varible names as in unix? For example,
rm(results*)
to remove all variable or function names that begin w/ results?
In R 2.2.0, there is the function 'glob2rx()' which can be used for this
purpose. As in
rm(list = ls(pattern = glob2rx(results*)))
-roger
Afshartous, David wrote:
All,
Is there are a wildcard in R for varible names as in unix? For example,
rm(results*)
to
On Mon, 2005-10-10 at 16:01 +0100, Prof Brian Ripley wrote:
On Mon, 10 Oct 2005, Marc Schwartz (via MN) wrote:
On Mon, 2005-10-10 at 10:37 -0400, Afshartous, David wrote:
All,
Is there are a wildcard in R for varible names as in unix? For example,
rm(results*)
to
FISCHER, Matthew [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
In a plot, can I specify pch to be a greek symbol? (I looked at
show.pch() in the Hmisc package but couldn't see the right symbols in
there).
If not, I guess I can get around this using text(x,y,expression()).
I'm
Hello all,
I am using an existing Fortran routine that takes a single character string
as argument. The routine echoes the argument that I provide. When working on
OS X 3.9 there seems to be no problem, ie the Fortran routine nicely echoes
my argument. However, I compiled the same package on a PC
For anyone who's looked at my previously posted problem I have managed
to solve the missing graph title by removing the main=Graph Title call
from my plot definition and adding a line defining the graph title as a
title call.
i.e. from
plot(Day, Ym1Imp, ylim=c(0,100), type=b, bty=l, main=Ym1
maintainer if I can figure out who he is. Any clues?
Try reading the docs! library(help='AnalyzeFMRI')
-- Bert Gunter
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
On Mon, 10 Oct 2005, Jason Horn wrote:
This is a question for any of you who use R.app (OS X). Is there any
way to resize the quartz plot window from within R? I know that you
can resize the window by dragging the corner of the window, and fro
the preferences panel. But is there a way to
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
Programming questions are appropriate for R-devel, as it says there.
The issue is OS-specific, but we don't see what your Fortran code is.
If you send a reproducible example to the R-devel list, people may be able
Jason,
?quartz lists the options, e.g.
quartz(width=6, height=7, pointsize=24)
All from the console.
A better alias for these questions is R-Sig-Mac (r-sig-
[EMAIL PROTECTED]).
Rob
On Oct 10, 2005, at 7:25 AM, Jason Horn wrote:
Hi all,
This is a question for any of you who use R.app
Hi,
If your limits were to be from -1 to +1 (instead of lower limit being 0),
your integral is:
pi * I_0(b)
Where I_0 is the modified Bessel's function of the zeroth order.
If it is from 0 to 1, then there is no closed form (the integrand is not
symmetric about 0). You must evaluate the
I need to extract identically named columns from several data frames in
a list. the column name is a variable (i.e. not known in advance). the
whole thing occurs within a function body. I'd like to use lapply with a
variable 'select' argument.
example:
tt - function (n) {
x -
Uwe Ligges a écrit :
[EMAIL PROTECTED] wrote:
Sorry for my lack of knowledge, but will the above trick really force
the regression line to pass through P ?
adding (0,0) in this new system of coordinates isn't it equivalent to
add P to the dataset in the original system ?
Well, you do not
On Mon, 10 Oct 2005, Real Miranda Rigoberto wrote:
I have a question referring to the calculation of variance estimation of
the survey package
I need to estimate the variance for different Domains but for a
stratified sampling desing in several stages. S?rndal et al (1992), CAP
10, makes
Thank you for your response. Sorry for insisting but I havent
understood if the data do not permit to include an individual effect or
if it is just the models I run.
We have several flowers and several fruits per individual plant. So
there is a correlation between the flowers/fruits of one
On Mon, 10 Oct 2005, joerg van den hoff wrote:
I need to extract identically named columns from several data frames in
a list. the column name is a variable (i.e. not known in advance). the
whole thing occurs within a function body. I'd like to use lapply with a
variable 'select' argument.
The problem is that subset looks into its parent frame but in this
case the parent frame is not the environment in tt but the environment
in lapply since tt does not call subset directly but rather lapply does.
Try this which is similar except we have added the line beginning
with environment
I am using R with Bioconductor to perform analyses on large datasets
using bootstrap methods. In an attempt to speed up my work, I have
inquired about using our local supercomputer and asked the administrator
if he thought R would run faster on our parallel network. I received the
following reply:
Hello all:
I frequently have glm models in which the residual variance is much
lower than the residual degrees of freedom (e.g. Res.Dev=30.5, Res.DF
= 82). Is it appropriate for me to use a quasipoisson error
distribution and test it with an F distribution? It seems to me that
I could
In general, R is not written in such a way that data remain in cache.
However, R can use optimized BLAS libraries, and these are. So if your
version of R is compiled to use an optimized BLAS library appropriate to
the machine (e.g., ATLAS, or Prof. Goto's Blas), AND a considerable
amount of
Martin Henry H. Stevens [EMAIL PROTECTED] writes:
Hello all:
I frequently have glm models in which the residual variance is much
lower than the residual degrees of freedom (e.g. Res.Dev=30.5, Res.DF
= 82). Is it appropriate for me to use a quasipoisson error
distribution and test it
Iain Gallagher wrote:
Hi. Sorry (esp to Uwe for the repeated messages!)
Here is the data and my code in full. Thanks for the
help.
Data.
Day Ym1Imp Ym1sham Semimp Semsham
0 5.785.781.221.36
1 44.36 42.116.26 18.83
3 38.39 14.66 18.02 2.86
5
Hello,
Has anyone ever written the R code that would extract data from the CPS
March Supplements?
If not, I'll give it a go.
--Bill
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
I am running R 2.1.1 on a Mac g5 under Mac OS 10.4.2.
I have an xyplot with a single conditioning variable (8 levels) .
Here is the code for the conditioning variable used in the formula
argument of xyplot:
factor(
drugauthoryear,
levels = c(
Error in do.call(pmax, lapply(cond, is.na)) :
symbol print-name too long
When I delete the labels vector, the code runs without difficulty.
Any thoughts on this error message?
Yes ... the labels are too long to be printed in the space available. Use
shorter labels. For example,
This error is a curious one.
When I remove the labels vector, the names of the levels are printed
by default in the strip without difficulty.
The names of the levels have lengths of 14 to 16 characters.
My labels had length 18 characters.
When I reduce the labels to length 10 characters (as
Hi,
I'm trying to ouput to a filled with a fixed precision:
eg. if I have data x=c(1.0,1.4,2.0), I want to be able to ouput the following
to a file:
1.00
1.40
2.00
I was wondering if there was a function to do this in R?
Thanks,
Richard
Richard Hedger
Hi,
I'm sort of a newbie to using R to deal with array data. I'm trying to
create a simple filtering function, which outputs only the rows of a
data frame that satisfies a specific criterion. I've set up an
iterative loop to apply the condition to each row. I can create a new
matrix and use
Well -- that **IS** curious. It sounds to me like a software bug or a typo
somewhere (are all your little 's OK?), but now you exceed my modest
expertise -- especially on a Mac (which apparently can be curious little
devils at times).
Deepayan -- where are you?
Cheers,
Bert
-Original
On Mon, 2005-10-10 at 19:50 -0400, Richard Hedger wrote:
Hi,
I'm trying to ouput to a filled with a fixed precision:
eg. if I have data x=c(1.0,1.4,2.0), I want to be able to ouput the following
to a file:
1.00
1.40
2.00
I was wondering if there was a
Have you received a reply to this post? I couldn't find one, and I
couldn't find a solution, even though one must exist. I can get the
substitute to work in main but not legend:
B - 2:3
eB - substitute(y==a*x^b, list(a=B[1], b=B[2]))
plot(1:2, 1:2, main=eB)
You should be
On Mon, 2005-10-10 at 17:04 -0700, Christina Yau wrote:
Hi,
I'm sort of a newbie to using R to deal with array data. I'm trying to
create a simple filtering function, which outputs only the rows of a
data frame that satisfies a specific criterion. I've set up an
iterative loop to apply the
Hi everyone.
I have a problem that I have been unable to determine either the best
way to proceed and why the methods I'm trying to use sometimes fail. I'm
using the pf() function in an optimization function to find a
noncentrality parameter that leads to a specific value at a specified
On 11/10/05 01:12, Earl F. Glynn wrote,:
FISCHER, Matthew [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
In a plot, can I specify pch to be a greek symbol? (I looked at
show.pch() in the Hmisc package but couldn't see the right symbols in
there).
If not, I guess I can get
I haven't look at your code but here a couple of things to try:
1. try using the square of the difference rather than the absolute value
as your objective so that your objective is differentiable.
2. your objective function may be relatively flat in which case it will be
difficult to get a
Dear userR,
With the following results, are they correct or acceptable?
x - c(1.4, 1.2, 2.8)
sum(x)
[1] 5.4
sum(x) == 5.4
[1] FALSE
(1.4 + 1.2 + 2.8) - 5.4
[1] -8.881784e-16
(1.4 + 1.2) - 2.6
[1] -4.440892e-16
2.6 - 1.5 - 1.1
[1] 0
version
_
platform i386-pc-mingw32
arch
See the R FAQ list, section 7. Why doesn't R think these numbers are equal?
Ted.
On 11/10/05 15:42, Joe wrote,:
Dear userR,
With the following results, are they correct or acceptable?
x - c(1.4, 1.2, 2.8)
sum(x)
[1] 5.4
sum(x) == 5.4
[1] FALSE
(1.4 + 1.2 + 2.8) - 5.4
[1] -8.881784e-16
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