Bill Shipley wrote:
Hello, and Happy New Year. My default working directory is getting very
cluttered. I know that I should be using a different working directory for
each project (I work in Windows), but do not know how to go about creating
different ones and moving back and forth between
Hello,
I've uploaded version 0.5-11 of RMySQL into CRAN, and it should be available
soon.
From the NEWS file:
Version 0.5-11
* Fixed a bug that would crash R with a failed mysql_real_connect().
* dbApply() is now working again (but still experimentally).
* Re-formatted the C code.
[0.5-9
Dear All,
can anyone explain this message
Error in load(matched2.RData) : Value of SET_STRING_ELT() must be a
'CHARSXP' not a 'builtin'?
Have I the possibility to retrieve the data?
TIA
Giovanni
--
dr. Giovanni Parrinello
Department of Biotecnologies
Medical Statistics Unit
University of
Claudia Tebaldi wrote:
Dear R-helpers,
I'm plotting geophysical data in the form of contours using
filled.contour. The display would be much more effective if the areas
with negative values could be color coded
by -- say -- cold colors in the range of blue to green, and conversely
the
You appear to have a corrupted character vector in your saved image.
Either R was internally corrupted when the image was saved or the file has
been corrupted. In neither case would the the data be intact.
On Fri, 5 Jan 2007, Giovanni Parrinello wrote:
Dear All,
can anyone explain this
Hi,
I'm looking for some lines of code that does the following:
I have a dataframe with 160 Columns and a number of rows (max 30):
Col1 Col2 Col3 ... Col 159 Col 160
Row 1 0 0 LD ... 0 VD
Row 2 HD 0 00 MD
Row 3 0 HD
Hy all,
I'm plotting graphs using plot() function, they are on X axes POSIX dates:
POSIXt oldClass POSIXct POSIXlt
I can't figure out why sometimes it prints the month and days and sometimes it
prints the unix timestamp.
It appens usually when the xlim is short like only some
Sorry i forgot to give the Rscript to execute with the datas in the archive
attached :
limite_x-structure(as.numeric(read.table(limite_x)),
class=c(POSIXt,POSIXct))
limite_y-as.numeric(read.table(limite_y))
test-as.numeric(read.table(y_values))
Hi
you could use also another approach in case of data frames
A - as.data.frame(A)
A0 - -A*log(A)
A0[is.na(A0)] - 0
which changes NaN's to zeroes
HTH
Petr
On 5 Jan 2007 at 16:38, talepanda wrote:
Date sent: Fri, 5 Jan 2007 16:38:05 +0900
From: talepanda [EMAIL
Dear R-helpers,
I need to compute probabilties of multinomial observations, eg by doing the
following:
y=sample(1:3,15,1)
prob=matrix(runif(45),15)
prob=prob/rowSums(prob)
diag(prob[,y])
However, my question is whether this is the most efficient way to do this.
In the call prob[,y] a whole
Hi
I am not sure if I understand how do you want to select unique items.
with
sapply(DF, function(x) !duplicated(x))
you can get data frame with TRUE when an item in particular column is
unique and FALSE in opposite. However then you need to choose which
rows to keep or discard
e.g.
maybe
prob[cbind(1:length(y), y)]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web:
Hello,
I'm using the gstat package within R for an automated procedure that
uses ordinary kriging.
I can see that there is a logical (singular) atrtibute of some
adjusted model semivariograms:
.- attr(*, singular)= logi TRUE
I cannot find documentation about the exact meaning and the
No attachment arrived: see the posting guide for handling attachments.
But in any case the posting guide says
If you are using an old version of R and think it does not work
properly, upgrade to the latest version and try that, before posting.
and your version of R is ancient. Note,
On Thu, 4 Jan 2007, Richard M. Heiberger wrote:
Get the RColorBrewer package from CRAN
Description: The packages provides palettes for drawing nice maps
shaded according to a variable.
The package vcd also offers the function diverge_hcl() that constructs
diverging palettes (based on
Hi Guys, it would be great if you could help me with a MLE problem in R.
I am trying to evaluate the maximum likelihood estimates of theta = (a1,
b1, a2, b2, P) which defines a mixture of a Poisson distribution and two
gamma prior distributions (where the Poisson means have a gamma
So maybe i will finaly succeed asking the right way,
Hy all,
I'm plotting graphs using plot() function, they are on X axes POSIX dates:
POSIXt oldClass POSIXct POSIXlt
I can't figure out why sometimes it prints the month and days and sometimes it
prints the unix timestamps.
Here is an
Franco,
You can provide lower and upper bounds on the parameters if you use optim
with method=L-BFGS-B.
Hth, Ingmar
From: francogrex [EMAIL PROTECTED]
Date: Fri, 5 Jan 2007 04:54:50 -0800 (PST)
To: r-help@stat.math.ethz.ch
Subject: [R] maximum likelihood estimation of 5 parameters
Hi
Liu, Delong (NIH/CIT) [C] liud2 at mail.nih.gov writes:
I want to extract estimated coeffiicents of each local polynomial at
given x from loess(), locfit(), or KernSmooth(). Can some experts
provide me with suggestions? Thanks.
Try
cars.lo - loess(dist ~ speed, cars)
str(cars.lo)
List
Dear all,
especially those of you that kindly provided suggestions yesterday,
I was not asking for cool palettes -- even if I now appreciate the
pointers -- but I was asking for a way to make the 0 level of a filled
contour plot correspond to the neutral color in the color scale
when the range
Liu, Delong (NIH/CIT) [C] liud2 at mail.nih.gov writes:
I want to extract estimated coeffiicents of each local polynomial at given x
from loess(), locfit(), or
KernSmooth(). Can some experts provide me with suggestions? Thanks.
Before you start on your own, also note Brian Ripleys recent
We can force the X axis into the format of our choice by
suppressing it in the plot and explicitly calling
axis.POSIXct with our choice of format
(see ?axis.POSIXct):
x - seq(from = as.POSIXct(2005-10-14),
to = as.POSIXct(2007-01-02),
by = day
y - as.numeric(x)
plot(y ~ x, xaxt =
Sorry, there was a missing parenthesis after the first statement.
Here is the code again:
x - seq(from = as.POSIXct(2005-10-14),
to = as.POSIXct(2007-01-02),
by = day)
y - as.numeric(x)
plot(y ~ x, xaxt = n)
at - x[seq(1, length(x), length = 5)]
axis.POSIXct(1, at, at, %y-%b-%d)
On
On 1/5/2007 9:08 AM, Claudia Tebaldi wrote:
Dear all,
especially those of you that kindly provided suggestions yesterday,
I was not asking for cool palettes -- even if I now appreciate the
pointers -- but I was asking for a way to make the 0 level of a filled
contour plot correspond to the
I am sorry to ask a trivial question but I am not a statistician.
When I need to compare more than two groups in a unbalanced design with
SAS system I use PROC GLM (like the example in data5.csv from Cody R.
Applied statistics and SAS programming language p.223). R glm gives
different results.
In R,
-- lm fits ordinary least squares linear models; this is likely to
be what you want if you were using PROC GLM.
-- glm fits generalized linear models (e.g. logit, Poisson, Gaussian,
etc.).
-- lmer and lme fit mixed models, similar to PROC MIXED
Cheers,
Hank
On Jan 5, 2007, at 10:05 AM,
Franco,
You can provide lower and upper bounds on the parameters if you use optim
with method=L-BFGS-B.
Hth, Ingmar
Thanks, but when I use L-BFGS-B it tells me that there is an error in
optim(start, f, method = method, hessian = TRUE, ...) : L-BFGS-B needs
finite values of 'fn'
--
View this
On Fri, 5 Jan 2007, francogrex wrote:
[quoting Ingmar Vissar without attribution, contrary to the posting
guide.]
Franco,
You can provide lower and upper bounds on the parameters if you use optim
with method=L-BFGS-B.
Hth, Ingmar
Thanks, but when I use L-BFGS-B it tells me that there is
Franco,
Is it possible that you have failed to provide the negative of loglikelihood
to optim, since optim, by default, minimizes a function? If you want to
do this withput redefining the log-likelihood, you should set fnscale= -1
(as hinted by Prof. Ripley). This would turn the problem into a
On Friday 05 January 2007 12:34, Petr Pikal wrote:
Hi
you could use also another approach in case of data
frames
A - as.data.frame(A)
A0 - -A*log(A)
A0[is.na(A0)] - 0
I think you meant A0[which(is.na(A0))] - 0
which changes NaN's to zeroes
HTH
Petr
Regards
Dear All,
I've read Thomas Lumley's fortune If the answer is parse() you should usually
rethink the question.. But I am not sure it that also applies (and why) to
other situations (Lumley's comment
http://tolstoy.newcastle.edu.au/R/help/05/02/12204.html
was in reply to accessing a list).
``MrJ Man'' wrote:
On Friday 05 January 2007 12:34, Petr Pikal wrote:
Hi
you could use also another approach in case of data
frames
A - as.data.frame(A)
A0 - -A*log(A)
A0[is.na(A0)] - 0
I think you meant A0[which(is.na(A0))] - 0
He most certainly DOES NOT mean this!
Ramon Diaz-Uriarte wrote:
Dear All,
I've read Thomas Lumley's fortune If the answer is parse() you should
usually
rethink the question.. But I am not sure it that also applies (and why) to
other situations (Lumley's comment
http://tolstoy.newcastle.edu.au/R/help/05/02/12204.html
was in
??
Or to add to what Peter Dalgaard said... (perhaps for the case of many more
functions)
Why eval(parse())? What's wrong with if then?
g - function(fpost,x){if(fpost==1)f.1 else f.2 }(x)
or switch() if you have more than 2 possible arguments? I think your remarks
reinforce the wisdom of
On Friday 05 January 2007 19:21, Peter Dalgaard wrote:
Ramon Diaz-Uriarte wrote:
Dear All,
I've read Thomas Lumley's fortune If the answer is parse() you should
usually rethink the question.. But I am not sure it that also applies
(and why) to other situations (Lumley's comment
On Friday 05 January 2007 19:35, Bert Gunter wrote:
??
Or to add to what Peter Dalgaard said... (perhaps for the case of many more
functions)
Why eval(parse())? What's wrong with if then?
g - function(fpost,x){if(fpost==1)f.1 else f.2 }(x)
or switch() if you have more than 2 possible
The other reason for considering which of the different approaches to use
would be performance:
f.1 - function(x) x+1
f.2 - function(x) x+2
system.time({
+ for (i in 1:10){
+ eval(parse(text=paste('f.', i%%2+1, sep='')))(i)
+ }
+ })
[1] 6.96 0.00 8.32 NA NA
The problem is with residuals in model3.
Fixed in 2.4.1 patched and later.
On Wed, 3 Jan 2007, Jarrod Hadfield wrote:
Hi,
I'm trying to compare models, one of which has all parameters fixed
using offsets. The log-likelihoods seem reasonble in all cases except
the model in which there are
Thank you for your example. I am now using it an example in the MMC
Mean--mean Multiple Comparisons plot in the HH package on CRAN. The
source for HH_1.17 is on CRAN in Austria as of this morning. The
Windows and MacOS binaries will be on CRAN in a few days and will then
propagate to the
IMHO, R is not good at really large-scale data mining, esp. when the
algorithm is complicated. The alternatives are
1. sampling your data; sometimes you really do not need that large
number of records and the accuracy might already be good enough when
you load less.
2. find an alternative
Hi r-user,
I am trying to save same objects in workspace, but the next error
message is show in line command,
save(area,cent,larg,mran,eddy,dia,vort,slan,file=stat.Rdata)
Erro en save.default(area, cent, larg, mran, eddy, dia, vort, slan, file
= stat.Rdata) :
5 arguments passed to
Using the inverse logistic transform to replace p by exp(xp)/(1+exp(xp)) allows
unconstrained fitting of xp. There may still be problems where xp tends to + or
- infinity depending on starting values.
francogrex [EMAIL PROTECTED] 01/05/07 11:54 PM
Hi Guys, it would be great if you could help
There is no 'save.default' function in R.
You were asked in the posting guide to report the results of
sessionInfo(). (It looks like an old version of R.oo might be the
culprit, and you might need to update.packages().)
On Fri, 5 Jan 2007, fernando espindola wrote:
Hi r-user,
I am trying
Ramon,
I prefer to use the list method for this type of thing, here are a couple of
reasons why (maybe you are more organized than me and would never do some of
the stupid things that I have, so these don't apply to you, but you can see
that the general suggestion applys to some of the rest of
Greg Snow wrote:
Personally I have never regretted trying not to underestimate my own future
stupidity.
Fortune!
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
I agree about sampling, but.. You can go a little further with your
hardware.
The defaults in R is to play nice and limit your allocation to half
the available RAM. Make sure you have a lot of disk swap space (at least
1G with 2G of RAM) and you can set your memory limit to 2G for R.
See
No I have not forgotten to use a negative fnscale to optimize, so as you
suggest I will post some parts of the code I am running to show you the
errors:
n
[1] 3 1 4 54 6 58 20 14 3 14 4 65 1 7 9 10 2 4
66
[20] 5 9 7 12 7 55 105 2 5 10 55 5 28
Peter Dalgaard wrote:
Greg Snow wrote:
Personally I have never regretted trying not to underestimate my own future
stupidity.
Fortune!
Seconded.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
On July 12, 2004 Spencer Graves wrote an email describing essentially
the same issue that I would like help on: calling the confint function
from within another homemade function. Because he provided many good
examples of the problem, I will not reproduce them here but will instead
refer readers
On Fri, 5 Jan 2007, Ramon Diaz-Uriarte wrote:
I see, this is direct way of dealing with the problem. However, you first need
to build the f list, and you might not know about that ahead of time. For
instance, if I build a function so that the only thing that you need to do to
use my function
On Fri, 2007-01-05 at 17:01 -0600, Inman, Brant A. M.D. wrote:
On July 12, 2004 Spencer Graves wrote an email describing essentially
the same issue that I would like help on: calling the confint function
from within another homemade function. Because he provided many good
examples of the
Franco,
It is up to the user of mle() to define a function that is numerically
well behaved - even near the boundaries - and/or to provide suitable
boundary constraints.
Using gamma() in situations in which the argument could be 500 (say) will
cause problems. It appears that the maximum of
I apologize for this post. I am new to R (two days) and I have tried and tried
to calculated confidence intervals for medians. Can someone help me?
Here is my data:
institution1
0.21
0.16
0.32
0.69
1.15
0.9
0.87
0.87
0.73
The first four observations compose group 1 and observations 5 through 9
Hello R-users,
I am using gls function in R to fit a model with certain correlation
structure.
The medol as:
fit.a-gls(y~1,data=test.data,correlation=corAR1(form=~1|aa),method=ML)
mu-summary(fit.a)$coefficient
With the toy data I made to test, the estimate of mu is exactly equal to
the
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