Re: [R] scatterplot of 100000 points and pdf file format
Prof Brian Ripley wrote: On Wed, 24 Nov 2004 [EMAIL PROTECTED] wrote: On 24-Nov-04 Witold Eryk Wolski wrote: Hi, I want to draw a scatter plot with 1M and more points and save it as pdf. This makes the pdf file large. So i tried to save the file first as png and than convert it to pdf. This looks OK if printed but if viewed e.g. with acrobat as document figure the quality is bad. Anyone knows a way to reduce the size but keep the quality? If you want the PDF file to preserve the info about all the 1M points then the problem has no solution. The png file will already have suppressed most of this (which is one reason for poor quality). I think you should give thought to reducing what you need to plot. Think about it: suppose you plot with a resolution of 1/200 points per inch (about the limit at which the eye begins to see rough edges). Then you have 4 points per square inch. If your 1M points are separate but as closely packed as possible, this requires 25 square inches, or a 5x5 inch (= 12.7x12.7 cm) square. And this would be solid black! Presumably in your plot there is a very large number of points which are effectively indistinguisable from other points, so these could be eliminated without spoiling the plot. I don't have an obviously best strategy for reducing what you actually plot, but perhaps one line to think along might be the following: 1. Multiply the data by some factor and then round the results to an integer (to avoid problems in step 2). Factor chosen so that the result of (4) below is satisfactory. 2. Eliminate duplicates in the result of (1). 3. Divide by the factor you used in (1). 4. Plot the result; save plot to PDF. As to how to do it in R: the critical step is (2), which with so many points could be very heavy unless done by a well-chosen procedure. I'm not expert enough to advise about that, but no doubt others are. unique will eat that for breakfast x - runif(1e6) system.time(xx - unique(round(x, 4))) [1] 0.55 0.09 0.64 0.00 0.00 length(xx) [1] 10001 ?table - reduces the data and ?image - shows it. And this is doing exactly what I need. (not my idea but one of Thomas Unternäher). Thanks Thomas. /E -- Dipl. bio-chem. Witold Eryk Wolski MPI-Moleculare Genetic Ihnestrasse 63-73 14195 Berlin tel: 0049-30-83875219 __(_ http://www.molgen.mpg.de/~wolski \__/'v' http://r4proteomics.sourceforge.net||/ \ mail: [EMAIL PROTECTED]^^ m m [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] what does order() stand for in an lme formula?
PD == Peter Dalgaard [EMAIL PROTECTED] on 24 Nov 2004 19:35:45 +0100 writes: PD Harry Athanassiou [EMAIL PROTECTED] PD writes: I'm a beginner in R, and trying to fit linear models with different intercepts per group, of the type y ~ A*x1 + B, where x1 is a numerical variable. I cannot understand whether I should use y1 ~ x1 +1 or y1 ~ order(x1) + 1 Although in the toy example included it makes a small difference, in models with many groups the models without order() converge slower if at all! PD Er? PD What gave you the idea of using order in the first PD place? To the best of my knowledge, order(x) is also in PD this context just a function, which for the nth PD observation returns the position of the nth largest PD observation in x. This is not likely to make sense as a PD predictor in a model. where on the other hand, rank(x1) may make sense and what Harry really intended to use. Martin Maechler, ETH Zurich __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Creating lists from matrices
Hello, I am using R 1.9.1 on Windows 2000 SP4. I have the following problem: Say I have a matrix, my.matrix [,1] [,2] [,3] [1,] A B C [2,] D E F [3,] G H I I would like to apply an operation to this matrix which returns a list my.list containing the following 3 elements, my.list [[1]] [1] A B C [[2]] [2] D E F [[3]] [3] G H I That is, each row of the original matrix is turned into a vector and these vectors are collected to a list. How do I do this? Thanks, Alexander __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Creating lists from matrices
Alexander Sokol [EMAIL PROTECTED] writes: [1] A B C [[2]] [2] D E F [[3]] [3] G H I That is, each row of the original matrix is turned into a vector and these vectors are collected to a list. How do I do this? split(my.matrix, row(my.matrix)) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Creating lists from matrices
my.matrix - matrix(LETTERS[1:9],3,3,byrow=TRUE) split(my.matrix, row(my.matrix)) $1 [1] A B C $2 [1] D E F $3 [1] G H I which even names the rows for you. On Thu, 25 Nov 2004, Alexander Sokol wrote: Hello, I am using R 1.9.1 on Windows 2000 SP4. I have the following problem: Say I have a matrix, my.matrix [,1] [,2] [,3] [1,] A B C [2,] D E F [3,] G H I I would like to apply an operation to this matrix which returns a list my.list containing the following 3 elements, my.list [[1]] [1] A B C [[2]] [2] D E F [[3]] [3] G H I That is, each row of the original matrix is turned into a vector and these vectors are collected to a list. How do I do this? Thanks, Alexander __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Danish characters i R2.0.1 vs R1.9.1 under winXP
On Thu, 25 Nov 2004, Jean Coursol wrote:
The same is true in french under linux.
No, it is not the same. Windows XP does this even in the Danish locale,
and that is a problem (in Windows).
Your problem is simply that you should be using a French locale to use
French characters. You should not expect French characters to be
recognised in a non-French locale, and if they were, that was a bug in R
1.9.1.
From ?Sys.setlocale
The locale describes aspects of the internationalization of a
program. Initially most aspects of the locale of R are set to
'C' (which is the default for the C language and reflects
North-American usage). R does set 'LC_CTYPE' and 'LC_COLLATE',
which allow the use of a different character set (typically ISO
Latin 1) and alphabetic comparisons in that character set
(including the use of 'sort')
Something changed
from 1.9.1 to 2.0.0.
Yes, that has already be explained in this thread, so please read earlier
replies.
To summarize: a bug has been corrected so R now works as it has always
been documented to do, print()ing only the printable characters of the
current locale. Unfortunately for German and Scandinavian locales (at
least), Windows XP does not correctly identify some of the characters in
their locales as used in the locale. As from 2.0.1 patched, we no longer
believe Windows, but we do still believe other OSes.
First, it is necessary to have .inputrc (in $HOME)
(or $INPUTRC defined) to enter and display 8-bits
characters under bash and R.
Enter from the console, yes, but not e.g. from a file.
#.inputrc (for readline library)
set input-meta on
set output-meta on
set convert-meta off
Then
Under R2.0.1, I have:
In what locale? It matters!
élément - é # error for object name
Error: syntax error
element - é
element
[1] \351 # different from R1.9.1 (=é)
Sys.setlocale('LC_ALL','fr_FR')
The setting you need is LC_CTYPE: see the help page.
[1] fr_FR
élément - é # OK for object name
élément
[1] é# OK for display
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595__
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[R] eval in correct frame?
I am trying, without success, to find out how to formulate correctly the
parameters of eval.
My code snippet looks like:
proutside - function(txt) {
cat(\n,txt,\n);
print(eval(parse(text = txt)))
}
vari - function(Ob) {
prininside - function(txt) {
cat(\n,txt,\n);
print(eval(parse(text = txt)))
}
prininside('inside'; 2*Ob)
proutside('outside'; 2*Ob)
}
Obs - matrix(rnorm(9),nrow=3,ncol=3)
vari(Obs)
'inside'; 2*Ob
[,1] [,2] [,3]
[1,] -1.566331 -1.98257 0.127522
[2,] 3.161932 -4.88416 2.355412
[3,] -0.763759 2.33552 2.165868
'outside'; 2*Ob
Error in eval(expr, envir, enclos) : Object Ob not found
My aim is to be able to use a function of type proutside as deep
inside a function within a function... as I wish. I suppose, that the
call of eval within proutside should have envir = parent.frame(),
enclos = , but I cannot figure that out correctly.
Thank for any help.
Christian
--
Dr.sc.math.Christian W. Hoffmann,
http://www.wsl.ch/staff/christian.hoffmann
Mathematics + Statistical Computing e-mail: [EMAIL PROTECTED]
Swiss Federal Research Institute WSL Tel: ++41-44-73922- -77 (office)
CH-8903 Birmensdorf, Switzerland -11(exchange), -15 (fax)
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Re: [R] eval in correct frame?
Christian Hoffmann [EMAIL PROTECTED] writes:
I am trying, without success, to find out how to formulate correctly
the parameters of eval.
My code snippet looks like:
proutside - function(txt) {
cat(\n,txt,\n);
print(eval(parse(text = txt)))
}
vari - function(Ob) {
prininside - function(txt) {
cat(\n,txt,\n);
print(eval(parse(text = txt)))
}
prininside('inside'; 2*Ob)
proutside('outside'; 2*Ob)
}
Obs - matrix(rnorm(9),nrow=3,ncol=3)
vari(Obs)
'inside'; 2*Ob
[,1] [,2] [,3]
[1,] -1.566331 -1.98257 0.127522
[2,] 3.161932 -4.88416 2.355412
[3,] -0.763759 2.33552 2.165868
'outside'; 2*Ob
Error in eval(expr, envir, enclos) : Object Ob not found
My aim is to be able to use a function of type proutside as deep
inside a function within a function... as I wish. I suppose, that the
call of eval within proutside should have envir = parent.frame(),
enclos = , but I cannot figure that out correctly.
I think you are looking for eval.parent() (in both cases; try using
prinside on something called txt)
--
O__ Peter Dalgaard Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
__
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Re: [R] Running R from CD?
On Mon, 22 Nov 2004, Prof Brian Ripley wrote: [...] BTW, I believe running R 2.0.x from a CD will be a lot slower than 1.9.1 because of lazy loading and frequent file accesses: that's a theoretical issue we intend to address for 2.1.0, but not one anyone has yet commented that it is a problem. I collected some data (under Windows XP). On a modern desktop, running R from a CD-R or from a USB 2.0 thumbdrive was perfectably acceptable, with startup times of about 5 secs and little delay when running. On a 2.5year old laptop with a USB 1.1 port (but the same thumbdrive) it took about 15secs to start and with frequent delays the first time an object was used -- I would not find that tolerable. The laptop's CD drive was slower than the desktop and there were delays when it powered down, but it was acceptable. This was less performance penalty than I was expecting, and less than I have seen on a high-latency network file system. So it looks as if all we can do is trade a slower startup time (by caching files) for removing hiatuses when running. (Caching the pkg.rdb and pkg.rdx files when a package is opened would probably only take up a little over 1Mb in a typical session.) Writing to the thumbdrive took about 20mins, as R has so many small files and the drive has a VFAT file system. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] substitute accents
I have an openoffice spreadsheet with a column of
character strings.
Some of them contain accents.
I want to read it in R so I have saved it as a csv
file using Western Europe (ISO-8859-1) character set
(the default, I've tried other sets but it doesn't
help).
R reads it fine with
CharMatrix-read.csv(test.csv,header=F,sep=,,as.is=TRUE);
Say I wan't to replace the 'o' with accent in the
first cell.
I've tried:
gsub('ó','o', CharMatrix[1,1])
But, It doesn't make any substitution
Trying to find a solution I input the character string
in R and do the substitution:
CharMatrix[1,1]-hóla
gsub('ó','o', CharMatrix[1,1])
And it works. I think the difference is that when I
now print the content of CharMatrix I get a \201
before the ó while I didn't get it with the openoffice
imported csv file.
I'm sure it is a problem with my understanding of how
accents can be specified. Can someone give me any
solutions / references?
Thanks,
M
_
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor0.0
year 2004
month10
day 04
language R
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RE: [R] help with error message
Before I receive a barrage of 'try looking in the help file' messages, I
have and to no avail. For a new user of R I would like to point out that
in order to be able to use the help files/manuals effectively one must
know the correct question and that only comes with using R!
Could someone please direct me to why I keep getting the following
error message
Error: subscript out of bounds
when the following code is run
z.score-function (group) {
for (t in levels(group$Country)){ # this will give
t countries
y-subset(group,factor(Country)==t) #particular
analyte over all countries equal to y
#calculate overall huber mean and sigma for a
particular analyte over all countries
ov.mu-hubers(group$X)$mu
ov.sigma-hubers(group$X)$s
#define arrays
#p.mu-array()
#p.sigma-array()
#z.value-array()
#calculate huber mean and sigma for given
analyte (defined by group) by selected country (y)
p.mu-hubers(y$X)$mu
#p.sigma-hubers(y$X)$s
#calculate z score for particular
analyte:country combination
#z.value-as.vector((p.mu[t]-ov.mu)/ov.sigma)
}
#data-list(mean=ov.mu,sd=ov.sigma)
return(p.mu)
}
# group entered as group=subset(sub2, factor(Analyte)==Cholesterol)
for example
#
#
#
#
#
Please ignore the commented out lines, these were put in for my own
use. The code gave the same error message with them removed.
Thankyou for your help
Mike Griffiths
Michael Griffiths, Ph.D.
Chemometrician
Training, Quality and Statistics Group
LGC Limited
Queens Road
Teddington
Middlesex, TW11 0LY, UK
Tel: +44 (0)20 8943 7352
Fax: +44 (0)20 8943 2767
e-mail: [EMAIL PROTECTED]
***
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Re: [R] substitute accents
Can you please tell us what locale you are working in?
This looks as if the problem might be the use of a UTF-8 locale, which R
does not currently support and which some Linux distros have made their
default. However, R does issue a warning -- so did you get one?
On Thu, 25 Nov 2004, Manuel Gutierrez wrote:
I have an openoffice spreadsheet with a column of
character strings.
Some of them contain accents.
I want to read it in R so I have saved it as a csv
file using Western Europe (ISO-8859-1) character set
(the default, I've tried other sets but it doesn't
help).
R reads it fine with
CharMatrix-read.csv(test.csv,header=F,sep=,,as.is=TRUE);
Say I wan't to replace the 'o' with accent in the
first cell.
I've tried:
gsub('ó','o', CharMatrix[1,1])
But, It doesn't make any substitution
Trying to find a solution I input the character string
in R and do the substitution:
CharMatrix[1,1]-hóla
gsub('ó','o', CharMatrix[1,1])
And it works. I think the difference is that when I
now print the content of CharMatrix I get a \201
before the ó while I didn't get it with the openoffice
imported csv file.
I'm sure it is a problem with my understanding of how
accents can be specified. Can someone give me any
solutions / references?
Thanks,
M
_
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor0.0
year 2004
month10
day 04
language R
__
Renovamos el Correo Yahoo!: ¡100 MB GRATIS!
Nuevos servicios, más seguridad
__
[EMAIL PROTECTED] mailing list
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595__
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RE: [R] help with error message - problem solved
Apologies to the listing, the problem was with the data set and not the
code
Thanks
Michael Griffiths, Ph.D.
Chemometrician
Training, Quality and Statistics Group
LGC Limited
Queens Road
Teddington
Middlesex, TW11 0LY, UK
Tel: +44 (0)20 8943 7352
Fax: +44 (0)20 8943 2767
e-mail: [EMAIL PROTECTED]
Michael Griffiths [EMAIL PROTECTED] 25/11/2004
11:20:56
Before I receive a barrage of 'try looking in the help file' messages,
I
have and to no avail. For a new user of R I would like to point out
that
in order to be able to use the help files/manuals effectively one must
know the correct question and that only comes with using R!
Could someone please direct me to why I keep getting the following
error message
Error: subscript out of bounds
when the following code is run
z.score-function (group) {
for (t in levels(group$Country)){ # this will
give
t countries
y-subset(group,factor(Country)==t) #particular
analyte over all countries equal to y
#calculate overall huber mean and sigma for a
particular analyte over all countries
ov.mu-hubers(group$X)$mu
ov.sigma-hubers(group$X)$s
#define arrays
#p.mu-array()
#p.sigma-array()
#z.value-array()
#calculate huber mean and sigma for given
analyte (defined by group) by selected country (y)
p.mu-hubers(y$X)$mu
#p.sigma-hubers(y$X)$s
#calculate z score for particular
analyte:country combination
#z.value-as.vector((p.mu[t]-ov.mu)/ov.sigma)
}
#data-list(mean=ov.mu,sd=ov.sigma)
return(p.mu)
}
# group entered as group=subset(sub2, factor(Analyte)==Cholesterol)
for example
#
#
#
#
#
Please ignore the commented out lines, these were put in for my own
use. The code gave the same error message with them removed.
Thankyou for your help
Mike Griffiths
Michael Griffiths, Ph.D.
Chemometrician
Training, Quality and Statistics Group
LGC Limited
Queens Road
Teddington
Middlesex, TW11 0LY, UK
Tel: +44 (0)20 8943 7352
Fax: +44 (0)20 8943 2767
e-mail: [EMAIL PROTECTED]
***
This email and any attachments are confidential. Any use,\ c...{{dropped}}
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[R] Error using glm with poisson family and identity link
Hi all I'm trying to use the function glm from the MASS package to do the following fit. fit - glm(FP ~ rand, data = tab, family = poisson(link = identity), subset = rand = 1) (FP is = 0) but I get the following error Error: no valid set of coefficients has been found:please supply starting values In addition: Warning message: NaNs produced in: log(x) in contrast if I fit a model without intercept fit - glm(FP ~ rand - 1, data = tab, family = poisson(link = identity), subset = rand = 1) everything goes fine. Now my guess is that the points naturally have a negative intercept so the error is produced because I'm using the poisson distribution for the y and negative values are of course not admitted. Am I right? Also if this is the cause, shouldn't the function always try to do the best fit given the parameters? I mean shouldn't it fit a model with intercept 0 anyway and report it as a bad fit? Thanks Federico __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R vs SPSS
Dear all, in last weeks you discussed about R vs SAS. I want to ask your opinion about a comparison between R and SPSS. I don't know this software, but some weeks ago I went to a presentation of this product. I found it really user-friendly with GUI (even if I'd prefer command line) and very usefull and simple to use in creation and managing tables, OLAP tecniques, pivot table. What you think about? Cordially Vito = Diventare costruttori di soluzioni Became solutions' constructors The business of the statistician is to catalyze the scientific learning process. George E. P. Box Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/palese/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] substitute accents
$ locale
LANG=en_GB
LC_CTYPE=en_GB
LC_NUMERIC=en_GB
LC_TIME=en_GB
LC_COLLATE=en_GB
LC_MONETARY=en_GB
LC_MESSAGES=en_GB
LC_PAPER=en_GB
LC_NAME=en_GB
LC_ADDRESS=en_GB
LC_TELEPHONE=en_GB
LC_MEASUREMENT=en_GB
LC_IDENTIFICATION=en_GB
LC_ALL=
$ locale charmap
ISO-8859-1
I have tried changing the locales with no difference.
Is this fine?
And, no, I didn't get any warning message.
My sistem is a debian sid under kde 3.3.
Thanks,
M
--- Prof Brian Ripley [EMAIL PROTECTED]
escribió:
Can you please tell us what locale you are working
in?
This looks as if the problem might be the use of a
UTF-8 locale, which R
does not currently support and which some Linux
distros have made their
default. However, R does issue a warning -- so did
you get one?
On Thu, 25 Nov 2004, Manuel Gutierrez wrote:
I have an openoffice spreadsheet with a column of
character strings.
Some of them contain accents.
I want to read it in R so I have saved it as a csv
file using Western Europe (ISO-8859-1) character
set
(the default, I've tried other sets but it doesn't
help).
R reads it fine with
CharMatrix-read.csv(test.csv,header=F,sep=,,as.is=TRUE);
Say I wan't to replace the 'o' with accent in the
first cell.
I've tried:
gsub('ó','o', CharMatrix[1,1])
But, It doesn't make any substitution
Trying to find a solution I input the character
string
in R and do the substitution:
CharMatrix[1,1]-hóla
gsub('ó','o', CharMatrix[1,1])
And it works. I think the difference is that when
I
now print the content of CharMatrix I get a \201
before the ó while I didn't get it with the
openoffice
imported csv file.
I'm sure it is a problem with my understanding of
how
accents can be specified. Can someone give me any
solutions / references?
Thanks,
M
_
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor0.0
year 2004
month10
day 04
language R
__
Renovamos el Correo Yahoo!: ¡100 MB GRATIS!
Nuevos servicios, más seguridad
__
[EMAIL PROTECTED] mailing list
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--
Brian D. Ripley,
[EMAIL PROTECTED]
Professor of Applied Statistics,
http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865
272861 (self)
1 South Parks Road, +44 1865
272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865
272595
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[R] MASS problem -- glmmPQL and anova
Hello: I am really stuck on this problem. Why do I get an error message with anova() when I compare these two equations? Hope someone can help. ANDREW fm1 - glmmPQL(choice ~ day + stereotypy, +random = ~ 1 | bear, data = learning, family = binomial) fm2 - glmmPQL(choice ~ day + envir + stereotypy, +random = ~ 1 | bear, data = learning, family = binomial) anova(fm1, fm2) Error in anova.lme(fm1, fm2) : Objects must inherit from classes gls, gnls lm,lmList, lme,nlme,nlsList, or nls anova(fm1) numDF denDF F-value p-value (Intercept) 1 2032 7.95709 0.0048 day 1 2032 213.98391 .0001 stereotypy 1 2032 0.42810 0.5130 anova(fm2) numDF denDF F-value p-value (Intercept) 1 2031 5.70343 0.0170 day 1 2031 213.21673 .0001 envir 1 2031 12.50388 0.0004 stereotypy 1 2031 0.27256 0.6017 -- Andrew R. Criswell, Ph.D. Graduate School, Bangkok University mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R vs SPSS
Hi, Le 25 nov. 04, à 13:15, Vito Ricci a écrit : command line) and very usefull and simple to use in I do not know R so much, nor SPSS. Then I appreciate SPSS, because tools are very practical to use. Every transformation, model analysis are easily made. In the other hand, let me say it doesn't run on Mac OS X 10.3 (only on Mac OS X 10.2), then the software editor didn't manage to update its product. R's communauty does. R is really made to make big computations on big servers, that's not SPSS cup of tea. Laurent __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R multithreading skills
Hi folks, is it possible to read more things about R behavior in multiprocessor / multihost environment ? Is there any distributed computation project associated to it ? Laurent __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Turning strings into expressions
Hello, I am running R 1.9.1 om Windows 2000 SP4. My problem is as follows: Say I have a dataframe my.frame with column names A and B. I have a string, my.string [1] A==1 B==2 And I would like to retrieve the subset corresponding to my.string, that is, from my.frame and my.string I would like to get the result of subset(my.frame,A==1 B==2) So I need to find a way to convert A==1 B==2 to A==1 B==2 I at first hoped that get() could do the job, but this does not work. Does anyone know how to do this? Thanks, Alexander __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Turning strings into expressions
On Thu, 25 Nov 2004 14:09:14 +0100, Alexander Sokol [EMAIL PROTECTED] wrote : Hello, I am running R 1.9.1 om Windows 2000 SP4. My problem is as follows: Say I have a dataframe my.frame with column names A and B. I have a string, my.string [1] A==1 B==2 And I would like to retrieve the subset corresponding to my.string, that is, from my.frame and my.string I would like to get the result of subset(my.frame,A==1 B==2) So I need to find a way to convert A==1 B==2 to A==1 B==2 I at first hoped that get() could do the job, but this does not work. Does anyone know how to do this? parse() does the conversion to an expression, but doesn't evaluate it. So you probably want eval(parse(text = A == 1 B == 2)) but you may want to set the envir argument to eval, to tell R where to go looking for A and B. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] (no subject)
Good morning, I'd like to know how to superimpose to a distribution of Pearson coefficient the Student cumulative distribution function. Thank you of helping me. Angela __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Turning strings into expressions
Hi Alexander, you could try: my.string - A==1 B==2 (my.frame - data.frame(A=sample(1:2, 20, TRUE), B=sample(1:2, 20, TRUE))) subset(my.frame, eval(parse(text=my.string))) 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: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Alexander Sokol [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, November 25, 2004 2:09 PM Subject: [R] Turning strings into expressions Hello, I am running R 1.9.1 om Windows 2000 SP4. My problem is as follows: Say I have a dataframe my.frame with column names A and B. I have a string, my.string [1] A==1 B==2 And I would like to retrieve the subset corresponding to my.string, that is, from my.frame and my.string I would like to get the result of subset(my.frame,A==1 B==2) So I need to find a way to convert A==1 B==2 to A==1 B==2 I at first hoped that get() could do the job, but this does not work. Does anyone know how to do this? Thanks, Alexander __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] (no subject)
Good morning, I tried to apply the ks test to a Student distribution by ks.test(input, pt, ncp = 0, df = 58) or ks.test(input, pt, df = 58) without success where input contains my data and 58 is the fredoom degree number. Why? Thank you, Angela __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Making legend() look like my plot()
Hello, I am using code like the following to create as simple plot... plot(x,y,type='b') lines(lowess(x,y),lwd=3,lty=3,col=2) I want to add a legend which shows lines looking exactly like those used in my plot, i.e. a thin black line with gaps taken up by circles (the default for type='b', and a thick dashed red line with no pch at all). I have two problems, 1) making the pch on the first like look like type = 'b' (gaps around pch) 2) surpressing a pch on for the second line Any help with these two problems would be greatly appreciated. Any archive of plots and code to browse which could help me visually find what I want and then copy the code? An online user contributable database of 'graphics in R' would be smashing. How come some smart people dont just let me do something like legend(xpos,ypos,legend=add) to add a legend to the current plot for all the relevant points and lines which have been added so far? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Re: how to remove time series trend in R?
Hi Terry, If I understood your problem you would estimate trend and seasonal (as sum of sin and cos) in a ts. If t is time, Y is your ts, T=f(t) is trend function of time (it could be linear, quadratic, etc. as better is for your data), e=errors/residuals Your model to fit will'be: Y(t)=T(t)+a*cos(2*pi*t/12)+b*sin(2*pi*t/12)+e(t) using lm() function to estimate a linear/polinomial trend and sin/cos seasonal: cos.t - cos(2*pi*t/12) sin.t - sin(2*pi*t/12) gfit-lm(y~t+cos.t+sin.t, data=yourdf) see this example: t-seq(1:48) t [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 [26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 y-10+5*t+0.5*cos(2*pi*t/12)+0.2*sin(2*pi*t/12)+rnorm(48) cos.t - cos(2*pi*t/12) sin.t - sin(2*pi*t/12) gfit-lm(y~t+cos.t+sin.t) summary(gfit) Call: lm(formula = y ~ t + cos.t + sin.t) Residuals: Min 1Q Median 3Q Max -2.10222 -0.62184 -0.09387 0.50586 2.74299 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 10.304660.30009 34.339 2e-16 *** t4.989870.01071 465.793 2e-16 *** cos.t0.302070.20604 1.4660.150 sin.t0.086990.20961 0.4150.680 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 1.008 on 44 degrees of freedom Multiple R-Squared: 0.9998, Adjusted R-squared: 0.9998 F-statistic: 7.525e+04 on 3 and 44 DF, p-value: 2.2e-16 I hope I helped you. Best Vito = Diventare costruttori di soluzioni Became solutions' constructors The business of the statistician is to catalyze the scientific learning process. George E. P. Box Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/palese/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] LDA with previous PCA for dimensionality reduction
Torsten Hothorn writes: as long as one does not use the information in the response (the class variable, in this case) I don't think that one ends up with an optimistically biased estimate of the error I would be a little careful, though. The left-out sample in the LDA-cross-validation, will still have influenced the PCA used to build the LDA on the rest of the samples. The sample will have a tendency to lie closer to the centre of the complete PCA than of a PCA on the remaining samples. Also, if the sample has a high leverage on the PCA, the directions of the two PCAs can be quite different. Thus, the LDA is built on data that fits better to the left-out sample than if the sample was a completely new sample. I have no proofs or numerical studies showing that this gives over-optimistic error rates, but I would not recommend placing the PCA outside the cross-validation. (The same for any resampling-based validation.) -- Bjørn-Helge Mevik __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R multiprocessor/multihost skills
On Thu, Nov 25, 2004 at 02:11:12PM +0100, Laurent Valdes wrote: Hi folks, is it possible to read more things about R behavior in multiprocessor / multihost environment ? Is there any distributed computation project associated to it ? Sure. I'd start with the paper 'Simple Parallel Statistical Computing in R' by Tony Rossini, Luke Tierney and Na Li: http://www.bepress.com/uwbiostat/paper193/ and the references therein. The underlying software (SNOW, Rmpi, Rpvm, ...) is readily available on CRAN, but you may have to do some work to get the communications libraries needed (lam, mpi, pvm, ...) built. One way to get a head start on deployment is to grab a Quantian dvd image: http://dirk.eddelbuettel.com/quantian which not only contains R alongside Snow, Rmpi, Rpvm for explicit parallelism using message parsing, but also provides a ready-to-use openMosix clustering environment where you can boot nodes 2, 3, ... right over the net off the first machine booted from the dvd. Hth, Dirk PS I amended your subject line as you really asked about multiprocessor and mulithost rather than multithreading (which is typically inside one cpu, and which R doesn't do, see http://developer.r-project.org). -- If your hair is standing up, then you are in extreme danger. -- http://www.usafa.af.mil/dfp/cockpit-phys/fp1ex3.htm __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] (no subject)
Angela Re wrote: Good morning, I tried to apply the ks test to a Student distribution by ks.test(input, pt, ncp = 0, df = 58) or ks.test(input, pt, df = 58) without success where input contains my data and 58 is the fredoom degree number. Why? Thank you, Angela __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Please do what the appended messages tells you to do: read the posting guide and learn to use a sensible subject line! The following works for me, so please also specify a reproducible example that shows what does not work input - rnorm(100) ks.test(input, pt, ncp = 0, df = 58) Uwe Ligges __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Automatic file reading
Sean Davis wrote:
If you simply want read all files in a given directory, you can do
something like:
fullpath = /home/andersm/tmp
filenames - dir(fullpath,pattern=*)
pair - sapply(filenames,function(x)
{read.table(paste(fullpath,'/',x,sep=))})
Slightly off-topic but it is more portable to use the file.path function
instead of paste when creating a file name.
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[R] Re: (no subject)
Angela Re wrote: Good morning, I tried to apply the ks test to a Student distribution by ks.test(input, pt, ncp = 0, df = 58) or ks.test(input, pt, df = 58) without success where input contains my data and 58 is the fredoom degree number. Why? Thank you, Angela It runs also for me: input-rt(100,58) ks.test(input, pt, ncp = 0, df = 58) One-sample Kolmogorov-Smirnov test data: input D = 0.0827, p-value = 0.5003 alternative hypothesis: two.sided Vito = Diventare costruttori di soluzioni Became solutions' constructors The business of the statistician is to catalyze the scientific learning process. George E. P. Box Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/palese/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] scatterplot of 100000 points and pdf file format
Hi Andy, On 25-Nov-04 Liaw, Andy wrote: From: [EMAIL PROTECTED] [...] X-round(rnorm(1e6),3);Y-round(rnorm(1e6),3) system.time(unique(X)) [1] 0.74 0.07 0.81 0.00 0.00 system.time(unique(cbind(X,Y))) [1] 350.81 4.56 356.54 0.00 0.00 Do you know if majority of that time is spent in unique() itself? If so, which method? What I see is: X-round(rnorm(1e6),3);Y-round(rnorm(1e6),3) system.time(unique(X), gcFirst=TRUE) [1] 0.25 0.01 0.26 NA NA system.time(unique(cbind(X,Y)), gcFirst=TRUE) [1] 101.80 0.34 104.61 NA NA system.time(dat - data.frame(x=X, y=Y), gcFirst=TRUE) [1] 10.17 0.00 10.24NANA system.time(unique(dat), gcFirst=TRUE) [1] 23.94 0.11 24.15NANA Andy I want to look into this a bit more systematically (I have an idea why 'unique' may be taking longer on the array from 'cbind' than on the dataframe), but I will be doing this on a much faster machine than I immediately have to hand, so will report results (if interesting) later. Meanwhile, I'm not sure what you mean by which method?, and I'm also wondering what gcFirst is about. Thanks, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 [NB: New number!] Date: 25-Nov-04 Time: 14:30:39 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R vs SPSS
Vito Ricci wrote: Dear all, in last weeks you discussed about R vs SAS. I want to ask your opinion about a comparison between R and SPSS. I don't know this software, but some weeks ago I went to a presentation of this product. I found it really user-friendly with GUI (even if I'd prefer command line) and very usefull and simple to use in creation and managing tables, OLAP tecniques, pivot table. What you think about? Cordially Vito What worries me about SPSS is that it often results in poor statistical practice. The defaults in dialog boxes are not very good in some cases, and like SAS, SPSS tends to lead users to make to many assumptions (linearity in regression being one of the key ones). -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Searching for a string in RSQLite
I'd like to search for a particular string in an SQLite database using RSQLite, but I'm running into problems constructing the query properly, because of embedded quotes and parens in the string. Is there a function that escapes these for me, or some other fixup that would let me do the queries below? In the real situation I don't have control over what strings get searched for. Example based on ?SQLite: library(RSQLite) m - dbDriver(SQLite) con - dbConnect(m, dbname = base.dbms) data(USArrests) dbWriteTable(con, USArrests, USArrests, overwrite = T) [1] TRUE state - Wyoming # this works fine: dbGetQuery(con, paste(SELECT * from USArrests where row_names=',state,',sep=)) row_names Murder Assault UrbanPop Rape 1 Wyoming6.8 161 60 15.6 # Buf if the search string contains characters that SQL interprets, I # get an error state - messy: ' ( dbGetQuery(con, paste(SELECT * from USArrests where row_names=',state,',sep=)) Error in sqliteExecStatement(con, statement) : RS-DBI driver: (error in statement: near (: syntax error) Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] logistic regression and 3PL model
Dear Mike, Pinheiro and Bates discuss a three-parameter logistic growth model in their Mixed Effects Models in S and S-PLUS, but as far as I know there's no direct way to fit the 3PL IRT model in R. It should be possible to fit such a model using one of the general optimisers in R, such as nlm() or optimise(), and I think that it would be a nice project to produce an IRT package for R. Regards, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Michael Lau Sent: Wednesday, November 24, 2004 10:26 PM To: [EMAIL PROTECTED] Subject: [R] logistic regression and 3PL model Hello colleagues, I am a novice with R and am stuck with an analysis I am trying to conduct. Any suggestions or feedback would be very much appreciated. I am analyzing a data set of psi (ESP) ganzfeld trials. The response variable is binary (correct/incorrect), with a 25% base rate. I've looked around the documentation and other online resources and cannot find how I can correct for that base rate when I conduct a logistic regression. I understand that the correction would be equivalent to the three parameter logistic model (3PL) in IRT but am unsure how to best fit it from a logistic regression in R. Thanks much, Mike Lau __ Michael Y. Lau, M.A. 118 Haggar Hall Department of Psychology University of Notre Dame Notre Dame, IN 46556 [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
On Thu, 25 Nov 2004, Federico Gherardini wrote: Hi all I'm trying to use the function glm from the MASS package to do the following It's in the stats package. fit. fit - glm(FP ~ rand, data = tab, family = poisson(link = identity), subset = rand = 1) (FP is = 0) but I get the following error Error: no valid set of coefficients has been found:please supply starting values In addition: Warning message: NaNs produced in: log(x) And did you follow the advice in the error message? in contrast if I fit a model without intercept fit - glm(FP ~ rand - 1, data = tab, family = poisson(link = identity), subset = rand = 1) everything goes fine. Now my guess is that the points naturally have a negative intercept so the error is produced because I'm using the poisson distribution for the y and negative values are of course not admitted. Am I right? We don't have your data, but it is plausible. Also if this is the cause, shouldn't the function always try to do the best fit given the parameters? I mean shouldn't it fit a model with intercept 0 anyway and report it as a bad fit? Well, I believe functions should do what they say on the box (and the help page), and not what some user hopes they might do by mind-reading. You do have a suitable set of starting values from the second fit, so why not just follow the rather explicit advice? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] R vs SPSS
Vito, I use SPSS mainly for descriptive analysis (tables, graphs, factor analysis..) and for data manipulation (you can see your data and verify/control each step of your manipulation), mainly exploring the analysis I need to develop in R (advanced clustering modelling, simulations..). SPSS huge worry : ITS VALUE.. Just actualize the annual fees .. If you have a lot of data manipulation and table/easy graphs production, you can consider it.. For statistical issues, spend the money into R trainings and R contribution Best regards Naji -Message d'origine- De : [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] la part de Frank E Harrell Jr Envoyé : jeudi 25 novembre 2004 15:57 À : Vito Ricci Cc : [EMAIL PROTECTED] Objet : Re: [R] R vs SPSS Vito Ricci wrote: Dear all, in last weeks you discussed about R vs SAS. I want to ask your opinion about a comparison between R and SPSS. I don't know this software, but some weeks ago I went to a presentation of this product. I found it really user-friendly with GUI (even if I'd prefer command line) and very usefull and simple to use in creation and managing tables, OLAP tecniques, pivot table. What you think about? Cordially Vito What worries me about SPSS is that it often results in poor statistical practice. The defaults in dialog boxes are not very good in some cases, and like SAS, SPSS tends to lead users to make to many assumptions (linearity in regression being one of the key ones). -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Problem with ODBC access to SQL database
I have the following problem in getting the sqlSave function from the example code in the R package RODBC to work with MySQL as ODBC server: - a new database is created, but no data is written to it - the example code works just fine when I use MS Access as ODBC server. --- offending code and output --- library(RODBC); channel - odbcConnect(opus); data(USArrests) # R 2.0.0 only sqlSave(channel, USArrests, rownames = state, addPK=TRUE) Error in sqlColumns(channel, tablename) : USArrests : table not found on channel Check case parameter in odbcConnect sqlSave(channel, USArrests, rownames = state, addPK=TRUE) Error in sqlSave(channel, USArrests, rownames = state, addPK = TRUE) : [RODBC] ERROR: Could not SQLExecDirect odbcGetErrMsg(channel) character(0) sqlTables(channel) TABLE_CAT TABLE_SCHEM TABLE_NAME TABLE_TYPE REMARKS 1 opus usarrests TABLE MySQL table - The next thing I did was to check from the MySQL commandline what the status of the database was. MySQL commandline query --- mysql use opus; Database changed mysql show tables; ++ | Tables_in_opus | ++ | usarrests | ++ 1 row in set (0.00 sec) mysql describe usarrests; +--+--+--+-+-+---+ | Field| Type | Null | Key | Default | Extra | +--+--+--+-+-+---+ | state| varchar(255) | | PRI | | | | Murder | double | YES | | NULL| | | Assault | int(11) | YES | | NULL| | | UrbanPop | int(11) | YES | | NULL| | | Rape | double | YES | | NULL| | +--+--+--+-+-+---+ 5 rows in set (0.01 sec) mysql select * from usarrests; Empty set (0.00 sec) - So: the table is created, but not filled. As noted, the example code works OK with MS access. The software versions I use are: - OS: Windows XP SP2 - R: 2.0.1; installed as binary - RODBC: latest version from CRAN - MySQL: 4.0.21-nt P.S. The problem has low priority for me bacause I can simply use Microsoft Access as RODBC server, but in future I would like to revert back to MySQL. [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] logistic regression and 3PL model
On Thu, 25 Nov 2004, John Fox wrote: Pinheiro and Bates discuss a three-parameter logistic growth model in their Mixed Effects Models in S and S-PLUS, but as far as I know there's no direct way to fit the 3PL IRT model in R. It should be possible to fit such a model using one of the general optimisers in R, such as nlm() or optimise(), and I optim(), not optimize() as there are at least two free parameters, I believe. think that it would be a nice project to produce an IRT package for R. As I understand it this is a logistic regression and not a logistic growth curve, the latter being fitted by least squares. For a known baseline (which is thus a 2-free PL model but what seems asked for here), a glm family can be constructed to allow glm() to do the fitting. This is model described at http://work.psych.uiuc.edu/irt/modeling_dich1.asp with c known to be 0.25. It would certainly be worth having an implementation of that in R, with c=0.5 being the most common case. It is quite straightforward to fit such models by direct optimization of the likelihood, and MASS4 p. 445 gives you a template for logistic regression that could easily be modified. -Original Message- To: [EMAIL PROTECTED] Subject: [R] logistic regression and 3PL model Hello colleagues, I am a novice with R and am stuck with an analysis I am trying to conduct. Any suggestions or feedback would be very much appreciated. I am analyzing a data set of psi (ESP) ganzfeld trials. The response variable is binary (correct/incorrect), with a 25% base rate. I've looked around the documentation and other online resources and cannot find how I can correct for that base rate when I conduct a logistic regression. I understand that the correction would be equivalent to the three parameter logistic model (3PL) in IRT but am unsure how to best fit it from a logistic regression in R. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R vs SPSS
Frank E Harrell Jr wrote: What worries me about SPSS is that it often results in poor statistical practice. The defaults in dialog boxes are not very good in some cases, and like SAS, SPSS tends to lead users to make to many assumptions (linearity in regression being one of the key ones). -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ My worry about SPSS is that it encourages people to do analysis and dataset manipulation 'on-the-fly', without leaving behind an audit trail that can be used to reconstruct the dataset and results. Certainly, SPSS has a 'paste' button which allows you to save a 'syntax' file of commands, but most users appear to ignore it. And post-hoc editing of graphs and tables cannot be saved thus (unless I'm missing out something here). -- Ronan M Conroy ([EMAIL PROTECTED]) Senior Lecturer in Biostatistics Royal College of Surgeons Dublin 2, Ireland +353 1 402 2431 (fax 2764) Just say no to drug reps http://www.nofreelunch.org/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Searching for a string in RSQLite
Single quotes in a string are escaped by putting two single quotes in a row. E.g., state - mess: '' ( Regards, -- David Duncan Murdoch wrote: I'd like to search for a particular string in an SQLite database using RSQLite, but I'm running into problems constructing the query properly, because of embedded quotes and parens in the string. Is there a function that escapes these for me, or some other fixup that would let me do the queries below? In the real situation I don't have control over what strings get searched for. Example based on ?SQLite: library(RSQLite) m - dbDriver(SQLite) con - dbConnect(m, dbname = base.dbms) data(USArrests) dbWriteTable(con, USArrests, USArrests, overwrite = T) [1] TRUE state - Wyoming # this works fine: dbGetQuery(con, paste(SELECT * from USArrests where row_names=',state,',sep=)) row_names Murder Assault UrbanPop Rape 1 Wyoming6.8 161 60 15.6 # Buf if the search string contains characters that SQL interprets, I # get an error state - messy: ' ( dbGetQuery(con, paste(SELECT * from USArrests where row_names=',state,',sep=)) Error in sqliteExecStatement(con, statement) : RS-DBI driver: (error in statement: near (: syntax error) Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] urgent
Hello, My name is Mrs. Jennifer Wilson i am a dying woman who have decided to donate what i have to you/ church.I am 59 years old and i was diagnosed for cancer for about 2 years ago,immediately after the death of my husband, who has left me everything he worked for. I have been touched by God to donate from what i have inherited from my late husband to the you for the god work of God,rather than allow my relatives to use my husband hard earned funds ungodly.Please pray,that the good Lord forgive me my sins.I have asked God to forgive me and i beleive he has because He is a merciful God. I will be going in for an operation in less than one hour. I decided to WILL/donate the sum of $1,500,000 (One million five hundred thousand dollars) to you for the good work of the lord, and also to help the motherless and less privilege and also for the assistance of the widows according to (JAMES 1:27). At the moment i cannot take any telephone calls right now due to the fact that my relatives are around me and my health status.I have adjusted my WILL and my lawyer is aware i have changed my will you and he will arrange the transfer of the funds from my account to you. I wish you all the best and may the good Lord bless you abundantly, and please use the funds well and always extend the good work to others. Contact my lawyer with this specified email [EMAIL PROTECTED] and tell him that i have WILLED ($1,500,000.00) to you and i have also notified him that i am WILLING that amount to you for a specific and good work.I know i dont know you but i have been directed to do this.Thanks and God bless. NB: I will appreciate your utmost confidentiality in this matter until the task is accomplished as I don't want anything that will Jeopardize my last wish. And Also I will be contacting with you by email as I don't want my relation or anybody to know because they are always around me. Regards, Jennifer Wilson __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] logistic regression and 3PL model
I don't know if I am missing something, but isn't there also a latent variable (trait) that must be integrated out using maybe Gauss-Hermite which might complicate a bit the calculations? So is this possible with `glm()'? Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Prof Brian Ripley [EMAIL PROTECTED] To: John Fox [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Thursday, November 25, 2004 5:13 PM Subject: RE: [R] logistic regression and 3PL model On Thu, 25 Nov 2004, John Fox wrote: Pinheiro and Bates discuss a three-parameter logistic growth model in their Mixed Effects Models in S and S-PLUS, but as far as I know there's no direct way to fit the 3PL IRT model in R. It should be possible to fit such a model using one of the general optimisers in R, such as nlm() or optimise(), and I optim(), not optimize() as there are at least two free parameters, I believe. think that it would be a nice project to produce an IRT package for R. As I understand it this is a logistic regression and not a logistic growth curve, the latter being fitted by least squares. For a known baseline (which is thus a 2-free PL model but what seems asked for here), a glm family can be constructed to allow glm() to do the fitting. This is model described at http://work.psych.uiuc.edu/irt/modeling_dich1.asp with c known to be 0.25. It would certainly be worth having an implementation of that in R, with c=0.5 being the most common case. It is quite straightforward to fit such models by direct optimization of the likelihood, and MASS4 p. 445 gives you a template for logistic regression that could easily be modified. -Original Message- To: [EMAIL PROTECTED] Subject: [R] logistic regression and 3PL model Hello colleagues, I am a novice with R and am stuck with an analysis I am trying to conduct. Any suggestions or feedback would be very much appreciated. I am analyzing a data set of psi (ESP) ganzfeld trials. The response variable is binary (correct/incorrect), with a 25% base rate. I've looked around the documentation and other online resources and cannot find how I can correct for that base rate when I conduct a logistic regression. I understand that the correction would be equivalent to the three parameter logistic model (3PL) in IRT but am unsure how to best fit it from a logistic regression in R. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
Dear Federico: Why do you use the identity link? That can produce situations with an average of (-2) Poisson defects per unit, for example. That's physical nonsense. Also, it seems essentially to be generating your error message. Also, have you considered the following: fit - glm(FP ~ offset(log(rand)), data = tab, family = poisson, subset = rand = 1) I haven't tried this, but it looks like this model is virtually equivalent to the one you wrote: FP ~ rand-1 with link = identity says estimate 'b' in PoissonMean = b*rand. FP ~ offset(log(rand)) with link = log (the default) says estimate 'a' in log(PoissonMean)-log(rand) = a. If I haven't made an error, then log(b) = a. In more general situations, if you really need the identity link, have you considered searching for good starting values, as Prof. Ripley suggested? You could build up to your final model by estimating simpler models and obtaining trial fits using the default log link? With those results and a little thought, you should be able to obtain reasonable starting values. hope this helps. spencer graves Prof Brian Ripley wrote: On Thu, 25 Nov 2004, Federico Gherardini wrote: Hi all I'm trying to use the function glm from the MASS package to do the following It's in the stats package. fit. fit - glm(FP ~ rand, data = tab, family = poisson(link = identity), subset = rand = 1) (FP is = 0) but I get the following error Error: no valid set of coefficients has been found:please supply starting values In addition: Warning message: NaNs produced in: log(x) And did you follow the advice in the error message? in contrast if I fit a model without intercept fit - glm(FP ~ rand - 1, data = tab, family = poisson(link = identity), subset = rand = 1) everything goes fine. Now my guess is that the points naturally have a negative intercept so the error is produced because I'm using the poisson distribution for the y and negative values are of course not admitted. Am I right? We don't have your data, but it is plausible. Also if this is the cause, shouldn't the function always try to do the best fit given the parameters? I mean shouldn't it fit a model with intercept 0 anyway and report it as a bad fit? Well, I believe functions should do what they say on the box (and the help page), and not what some user hopes they might do by mind-reading. You do have a suitable set of starting values from the second fit, so why not just follow the rather explicit advice? -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Avoiding for-loops
Hello R-users,
I have a symmetric matrix of numerical values and I
want to obtain those values in the upper or lower
triangle of the matrix in a vector. I tried to do the
job by using two for-loops but it doens't seem to be a
clever way, and I'd like to know a more efficient code
for a large matrix of thousands of rows and columns.
Below is my code for your reference.
Thanks a lot.
John
# mtx.sym is a symmetric matrix
my.ftn - function(size_mtx, mtx) {
+ my.vector - c()
+ for ( i in 1:size_mtx ) {
+ cat(.)
+ for ( j in 1:size_mtx ) {
+ if ( upper.tri(mtx)[i,j] ) {
+ my.vector - c(my.vector, mtx[i,j])
+ }}}
+ cat(\n)
+ }
# if I have a matrix, mtx.sym, of 100x100
my.ftn(100, mtx.sym)
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RE: [R] Searching for a string in RSQLite
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Duncan Murdoch Sent: 25 November 2004 15:38 To: [EMAIL PROTECTED] Subject: [R] Searching for a string in RSQLite I'd like to search for a particular string in an SQLite database using RSQLite, but I'm running into problems constructing the query properly, because of embedded quotes and parens in the string. Is there a function that escapes these for me, or some other fixup that would let me do the queries below? In the real situation I don't have control over what strings get searched for. Example based on ?SQLite: library(RSQLite) m - dbDriver(SQLite) con - dbConnect(m, dbname = base.dbms) data(USArrests) dbWriteTable(con, USArrests, USArrests, overwrite = T) [1] TRUE state - Wyoming # this works fine: dbGetQuery(con, paste(SELECT * from USArrests where row_names=',state,',sep=)) row_names Murder Assault UrbanPop Rape 1 Wyoming6.8 161 60 15.6 # Buf if the search string contains characters that SQL interprets, I # get an error state - messy: ' ( dbGetQuery(con, paste(SELECT * from USArrests where row_names=',state,',sep=)) Error in sqliteExecStatement(con, statement) : RS-DBI driver: (error in statement: near (: syntax error) Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html The normal character for escaping the next character to prevent it being interpreted in SQL (including SQLite) is the backslash (i.e. \). Unless, of course, I'm not understanding the precise nature of your request. Regards Mike __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Turning strings into expressions
Dimitris Rizopoulos [EMAIL PROTECTED] writes: Hi Alexander, you could try: my.string - A==1 B==2 (my.frame - data.frame(A=sample(1:2, 20, TRUE), B=sample(1:2, 20, TRUE))) subset(my.frame, eval(parse(text=my.string))) Hmm, considering the nonstandard evaluation that is going on inside subset(), I think I'd rather try my.frame - data.frame(A=sample(1:2, 20, TRUE), B=sample(1:2, 20,TRUE)) my.string - A==1 B==2 l - as.list(parse(text=my.string)) names(l)-sub eval(substitute(subset(my.frame, sub), l)) A B 18 1 2 (or perhaps l - list(sub=parse(text=my.string)[[1]]) is less cryptic). Point being that this way you'll literally evaluate substitute(subset(my.frame, sub), l) subset(my.frame, A == 1 B == 2) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Searching for a string in RSQLite
You may find dQuote() and sQuote() to be helpful, but a better
Ooops, dQuote() and sQuote() won't be of much use as they escape
quotes with quotes.
A regular expression should do the trick: gsub(', ', Hi
y'all). (Note that this looks like it has too many backslashes, but
this is just the way R prints escaped strings, use str() to see the
unescaped string)
Hadley
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Spam {Re: [R] urgent}
I'm sorry for this spam that astonishingly came through to the
mailing list.
It seems these guys have been exercising against known spam
filters and achieved more than in the past.
Also, recent versions of our spamfilter have been tuned such as
to rather produce a few false negatives {spam not detected}
with hardly ever any false positive {non-spam not delivered to recipient},
which makes a lot of sense.
We (I and local e-mail administrators) do keep an eye on this,
me spending a little time for manual tuning, but we do not
want to allocate too much time for this.
PLEASE do not reply to this e-mail (at least not to R-help!).
It is *not* relevant to R and not worth the time (also since too
many people think they know what they are talking about :-).
If you want, reply to me privately.
Martin Maechler, ETH Zurich
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Re: [R] scatterplot of 100000 points and pdf file format
(Ted Harding) [EMAIL PROTECTED] writes: I want to look into this a bit more systematically (I have an idea why 'unique' may be taking longer on the array from 'cbind' than on the dataframe), Just look inside the functions. One is pasting columns together, the other is using a paste() construct inside an apply() function. So with two columns by 1e6 rows, one is doing one large paste and the other a million small ones. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Avoiding for-loops
lower triangle can be obtained by
A[row(A)col(A)]
url:www.econ.uiuc.edu/~rogerRoger Koenker
email [EMAIL PROTECTED] Department of Economics
vox:217-333-4558University of Illinois
fax:217-244-6678Champaign, IL 61820
On Nov 25, 2004, at 11:15 AM, John wrote:
Hello R-users,
I have a symmetric matrix of numerical values and I
want to obtain those values in the upper or lower
triangle of the matrix in a vector. I tried to do the
job by using two for-loops but it doens't seem to be a
clever way, and I'd like to know a more efficient code
for a large matrix of thousands of rows and columns.
Below is my code for your reference.
Thanks a lot.
John
# mtx.sym is a symmetric matrix
my.ftn - function(size_mtx, mtx) {
+ my.vector - c()
+ for ( i in 1:size_mtx ) {
+ cat(.)
+ for ( j in 1:size_mtx ) {
+ if ( upper.tri(mtx)[i,j] ) {
+ my.vector - c(my.vector, mtx[i,j])
+ }}}
+ cat(\n)
+ }
# if I have a matrix, mtx.sym, of 100x100
my.ftn(100, mtx.sym)
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Re: [R] Avoiding for-loops
John wrote:
Hello R-users,
I have a symmetric matrix of numerical values and I
want to obtain those values in the upper or lower
triangle of the matrix in a vector. I tried to do the
job by using two for-loops but it doens't seem to be a
clever way, and I'd like to know a more efficient code
for a large matrix of thousands of rows and columns.
Below is my code for your reference.
See ?upper.tri
Uwe Ligges
Thanks a lot.
John
# mtx.sym is a symmetric matrix
my.ftn - function(size_mtx, mtx) {
+ my.vector - c()
+ for ( i in 1:size_mtx ) {
+ cat(.)
+ for ( j in 1:size_mtx ) {
+ if ( upper.tri(mtx)[i,j] ) {
+ my.vector - c(my.vector, mtx[i,j])
+ }}}
+ cat(\n)
+ }
# if I have a matrix, mtx.sym, of 100x100
my.ftn(100, mtx.sym)
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Re: [R] Avoiding for-loops
On Thursday 25 November 2004 11:15, John wrote:
Hello R-users,
I have a symmetric matrix of numerical values and I
want to obtain those values in the upper or lower
triangle of the matrix in a vector. I tried to do the
job by using two for-loops but it doens't seem to be a
clever way, and I'd like to know a more efficient code
for a large matrix of thousands of rows and columns.
Below is my code for your reference.
Try
mtx[lower.tri(mtx)]
which should give you the same order as your code (untested).
HTH,
Deepayan
Thanks a lot.
John
# mtx.sym is a symmetric matrix
my.ftn - function(size_mtx, mtx) {
+ my.vector - c()
+ for ( i in 1:size_mtx ) {
+ cat(.)
+ for ( j in 1:size_mtx ) {
+ if ( upper.tri(mtx)[i,j] ) {
+ my.vector - c(my.vector, mtx[i,j])
+ }}}
+ cat(\n)
+ }
# if I have a matrix, mtx.sym, of 100x100
my.ftn(100, mtx.sym)
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[R] arrows in histograms
Dear all, I have a null frequency histogram of pairwise associations values. How can I put an arrow on the histogram to indicate the critical value obtained from the null distribution? [In this case, quantile 0.05]. Thanks -- Rogério R. Silva MZUSP http://www.mz.usp.br Linux User # 354364 Linux counter http://counter.li.org __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
Spencer Graves [EMAIL PROTECTED] writes: Dear Federico: Why do you use the identity link? That can produce situations with an average of (-2) Poisson defects per unit, for example. That's physical nonsense. So is _not_ using the identity link when the model is manifestly additive on the identity scale. E.g. calibrating differential spectrofluorometry with photon counters recording linear combinations of intensities at different wavelengths. I've bumped into similar situations before (binomial(link=identity), I think it was then) and the glm.fit algorithm could use improvement in dealing with the parameter constraints in these cases. With the standard IRLS algorithm, if the maximum is on the boundary, you basically hit a random point on the boundary and get stuck there with a search direction pointing out of the valid region. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Avoiding for-loops
On 25-Nov-04 John wrote: Hello R-users, I have a symmetric matrix of numerical values and I want to obtain those values in the upper or lower triangle of the matrix in a vector. I tried to do the job by using two for-loops but it doens't seem to be a clever way, and I'd like to know a more efficient code for a large matrix of thousands of rows and columns. The two functions uuper.tri and lower.tri do just this job! Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 [NB: New number!] Date: 25-Nov-04 Time: 18:54:20 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
Hi, Peter: What do you do in such situations? Sundar Dorai-Raj and I have extended glm concepts to models driven by a sum of k independent Poissons, with the a linear model for log(defectRate[i]) for each source (i = 1:k). To handle convergence problems, etc., I think we need to use informative Bayes, but we're not there yet. In any context where things are done more than once [which covers most human activities], informative Bayes seems sensible. A related question comes with data representing the differences between Poisson counts, e.g., with d[i] = X[i]-X[i-1] = the number of new defects added between steps i-1 and i in a manufacturing process. Most of the time, d[i] is nonnegative. However, in some cases, it can be negative, either because of metrology errors in X[i] or because of defect removal between steps i-1 and i. Comments? Best Wishes, Spencer Graves Peter Dalgaard wrote: Spencer Graves [EMAIL PROTECTED] writes: Dear Federico: Why do you use the identity link? That can produce situations with an average of (-2) Poisson defects per unit, for example. That's physical nonsense. So is _not_ using the identity link when the model is manifestly additive on the identity scale. E.g. calibrating differential spectrofluorometry with photon counters recording linear combinations of intensities at different wavelengths. I've bumped into similar situations before (binomial(link=identity), I think it was then) and the glm.fit algorithm could use improvement in dealing with the parameter constraints in these cases. With the standard IRLS algorithm, if the maximum is on the boundary, you basically hit a random point on the boundary and get stuck there with a search direction pointing out of the valid region. -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Making legend() look like my plot()
Is this an impossible task? How about just problem 2 below, having one pch in one legend entry, but no pch in the second? On Thu, 25 Nov 2004, Dan Bolser wrote: Hello, I am using code like the following to create as simple plot... plot(x,y,type='b') lines(lowess(x,y),lwd=3,lty=3,col=2) I want to add a legend which shows lines looking exactly like those used in my plot, i.e. a thin black line with gaps taken up by circles (the default for type='b', and a thick dashed red line with no pch at all). I have two problems, 1) making the pch on the first like look like type = 'b' (gaps around pch) 2) surpressing a pch on for the second line Any help with these two problems would be greatly appreciated. Any archive of plots and code to browse which could help me visually find what I want and then copy the code? An online user contributable database of 'graphics in R' would be smashing. How come some smart people dont just let me do something like legend(xpos,ypos,legend=add) to add a legend to the current plot for all the relevant points and lines which have been added so far? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] arrows in histograms
On 25-Nov-04 Rogerio Rosa da Silva wrote: Dear all, I have a null frequency histogram of _pairwise associations values. How can I put _an arrow on the histogram to indicate the critical value obtained from the null distribution? [In this case, quantile 0.05]. ?Something like q05 - qnorm(0.95) ## or qnorm(0.05) -- not sure which you really want hist(x,xlim=c(-5,5)) arrows(q05,400,q05,0,length=0.1) text(q05+0.1,400,5 per cent point,adj=0) Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 [NB: New number!] Date: 25-Nov-04 Time: 19:46:15 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R-2.0.1 reinstall non-CRAN pkg
I am trying to upgrade to R-2.0.1 from R-1.9 on a Mac running OS X 10.3. I have some simple packages I wrote myself that have to be reinstalled to be recognized as valid packages. I have been using them for a while on earlier versions of R, so didn't expect to have any problems. I am probably going about this the wrong way? I simply used R CMD build mypkgdir and then R CMD install mypkgdir.tar.gz the package installs without any error messages. however, library(mypkgname) still generates spiteful Error in library(mypkgname) : 'mypkgname' is not a valid package -- installed 2.0.0? messages. My apologies if answers to this kind of question have already been posted. I have looked over the archived r-help threads for the last couple of months. Best Anthony Westerling __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Making legend() look like my plot()
Dan Bolser wrote: Is this an impossible task? How about just problem 2 below, having one pch in one legend entry, but no pch in the second? Please be at least a little bit patient! This is not a hotline! People are not working 24 hours a day just to answer your questions at once - they are answering questions on a voluntary basis! answer 1) is not straightforward, but you might want to use one of fillable symbols mentioned in ?points, e.g. number 21 answer 2) pch = c(1, NA) should do the trick. legend(., pch=c(21,NA), lwd=c(1,3), lty=c(1,3), pt.bg=white, col=1:2) Uwe Ligges On Thu, 25 Nov 2004, Dan Bolser wrote: Hello, I am using code like the following to create as simple plot... plot(x,y,type='b') lines(lowess(x,y),lwd=3,lty=3,col=2) I want to add a legend which shows lines looking exactly like those used in my plot, i.e. a thin black line with gaps taken up by circles (the default for type='b', and a thick dashed red line with no pch at all). I have two problems, 1) making the pch on the first like look like type = 'b' (gaps around pch) 2) surpressing a pch on for the second line Any help with these two problems would be greatly appreciated. Any archive of plots and code to browse which could help me visually find what I want and then copy the code? An online user contributable database of 'graphics in R' would be smashing. How come some smart people dont just let me do something like legend(xpos,ypos,legend=add) to add a legend to the current plot for all the relevant points and lines which have been added so far? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
Spencer Graves [EMAIL PROTECTED] writes: Hi, Peter: What do you do in such situations? Sundar Dorai-Raj and I have extended glm concepts to models driven by a sum of k independent Poissons, with the a linear model for log(defectRate[i]) for each source (i = 1:k). To handle convergence problems, etc., I think we need to use informative Bayes, but we're not there yet. In any context where things are done more than once [which covers most human activities], informative Bayes seems sensible. A related question comes with data representing the differences between Poisson counts, e.g., with d[i] = X[i]-X[i-1] = the number of new defects added between steps i-1 and i in a manufacturing process. Most of the time, d[i] is nonnegative. However, in some cases, it can be negative, either because of metrology errors in X[i] or because of defect removal between steps i-1 and i. Comments? I haven't got all that much experience with it, but obviously, the various algorithms for constrained optimization (box- or otherwise) at least allow you to find a proper maximum likelihood estimator. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
Hi, Peter: Thanks for the comment and reply. I generally avoid constrained optimizers for three reasons: 1. My experience with them has included many cases where the optimizer would stop with an error when testing parameter values that violate the constraints. If I transform the parameter space to remove the constraints, that never happens. The constrained optimizers in R 2.0.1 may not exhibit this behavior, but I have not checked. 2. In a few cases, I've plotted the log(likelihood) vs. parameter values using various transformations. When I've done that, I typically found that the most nearly parabolic performance used unconstrained parameterizations. This makes asymptotic normality more useful and increases the accuracy of simple, approximate sequential Bayesian procedures. 3. When I think carefully about a particular application, I often find a rationale for claiming that a certain unconstrained parameterization provides a better description of the application. For example, interest income on investments is essentially additive on the log scale. Similarly, the concept of materiality in Accounting is closer to being constant in log space: One might look for an error of a few Euros in the accounts of a very small business, but in auditing some major government accounts, errors on the order of a few Euros might not be investigated. Also, measurement errors with microvolts are much smaller than with megavolts; expressing the measurements in decibels (i.e., on the log scale) makes the measurement errors more nearly comparable. Thanks again for your comments. Best Wishes, Spencer Graves Peter Dalgaard wrote: Spencer Graves [EMAIL PROTECTED] writes: Hi, Peter: What do you do in such situations? Sundar Dorai-Raj and I have extended glm concepts to models driven by a sum of k independent Poissons, with the a linear model for log(defectRate[i]) for each source (i = 1:k). To handle convergence problems, etc., I think we need to use informative Bayes, but we're not there yet. In any context where things are done more than once [which covers most human activities], informative Bayes seems sensible. A related question comes with data representing the differences between Poisson counts, e.g., with d[i] = X[i]-X[i-1] = the number of new defects added between steps i-1 and i in a manufacturing process. Most of the time, d[i] is nonnegative. However, in some cases, it can be negative, either because of metrology errors in X[i] or because of defect removal between steps i-1 and i. Comments? I haven't got all that much experience with it, but obviously, the various algorithms for constrained optimization (box- or otherwise) at least allow you to find a proper maximum likelihood estimator. -- Spencer Graves, PhD, Senior Development Engineer O: (408)938-4420; mobile: (408)655-4567 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Making legend() look like my plot()
On Thu, 25 Nov 2004, Uwe Ligges wrote: Dan Bolser wrote: Is this an impossible task? How about just problem 2 below, having one pch in one legend entry, but no pch in the second? Please be at least a little bit patient! This is not a hotline! People are not working 24 hours a day just to answer your questions at once - they are answering questions on a voluntary basis! answer 1) is not straightforward, but you might want to use one of fillable symbols mentioned in ?points, e.g. number 21 answer 2) pch = c(1, NA) should do the trick. legend(., pch=c(21,NA), lwd=c(1,3), lty=c(1,3), pt.bg=white, col=1:2) Ahhh... I tried pch=c(1,NULL), pt.bg='white' I couldn't work out what was going on.. thanks very much for the info Uwe Ligges On Thu, 25 Nov 2004, Dan Bolser wrote: Hello, I am using code like the following to create as simple plot... plot(x,y,type='b') lines(lowess(x,y),lwd=3,lty=3,col=2) I want to add a legend which shows lines looking exactly like those used in my plot, i.e. a thin black line with gaps taken up by circles (the default for type='b', and a thick dashed red line with no pch at all). I have two problems, 1) making the pch on the first like look like type = 'b' (gaps around pch) 2) surpressing a pch on for the second line Any help with these two problems would be greatly appreciated. Any archive of plots and code to browse which could help me visually find what I want and then copy the code? An online user contributable database of 'graphics in R' would be smashing. How come some smart people dont just let me do something like legend(xpos,ypos,legend=add) to add a legend to the current plot for all the relevant points and lines which have been added so far? __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] support vector machine
Hi Everyone Thanks to those who responded last time. I am still having problems. I really want to find one of those tutorials on how to use svm() so I can then get going using it myself. Issues are which kernel to choose, how to tune the parameters. If anyone know of a tutorial please let me know. Stephen [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] help with glmmPQL
Hello: Will someone PLEASE help me with this problem. This is the third time I've posted it. When I appply anova() to two equations estimated using glmmPQL, I get a complaint, anova(fm1, fm2) Error in anova.lme(fm1, fm2) : Objects must inherit from classes gls, gnls lm,lmList, lme,nlme,nlsList, or nls The two equations I estimated are these: fm1 - glmmPQL(choice ~ day + stereotypy, +random = ~ 1 | bear, data = learning, family = binomial) fm2 - glmmPQL(choice ~ day + envir + stereotypy, +random = ~ 1 | bear, data = learning, family = binomial) Individually, I get results from anova(): anova(fm1) numDF denDF F-value p-value (Intercept) 1 2032 7.95709 0.0048 day 1 2032 213.98391 .0001 stereotypy 1 2032 0.42810 0.5130 anova(fm2) numDF denDF F-value p-value (Intercept) 1 2031 5.70343 0.0170 day 1 2031 213.21673 .0001 envir 1 2031 12.50388 0.0004 stereotypy 1 2031 0.27256 0.6017 I did look through the archives but didn't finding anything relevant to my problem. Hope someone can help. ANDREW _ platform i586-mandrake-linux-gnu arch i586 os linux-gnu system i586, linux-gnu status major2 minor0.0 year 2004 month10 day 04 language R -- Andrew R. Criswell, Ph.D. Graduate School, Bangkok University __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Testing for S4 objects
Dear r-help list members, Is there a way to test whether an object is an S4 object? The best that I've been able to come up with is isS4object - function(object) !(is.null(slotNames(object))) which assumes that an S4 object has at least one slot. I think this is safe, but perhaps I'm missing something. Thanks, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to expand the size limit of a vector?
Hi everybody on this list, Could somebody please tell me how to expand the size limit of a vector in R? In my simulation I get the following error message: x=kronecker(diag(1,100),matrix(1,100,100)) Error: cannot allocate vector of size 781250 Kb In addition: Warning message: Reached total allocation of 511Mb: see help(memory.size) Thanks in advance Yours sincerely, Guoqi Qian - Guoqi Qian Department of Statistics La Trobe University Bundoora, VIC 3086 AUSTRALIA Email:[EMAIL PROTECTED] Tel: +61 3 9479 2609 Fax: +61 3 9479 2466 http://www.latrobe.edu.au/www/statistic/ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R vs SPSS
As I started out using SPSS when there was no GUI (in fact, no interactive interface at all), I automatically open up the syntax editing window when I have to use it. It's a workable text editor, you can run all or part of the code at will, and build up a code file in much the same way as R. On the other hand, it does encourage the user who has not taken Pope to heart (A little learning...) to put their data through a high-powered analysis while convincing themselves that they know what they are doing. I confess to having done it more than once in the past. It was when I began reviewing other researcher's papers, and thinking 'This guy didn't know what he was doing.' and then, 'And you've done it too, brother.' that I resolved to be more circumspect. Jim __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Response Surface
Hi. I'm a student at Simon Fraser University in British Columbia, Canada. I can't for the life of me figure out how to plot a 3D surface (A 3D response surface to be more specific) in R. I found your email address on a web board, and saw someone mention wireframe(), but using the help in R yielded no results. Any suggestions? Thanks. Dean Vrecko __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Response Surface
On Thursday 25 November 2004 22:35, [EMAIL PROTECTED] wrote: Hi. I'm a student at Simon Fraser University in British Columbia, Canada. I can't for the life of me figure out how to plot a 3D surface (A 3D response surface to be more specific) in R. I found your email address on a web board, and saw someone mention wireframe(), but using the help in R yielded no results. Any suggestions? Read help(persp) and run example(persp) demo(persp) Deepayan __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Response Surface
wireframe() is available in the package lattice. --Matt -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Thursday, November 25, 2004 20:36 PM To: [EMAIL PROTECTED] Subject: [R] Response Surface Hi. I'm a student at Simon Fraser University in British Columbia, Canada. I can't for the life of me figure out how to plot a 3D surface (A 3D response surface to be more specific) in R. I found your email address on a web board, and saw someone mention wireframe(), but using the help in R yielded no results. Any suggestions? Thanks. Dean Vrecko __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Response Surface
type ?wireframe rather than wireframe() Tom Mulholland -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Friday, 26 November 2004 12:36 PM To: [EMAIL PROTECTED] Subject: [R] Response Surface Hi. I'm a student at Simon Fraser University in British Columbia, Canada. I can't for the life of me figure out how to plot a 3D surface (A 3D response surface to be more specific) in R. I found your email address on a web board, and saw someone mention wireframe(), but using the help in R yielded no results. Any suggestions? Thanks. Dean Vrecko __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] help with glmmPQL
For better or worse, it's holidays in the states. Very amusing for me being in a non-Thanksgiving celebrating country. In addition, it's not a problem. The complaint is valid. Probably no one has coded up the right solution yet for comparison. I can't recall if one would want those statistics for a binomial random effects model, but I do recall some issues with model comparison in that setting, though they are a bit dated (say, 2 years or so). On Fri, 26 Nov 2004 09:31:40 +0700, Andrew Criswell [EMAIL PROTECTED] wrote: Hello: Will someone PLEASE help me with this problem. This is the third time I've posted it. When I appply anova() to two equations estimated using glmmPQL, I get a complaint, anova(fm1, fm2) Error in anova.lme(fm1, fm2) : Objects must inherit from classes gls, gnls lm,lmList, lme,nlme,nlsList, or nls The two equations I estimated are these: fm1 - glmmPQL(choice ~ day + stereotypy, +random = ~ 1 | bear, data = learning, family = binomial) fm2 - glmmPQL(choice ~ day + envir + stereotypy, +random = ~ 1 | bear, data = learning, family = binomial) Individually, I get results from anova(): anova(fm1) numDF denDF F-value p-value (Intercept) 1 2032 7.95709 0.0048 day 1 2032 213.98391 .0001 stereotypy 1 2032 0.42810 0.5130 anova(fm2) numDF denDF F-value p-value (Intercept) 1 2031 5.70343 0.0170 day 1 2031 213.21673 .0001 envir 1 2031 12.50388 0.0004 stereotypy 1 2031 0.27256 0.6017 I did look through the archives but didn't finding anything relevant to my problem. Hope someone can help. ANDREW _ platform i586-mandrake-linux-gnu arch i586 os linux-gnu system i586, linux-gnu status major2 minor0.0 year 2004 month10 day 04 language R -- Andrew R. Criswell, Ph.D. Graduate School, Bangkok University __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- best, -tony --- A.J. Rossini [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] barplot(2?) with CI from a zero reference line
Dear R Users, (and dear Marc) First of all many thanks for the answers to my previous questions. I would like to barplot the mean percent change of a variate with it's CI. Bars should start from the zero reference line to height (in barplot2). Is there a way to tweak barplot2, for example, to do that ? I have tried to see what the function was but unlike other functions was not able to list it by barplot2. Is it because it is called through UseMethods ? Thanks for any help. Jean-Louis __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] support vector machine
http://www.maths.lth.se/help/R/.R/library/e1071/doc/svmdoc.pdf regards, christian stephenc wrote: Hi Everyone Thanks to those who responded last time. I am still having problems. I really want to find one of those tutorials on how to use svm() so I can then get going using it myself. Issues are which kernel to choose, how to tune the parameters. If anyone know of a tutorial please let me know. Stephen [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error using glm with poisson family and identity link
On Thu, 25 Nov 2004, Spencer Graves wrote: I generally avoid constrained optimizers for three reasons: 1. My experience with them has included many cases where the optimizer would stop with an error when testing parameter values that violate the constraints. If I transform the parameter space to remove the constraints, that never happens. The constrained optimizers in R 2.0.1 may not exhibit this behavior, but I have not checked. They do not, and OTOH if the MLE really does lie on the boundary (and here it may well, with one data point fitted with mean zero) transformation will often not find a good solution. Constrained optimization is a hard problem, good methods are very complex and good code to implement them is usually expensive. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to expand the size limit of a vector?
On Fri, 26 Nov 2004, Guoqi Qian wrote: Hi everybody on this list, Could somebody please tell me how to expand the size limit of a vector in R? You must be using Windows, without telling us. The rw-FAQ tells you the answer, as well as the help page that message refers to. It's better for you that you learn to use these resources than ask 2000 people to read them for you. You may also need to get a bigger computer, since you are trying to create a single object bigger than your computer's RAM memory and a high proportion of the address space of a Windows machine. It might be more sensible to ask your more senior colleagues for advice on your statistical computing and see if you can avoid this computation. In my simulation I get the following error message: x=kronecker(diag(1,100),matrix(1,100,100)) Error: cannot allocate vector of size 781250 Kb In addition: Warning message: Reached total allocation of 511Mb: see help(memory.size) PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html which asks you to read the rw-FAQ before posting. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] barplot(2?) with CI from a zero reference line
I didn't know how to do this but I knew it had to been asked about. Try getS3method(barplot2,default) Make sure you've loaded gplots. I guessed default, but I wonder how you would find out the class if it had been something else. I guess that's something to work on when I'm next twiddling my thumbs. Tom Mulholland -Original Message- From: Jean-Louis Abitbol [mailto:[EMAIL PROTECTED] Sent: Friday, 26 November 2004 2:56 PM To: [EMAIL PROTECTED] Subject: [R] barplot(2?) with CI from a zero reference line Dear R Users, (and dear Marc) First of all many thanks for the answers to my previous questions. I would like to barplot the mean percent change of a variate with it's CI. Bars should start from the zero reference line to height (in barplot2). Is there a way to tweak barplot2, for example, to do that ? I have tried to see what the function was but unlike other functions was not able to list it by barplot2. Is it because it is called through UseMethods ? Thanks for any help. Jean-Louis __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
