Re: [R] (robust) mixed-effects model with covariate
What do you get from the following: max(with(df1, table(Subj, Time)))? With cases like this, lme gives an answer with a bogus distinction between variance components for time and residuals. I don't know about aov or JMP, but I know that varcomp in S-Plus also produces garbage answers in such cases as well. With mixed models, there seem to be many ways to specify models with random effects that are not estimable. Some some standard software (like varcomp in S-Plus or lme) does not (adequately?) test for these situations, with the result that the algorithm sometimes returns answers that are not correct, at least with the distinction between time and residual. I don't know if this applies to your example since it is not self contained. Your code raises other question. For example, what is the class of Time in df1? Might it be treated differently between aov and lme? How many levels does Group have, etc.? Hope this helps. Spencer Graves p.s. If you'd like more help from this listserve, please submit another post that includes commented, minimal, self-contained, reproducible code, as suggested in the posting guide www.R-project.org/posting-guide.html. Giuseppe Pagnoni wrote: Dear Thilo, many thanks for your reply. I realized that there was an error in my formula which should have been: aov(y ~ Group * (Time + Age) + Error (Subj/Time), data=df1) or alternatively: lme(RVP.A ~ Group*(Time+Age), random = ~ 1|Subj/Time,data=df1) but I get different results in each case, and different still from the results of another stat program (JMP). The problem is that I am not sure which one (if one indeed is) correct! Also, in the model you proposed: lme(y~Group*Time, random ~ age | Subj, data = df1) it appears that age is not between the effects of interests, so I do not get an estimate of the significance of the Age or the Age*Group effect. I have Pinheiro Bates, and I read the first chapter but it didn't seem to provide an example analogous to my case. Also, it looks like it would take me some months to study the book thoroughly and frankly that seems a bit excessive for such a (apparently?) simple problem I was hoping somebody would magically provide the correct syntax :-) ! thanks again anyway for your help best regards giuseppe Thilo Kellermann wrote: On Monday 24 July 2006 20:16, Giuseppe Pagnoni wrote: Dear all, First of all I apologize if you received this twice: I was checking the archive and I noticed that the text was scrubbed from the message, probably due to some setting in my e-mail program. I am unsure about how to specify a model in R and I thought of asking some advice to the list. I have two groups (Group= A, B) of subjects, with each subject undertaking a test before and after a certain treatment (Time= pre, post). Additionally, I want to enter the age of the subject as a covariate (the performance on the test is affected by age), and I also want to allow different slopes for the effect of age in the two groups of subjects (age might affect the performance of the two groups differentially). Is the right model to use something like the following? aov (y ~ Group*Time + Group*Age + Error(Subj/Group), data=df1 ) (If I enter that command, within summary, I get the following: Error() model is singular in: aov(y ~ Group * Time + Group * Age + Error(Subj/Group), data = df1)) try: aov(y~Group*Time*Age + Error(Subj*Time*Age), data = df1) which specifies an ANOVA (but not with mixed effects) with three main effects and all interaction terms plus an error term that is independent between groups (!) and relates to within subjects variability. For a real mixed effects analysis you should use the (n)lme function from the nlme package and one possible model could look like this: lme(y~Group*Time, random ~ age | Subj, data = df1) but the exact specification depends on your assumptions, in which it is possible to specify two or three models and compare their fits with anova(). For more information on mixed effects you should consult: Jose C. Pinheiro Douglas M. Bates (2000) Mixed-Effects Models in S and S-PLUS. Springer, New York. Good luck, Thilo As a second question: I have an outlier in one of the two groups. The outlier is not due to a measurement error but simply to the performance of the subject (possibly related to his medical history, but I have no way to determine that with certainty). This subject is signaled to be an outlier within its group: averaging the pre and post values for the performance of the subjects in his group, the Grubbs test yields a probability of 0.002 for the subject to be an outlier (the subject is marked as a significant outlier also if I perform the test separately on the pre and the post data). If I remove this subject from its group, I
[R] run self written functions
Hello,
I'm not sure if I'm in the right place with my question...
I'm running R on Windows and wrote a function and saved it as .R file.
It looks like this:
bmi - function(weight, height) {
bmi - weight / height^2
bmi
}
If I want to use this function, I have to mark everything and then press
Ctrl-R. But then everything single line is executed on the command line,
which means that I will loose my history when the code becomes longer.
Further, I wonder if there is any way to do some output for control
within the function (or any other possibilities to debug in a way).
Maybe, I have chosen a completely wrong way? I only want to make it easy
to create some graphical visualizations of data which will be read in by
csv. files, has to be converted and then displayed depending on some
displaying parameters.
Ciao,
Antje
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[R] From 2.2.1 to 2.3
Hello everyone. Currently I am running R 2.2.1 (windows), and I will like to update to 2.3. I wanted to ask if it is possible to update without having to removed 2.2.1; or do I first need to delete 2.2.1? Thank you very much. Tony [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] spatstat 1.9-4
Version 1.9-4 of package 'spatstat' has been sent to CRAN. It includes new code for perfect simulation of point processes and various improvements. The release notes are available at http://www.spatstat.org/spatstat/current/spatstatRELEASE-NOTES-1.9-4 Adrian Baddeley ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Math elements in panel headers of lattice plots?
Hi, I would like to put a math expression in the header of a panel. Acctually, I need to substitute the 'a' to a script-a in a transcription contained in the factor by which the data set is divided. So, /spak/ should be /spAk/ where A should be a script-a. I guessed that math expressions would be the way to go on this.. Is this possible? /Fredrik -- Give up learning, and put an end to your troubles. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] From 2.2.1 to 2.3
On Wed, 2 Aug 2006, [EMAIL PROTECTED] wrote: Hello everyone. Currently I am running R 2.2.1 (windows), and I will like to update to 2.3. I wanted to ask if it is possible to update without having to removed 2.2.1; or do I first need to delete 2.2.1? There is no such version as `2.3', as the posting guide points out. Updating is discussed in detail the rw-FAQ, to which the posting guide referred you. Thank you very much. Tony [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do! -- 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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] From 2.2.1 to 2.3
http://cran.r-project.org/bin/windows/base/rw-FAQ.html#What_0027s-the-best-way-to-upgrade_003f [EMAIL PROTECTED] wrote: Hello everyone. Currently I am running R 2.2.1 (windows), and I will like to update to 2.3. I wanted to ask if it is possible to update without having to removed 2.2.1; or do I first need to delete 2.2.1? Thank you very much. Tony [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.spss and umlaut
Hello Am Mittwoch, 2. August 2006 17.11 schrieb Thomas Lumley: This sounds like a conflict between encodings -- eg if R is assuming UTF-8 and the file is encoding in Latin-1 then the sequence U+00FC : LATIN SMALL LETTER U WITH DIAERESIS U+0072 : LATIN SMALL LETTER R is coded as FC72 in the file, which is an illegal byte sequence in UTF-8. Hex: 74 65 20 66 fc 72 20 61 6c 6c 65 53 45 2f 31 36 Text: t e f ü r a l l e S E / 1 6 The underlying C code (being written in the US quite a long time ago) doesn't know about encodings, and I don't know what the rules are in SPSS for valid characters (I suspect that in these old portable file formats it probably just reads and writes bytes, leaving it up to the OS to interpret them. But why stopp the C code reading? Is / not the endmark of the string? What is the problem, if I chance that in the source? You could try running R in a non-UTF-8 locale to see if it helps. I think my local is non-UTF-8 (de_CH, isolatin). How can I check that, and set an other temporary? A dirty hack like this: sed s/ä/ae/g | sed s/ö/oe/g | sed s/ü/ue/g | sed s/Ä/Ae/g | sed s/Ö/Oe/g | sed s/Ü/Ue/g didn't work (file 'projets_non_umlaut.por' is not in any supported SPSS format). Thomas If anyone has definitive information about how SPSS represents strings and decides on valid characters that might be useful too. -thomas library(foreign) spssdaten - read.spss(projets.por) attr(spssdaten$PROJETX, value.labels)[1:20] Bg Stammzellenforschung Bb 863 862 Bb Neugestaltung des Finanzausgleichs 861 854 EV Postdienste f Bb 853 852 Bb Bg Steuerpaket 851 843 Bb Anhebung der Mehrwertsteuer s 11. AHV-Revision 842 841 Volkinitiative Lebenslange Verwahrung 833 832 Gegenentwurf zur Avanti EV Lehrstellen-Initiative 831 824 EV Moratorium Plus EV Strom ohne Atom 823 822 EV Ja zu fairen Mieten EV Gleiche Rechte f 821 815 EV GesundheitsinitiativeEV Sonntags-Initiative 814 813 The SPSS-File is okay: system(cat projets.por |grep Postdienste) echtserwerb 3. GenerationSD/N/EV Postdienste für alleSE/16/Änderrung Bg EOG Mut How can I read the SPSS-File with the Umlaut? Bye Thomas Kuster R: 2.1.0 (2005-04-18) OS: Debian Linux, 2.6.10-isgee-neptun-1 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] run self written functions
Hello
Am Donnerstag, 3. August 2006 09.36 schrieb Antje:
Hello,
I'm not sure if I'm in the right place with my question...
I'm running R on Windows and wrote a function and saved it as .R file.
It looks like this:
bmi - function(weight, height) {
bmi - weight / height^2
bmi
}
If I want to use this function, I have to mark everything and then press
Ctrl-R. But then everything single line is executed on the command line,
which means that I will loose my history when the code becomes longer.
Further, I wonder if there is any way to do some output for control
within the function (or any other possibilities to debug in a way).
source(your-r-file.R)
mybmi - bmi(180, 70)
Maybe, I have chosen a completely wrong way? I only want to make it easy
to create some graphical visualizations of data which will be read in by
csv. files, has to be converted and then displayed depending on some
displaying parameters.
look here (directory R):
http://tomix.homelinux.org/~thomas/eth/5_semester/semesterarbeit_WS_2005_2006/
auswertung.R is the main file.
Ciao,
Antje
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Re: [R] Summary method needed?
At 21:04 02.08.2006 +0100, Prof Brian Ripley wrote: On Wed, 2 Aug 2006, Christian Hennig wrote: Thank you Brian! I'm updating my fpc package at the moment and will add some new functions. I learned that there should be print and summary methods for the key functions. for 'classes', I think. Yes. But in some cases the print method will make use of more or less all the output information of the function. Is there any reason to implement a summary method in these cases? Would a more concise print() method be useful? If so the existing print() could become summary(). :-) What I initially did some years ago was to write summary methods to print out the required informations. Then M. Maechler told me that this is not the purpose of a summary method and I should write a print.summary method for this. Now I realise that I actually just want to print, and I don't really need the extra synopsis to be done by summary(). Now is there any recommendation on this? My intuition would be to write a print, but not a summary method. That sounds fine for your purposes. Maybe I am wrong, but as far as I see, print() has the disadvantage that it has to return x and must not return the summarized results as an object. You remember the difficulties with print.survfit. Instead it seems to be allowed that summary does not only print but also return summarized results. Is that right? Greetings, Heinz -- 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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] geodesic distance
Hi,
has anyone ever seen implemented in R the following geodesic
distance between positive definite pxp matrices A and B?
d(A,B) = \sum_{i=1}^p (\log \lambda_i)^2
were \lambda is the solution of det(A -\lambda B) = 0
thanks
stefano
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Re: [R] Summary method needed?
HeinzT == Heinz Tuechler [EMAIL PROTECTED]
on Thu, 03 Aug 2006 09:39:35 +0100 writes:
HeinzT At 21:04 02.08.2006 +0100, Prof Brian Ripley wrote:
On Wed, 2 Aug 2006, Christian Hennig wrote:
Thank you Brian!
I'm updating my fpc package at the moment and will add some new
functions.
I learned that there should be print and summary methods for the key
functions.
for 'classes', I think.
Yes.
But in some cases the print method will make use of
more or less all the output information of the
function. Is there any reason to implement a summary
method in these cases?
Would a more concise print() method be useful? If so
the existing print() could become summary().
:-)
What I initially did some years ago was to write summary
methods to print out the required informations. Then
M. Maechler told me that this is not the purpose of a
summary method and I should write a print.summary method
for this. Now I realise that I actually just want to
print, and I don't really need the extra synopsis to
be done by summary().
Now is there any recommendation on this? My intuition
would be to write a print, but not a summary method.
That sounds fine for your purposes.
HeinzT Maybe I am wrong, but as far as I see, print() has
HeinzT the disadvantage that it has to return x and must
HeinzT not return the summarized results as an object. You
HeinzT remember the difficulties with print.survfit.
HeinzT Instead it seems to be allowed that summary does not
HeinzT only print but also return summarized results.
HeinzT Is that right?
yes, that's right, But let me tell more on the story:
I think I had recommended summary.FOO() and print.summary.FOO()
to Christian because he *did* compute a few `useful' things on
his FOO object. In such a situation, we (R-core) recommend and
usually (very rarely not; for back-compatibility reasons only) implement
the following:
print.FOO: gives a one (sometimes two) paragraph overview of
the fitted model object, and
--yes-- *should* return its *unchanged* argument invisibly.
summary.FOO: computes more interesting things from the original
FOO object, does *NOT print* anything (explicitly)
and returns an object of class summary.FOO.
print.summary.FOO: now prints (an overview of) the summary.FOO
object and -- since it's a print() method -- also
returns its argument unchanged and invisibly.
Examples of the above, can be inspected e.g., for
'FOO' in { lm, glm, aov, nls, loess, princomp, prcomp, .. }
For Joe Average User, of course it *looks* like
summary( FOO ) would print , but that's just because it
returns a summary.FOO object which is auto-printed subsequently
-- unless it's assigned or used in another expression.
My reasons for *not* providing a summary.* and print.summary.*
method would have to be either *both* of 1) and 2) *or* 3) :
1) print(obj) gives enough information
2) the 'obj' already contains the interesting quantities, or
these are either trivially computable from the contents of 'obj'
or are already provided by other FOO methods, e.g., vcov.FOO().
3) lack of time and/or motivation; lazyness
Martin Maechler, ETH Zurich
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[R] NLME: Problem with plotting ranef vs a factor
Hi I am following the model building strategy that is outlined in the Pinheiro and Bates book wrt including covariates but am having a problem with the plot. Basically I am using 4 covariates (1 of them is continuous) and 3 of them are fine but the 4th one is being shown as a scatterplot despite the fact that it is a factor. I have explicitly declared this to be a factor (pcat-as.factor(pcat)) and have also checked by using the is.factor and the levels command that it is a factor. Yet despite this the plot command is not recognising it as a factor. Here is more information about my problem: I am reading in the data by: Data-read.csv(Data1_93_2.csv,header=T) attach(Data) Data1_93-transform(Data,log2game=log2(gamedens+1)) pcat-as.factor(pcat) Data1_93-groupedData(log2game ~ day | subjectno, data=Data1_93) detach(Data) Here is the code to check that the covariate called pcat is indeed a factor: levels(pcat) [1] 1 2 3 is.factor(pcat) [1] TRUE and then after the model is fitted I extract the random effects: D1C2.ran - ranef(mod11.103nlme,augFrame=T) and here is an extract from the object: C R day gamedens pcat site mutcat1 pdens0 log2game NA02_259 -1.016987007 0.0162825099 15.75000 23.51 Namaacha Mixed 15018 3.761099 NA02_073 -0.939355374 0.0132589702 10.5 23.750001 Namaacha Resistant6170 3.675543 M00_12 -0.775048474 0.0047124742 10.5 25.01 Mpumulanga Sensitive 17525 3.768326 M00_93 -0.555801118 0.0053872868 14.0 37.52 Mpumulanga Sensitive 332000 4.254319 NA02_053 -0.327990343 -0.0037659864 6.0 39.250001 Namaacha Resistant 65529 4.292481 Note that this output also seems to indicate that pcat is a factor as it is summarised correctly. I then generate plots for my random effects: plot(D1C2.ran,form= C ~site+mutcat2+pcat+pdens0) and the problem is that the panel for my random effects vs pcat is displayed as a scatterplot rather than as a boxplot. I am getting told to check warnings and these warnings look like: Warning messages: 1: at 0.99 2: radius 0.0001 3: all data on boundary of neighborhood. make span bigger 4: pseudoinverse used at 0.99 5: neighborhood radius 0.01 6: reciprocal condition number -1.#IND 7: zero-width neighborhood. make span bigger I do not get these warnings if I exclude the problematic variable pcat so must be something to do with this. Any ideas? Many thanks Greg [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about dll crashing R
I have ported some R code to C to make it faster.
I can perform .Call(foobar,) once and it
works fine. Absolutely correct answer.
If I put a loop inside foobar and run the main code
routine more than 100 times, it crashes R.
Or if I call .Call(foobar) seperately more than
two tims it crashes R.
For the most part I am doing matirx multiplies using
EXP matrixprod (SEXP x , SEXP y )
{ int nrx , ncx , nry , ncy , mode;
SEXP xdims , ydims , ans ;
char *transa = N , *transb = N ;
double one = 1.0 , zero = 0.0 ;
xdims = getAttrib (x , R_DimSymbol ) ;
ydims = getAttrib (y , R_DimSymbol ) ;
mode = REALSXP;
nrx = INTEGER( xdims ) [ 0 ] ;
ncx = INTEGER( xdims ) [ 1 ] ;
nry = INTEGER( ydims ) [ 0 ] ;
ncy = INTEGER( ydims ) [ 1 ] ;
PROTECT( ans = allocMatrix (mode, nrx , ncy ) ) ;
F77_CALL(dgemm) ( transa , transb , nrx , ncy , ncx
, one ,
REAL( x ) , nrx , REAL( y ) , nry , zero ,
REAL( ans ) , nrx ) ;
UNPROTECT( 1 ) ;
return( ans ) ;
}
I am also generating random multiavriate normals using
the (not pretty) code
*
1) Generate P independent standard normal deviates -
Ei ~ N(0,1)
2) Using Cholesky decomposition find A s.t.
trans(A)*A = COVM
3) trans(A)E + MEANV ~ N(MEANV,COVM)
*/
SEXP mvntest (SEXP mean, SEXP cov, SEXP temp)
{ int nrx , ncx , nry , ncy ,info,mode;
SEXP xdims , ydims , ans;
int i,j, one=1;
info = 1;
xdims = getAttrib (mean , R_DimSymbol ) ;
ydims = getAttrib (cov , R_DimSymbol ) ;
mode = REALSXP;
nrx = INTEGER( xdims ) [ 0 ] ;
ncx = INTEGER( xdims ) [ 1 ] ;
nry = INTEGER( ydims ) [ 0 ] ;
ncy = INTEGER( ydims ) [ 1 ] ;
/* create the upper trianglular matrix A */
/* such that t(A) %*% A = Sigma */
GetRNGstate();
F77_CALL(dpofa) ( REAL( cov ), nry , ncy , info);
Rprintf(Info = %d\n,info);
for(i=0;inry;i++)
for(j=0;ji;j++)
REAL(cov)[i+j*ncy] = 0.0;
PROTECT( ans = allocMatrix (mode, nrx , one ) ) ;
for(i=0;inry;i++)
REAL(temp)[i] = rnorm(0,1);
ans = tmatrixprod(cov,temp);
for(i=0;inry;i++)
REAL(ans)[i] = REAL(ans)[i]+REAL(mean)[i];
UNPROTECT( 1 ) ;
PutRNGstate();
return( ans ) ;
}
I have a feeling I am messing up memory usage
somewhere but haven't a clue. Do I need to do garbage
collecting inside the C program ?The fact that the
code
runs a few times before R crashes is driving me nuts.
I send most of what I need into the C routine from R,
so I am not creating that many SEXP objects within the
program.
Any hints or ideas ?
Thanks!
Benn
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Re: [R] best way to calculate per-parameter differences in across-subject means
Thanks, I'll look at that.
In the meantime, the code below is what I came up with myself. It does
what I want
# SelectCases(dat,crit) == subset(dat, crit, drop=FALSE)
SENSICK.AvScores- function( dat=SENSICK.items.tr )
{
n-nlevels(dat$Symptom)
data.frame(
Patient=c( rep(1,n), rep(0, 3*n)),
WasSick=c( rep(1,2*n), rep(NA,2*n)),
StrictSick=c( rep(NA,2*n), rep(-1,n), rep(1,n)),
Symptom=rep(levels(dat$Symptom),4),
AvScore=c(
with( SelectCases(dat, 'Patient==1 WasSick==1'),
tapply(Score,
Symptom , mean) ),
with( SelectCases(dat, 'Patient==0 WasSick==1'),
tapply(Score,
Symptom , mean) ),
with( SelectCases(dat, 'Patient==0 StrictSick==-1'),
tapply(Score, Symptom , mean) ),
with( SelectCases(dat, 'Patient==0 StrictSick==1'),
tapply(Score,
Symptom , mean) )
)
)
}
AvScores-SENSICK.AvScores
with( AvScores, (barchart( AvScore[Patient==1] - AvScore[Patient==0
WasSick==1]) ~ Symptom, scales=list( rot=c(45,0)) )
-- Dieter wrote:
Maybe it's a bit more than you want, but possibly you are happy with it: see the
example under TukeyHSD.
summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks))
TukeyHSD(fm1, tension, ordered = TRUE)
plot(TukeyHSD(fm1, tension))
Dieter
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Re: [R] mult comp significance
I read two questions: physical meaning and how to display only significant differences. PHYSICAL MEANING A p value is the probability of obtaining by chance alone a result at least as extreme as what we observe. A simultaneous or adjusted p value is the probability that at least one of the multiple tests was at least as extreme as what observed. A (1-alpha) confidence interval or region is a random set that includes the true but unknown value with probability at least (1-alpha). A set of (1-alpha) simultaneous confidence intervals is a random set in the multidimensional space that includes the true but unknown values of all parameter comparisons with probability at least (1-alpha). SELECTING ONLY SIGNIFICANT DIFFERENCES Consider the following modification of the first example in the simint help page: library(multcomp) data(recovery) RecInts - simint(minutes~blanket, data=recovery, type=Dunnett, conf.level=0.9, alternative=less,eps=0.0001) To see the structure of RecInts, look at str(RecInts). This indicates that RecInts is a list, and one of its components is a named vector called p.value.adj. This suggests we try the following: with(RecInts, p.value.adj[p.value.adj0.05]) blanketb2-blanketb0 5.484862e-05 Hope this helps. Spencer Graves Nair, Murlidharan T wrote: This has a stats question and a R question. I am sure there are many core statisticians here how would know the answer to this simple question. In determining the significant comparisons using the methods in multcomp, the ones that are designated as significant are the ones that do not intersect the zero line. What is the physical meaning of this and why are those considered significant? I can sort those out and pick out by their adjusted pvalues to pick the top ones correct? Is there an method the multcomp that will output only the significant ones? Thanks ../Murli [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tcltk package
On Tuesday 01 August 2006 19:24, John McHenry wrote: [...] Yes, I built R myself. I couldn't find a debian package for R 2.3.1. The latest available is 2.2.1. Oh, but there is... right on CRAN. For Dapper just add this line to your sources.list: deb http://cran.R-project.org/bin/linux/ubuntu/ dapper/ This repository has lots of other packages compiled for Ubuntu, feel free to take a look. HTH, Adrian -- Adrian DUSA Romanian Social Data Archive 1, Schitu Magureanu Bd 050025 Bucharest sector 5 Romania Tel./Fax: +40 21 3126618 \ +40 21 3120210 / int.101 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Default first argument in assignment function possible?
Dear All,
is there a possibility to provide a default for the first argument in an
assignment function?
I could not find out if and how. You see in the example below that not
explicitly stating the first argument leads to errors.
Thanks,
Heinz Tüchler
### example of assignment function
'testfun-' - function(x=a, y=b, value)
{ x - x+ y+ value
print(x)
x }
a - 0; b - 1; c - 2
testfun(c, 2) - 2 # result: 6
c - 2
testfun(c ) - 2 # result: 5
testfun( , 2) - 2 # Error: argument is missing, with no default
testfun(y=2) - 2 # Error: target of assignment expands to non-language
object
testfun() - 2 # Error: invalid (NULL) left side of assignment
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status Patched
major 2
minor 3.1
year 2006
month 07
day23
svn rev38687
language R
version.string Version 2.3.1 Patched (2006-07-23 r38687)
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Re: [R] NLME: Problem with plotting ranef vs a factor
Greg, be careful using attach() and detach(). From the syntax snippets you showed it seems that you did create an object pcat (factor variable), but you did not change the respective variable in your data frame. Try to remove pcat and see what happens do the results of lme()! Dirk __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] random effects with lmer() and lme(), three random factors
In theory, 'lme' may be able to handle crossed random effects, but I don't know how to do it. I've used 'lmer' for such models, and your 'lmer' code looks plausible to me. Therefore, I would be inclined to believe the 'lmer' results. To remove any lingering doubt, I suggest you construct Monte Carlo simulations where you know the structure. Then feed the simulations to 'lme' and 'lmer' and see what you get. Hope this helps. Spencer Graves Xianqun (Wilson) Wang wrote: Hi, all, I have a question about random effects model. I am dealing with a three-factor experiment dataset. The response variable y is modeled against three factors: Samples, Operators, and Runs. The experimental design is as follow: 4 samples were randomly chosen from a large pool of test samples. Each of the 4 samples was analyzed by 4 operators, randomly selected from a group of operators. Each operator independently analyzed same samples over 5 runs (runs nested in operator). I would like to know the following things: (1) the standard deviation within each run; (2) the standard deviation between runs; (3) the standard deviation within operator (4) the standard deviation between operator. With this data, I assumed the three factors are all random effects. So the model I am looking for is Model: y = Samples(random) + Operator(random) + Operator:Run(random) + Error(Operator) + Error(Operator:Run) + Residuals I am using lme function in nlme package. Here is the R code I have 1. using lme: First I created a grouped data using gx - groupedData(y ~ 1 | Sample, data=x) gx$dummy - factor(rep(1,nrow(gx))) then I run the lme fm- lme(y ~ 1, data=gx, random=list(dummy=pdBlocked(list(pdIdent(~Sample-1), pdIdent(~Operator-1), pdIdent(~Operator:Run-1) finally, I use VarCorr to extract the variance components vc - VarCorr(fm) Variance StdDev Operator:Run 1.595713e-10:20 1.263215e-05:20 Sample 5.035235e+00: 4 2.243933e+00: 4 Operator 5.483145e-04: 4 2.341612e-02: 4 Residuals8.543601e-02: 1 2.922944e-01: 1 2.Using lmer in Matrix package: fm - lmer(y ~ (1 | Sample) + (1 | Operator) + (1|Operator:Run), data=x) summary(fm) Linear mixed-effects model fit by REML Formula: H.I.Index ~ (1 | Sample.Name) + (1 | Operator) + (1 | Operator:Run) Data: x AIC BIClogLik MLdeviance REMLdeviance 96.73522 109.0108 -44.36761 90.80064 88.73522 Random effects: Groups NameVariance Std.Dev. Operator:Run (Intercept) 4.2718e-11 6.5359e-06 Operator (Intercept) 5.4821e-04 2.3414e-02 Sample (Intercept) 5.0352e+00 2.2439e+00 Residual 8.5436e-02 2.9229e-01 number of obs: 159, groups: Operator:Run, 20; Operator, 4; Sample.Name, 4 Fixed effects: Estimate Std. Error t value (Intercept) 0.0020818 1.1222683 0.001855 There is a difference between lmer and lme is for the factor Operator:Run. I cannot find where the problem is. Could anyone point me out if my model specification is correct for the problem I am dealing with. I am pretty new user to lme and lmer. Thanks for your help! Wilson Wang [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] tcl/tk bind destroy event
Perhaps Destroy key is unknown by Tcl; it is not in the
Event modifiers table in Welch Book...
But try with Control-L or Shift-Control_L, it runs
(but not with Control_L-Shift ??).
Use xmodmap to see the current mappings from keys to modifiers.
Jean Coursol
On Wed, 2 Aug 2006, Franco Mendolia wrote:
Hello!
I want to create a messagebox whenever the user wants to destroy the window
(e.g. Alt-F4 or the 'x' in the right top corner) and ask if a modified file
should be saved or not.
If 'cancel' is chosen then nothing should happen and the windows still should
be
existing. This doesn't work. When I press cancel the window will be destroyed
although.
I also implemented a menu item 'Quit' where I show the same messagebox and
there
it works fine.
How can I make it work or is there another method to do this? I'm very new to
R
and tcl/tk.
Here is part of my code:
exitProg - function()
{
returnVal - tkmessageBox(title=Question,
message=Save modified file?,
icon=question, type=yesnocancel, default=yes)
returnVal - as.character(returnVal)
if( returnVal == yes )
{
# save file
value - saveFile()
# destroy window when save was successfull
if( value == 1 )
tkdestroy(mw)
}
if( returnVal == no )
{
tkdestroy(mw)
}
if( returnVal == cancel )
{
# do nothing
cat(Cancel was pressed.\n)
}
}
# bind the destroy event in order to show a message box
tkbind(mw,Destroy,exitProg)
# menu item which works fine
tkadd(fileMenu, command, label=Quit, command=exitProg)
Thank you.
Franco Mendolia
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[R] between-within anova: aov and lme
I have 2 questions on ANOVA with 1 between subjects factor and 2 within factors. 1. I am confused on how to do the analysis with aov because I have seen two examples on the web with different solutions. a) Jon Baron (http://www.psych.upenn.edu/~baron/rpsych/rpsych.html) does 6.8.5 Example 5: Stevens pp. 468 - 474 (one between, two within) between: gp within: drug, dose aov(effect ~ gp * drug * dose + Error(subj/(dose*drug)), data=Ela.uni) b) Bill Venables answered a question on R help as follows. - factor A between subjects - factors B*C within subjects. aov(response ~ A*B*C + Error(subject), Kirk) An alternative formula would be response ~ A/(B*C) + Error(subject), which would only change things by grouping together some of the sums of squares. --- SO: which should I do aov(response ~ A*B*C + Error(subject), Kirk) aov(response ~ A/(B*C) + Error(subject), Kirk) aov(response ~ A*B*C + Error(subject/(B*C)), Kirk) 2. How would I do the analysis in lme()? Something like lme(response~A*B*C,random=~1|subject/(B*C))??? Thanks very much for any help! Bill Simpson __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] tcl/tk bind destroy event
Hi!
Perhaps Destroy key is unknown by Tcl; it is not in the
Event modifiers table in Welch Book...
I think the Destroy key is known, because when destroying the window with
Alt-F4
or the littel x in the topcorner my function exitProg is executed.
What I actually search for is a possibility to let the window not be destroyed
when user closes the window by Alt-F4 or the little x.
Franco Mendolia
exitProg - function()
{
returnVal - tkmessageBox(title=Question,
message=Save modified file?,
icon=question, type=yesnocancel, default=yes)
returnVal - as.character(returnVal)
if( returnVal == yes )
{
# save file
value - saveFile()
# destroy window when save was successfull
if( value == 1 )
tkdestroy(mw)
}
if( returnVal == no )
{
tkdestroy(mw)
}
if( returnVal == cancel )
{
# do nothing
cat(Cancel was pressed.\n)
}
}
# bind the destroy event in order to show a message box
tkbind(mw,Destroy,exitProg)
# menu item which works fine
tkadd(fileMenu, command, label=Quit, command=exitProg)
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[R] Invoking R on UNIX Command line from a PERL CGI
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Re: [R] tcl/tk bind destroy event
Jean Coursol [EMAIL PROTECTED] writes:
Perhaps Destroy key is unknown by Tcl; it is not in the
Event modifiers table in Welch Book...
But try with Control-L or Shift-Control_L, it runs
(but not with Control_L-Shift ??).
I don't think that is it. As I read it, Tk catches the Destroy event
alright, but the problem is that it is too late: the window
destruction is already in motion at the time and there's no way to
undestroy it from within binding. I think you need to create a wm
protocol handler for WM_DELETE_WINDOW. As in
x - tktoplevel()
tcl(wm, protocol, x, WM_DELETE_WINDOW, quote(cat(I'm staying!\n)))
--- now try to destroy the window using window controls ---
tcl(wm, protocol, x, WM_DELETE_WINDOW, )
--- things should be back to normal now ---
Use xmodmap to see the current mappings from keys to modifiers.
Jean Coursol
On Wed, 2 Aug 2006, Franco Mendolia wrote:
Hello!
I want to create a messagebox whenever the user wants to destroy the window
(e.g. Alt-F4 or the 'x' in the right top corner) and ask if a modified
file
should be saved or not.
If 'cancel' is chosen then nothing should happen and the windows still
should be
existing. This doesn't work. When I press cancel the window will be
destroyed
although.
I also implemented a menu item 'Quit' where I show the same messagebox and
there
it works fine.
How can I make it work or is there another method to do this? I'm very new
to R
and tcl/tk.
Here is part of my code:
exitProg - function()
{
returnVal - tkmessageBox(title=Question,
message=Save modified file?,
icon=question, type=yesnocancel, default=yes)
returnVal - as.character(returnVal)
if( returnVal == yes )
{
# save file
value - saveFile()
# destroy window when save was successfull
if( value == 1 )
tkdestroy(mw)
}
if( returnVal == no )
{
tkdestroy(mw)
}
if( returnVal == cancel )
{
# do nothing
cat(Cancel was pressed.\n)
}
}
# bind the destroy event in order to show a message box
tkbind(mw,Destroy,exitProg)
# menu item which works fine
tkadd(fileMenu, command, label=Quit, command=exitProg)
Thank you.
Franco Mendolia
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and provide commented, minimal, self-contained, reproducible code.
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] tcl/tk bind destroy event
I don't think that is it. As I read it, Tk catches the Destroy event alright, but the problem is that it is too late: the window destruction is already in motion at the time and there's no way to undestroy it from within binding. I think you need to create a wm protocol handler for WM_DELETE_WINDOW. As in x - tktoplevel() tcl(wm, protocol, x, WM_DELETE_WINDOW, quote(cat(I'm staying!\n))) --- now try to destroy the window using window controls --- tcl(wm, protocol, x, WM_DELETE_WINDOW, ) --- things should be back to normal now --- Yeah! That's it! Thanks! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with factor in list (same name for diffrent category)
Hello I try to ignore the problems with the umlaut. I read the file in with: library(foreign) daten - read.spss(filename) Then I want select a Vorlage (vote), but if I want select for example [6] and not 6 and 7 how can I select it? The real name is a other problem, but I can solve this via date. levels(daten$PROJETX) [1] Bg Stammzellenforschung [2] Bb [3] Bb Neugestaltung des Finanzausgleichs [4] [5] EV Postdienste f [6] Bb [7] Bb [8] Bg Steuerpaket [9] Bb Anhebung der Mehrwertsteuer s [10] 11. AHV-Revision [11] Volkinitiative Lebenslange Verwahrung [12] . . . Select with: pos - daten$PROJETX==11. AHV-Revision summary(pos) Mode FALSETRUE logical 2003731002 This is okay. pos - daten$PROJETX==(levels(daten$PROJETX)[6]) summary(pos) Mode FALSETRUE logical 1949046471 Too many TRUE's. Thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vector DF [a, b] to replace values :was Finding the position of a variable in a data.frame
--- jim holtman [EMAIL PROTECTED] wrote: ?which which(Df = 50, arr.ind=T) row col 5 5 4 This works very nicely as has some other suggestions on how to replace a value. Assuming that I have more than one correction to make where Df = 50, can I use vectors in the Df[] to do this. My attempt shows that I can use vectors but it appears thatthere is something wrong with my logic Eample cat - c( 3,5,6,8,0) dog - c(3,5,3,6, 0) rat - c (5, 5, 4, 9, 51) bat - c( 12, 42, 45, 32, 54) Df - data.frame(cbind(cat, dog, rat, bat)) post - which(Df = 50, arr.ind=T) # find values .= 50 post correction - c(77, 88) # new correct values row - post[ ,1] # vector for row number col - post[ ,2] # vector for column number Df[row,col] - correction Df ---result- cat dog rat bat 1 3 3 5 12 2 5 5 5 42 3 6 3 4 45 4 8 6 9 32 5 0 0 88 88 I am replacing both instances with 88 which is not correct Thanks On 8/2/06, John Kane [EMAIL PROTECTED] wrote: Simple problem but I don't see the answer. I'm trying to clean up some data I have 120 columns in a data.frame. I have one value in a column named blaw that I want to change. How do I find the coordinates. I can find the row by doing a subset on the data.frame but how do I find out here blaw is in columns without manually counting them or converting names(Df) to a list and reading down the list. Simple example cat - c( 3,5,6,8,0) dog - c(3,5,3,6, 0) rat - c (5, 5, 4, 9, 0) bat - c( 12, 42, 45, 32, 54) Df - data.frame(cbind(cat, dog, rat, bat)) Df subset(Df, bat = 50) results cat dog rat bat 5 0 0 0 54 Thus I know that my target is in row 5 but how do I figure out where 'bat' is? All I want to do is be able to say Df[5,4] - 100 Is there some way to have function(bat) return the column number: some kind of a colnum() function? I had thought that I had found somthing in library(gdata) matchcols but no luck. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to use the EV AND condEV from BMA's results?
Dear friends, In R, the help of bic.glm tells the difference between postmean(the posterior mean of each coefficient from model averaging) and condpostmean(the posterior mean of each coefficient conditional on the variable being included in the model), But it's still unclear about the results explanations, and the artile of Rnews in 2005 on BMA still don't give more detail on it. Suppose my results of logistic regression analyzed by bic.glm (BMA) as follows:(dataset is birthwt(MASS) and i include the interaction) p!=0 EV SD condEV cond SDmodel 1 model 2 model 3 model 4model 5 Intercept 100 0.1841 1.2204 0.1841.2201.017 1.175-0.853-1.057 0.532 age17.8 -0.0113 0.0285 -0.0630.036 . . . . -0.071 lwt 50.0 -0.0079 0.0093 -0.016 0.007 -0.017- 0.017 . . . smokeTRUE 9.5 0.0469 0.1798 0.4960.345 . . . . . ptdTRUE 99.41.5161 0.4751 1.526 0.461 1.407 1.596 1.732 1.463 1.608 htTRUE54.4 0.9477 1.02691.742 0.7441.894 1.930 . . . uiTRUE13.30.0976 0.2987 0.7310.453 . . . . . ftv 12.3 .1-0.0257 0.5117 -0.209 2.438. . -0.867 . . .2+0.7470 2.1277 6.0813.371. .6.024 . . age.ftv1 33.7 -0.0136 0.0278 -0.0400.035 . - 0.036 . . . age.ftv2. 15.9 -0.0340 0.0950 -0.2140.135 . . -0.271 . . smokeTRUE.uiTRUE 2.4 0.0103 0.12090.422 0.652. . . . . nVar3 4 3 1 2 post prob 0.117 0.086 0.083 0.061 0.044 1. how should I write my final logistic model? 2. Which parameter estimation should be used, condEV OR EV? How should I use the two different parameter estimations correctly? Thanks for your precious time! -- Kind Regards, Zhi Jie,Zhang [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vector DF [a, b] to replace values :was Finding the position of a variable in a data.frame
John Kane [EMAIL PROTECTED] writes: --- jim holtman [EMAIL PROTECTED] wrote: ?which which(Df = 50, arr.ind=T) row col 5 5 4 This works very nicely as has some other suggestions on how to replace a value. Assuming that I have more than one correction to make where Df = 50, can I use vectors in the Df[] to do this. My attempt shows that I can use vectors but it appears thatthere is something wrong with my logic Eample cat - c( 3,5,6,8,0) dog - c(3,5,3,6, 0) rat - c (5, 5, 4, 9, 51) bat - c( 12, 42, 45, 32, 54) Df - data.frame(cbind(cat, dog, rat, bat)) post - which(Df = 50, arr.ind=T) # find values .= 50 post correction - c(77, 88) # new correct values row - post[ ,1] # vector for row number col - post[ ,2] # vector for column number Df[row,col] - correction Df ---result- cat dog rat bat 1 3 3 5 12 2 5 5 5 42 3 6 3 4 45 4 8 6 9 32 5 0 0 88 88 I am replacing both instances with 88 which is not correct You're being bitten by two issues here: One is that Df[row,col] does not vectorize in parallel in the two indices: Df[1:2,1:2] has *four* elements, not two. Another is that data frames are not matrices: you can have different types of values in different columns, so you cannot expect to pick an arbitrary set of elements and assign a vector into it (well, it doesn't work, anyway). This sort of stuff works much easier with matrices, where we also have the wonderful feature of indexing with a matrix: M - cbind(cat, dog, rat, bat) M[post]- correction M cat dog rat bat [1,] 3 3 5 12 [2,] 5 5 5 42 [3,] 6 3 4 45 [4,] 8 6 9 32 [5,] 0 0 77 88 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] bringToTop without focus?
Hi all who know R on Windows, Quick question: Is there a way to do bringToTop(stay=TRUE) without giving focus? I would like to pop a graph window but I would like to preserve focus in the window which I was in before. Thanks for any lead, Chris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bringToTop without focus?
On 8/3/2006 9:13 AM, [EMAIL PROTECTED] wrote:
Hi all who know R on Windows,
Quick question: Is there a way to do bringToTop(stay=TRUE) without giving
focus? I would like to pop a graph window but I would like to preserve
focus in the window which I was in before.
No, there's no way to do that. The closest is
{ bringToTop(stay=TRUE); bringToTop(-1) }
but that only works if the previous window was the console.
Duncan Murdoch
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[R] Looking for transformation to overcome heterogeneity of variances
Dear All My data consists in 96 groups, each one with 10 observations. Levene's test suggests that the variances are not equal, and therefore I have tried to apply the classical transformations to have homocedasticity in order to be able to use ANOVA. Unfortunately, no transformation that I have used transforms my data into data with homocedasticity. The histogram of variances is at http://phhs80.googlepages.com/hist1.png Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.spss and umlaut
On Thu, 3 Aug 2006, Thomas Kuster wrote: Hello Am Mittwoch, 2. August 2006 17.11 schrieb Thomas Lumley: This sounds like a conflict between encodings -- eg if R is assuming UTF-8 and the file is encoding in Latin-1 then the sequence U+00FC : LATIN SMALL LETTER U WITH DIAERESIS U+0072 : LATIN SMALL LETTER R is coded as FC72 in the file, which is an illegal byte sequence in UTF-8. Hex: 74 65 20 66 fc 72 20 61 6c 6c 65 53 45 2f 31 36 Text: t e f ? r a l l e S E / 1 6 Ok, so that looks like Latin-1 encoding in the file The underlying C code (being written in the US quite a long time ago) doesn't know about encodings, and I don't know what the rules are in SPSS for valid characters (I suspect that in these old portable file formats it probably just reads and writes bytes, leaving it up to the OS to interpret them. But why stopp the C code reading? Is / not the endmark of the string? What is the problem, if I chance that in the source? You haven't shown anything that indicates that the C code stopped reading. More likely R just stops displaying when it gets to an illegal byte sequence. You could use nchar() to count the bytes in the string to find out. You could try running R in a non-UTF-8 locale to see if it helps. I think my local is non-UTF-8 (de_CH, isolatin). How can I check that, and set an other temporary? You can use charToRaw() to see what R thinks the byte sequence is for a word with a u-umlaut. Sys.setlocale() will let you change the locale, but your locale does look non-UTF-8. This is all guesswork since we can't see the file. -thomas__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for transformation to overcome heterogeneity of variances
Paul Smith [EMAIL PROTECTED] writes: Dear All My data consists in 96 groups, each one with 10 observations. Levene's test suggests that the variances are not equal, and therefore I have tried to apply the classical transformations to have homocedasticity in order to be able to use ANOVA. Unfortunately, no transformation that I have used transforms my data into data with homocedasticity. The histogram of variances is at http://phhs80.googlepages.com/hist1.png Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? Not based on that information. Try the following instead: fit - lm(y~g) par(mfrow=c(2,2)); plot(fit) -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] get() in sapply() in with()
Dear All,
applying some function within a with() function I wanted to use also
sapply() and get() to form a data.frame, but did not succede.
Below is a simplified example.
It is possible to use sapply() within a with() function, it is also
possible to use get() within a with() function, but when I try to use get
within sapply within with I arrive at Error in get(x, envir, mode,
inherits) : variable v5 was not found.
Is there a solution?
Thanks,
Heinz Tüchler
## example
df1 - data.frame(v5=16:20, v6=21:25, v7=I(letters[16:20]), v8=letters[16:20])
with(df1, sapply(c('v5', 'v6'), get) ) ## Error, see next line
## Error in get(x, envir, mode, inherits) : variable v5 was not found
with(df1, sapply(list(v5, v6), mean) ) # does work
with(df1, get('v5') ) # does work
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status Patched
major 2
minor 3.1
year 2006
month 07
day23
svn rev38687
language R
version.string Version 2.3.1 Patched (2006-07-23 r38687)
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Re: [R] Looking for transformation to overcome heterogeneity of variances
On 03 Aug 2006 15:45:10 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote: My data consists in 96 groups, each one with 10 observations. Levene's test suggests that the variances are not equal, and therefore I have tried to apply the classical transformations to have homocedasticity in order to be able to use ANOVA. Unfortunately, no transformation that I have used transforms my data into data with homocedasticity. The histogram of variances is at http://phhs80.googlepages.com/hist1.png Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? Not based on that information. Try the following instead: fit - lm(y~g) par(mfrow=c(2,2)); plot(fit) Thanks, Peter. By 'g', you mean factor1* factor2*factor3*factor4 ? Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] efficient way to make NAs of empty cells in a factor (or character)
Dear all, I have some csv-files (originating from Excel-files) containing empty cells. In my example file I have four variables of different classes, each with some empty cells in the original csv-file: test - read.csv2(test.csv, dec=.) test id id2 x y 1 a 1 NA 2 b e NA 2.2 3 f 3 3.3 4 c g 4 4.4 class(test$id) [1] factor class(test$id2) [1] factor class(test$x) [1] integer class(test$y) [1] numeric In the help text of read.csv2 you can read 'Blank fields are also considered to be missing values in logical, integer, numeric and complex fields.'. Thus, empty cells in a factor (or a character I assume) is not considered as missing values but an own level: is.na(test$id) [1] FALSE FALSE FALSE FALSE levels(test$id) [1] a b c When I work with my real (larger) dataset I would like to use functions like 'is.na' and '!is.na' on factors. Now I wonder if there is an R alternativ to do 'search (for empty cells) and replace (with NA)' in Excel? I have tried a modification of Uwe Ligges suggestion on missing value posted 2 Aug: is.na(test[test==]) - TRUE ...but it did not work on the data set: Error in [-.data.frame(`*tmp*`, test == , value = c(NA, NA, NA, NA : rhs is the wrong length for indexing by a logical matrix However it worked fine when applied to a single vector: is.na(test$id[test$id==]) - TRUE test$id [1] abNA c Levels: a b c is.na(test$id) [1] FALSE FALSE TRUE FALSE Is there a more efficient way to fill empty cells in all my factors in R or should I just do it in advance in Excel by 'search and replace'? Thanks in advance! -- Henrik Pärn Department of Biology NTNU 7491 Trondheim Norway +47 735 96282 (office) +47 909 89 255 (mobile) +47 735 96100 (fax) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bringToTop without focus?
Thanks for fast response.
'Had another look at the problem and found a partial but nice solution which
will solve my present problem and might be useful to others.
Here is the context:
In fact I have all my data and stuff in Excel and I use R(D)COM with Rexcel to
communicate with R and do some of the work there. One of the things I do is to
create a graph in R and to display it. To make sure it stays visible, I use
bringToTop(stay=TRUE) from R. But then I would like to keep working in Excel
without having to re-activate the workbook window. This is where the problem
originated. Within excel there is a somewhat crude solution:
Submit the code which makes the graph via a VBA procedure (macro) and end it
with SendKeys %{TAB}. Here is an example:
Sub Refilter
Range(Alldata).AdvancedFilter Action:=xlFilterCopy, CriteriaRange:=Range( _
Criteria), CopyToRange:=Range(Extractrange), Unique:=False
SendCommands Range(DiagPlot)
SendKeys %{TAB}
End sub
Here DiagPlot is a range which contains the code to construct the plot such
as plot(x,y).
The VBA macro should be run when the worksheet of interest has focus. The call
SendKeys %{TAB} simulates pressing ALT TAB which goes back to the previous
window on the stack. This was the one which had focus when bringToTop. There we
go.
Have a nice day,
Chris.
Morale: if impossible / change the point of view
-Original Message-
From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
Sent: Thursday, 03 August, 2006 3:18 PM
To: Ritter, Christian C GSMCIL-GSTMS/2
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] bringToTop without focus?
On 8/3/2006 9:13 AM, [EMAIL PROTECTED] wrote:
Hi all who know R on Windows,
Quick question: Is there a way to do bringToTop(stay=TRUE) without giving
focus? I would like to pop a graph window but I would like to preserve
focus in the window which I was in before.
No, there's no way to do that. The closest is
{ bringToTop(stay=TRUE); bringToTop(-1) }
but that only works if the previous window was the console.
Duncan Murdoch
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and provide commented, minimal, self-contained, reproducible code.
[R] Vectorizing a for loop
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
Thanks!
-- Dan Gerlanc
Williams College
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and provide commented, minimal, self-contained, reproducible code.
[R] questions on plotting dedrograms
Hi, i've two questions concerning the plot of a dendrogram. first, i use hclust for clustering and if i plot the dendrogram, then the maximal height is the maximal dissimilarity found in my data. but i want to have a arbitary maximal height. for example if the maximal dissimilarity in my data is 50 and i want a height of 100, the plot should be compressed by 1/2 and the line to the root node should be as long as the rest of the plot. How can i do that ? (i want to use it for comparing 2 clusterings with the same maximal possible dissimilarity) The other question is, how can i plot the dendrogram horizontal, but with the root node to the right ? thx for your efford alexander scheidler __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] questions on plotting dedrograms
Hi, i've two questions concerning the plot of a dendrogram. first, i use hclust for clustering and if i plot the dendrogram, then the maximal height is the maximal dissimilarity found in my data. but i want to have a arbitary maximal height. for example if the maximal dissimilarity in my data is 50 and i want a height of 100, the plot should be compressed by 1/2 and the line to the root node should be as long as the rest of the plot. How can i do that ? (i want to use it for comparing 2 clusterings with the same maximal possible dissimilarity) The other question is, how can i plot the dendrogram horizontal, but with the root node to the right ? thx for your efford alexander scheidler __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient way to make NAs of empty cells in a factor (orcharacter)
try to use the 'na.strings' argument of read.csv(), e.g., test - read.csv(test.csv, na.strings = ) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Henrik Parn [EMAIL PROTECTED] To: R-help r-help@stat.math.ethz.ch Sent: Thursday, August 03, 2006 3:46 PM Subject: [R] efficient way to make NAs of empty cells in a factor (orcharacter) Dear all, I have some csv-files (originating from Excel-files) containing empty cells. In my example file I have four variables of different classes, each with some empty cells in the original csv-file: test - read.csv2(test.csv, dec=.) test id id2 x y 1 a 1 NA 2 b e NA 2.2 3 f 3 3.3 4 c g 4 4.4 class(test$id) [1] factor class(test$id2) [1] factor class(test$x) [1] integer class(test$y) [1] numeric In the help text of read.csv2 you can read 'Blank fields are also considered to be missing values in logical, integer, numeric and complex fields.'. Thus, empty cells in a factor (or a character I assume) is not considered as missing values but an own level: is.na(test$id) [1] FALSE FALSE FALSE FALSE levels(test$id) [1] a b c When I work with my real (larger) dataset I would like to use functions like 'is.na' and '!is.na' on factors. Now I wonder if there is an R alternativ to do 'search (for empty cells) and replace (with NA)' in Excel? I have tried a modification of Uwe Ligges suggestion on missing value posted 2 Aug: is.na(test[test==]) - TRUE ...but it did not work on the data set: Error in [-.data.frame(`*tmp*`, test == , value = c(NA, NA, NA, NA : rhs is the wrong length for indexing by a logical matrix However it worked fine when applied to a single vector: is.na(test$id[test$id==]) - TRUE test$id [1] abNA c Levels: a b c is.na(test$id) [1] FALSE FALSE TRUE FALSE Is there a more efficient way to fill empty cells in all my factors in R or should I just do it in advance in Excel by 'search and replace'? Thanks in advance! -- Henrik Pärn Department of Biology NTNU 7491 Trondheim Norway +47 735 96282 (office) +47 909 89 255 (mobile) +47 735 96100 (fax) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing a for loop
Hi,
Try this:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
expanded -expand.grid(rows,columns)
abslist -abs(ex$Var1-ex$Var2)
res - matrix(abslist,nrow = length(rows), ncol = length(columns))
Neurorox
From: Daniel Gerlanc [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] Vectorizing a for loop
Date: Thu, 3 Aug 2006 10:10:46 -0400
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
Thanks!
-- Dan Gerlanc
Williams College
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing a for loop
On Thu, 3 Aug 2006, Daniel Gerlanc wrote:
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
abs(outer(rows,columns,-))
-thomas
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing a for loop
Hi
outer(rows, columns, function(x,y) abs(x-y))
shall do it.
HTH
Petr
On 3 Aug 2006 at 10:10, Daniel Gerlanc wrote:
Date sent: Thu, 3 Aug 2006 10:10:46 -0400
From: Daniel Gerlanc [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] Vectorizing a for loop
Send reply to: [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x
length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
Thanks!
-- Dan Gerlanc
Williams College
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
Petr Pikal
[EMAIL PROTECTED]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing a for loop
Sorry,
I used ex instead of expanded (was working name).
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
expanded -expand.grid(rows,columns)
abslist -abs(expanded$Var1-expanded$Var2)
res - matrix(abslist,nrow = length(rows), ncol = length(columns))
Désolé encore
Neuro
From: Daniel Gerlanc [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] Vectorizing a for loop
Date: Thu, 3 Aug 2006 10:10:46 -0400
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
Thanks!
-- Dan Gerlanc
Williams College
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing a for loop
On Thu, 2006-08-03 at 10:10 -0400, Daniel Gerlanc wrote:
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
Thanks!
See ?outer
outer(rows, columns, function(x, y) abs(x - y))
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]89 10 11 12
[3,]789 10 11
[4,]6789 10
[5,]56789
HTH,
Marc Schwartz
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing a for loop
res-outer(rows,columns,FUN=function(x,y) abs(x-y))
will help you.
Am Thu, 03 Aug 2006 16:10:46 +0200 schrieb Daniel Gerlanc
[EMAIL PROTECTED]:
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a function
included in the recommended packages that does this?
Thanks!
-- Dan Gerlanc
Williams College
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--
Universität Hamburg
Institut für Statistik und Ökonometrie
Dipl.-Wi.-Math. Eik Vettorazzi
Von-Melle-Park 5
20146 Hamburg
Tel.: +49 40-42838-3540
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Re: [R] Vectorizing a for loop
try this:
abs(outer(rows, columns, -))
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Daniel Gerlanc [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Thursday, August 03, 2006 4:10 PM
Subject: [R] Vectorizing a for loop
Hello all,
Consider the following problem:
There are two vectors:
rows - c(1, 2, 3, 4, 5)
columns - c(10, 11, 12, 13, 14)
I want to create a matrix with dimensions length(rows) x
length(columns):
res - matrix(nrow = length(rows), ncol = length(columns))
If i and j are the row and column indexes respectively, the
values
of the cells are abs(rows[i] - columns[j]). The resultant matrix
follows:
[,1] [,2] [,3] [,4] [,5]
[1,]9 10 11 12 13
[2,]8910 11 12
[3,]78 9 10 11
[4,]67 8 9 10
[5,]56 7 89
This matrix may be generated by using a simple for loop:
for(i in 1:length(rows)){
for(j in 1:length(columns)){
res[i,j] - abs(rows[i] - columns[j])
}
}
Is there a quicker, vector-based approach for doing this or a
function
included in the recommended packages that does this?
Thanks!
-- Dan Gerlanc
Williams College
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Re: [R] Looking for transformation to overcome heterogeneity of variances
On 03 Aug 2006 16:32:38 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote: My data consists in 96 groups, each one with 10 observations. Levene's test suggests that the variances are not equal, and therefore I have tried to apply the classical transformations to have homocedasticity in order to be able to use ANOVA. Unfortunately, no transformation that I have used transforms my data into data with homocedasticity. The histogram of variances is at http://phhs80.googlepages.com/hist1.png Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? Not based on that information. Try the following instead: fit - lm(y~g) par(mfrow=c(2,2)); plot(fit) Thanks, Peter. By 'g', you mean factor1* factor2*factor3*factor4 If that defines your 96 groups, yes. Thanks, Peter. The result of fit - lm(tardiness ~ interaction(factor1,factor2,factor3,factor4)) par(mfrow=c(2,2)); plot(fit) Warning message: X11 used font size 8 when 7 was requested is at http://phhs80.googlepages.com/2transform1.png Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for transformation to overcome heterogeneity of variances
Paul Smith [EMAIL PROTECTED] writes: On 03 Aug 2006 15:45:10 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote: My data consists in 96 groups, each one with 10 observations. Levene's test suggests that the variances are not equal, and therefore I have tried to apply the classical transformations to have homocedasticity in order to be able to use ANOVA. Unfortunately, no transformation that I have used transforms my data into data with homocedasticity. The histogram of variances is at http://phhs80.googlepages.com/hist1.png Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? Not based on that information. Try the following instead: fit - lm(y~g) par(mfrow=c(2,2)); plot(fit) Thanks, Peter. By 'g', you mean factor1* factor2*factor3*factor4 If that defines your 96 groups, yes. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] get() in sapply() in with()
On Thu, 3 Aug 2006, Heinz Tuechler wrote:
Dear All,
applying some function within a with() function I wanted to use also
sapply() and get() to form a data.frame, but did not succede.
Below is a simplified example.
It is possible to use sapply() within a with() function, it is also
possible to use get() within a with() function, but when I try to use get
within sapply within with I arrive at Error in get(x, envir, mode,
inherits) : variable v5 was not found.
Is there a solution?
This isn't what you want to hear, but the solution is probably to not use
get(). How to not use get() will depend on what your problem really is.
## example
df1 - data.frame(v5=16:20, v6=21:25, v7=I(letters[16:20]), v8=letters[16:20])
with(df1, sapply(c('v5', 'v6'), get) ) ## Error, see next line
## Error in get(x, envir, mode, inherits) : variable v5 was not found
get() looks in the environment it was called from, which is the
environment inside lapply(), whose parent is the environment of
the base package. There is no v5 there.
with(df1, sapply(list(v5, v6), mean) ) # does work
This works because list(v5,v6) is evaluated in df1.
with(df1, get('v5') ) # does work
This works because get() looks in the environment it was called from,
which is df1.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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[R] bullseye or polar display of circular data
I have data for several rings of a left heart chamber, and which I would like to display in concentric rings, with color-encoding of the values. Each ring corresponds to one slice through the heart, and the rings correspond to positions from the base to the apex of the heart as you move from the outermost ring to the innermost one. The data have a circular pattern. These types of displays are referred to as bullseye displays in the nuclear medicine literature. Does any reader of these messages know of a R function/package that offers this functionality? Also I noticed that in some contexts you can define a circular attribute for your data. Are there plot routines for such circular data? thank you! Michael Jerosch-Herold __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about dll crashing R
On Thu, 3 Aug 2006, Benn Fine wrote:
I have ported some R code to C to make it faster.
I can perform .Call(foobar,) once and it
works fine. Absolutely correct answer.
If I put a loop inside foobar and run the main code
routine more than 100 times, it crashes R.
Or if I call .Call(foobar) seperately more than
two tims it crashes R.
snip
SEXP mvntest (SEXP mean, SEXP cov, SEXP temp)
{ int nrx , ncx , nry , ncy ,info,mode;
SEXP xdims , ydims , ans;
int i,j, one=1;
info = 1;
xdims = getAttrib (mean , R_DimSymbol ) ;
ydims = getAttrib (cov , R_DimSymbol ) ;
mode = REALSXP;
nrx = INTEGER( xdims ) [ 0 ] ;
ncx = INTEGER( xdims ) [ 1 ] ;
nry = INTEGER( ydims ) [ 0 ] ;
ncy = INTEGER( ydims ) [ 1 ] ;
/* create the upper trianglular matrix A */
/* such that t(A) %*% A = Sigma */
GetRNGstate();
F77_CALL(dpofa) ( REAL( cov ), nry , ncy , info);
Rprintf(Info = %d\n,info);
for(i=0;inry;i++)
for(j=0;ji;j++)
REAL(cov)[i+j*ncy] = 0.0;
PROTECT( ans = allocMatrix (mode, nrx , one ) ) ;
for(i=0;inry;i++)
REAL(temp)[i] = rnorm(0,1);
ans = tmatrixprod(cov,temp);
^
Here you are returning a new SEXP from tmatrixprod but not protecting it.
Remember, PROTECT() operates on the *value* of its argument, so it
protects the thing that ans points to. When ans points to a new thing, it
is still the old thing that is protected.
-thomas
for(i=0;inry;i++)
REAL(ans)[i] = REAL(ans)[i]+REAL(mean)[i];
UNPROTECT( 1 ) ;
PutRNGstate();
return( ans ) ;
}
I have a feeling I am messing up memory usage
somewhere but haven't a clue. Do I need to do garbage
collecting inside the C program ?The fact that the
code
runs a few times before R crashes is driving me nuts.
I send most of what I need into the C routine from R,
so I am not creating that many SEXP objects within the
program.
Any hints or ideas ?
Thanks!
Benn
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Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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Re: [R] get() in sapply() in with()
Thank you, Thomas!
It helps a lot to know that something is impossible and that I have to look
for a different kind of solution.
Heinz
At 07:33 03.08.2006 -0700, Thomas Lumley wrote:
On Thu, 3 Aug 2006, Heinz Tuechler wrote:
Dear All,
applying some function within a with() function I wanted to use also
sapply() and get() to form a data.frame, but did not succede.
Below is a simplified example.
It is possible to use sapply() within a with() function, it is also
possible to use get() within a with() function, but when I try to use get
within sapply within with I arrive at Error in get(x, envir, mode,
inherits) : variable v5 was not found.
Is there a solution?
This isn't what you want to hear, but the solution is probably to not use
get(). How to not use get() will depend on what your problem really is.
## example
df1 - data.frame(v5=16:20, v6=21:25, v7=I(letters[16:20]),
v8=letters[16:20])
with(df1, sapply(c('v5', 'v6'), get) ) ## Error, see next line
## Error in get(x, envir, mode, inherits) : variable v5 was not found
get() looks in the environment it was called from, which is the
environment inside lapply(), whose parent is the environment of
the base package. There is no v5 there.
with(df1, sapply(list(v5, v6), mean) ) # does work
This works because list(v5,v6) is evaluated in df1.
with(df1, get('v5') ) # does work
This works because get() looks in the environment it was called from,
which is df1.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient way to make NAs of empty cells in a factor (or character)
Hi try to set na.strings = in calling read.csv2. Works for me is.na(read.delim(clipboard, na.strings=)$mono) [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE read.delim(clipboard, na.strings=)$mono [1] hruby hruby jemny jemny nejhrubsi nejhrubsi standard standard NA Levels: hruby jemny nejhrubsi standard or you can try test[(test==)] - NA HTH Petr On 3 Aug 2006 at 15:46, Henrik Parn wrote: Date sent: Thu, 03 Aug 2006 15:46:32 +0200 From: Henrik Parn [EMAIL PROTECTED] Organization: NTNU To: R-help r-help@stat.math.ethz.ch Subject:[R] efficient way to make NAs of empty cells in a factor (or character) Send reply to: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] Dear all, I have some csv-files (originating from Excel-files) containing empty cells. In my example file I have four variables of different classes, each with some empty cells in the original csv-file: test - read.csv2(test.csv, dec=.) test id id2 x y 1 a 1 NA 2 b e NA 2.2 3 f 3 3.3 4 c g 4 4.4 class(test$id) [1] factor class(test$id2) [1] factor class(test$x) [1] integer class(test$y) [1] numeric In the help text of read.csv2 you can read 'Blank fields are also considered to be missing values in logical, integer, numeric and complex fields.'. Thus, empty cells in a factor (or a character I assume) is not considered as missing values but an own level: is.na(test$id) [1] FALSE FALSE FALSE FALSE levels(test$id) [1] a b c When I work with my real (larger) dataset I would like to use functions like 'is.na' and '!is.na' on factors. Now I wonder if there is an R alternativ to do 'search (for empty cells) and replace (with NA)' in Excel? I have tried a modification of Uwe Ligges suggestion on missing value posted 2 Aug: is.na(test[test==]) - TRUE ...but it did not work on the data set: Error in [-.data.frame(`*tmp*`, test == , value = c(NA, NA, NA, NA : rhs is the wrong length for indexing by a logical matrix However it worked fine when applied to a single vector: is.na(test$id[test$id==]) - TRUE test$id [1] abNA c Levels: a b c is.na(test$id) [1] FALSE FALSE TRUE FALSE Is there a more efficient way to fill empty cells in all my factors in R or should I just do it in advance in Excel by 'search and replace'? Thanks in advance! -- Henrik Pärn Department of Biology NTNU 7491 Trondheim Norway +47 735 96282 (office) +47 909 89 255 (mobile) +47 735 96100 (fax) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] run self written functions
Antje niederlein-rstat at yahoo.de writes:
I'm not sure if I'm in the right place with my question...
I'm running R on Windows and wrote a function and saved it as .R file.
It looks like this:
bmi - function(weight, height) {
bmi - weight / height^2
bmi
}
If I want to use this function, I have to mark everything and then press
Ctrl-R. But then everything single line is executed on the command line,
which means that I will loose my history when the code becomes longer.
I suggest that you forget the history function, it tends to accumulate all the
nonsense we make. Use the following method instead: Keep a RGui (assuming
Windows) on the left half of your screen, and an editor (Tinn-R, I suggest, if
you are no emacian) on the right half. Write everything you plan to keep in the
editor, and send it to RGui using the menu items of the editor provided. That
way, you end up with a nice little program that you can take home and re-run
tomorry, when your BMI has changed (I hope it stays constant).
R is a bit different from other programming languages, it does not need the
assignment to the function name. The following is the same as your function.
bmi - function(weight, height) weight / height^2
Dieter
(TÜ)
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[R] levels of an array (strings and numbers)
Reading a csv file (db.csv) as: yes,full no,full no,empty I can use the command 'levels' to extract the different values appearing for each column as follows: d-read.csv(db.csv) levels(d[,2]) which returns [1] empty full while, doing the same with a numerical csv file as: 1,6 0,6 0,7 the same instruction returns an empty output instead of 6 7 Any idea about that? Thanks in advance, Alessandro -- Alessandro Antonucci Dalle Molle Institute for Artificial Intelligence (IDSIA) at Idsiae-mail: [EMAIL PROTECTED] Galleria 2 web: idsia.ch/~alessandro Via Cantonale mobile: +39 339-567-23-28 CH-6928 tel: +41 58-666-66-70 Manno - Lugano fax: +41 58-666-66-61 Switzerland skype: alessandro.antonucci __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bullseye or polar display of circular data
On Thu, 2006-08-03 at 07:46 -0700, Michael Jerosch-Herold wrote: I have data for several rings of a left heart chamber, and which I would like to display in concentric rings, with color-encoding of the values. Each ring corresponds to one slice through the heart, and the rings correspond to positions from the base to the apex of the heart as you move from the outermost ring to the innermost one. The data have a circular pattern. These types of displays are referred to as bullseye displays in the nuclear medicine literature. Does any reader of these messages know of a R function/package that offers this functionality? Also I noticed that in some contexts you can define a circular attribute for your data. Are there plot routines for such circular data? thank you! Michael Jerosch-Herold You might want to take a look at the 'circular' or 'CircStats' packages on CRAN: http://cran.us.r-project.org/src/contrib/Descriptions/circular.html http://cran.us.r-project.org/src/contrib/Descriptions/CircStats.html There are some examples of plots generated using the packages in the R Graphics Gallery here: http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=121 http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=97 HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vector DF [a, b] to replace values :was Finding the position of a variable in a data.frame
--- Peter Dalgaard [EMAIL PROTECTED] wrote: John Kane [EMAIL PROTECTED] writes: --- jim holtman [EMAIL PROTECTED] wrote: ?which which(Df = 50, arr.ind=T) row col 5 5 4 This works very nicely as has some other suggestions on how to replace a value. Assuming that I have more than one correction to make where Df = 50, can I use vectors in the Df[] to do this. My attempt shows that I can use vectors but it appears thatthere is something wrong with my logic Eample cat - c( 3,5,6,8,0) dog - c(3,5,3,6, 0) rat - c (5, 5, 4, 9, 51) bat - c( 12, 42, 45, 32, 54) Df - data.frame(cbind(cat, dog, rat, bat)) post - which(Df = 50, arr.ind=T) # find values .= 50 post correction - c(77, 88) # new correct values row - post[ ,1] # vector for row number col - post[ ,2] # vector for column number Df[row,col] - correction Df ---result- cat dog rat bat 1 3 3 5 12 2 5 5 5 42 3 6 3 4 45 4 8 6 9 32 5 0 0 88 88 I am replacing both instances with 88 which is not correct You're being bitten by two issues here: One is that Df[row,col] does not vectorize in parallel in the two indices: Df[1:2,1:2] has *four* elements, not two. It took a moment to see it but it's rather obvious now. I'm still fuzzy on the differences in behaviour between a data frame and a matrix. Another is that data frames are not matrices: you can have different types of values in different columns, so you cannot expect to pick an arbitrary set of elements and assign a vector into it (well, it doesn't work, anyway). This sort of stuff works much easier with matrices, where we also have the wonderful feature of indexing with a matrix: M - cbind(cat, dog, rat, bat) M[post]- correction M cat dog rat bat [1,] 3 3 5 12 [2,] 5 5 5 42 [3,] 6 3 4 45 [4,] 8 6 9 32 [5,] 0 0 77 88 Which of course I, stupidly thought that I was doing in the data frame. Thank you very much. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about dll crashing R
On Thu, 3 Aug 2006, Thomas Lumley wrote:
On Thu, 3 Aug 2006, Benn Fine wrote:
I have ported some R code to C to make it faster.
I can perform .Call(foobar,) once and it
works fine. Absolutely correct answer.
If I put a loop inside foobar and run the main code
routine more than 100 times, it crashes R.
Or if I call .Call(foobar) seperately more than
two tims it crashes R.
snip
SEXP mvntest (SEXP mean, SEXP cov, SEXP temp)
{ int nrx , ncx , nry , ncy ,info,mode;
SEXP xdims , ydims , ans;
int i,j, one=1;
info = 1;
xdims = getAttrib (mean , R_DimSymbol ) ;
ydims = getAttrib (cov , R_DimSymbol ) ;
mode = REALSXP;
nrx = INTEGER( xdims ) [ 0 ] ;
ncx = INTEGER( xdims ) [ 1 ] ;
nry = INTEGER( ydims ) [ 0 ] ;
ncy = INTEGER( ydims ) [ 1 ] ;
/* create the upper trianglular matrix A */
/* such that t(A) %*% A = Sigma */
GetRNGstate();
F77_CALL(dpofa) ( REAL( cov ), nry , ncy , info);
Rprintf(Info = %d\n,info);
for(i=0;inry;i++)
for(j=0;ji;j++)
REAL(cov)[i+j*ncy] = 0.0;
PROTECT( ans = allocMatrix (mode, nrx , one ) ) ;
for(i=0;inry;i++)
REAL(temp)[i] = rnorm(0,1);
ans = tmatrixprod(cov,temp);
^
Here you are returning a new SEXP from tmatrixprod but not protecting it.
Remember, PROTECT() operates on the *value* of its argument, so it
protects the thing that ans points to. When ans points to a new thing, it
is still the old thing that is protected.
and the old value of 'ans' is never used.
for(i=0;inry;i++)
REAL(ans)[i] = REAL(ans)[i]+REAL(mean)[i];
UNPROTECT( 1 ) ;
PutRNGstate();
return( ans ) ;
but the next two lines do not do any allocations. However, PutRNGstate
does, so you need to UNPROTECT *after* that call.
While we are at it, REAL(ans) is a function call, and you will do better
to assign the value once outside the loop.
I have a feeling I am messing up memory usage
somewhere but haven't a clue. Do I need to do garbage
collecting inside the C program ?The fact that the
code
runs a few times before R crashes is driving me nuts.
I send most of what I need into the C routine from R,
so I am not creating that many SEXP objects within the
program.
Any hints or ideas ?
See the debugging sections in `Writing R Extensions'. Using
gctorture(TRUE) and valgrind (if possible) will find these things faster.
--
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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for transformation to overcome heterogeneity ofvariances
I know I'm coming late to this, but ... Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? It is not usually useful to worry about this. In my experience, the gain in efficiency from using an essentially ideal weighted analysis vs. an approximate unweighted one is usually small and unimportant (transformation to simplify a model is another issue ...). Of far greater importance usually is the loss in efficiency due to the presence of a few unusual extreme values; have you carefully checked to make sure that none of the large sample variances you have are due merely to the presence of a small number of highly discrepant values? -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.spss and umlaut
I have gone and looked at the code for reading SPSS portable files, and
the file format appears to specify that you cannot read many legal
characters.
Part of the header information in the file format is a 256-byte
translation table apparently designed for translating between character
representations. It can mark characters as untranslateable, and the
code for reading character strings replaces untranslateable characters
with NULs.
In the example file in the foreign package the only translatable
characters are the ASCII alphanumeric characters and
.(+0[]!$*);^-/|,%_?`:#@'=~{}\
So, it looks as though your SPSS portable file may be marking character
code FC as untranslatable. This is easy to check -- look at the start of
the file and find the sequence ABCDEF..., which is in the middle of the
translation table. See if u-umlaut is in the table. It might even work to
modify the translation table to allow the accented characters.
It looks as though SPSS .sav files don't have this limitation.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
[EMAIL PROTECTED] University of Washington, Seattle
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[R] Help Building packages for windows
Hello all, I've tried to build my linux source package with the following commands and then run it through the RCrossBuild package with R 2.3.1, but I get errors about how the built field is incorrect and that my Rd file has a non-empty \name. I've looked through the Rd file and it seems to not have any empty \name values, but I'm not sure how to tell if any are 'missing'. The built field is: Built: R 2.3.1; ; 2006-08-02 21:42:59; unix R CMD INSTALL --build pkgName R CMD INSTALL --build --use-zip pkgName R CMD BUILD --use-zip pkgName R CMD BUILD pkgName An RCHECK on the package gives back the warning that subdirectory data contains no data sets, despite the presence of Rdata.rdb, Rdata.rds, and Rdata.rdx in the directory. This package is an annotation package for an array and can be used fine in *nix R. If I try to install the package in R for windows it gives me a bunch of errors with package names that look like random strings of characters. Any help would be appreciated. Thanks, Denise sessionInfo() Version 2.3.1 (2006-06-01) i686-pc-linux-gnu attached base packages: [1] methods stats graphics grDevices utils datasets [7] base other attached packages: pkgName 2.3 Linux 2.4.20-31.9 #1 Tue Apr 13 17:38:16 EDT 2004 i686 athlon i386 GNU/Linux __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] levels of an array (strings and numbers)
Alessandro Antonucci wrote: Reading a csv file (db.csv) as: yes,full no,full no,empty I can use the command 'levels' to extract the different values appearing for each column as follows: d-read.csv(db.csv) levels(d[,2]) which returns [1] empty full while, doing the same with a numerical csv file as: 1,6 0,6 0,7 R (i.e. read.csv) assumes this is numeric rather than categorical data. Hence specify the class of each variable in read.cvs' argument colClasses. Uwe Ligges the same instruction returns an empty output instead of 6 7 Any idea about that? Thanks in advance, Alessandro __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] DOE in R
see in line Petr Pikal wrote: See ?aov ?lm something like lm(result~yourfactor1+yourfactor2+yourfactor3+yourfactor5, data=yourdataframe) or lm(result~(yourfactor1+yourfactor2+yourfactor3+yourfactor5)^2, data=yourdataframe) if you want all interactions. Make sure all yourfactors are of class factor. Hope this helps. Spencer Graves or aov(result~yourfactor1+yourfactor2+yourfactor3+yourfactor5, data=yourdataframe) but the exact structure of lm or aov construction depends on what you want to test. HTH Petr On 28 Jul 2006 at 21:43, [EMAIL PROTECTED] wrote: Date sent:Fri, 28 Jul 2006 21:43:38 -0700 From: [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject: [R] DOE in R Send reply to:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] Hi. I'm a student in a graduate program at Simon Fraser University in Canada. I am trying to run a simple screening experiment with some simulated data. I simply want to do an ANOVA of an experiemnt with 5 factors (4 have 2 levels, the last has 3 levels) and 48 runs (ie, full factorial). The thing is that I have multiple observations for each level combination (run). So, 1) How do I do the anova based on the setup above? and 2) More importantly, because of convergence issues for my simulations, I will likely have an unequal number of observations for the 48 runs. How can I do this? Seems like a straightforward enough situation. I am trying to avaoid writing my own C code to do the analysis since I am working under some pretty tight time constraints. ANy help would be appreciated. Thanks very much. Dean Vrecko __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for transformation to overcome heterogeneity ofvariances
Peter You question is difficult to answer without more information about the distribution of your residuals. Different residual patterns call for different transformations to stabilize the variance. One very common form of heterocedasticity is increasing variance with increasing values of an independent predictor, i.e. the variance of the residuals of y=x increase as x increases. In this case a log transformation of some, or all, of the independent variables of the helps. Please also note the comment by Bert Gunter (included below) in which some important points are raised, particularly about extreme values. If you want more help, please describe the pattern of your residuals. John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC, University of Maryland School of Medicine Claude D. Pepper OAIC, University of Maryland Clinical Nutrition Research Unit, and Baltimore VA Center Stroke of Excellence University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) [EMAIL PROTECTED] Berton Gunter [EMAIL PROTECTED] 8/3/2006 11:56:28 AM I know I'm coming late to this, but ... Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? It is not usually useful to worry about this. In my experience, the gain in efficiency from using an essentially ideal weighted analysis vs. an approximate unweighted one is usually small and unimportant (transformation to simplify a model is another issue ...). Of far greater importance usually is the loss in efficiency due to the presence of a few unusual extreme values; have you carefully checked to make sure that none of the large sample variances you have are due merely to the presence of a small number of highly discrepant values? -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.spss 'error reading system-file header'
Thank you for your answer. I did try this but got similar errors. I tried all the other spss-specific formats with same result. A way to get access to the data is: save in sas-xport format (though all labels are lost) then save in another sas-format and asking spss to also save value labels in a sas-format file which can then be edited to extract both value labels and variable labels. These methods are quite laborious and quite complicated, though, and you need access to spss. Ulrich Keller wrote: Question 2: Try saving the data as an SPSS portable file. I never had trouble reading these in R. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.spss 'error reading system-file header'
Thank you Michael I think you point to the real cause of the problem. I solved my own immediate problem by using StatTransfer to transfer from sav to sav. My real concern is that the read.spss() function will become obsolete. Most of the data I have received in the last year have had those problems (and increasingly so), now they have become so serious that the import process fails all together. Therefore I believe it is important to do something about Michael Bibo wrote: I don't know if this is relevant in your particular case, but the error messages you quote are precisely what you get if the SPSS .sav file has been created with the SPSS Data Entry product. If this is the case, it is covered by section 3.1 of the R Data Import/Export document. If that is the problem, and if you don't have access to SPSS Data Entry, I could do a data export for you. If I can help, please respond on both my work email (below) and [EMAIL PROTECTED], as I'm not sure which one I would see first. Michael Bibo Research Officer Projects and Research Service West Moreton Community Health Services West Moreton Health Service District Queensland Health [EMAIL PROTECTED] Ph. +61 7 3817 2400 P.O. Box 878 Ipswich, 4305 Australia This e-mail, including any attachments sent with it, is confidential and for the sole use of the intended recipient(s). This confidentiality is not waived or lost if you receive it and you are not the intended recipient(s), or if it is transmitted/ received in error. Any unauthorised use, alteration, disclosure, distribution or review of this e-mail is prohibited. It may be subject to a statutory duty of confidentiality if it relates to health service matters. If you are not the intended recipient(s), or if you have received this e-mail in error, you are asked to immediately notify the sender by telephone or by return e-mail. You should also delete this e-mail message and destroy any hard copies produced. * This email, including any attachments sent with it, is confidential and for the sole use of the intended recipient(s). This confidentiality is not waived or lost, if you receive it and you are not the intended recipient(s), or if it is transmitted/ received in error. Any unauthorised use, alteration, disclosure, distribution or review of this email is strictly prohibited. The information contained in this email, including any attachment sent with it, may be subject to a statutory duty of confidentiality if it relates to health service matters. If you are not the intended recipient(s), or if you have received this email in error, you are asked to immediately notify the sender by telephone collect on Australia +61 1800 198 175 or by return email. You should also delete this email, and any copies, from your computer system network and destroy any hard copies produced. If not an intended recipient of this email, you must not copy, distribute or take any action(s) that relies on it; any form of disclosure, modification, distribution and/or publication of this email is also prohibited. Although Queensland Health takes all reasonable steps to ensure this email does not contain malicious software, Queensland Health does not accept responsibility for the consequences if any person's computer inadvertently suffers any disruption to services, loss of information, harm or is infected with a virus, other malicious computer programme or code that may occur as a consequence of receiving this email. Unless stated otherwise, this email represents only the views of the sender and not the views of the Queensland Government. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] gsummary
Could someone give me a hand with the format of the gsummary function? Basically, I have a large set of xyz coordinates generated by LiDAR data (37 million points) and I am trying to derive various summary statistics on the z-coordinates by a grid cell. I wrote a function to do this by creating factors from the x- and y- coordinates and then using gsummary. However, I want the function to calculate BOTH the max z-value in a grid cell and the number of z-values in that cell. gsummary is supposed to be able to use a list a functions as input to FUN, but I am having trouble coding it correctly. Here is what I came up to so far. gr-gsummary(xyz.packet,FUN=list(numeric=max,numeric=length),omitGroupingFactor=T,form=~X/Y) The help for the function gives no examples using multiple functions as a FUN value, so the help pages haven't been overly useful in this case. Thanks in advance, Mike Mike R. Saunders Forest Biometrician Cooperative Forest Research Unit University of Maine 5755 Nutting Hall Orono, ME 04469-5755 207-581-2763 (O) 207-581-2833 (F) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fitting a model with the nlme package
Dear all, I am analyzing some data that requires a mixed model. I have been reading Pinheiro and Bates' book, but cannot find the notation to fit the following model: Suppose I have the dataset below. Here I am fitting variable p as a fixed effect, variable h as a random effect and variable t as nested within h. I would like to include the variable j as well as an independent (non-nested) random effect. I can't find the notation in the book to tell R to do this. Does anybody know? library(nlme) h-c(1,1,1,2,2,2,3,3,3) j-c(2,3,8,3,4,3,9,5,4) t-c(1,2,3,1,2,3,1,2,3) f-c(1,2,1,2,1,2,1,2,1) p-c(45,32,35,12,23,12,2,9,12) set-data.frame(p,h,j,t) out-lme(p~ -1 + f,data=set, random=~1|h/t) Thanks a alot, frank. -- unladen european swallow __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] gsummary
The gsummary function was intended to apply a summary function to each column in each of the groups of observations. The (implicit) definition of a summary function is that it produces a scalar from a vector. (See the groupings of functions in Table A.1, p. 472 of Statistical Models in S, a.k.a. The White Book.) I don't think you need to use gsummary for your application. You may be able to use tapply or a combination of split and lapply instead. On 8/3/06, Mike Saunders [EMAIL PROTECTED] wrote: Could someone give me a hand with the format of the gsummary function? Basically, I have a large set of xyz coordinates generated by LiDAR data (37 million points) and I am trying to derive various summary statistics on the z-coordinates by a grid cell. I wrote a function to do this by creating factors from the x- and y- coordinates and then using gsummary. However, I want the function to calculate BOTH the max z-value in a grid cell and the number of z-values in that cell. gsummary is supposed to be able to use a list a functions as input to FUN, but I am having trouble coding it correctly. Here is what I came up to so far. gr-gsummary(xyz.packet,FUN=list(numeric=max,numeric=length),omitGroupingFactor=T,form=~X/Y) The help for the function gives no examples using multiple functions as a FUN value, so the help pages haven't been overly useful in this case. Thanks in advance, Mike Mike R. Saunders Forest Biometrician Cooperative Forest Research Unit University of Maine 5755 Nutting Hall Orono, ME 04469-5755 207-581-2763 (O) 207-581-2833 (F) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for transformation to overcome heterogeneity ofvariances
[Resending -- recipient list length issue] John Sorkin [EMAIL PROTECTED] writes: Peter Erm, that was Paul's question, not mine! If you want to help, please look at the pattern of residuals which he put up on the web on my request You question is difficult to answer without more information about the distribution of your residuals. Different residual patterns call for different transformations to stabilize the variance. One very common form of heterocedasticity is increasing variance with increasing values of an independent predictor, i.e. the variance of the residuals of y=x increase as x increases. In this case a log transformation of some, or all, of the independent variables of the helps. Please also note the comment by Bert Gunter (included below) in which some important points are raised, particularly about extreme values. If you want more help, please describe the pattern of your residuals. John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC, University of Maryland School of Medicine Claude D. Pepper OAIC, University of Maryland Clinical Nutrition Research Unit, and Baltimore VA Center Stroke of Excellence University of Maryland School of Medicine Division of Gerontology Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) [EMAIL PROTECTED] Berton Gunter [EMAIL PROTECTED] 8/3/2006 11:56:28 AM I know I'm coming late to this, but ... Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? It is not usually useful to worry about this. In my experience, the gain in efficiency from using an essentially ideal weighted analysis vs. an approximate unweighted one is usually small and unimportant (transformation to simplify a model is another issue ...). Of far greater importance usually is the loss in efficiency due to the presence of a few unusual extreme values; have you carefully checked to make sure that none of the large sample variances you have are due merely to the presence of a small number of highly discrepant values? -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to access a column by its label?
Hi all, Is there any way to access a column of a data frame by its label (title) rather than its column index? For example, I'd like to be able to select animals[,weight] rather than animals[,3], if the third column of the animals data frame has the label weight. Thank you! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.spss 'error reading system-file header'
On Thu, 3 Aug 2006, Finn Sand? wrote: My real concern is that the read.spss() function will become obsolete. Most of the data I have received in the last year have had those problems (and increasingly so), now they have become so serious that the import process fails all together. Therefore I believe it is important to do something about Well, we would be happy if someone did something about it. It still reads all the files it used to read (which includes all the SPSS files I have ever encountered in my work). As has been pointed out several times, PSPP now has newer code to read SPSS files than the code in the foreign package, and that could be adapted. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fitting a model with the nlme package
On 8/3/06, Frank Johannes [EMAIL PROTECTED] wrote: Dear all, I am analyzing some data that requires a mixed model. I have been reading Pinheiro and Bates' book, but cannot find the notation to fit the following model: Suppose I have the dataset below. Here I am fitting variable p as a fixed effect, variable h as a random effect and variable t as nested within h. I would like to include the variable j as well as an independent (non-nested) random effect. I can't find the notation in the book to tell R to do this. Does anybody know? library(nlme) h-c(1,1,1,2,2,2,3,3,3) j-c(2,3,8,3,4,3,9,5,4) t-c(1,2,3,1,2,3,1,2,3) f-c(1,2,1,2,1,2,1,2,1) p-c(45,32,35,12,23,12,2,9,12) set-data.frame(p,h,j,t) out-lme(p~ -1 + f,data=set, random=~1|h/t) It is not easy to fit mixed models with non-nested grouping factors for the random effects using lme. I would recommend using lmer from the lme4/Matrix package instead. Even with lmer you can't fit the model you want to fit because you have 9 levels of the interaction h:t and only 9 observations. The version of lmer in the soon-to-be-released Matrix_0.995-13 even produces a warning about this. library(lme4) Loading required package: Matrix Loading required package: lattice Loading required package: lattice set - data.frame(h = factor(c(1,1,1,2,2,2,3,3,3)), + j = factor(c(2,3,8,3,4,3,9,5,4)), + t = factor(c(1,2,3,1,2,3,1,2,3)), + f = c(1,2,1,2,1,2,1,2,1), + p = c(45,32,35,12,23,12,2,9,12)) (out - lmer(p ~ f - 1 + (1|h/t) + (1|j), set)) Linear mixed-effects model fit by REML Formula: p ~ f - 1 + (1 | h/t) + (1 | j) Data: set AIC BIClogLik MLdeviance REMLdeviance 39.21916 40.00806 -15.60958 36.17505 31.21916 Random effects: Groups NameVariance Std.Dev. t:h (Intercept) 1.7053e-22 1.3059e-11 j(Intercept) 1.0067e+02 1.0033e+01 h(Intercept) 1.2655e+02 1.1249e+01 Residual 3.4106e-13 5.8400e-07 number of obs: 9, groups: t:h, 9; j, 6; h, 3 Fixed effects: Estimate Std. Error t value f 12.000 4.899 2.4495 Warning messages: 1: Estimated variance for group(s) t:h is zero in: LMEoptimize-(`*tmp*`, value = list(maxIter = 200, tolerance = 1.49011611938477e-08, 2: nlminb returned message singular convergence (7) in: LMEoptimize-(`*tmp*`, value = list(maxIter = 200, tolerance = 1.49011611938477e-08, __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to access a column by its label?
?colnames hih __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to access a column by its label?
On Thu, 2006-08-03 at 14:44 -0400, Neil McLeod wrote: Hi all, Is there any way to access a column of a data frame by its label (title) rather than its column index? For example, I'd like to be able to select animals[,weight] rather than animals[,3], if the third column of the animals data frame has the label weight. Thank you! You answered your own question...animals[,weight] You can also do: animals$weight or animals[[weight]] or subset(animals, select = weight) See ?Extract and ?subset for more information. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot facet label font size
This works OK, but there is some extra spacing between the panels, the top axis and the strip on the top, and the left labels and panel. How can I remove these extra spaces? I've tried changing various layout.widths settings with no luck. It seems the spaces are calculated based on the number of conditioning variables, in this case 2 (sex+smoker). Thanks in advance... -Sam -Original Message- From: Gabor Grothendieck [mailto:[EMAIL PROTECTED] Sent: Wednesday, August 02, 2006 6:04 PM To: Walker, Sam Cc: r-help@stat.math.ethz.ch Subject: Re: [R] ggplot facet label font size On 8/2/06, Walker, Sam [EMAIL PROTECTED] wrote: How do I change the font size in the facet labels along the edges of the plot? For example (from the ggplot help file): p-ggplot(tips, sex ~ smoker, aesthetics=list(x=tip/total_bill)) gghistogram(p) In this plot, the facet labels are smoker: No, smoker: Yes, sex: Female, sex: Male. What command can I use to reduce the font size of these labels? In lattice terminology, cex is used to scale these strip labels. But I couldn't find the equivalent in ggplot. The reason I'm asking is I have a 9x7 array of plots which I've been plotting with lattice. I wanted to use ggplot because I like having the labels on the edge of the plots Note that lattice can do that by using custom strip functions: library(ggplot) # data resides here library(lattice) my.strip - function(which.given, which.panel, ...) if (which.given == 1 which.panel[2] == 2) strip.default(which.given, which.panel, ...) my.strip.left - function(which.given, which.panel, ..., horizontal) if (which.given == 2 which.panel[1] == 1) strip.default(which.given, which.panel, horizontal = FALSE, ...) histogram(~ tip/total_bill | sex + smoker, tips, strip = my.strip, strip.left = my.strip.left, par.settings = list(add.text = list(cex = 0.7))) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for transformation to overcome heterogeneity of variances
Paul: It is too bad that most peoples statistical thought processes lead them to thinking heterogeneity is something to overcome so that a simple test of differences in means with ANOVA can be made. If you have that much heterogeneity among 96 groups (hard for me to imagine otherwise), perhaps the distributional heterogeneity rather than simple shifts in means is the more important effect. You might try using omnibus tests of distributional differences (eg., MRPP, coverage tests, etc.) or compare multiple quantiles (e.g., with quantile regression) since you've already admitted that the group distributions differ by more than just a shift in means. Heterogeneous variances among groups immediately implies that there is not a single parameter describing changes in distributions among groups. Focusing on just a comparison of means, while traditional and analytically expedient, is unlikely to be very enlightening. You could of course, weight each group inversely by its variance to achieve a weighted comparison of means. But doing this just makes it so that you've made a valid test on only one of the parameters characterizing distributional differences. A better analysis but still not as enlightening as possible. My 2 pence. Brian Brian S. Cade U. S. Geological Survey Fort Collins Science Center 2150 Centre Ave., Bldg. C Fort Collins, CO 80526-8818 email: [EMAIL PROTECTED] tel: 970 226-9326 Paul Smith [EMAIL PROTECTED] Sent by: [EMAIL PROTECTED] 08/03/2006 07:33 AM To r-help@stat.math.ethz.ch cc Subject [R] Looking for transformation to overcome heterogeneity of variances Dear All My data consists in 96 groups, each one with 10 observations. Levene's test suggests that the variances are not equal, and therefore I have tried to apply the classical transformations to have homocedasticity in order to be able to use ANOVA. Unfortunately, no transformation that I have used transforms my data into data with homocedasticity. The histogram of variances is at http://phhs80.googlepages.com/hist1.png Is someone able to suggest to me a transformation to overcome the problem of heterocedasticity? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] meta characters in file path
Hi, I need to read in some files. The file names contain come meta characters such as @, #, and white spaces etc, In read.csv, file= option, is there any way that one can make the function to recognize a file path with those characters? Thanks Johnny [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] meta characters in file path
What is the problem you are having? Seems to work fine for me running under Windows2000: write.table(data.frame(a=1:3,b=4:6), file=@# x.csv, sep=,) read.csv(file=@# x.csv) a b 1 1 4 2 2 5 3 3 6 sessionInfo() Version 2.3.1 (2006-06-01) i386-pc-mingw32 attached base packages: [1] methods stats graphics grDevices utils datasets [7] base other attached packages: XML 0.99-8 Li,Qinghong,ST.LOUIS,Molecular Biology wrote: Hi, I need to read in some files. The file names contain come meta characters such as @, #, and white spaces etc, In read.csv, file= option, is there any way that one can make the function to recognize a file path with those characters? Thanks Johnny [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fitting a model with the nlme package
(out - lmer(p ~ f - 1 + (1|h/t) + (1|j), set)) Doug: It seems the nesting syntax is handled a bit differently. Is (1|h/t) equivalent to the old lme nesting syntax? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Weibull distribution
On 7/21/06, Thomas Lumley [EMAIL PROTECTED] wrote:
On Fri, 21 Jul 2006, Valentin Dimitrov wrote:
Dear Leaf,
I modified your code as follows:
gamma.fun - function(mu,sd,start=100)
{
f.fn - function(alpha)
{abs(sd^2-mu^2/(gamma(1+1/alpha))^2*(gamma(1+2/alpha)-(gamma(1+1/alpha))^2))}
alpha - optim(start, f.fn)
beta - mu/gamma(1+1/alpha$par)
return(list=c(a=alpha$par,b=beta));
}
Now it works properly.
First, I added an abs(). You tried to solve an
equation by means of the R-function optim(), which
finds a minimum. That's why you can find the solution
of f(x)=a through minimization of abs(f(x)-a).
Second, I deleted the optim-method BFGS from the
optim() function, because it is not appropriate in
this case.
optim() is not appropriate at all in this case -- its help page says to
use optimize() for one-dimensional problems.
Just to clarify: The help page says:
'optim' will work with one-dimensional 'par's, but the default
method does not work well (and will warn). Use 'optimize' instead.
In other words, if you for instance use the 'BFGS' method, optim is
perfectly OK for one-dimensional problems.
In fact, in one dimension there isn't any need to resort to optimization
when you really want root-finding, and uniroot() is more appropriate than
optimize().
One reason to use optim instead of uniroot or optimize is that you
need not specify a finite interval that covers the solution.
Göran
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[R] geodesic distance (solved)
I found the answer to my problem. stefano __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot facet label font size
Hi Sam, How do I change the font size in the facet labels along the edges of the plot? Unfortunately, you can't currently change the size of those fonts. However, it is on my todo list (as well as completely custom strip functions) and should be available in the near future. One thing you could do is have a look at ggopt, where you can at least change the strip text, if not the size. Regards, Hadley On 8/2/06, Walker, Sam [EMAIL PROTECTED] wrote: For example (from the ggplot help file): p-ggplot(tips, sex ~ smoker, aesthetics=list(x=tip/total_bill)) gghistogram(p) In this plot, the facet labels are smoker: No, smoker: Yes, sex: Female, sex: Male. What command can I use to reduce the font size of these labels? In lattice terminology, cex is used to scale these strip labels. But I couldn't find the equivalent in ggplot. The reason I'm asking is I have a 9x7 array of plots which I've been plotting with lattice. I wanted to use ggplot because I like having the labels on the edge of the plots, but the label font size is too large and exceeding the size of the label box. Thanks in advance... -Sam __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding the position of a variable in a data.frame
You don't need to find out the column index. This works: Df[5,'bat'] - 100 -Don At 5:01 PM -0400 8/2/06, John Kane wrote: Simple problem but I don't see the answer. I'm trying to clean up some data I have 120 columns in a data.frame. I have one value in a column named blaw that I want to change. How do I find the coordinates. I can find the row by doing a subset on the data.frame but how do I find out here blaw is in columns without manually counting them or converting names(Df) to a list and reading down the list. Simple example cat - c( 3,5,6,8,0) dog - c(3,5,3,6, 0) rat - c (5, 5, 4, 9, 0) bat - c( 12, 42, 45, 32, 54) Df - data.frame(cbind(cat, dog, rat, bat)) Df subset(Df, bat = 50) results cat dog rat bat 5 0 0 0 54 Thus I know that my target is in row 5 but how do I figure out where 'bat' is? All I want to do is be able to say Df[5,4] - 100 Is there some way to have function(bat) return the column number: some kind of a colnum() function? I had thought that I had found somthing in library(gdata) matchcols but no luck. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tcltk package
Hi Adrian, Thanks for the tip. I re-installed and everything seems to work just fine. Thanks, Jack. Adrian DUSA [EMAIL PROTECTED] wrote: On Tuesday 01 August 2006 19:24, John McHenry wrote: [...] Yes, I built R myself. I couldn't find a debian package for R 2.3.1. The latest available is 2.2.1. Oh, but there is... right on CRAN. For Dapper just add this line to your sources.list: deb http://cran.R-project.org/bin/linux/ubuntu/ dapper/ This repository has lots of other packages compiled for Ubuntu, feel free to take a look. HTH, Adrian -- Adrian DUSA Romanian Social Data Archive 1, Schitu Magureanu Bd 050025 Bucharest sector 5 Romania Tel./Fax: +40 21 3126618 \ +40 21 3120210 / int.101 - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding the position of a variable in a data.frame
--- Don MacQueen [EMAIL PROTECTED] wrote: You don't need to find out the column index. This works: Df[5,'bat'] - 100 -Don Thanks, I'd tried Df[5, bat] - 100 :( I never thought of the ' ' being needed. At 5:01 PM -0400 8/2/06, John Kane wrote: Simple problem but I don't see the answer. I'm trying to clean up some data I have 120 columns in a data.frame. I have one value in a column named blaw that I want to change. How do I find the coordinates. I can find the row by doing a subset on the data.frame but how do I find out here blaw is in columns without manually counting them or converting names(Df) to a list and reading down the list. Simple example cat - c( 3,5,6,8,0) dog - c(3,5,3,6, 0) rat - c (5, 5, 4, 9, 0) bat - c( 12, 42, 45, 32, 54) Df - data.frame(cbind(cat, dog, rat, bat)) Df subset(Df, bat = 50) results cat dog rat bat 5 0 0 0 54 Thus I know that my target is in row 5 but how do I figure out where 'bat' is? All I want to do is be able to say Df[5,4] - 100 Is there some way to have function(bat) return the column number: some kind of a colnum() function? I had thought that I had found somthing in library(gdata) matchcols but no luck. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Ambitious newbie with some ongoing Q's
I'm new to the list and I've been playing about with R for some months now, mostly using the power analysis routines including the pwr package. I'm currently looking at a project which will require a repeated-measures MANCOVA. I've been reviewing the files available at CRAN, including http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_repms.html and http://tolstoy.newcastle.edu.au/R/help/03b/7663.html which describe repeated-measures ANOVA Is making this a repeated-measures MANCOVA as simple as adding a covariate to the fixed model and then making the Y and covariate variables matrices? If not, how do I do it? Can I do it? And if I can do it, can I also do a power analysis? (OK, I'm a greedy little newbie...) Thanks in advance. -- I can answer any question. I don't know is an answer. I don't know yet is a better answer. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random structure of nested design in lme
I'm not familiar with 'aov', but I have two observations that might help you: 1. UNESTIMABLE VARIANCE COMPONENT The variance component 'soiltype' is not estimable in your 'lme' model: lme(NA.1~soiltype*habitat,random=~1|destination/soiltype) That's because each level of 'soiltype' occurs only once within each level of 'destination' in the self-contained example you provided below. To confirm this, I deleted 'soiltype' from this model: fit.lme - lme(response~soiltype*habitat, random=~1|destination/origin) fit.lme0 - lme(response~soiltype*habitat, random=~1|destination) The answers seemed to be identical except for one thing: VarCorr(fit.lme) Variance StdDev destination = pdLogChol(1) (Intercept) 0.004149471 0.06441639 origin = pdLogChol(1) (Intercept) 0.060968550 0.24691810 Residual 0.007265180 0.08523603 VarCorr(fit.lme0) destination = pdLogChol(1) VarianceStdDev (Intercept) 0.004149471 0.06441639 Residual0.068233730 0.26121587 The Residual variance in fit.lme0 equals the sum of origin and Residual variances in fit.lme. It would help if 'lme' checked for situations like this and either refused to run or dropped inestimable variance components. However, it's possible that there are so many ways that variance components can be inestimable that it's just not feasible to check for them all. (The function 'varcomp' in S-Plus 6.2 has the same problem.) 2. CROSSED OR NESTED? Are 'destination' and 'origin' crossed or nested in your 'aov' model: aov(response~soiltype*habitat+Error(destination+origin)) I have not used 'aov', and I don't think I should take the time now to try to figure this out. However, this model specification suggests to me that 'destination' and 'origin' might be crossed not nested. (The difference is the 'destination:origin' interaction: If 'destination+origin' is crossed, their interaction is used as the error term; otherwise, it looks to me like you have a saturated model.) By contrast, 'destination/origin' in lme is 'nested', which means that the variance component for 'origin' is in essence the crossed term and the interaction combined. I believe there is a way to estimate crossed random effects using 'lme', but I don't understand how. Fortunately, we can do it using 'lmer' in the 'lme4' and 'Matrix' packages. Because of potential conflicts between 'nlme' and 'lme4', I always quit R and restart when I switch from one to another. The following will then fit something using 'lmer' that looks like it might match your 'aov' fit: library(lme4) fit.lme4 - lmer( response~soiltype*habitat +(1|destination)+(1|origin), Dat0) where Dat0 is a data.frame with columns 'response', 'soiltype', 'habitat', 'destination' and 'origin'. I don't know 'aov' well enough to determine easily if the results from this 'lmer' fit match those from 'aov', but I hope this helps. Spencer Graves ESCHEN Rene wrote: Spencer, Thank you for the kind and elaborate reply to my previous post. I did consider the option you suggested and many variations. Depending on the order of the random factors, lme will either give the same output as the aov model for soiltype or for habitat, but not both in the same model. The closest I came was anova(lme(NA.1~soiltype*habitat,random=~1|destination/soiltype)) However, it apppears that in this case the interaction is tested at the same level as soiltype. In this post, a small sample dataset with a brief explanation of the meaning of the different column titles is included below. Also, I included both the aov model and the lme model. Hopefully, this will help to get closer to a solution to my problem. Best regards, René Eschen. ___ #Small sample dataset # data=read.table(Sample dataset.csv,header=T) require(nlme) soiltype=factor(soiltype) habitat=factor(habitat) destination=factor(destination) origin=factor(origin) summary(aov(response~soiltype*habitat+Error(destination+origin))) anova(lme(response~soiltype*habitat,random=~1|destination/origin)) # #habitat type is either 'arable' or 'grassland' #destination indicates what site the soil was transplanted into, and is considered a random factor within habitat type #soiltype is either 'arable' or 'grassland' #origin indicates what site the soil was taken from, and is considered a random factor within soil type #response is the response variable, typically some plant parameter such as growth rate or number of leaves, but in this example it is a random number between 0 and 1. # habitat destination soiltype originresponse 1 1 1 1 0.63 1 2 1 1 0.76 1 3 1 1 0.14 2 4 1 1 0.27 2 5 1
Re: [R] Finding the position of a variable in a data.frame
On 8/3/06, John Kane [EMAIL PROTECTED] wrote: --- Don MacQueen [EMAIL PROTECTED] wrote: You don't need to find out the column index. This works: Df[5,'bat'] - 100 -Don Thanks, I'd tried Df[5, bat] - 100 :( I never thought of the ' ' being needed. Right -- the quotes are not needed if you use $ but they are needed if you use [. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Building a random walk vector
I'm studying R in my free time. I want to build a vector where each element is equal to the element before it in the sequence plus some random tweak. In python, I would write: vec = [100] * 50 # make a 50-element list with each element set to 100 from random import randint for i, v in enumerate(vec): if i is not 0: # if we're not on the first element vec[i] = vec[i-1] + randint(-2, 2) I suspect R has some fancier way of doing this. How to? TIA -- A better way of running series of SAS programs: http://overlook.homelinux.net/wilsonwiki/SasAndMakefiles __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Questions about sweave...
Evening all:
I'm taking a little time to experiment with R, Sweave, and Miktex/LaTex but
I've run up against some problems and -well- I hope that there are some on
the list who might have some suggestions. This will be kind of wordy as I
will include the complete files involved as I'm just not sure what I'm
looking for. Apologies at the outset.
I created the source file (example.Snw):
\documentclass[12pt,letterpaper]{article}
\title{Sweave Example 1}
\author{Friedrich Leisch}
\begin{document}
\maketitle
In this example we embed parts of the examples from the
\texttt{kruskal.test} help page into a \LaTeX{} document:
=
data(airquality)
library(stats)
kruskal.test(Ozone ~ Month , data=airquality)
@
which shows that the location parameter of the Ozone distribution varies
significantly from month to month. Finally we include a boxplot of the data:
\begin{center}
fig=TRUE , echo=FALSE=
boxplot(Ozone ~ Month , data=airquality)
@
\end{center}
\end{document}
which R seems to accept gracefully to produce the tex file (example.tex):
\documentclass[12pt,letterpaper]{article}
\title{Sweave Example 1}
\author{Friedrich Leisch}
\usepackage{C:/PROGRA~1/R/R-23~1.1/share/texmf/Sweave}
\begin{document}
\maketitle
In this example we embed parts of the examples from the
\texttt{kruskal.test} help page into a \LaTeX{} document:
\begin{Schunk}
\begin{Sinput}
data(airquality)
library(stats)
kruskal.test(Ozone ~ Month, data = airquality)
\end{Sinput}
\begin{Soutput}
Kruskal-Wallis rank sum test
data: Ozone by Month
Kruskal-Wallis chi-squared = 29.2666, df = 4, p-value = 6.901e-06
\end{Soutput}
\end{Schunk}
which shows that the location parameter of the Ozone distribution varies
significantly from month to month. Finally we include a boxplot of the data:
\begin{center}
\includegraphics{example-002}
\end{center}
\end{document}
but when I try to run Latex on it I get errors (example.log):
This is e-TeX, Version 3.141592-2.2 (MiKTeX 2.4) (preloaded format=latex
2006.8.3) 3 AUG 2006 20:57
entering extended mode
**example.tex
(example.tex
LaTeX2e 2005/12/01
Babel v3.8g and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, french, ukenglish, loaded.
(C:\texmf\tex\latex\base\article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(C:\texmf\tex\latex\base\size12.clo
File: size12.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
)
! Missing \endcsname inserted.
to be read again
\protect
l.5 \begin
{document}
The control sequence marked to be read again should
not appear between \csname and \endcsname.
! LaTeX Error: Missing \begin{document}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H return for immediate help.
...
l.5 \begin
{document}
You're in trouble here. Try typing return to proceed.
If that doesn't work, type X return to quit.
! Extra \endcsname.
[EMAIL PROTECTED] [EMAIL PROTECTED] -h@@k\endcsname
[EMAIL PROTECTED] \let
\CurrentOptio...
l.5 \begin
{document}
I'm ignoring this, since I wasn't doing a \csname.
! Missing \endcsname inserted.
to be read again
\protect
l.5 \begin
{document}
The control sequence marked to be read again should
not appear between \csname and \endcsname.
! Extra \endcsname.
[EMAIL PROTECTED]@aded ...er \ifx \csname [EMAIL PROTECTED]
\relax \expandafter
[EMAIL PROTECTED]
l.5 \begin
{document}
I'm ignoring this, since I wasn't doing a \csname.
! Missing \endcsname inserted.
to be read again
\protect
l.5 \begin
{document}
The control sequence marked to be read again should
not appear between \csname and \endcsname.
! Extra \endcsname.
[EMAIL PROTECTED]@ptions ...xdef \csname [EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
l.5 \begin
{document}
I'm ignoring this, since I wasn't doing a \csname.
! Missing \endcsname inserted.
to be read again
\protect
l.5 \begin
{document}
The control sequence marked to be read again should
not appear between \csname and \endcsname.
! Missing \endcsname inserted.
to be read again
\protect
l.5 \begin
{document}
The control sequence marked to be read again should
not appear between \csname and \endcsname.
! Extra \endcsname.
argument ...e/texmf/[EMAIL PROTECTED] \endcsname
,
l.5 \begin
{document}
I'm ignoring this, since I wasn't doing a \csname.
! Missing \endcsname inserted.
to be read again
[R] gnlsControl
When I run gnls I get the error:
Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal, data = xy, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976563
My first thought was to decrease minFactor but gnlsControl does not contain
minFactor nor nlsMinFactor (see below). It does however contain nlsMaxIter
and nlsTol which I assume are the analogs of maxiter and tol in nls.control.
I would be happy to hear from anyone who has an idea on what parameters in
gnlsControl to change to get convergence.
Cheers
Dan Coleman
Genentech Inc.
gnlsControl
function (maxIter = 50, nlsMaxIter = 7, msMaxIter = 50, minScale = 0.001,
tolerance = 1e-06, nlsTol = 0.001, msTol = 1e-07, msScale = lmeScale,
returnObject = FALSE, msVerbose = FALSE, apVar = TRUE, .relStep =
(.Machine$double.eps)^(1/3),
nlmStepMax = 100, opt = c(nlminb, optim), optimMethod = BFGS,
minAbsParApVar = 0.05)
{
list(maxIter = maxIter, nlsMaxIter = nlsMaxIter, msMaxIter = msMaxIter,
minScale = minScale, tolerance = tolerance, nlsTol = nlsTol,
msTol = msTol, msScale = msScale, returnObject = returnObject,
msVerbose = msVerbose, apVar = apVar, nlmStepMax = nlmStepMax,
opt = match.arg(opt), optimMethod = optimMethod, .relStep =
.relStep,
minAbsParApVar = minAbsParApVar)
}
nls.control
$maxiter
[1] 50
$tol
[1] 1e-05
$minFactor
[1] 0.0009765625
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building a random walk vector
On 8/3/2006 9:17 PM, Matthew Wilson wrote: I'm studying R in my free time. I want to build a vector where each element is equal to the element before it in the sequence plus some random tweak. In python, I would write: vec = [100] * 50 # make a 50-element list with each element set to 100 from random import randint for i, v in enumerate(vec): if i is not 0: # if we're not on the first element vec[i] = vec[i-1] + randint(-2, 2) I suspect R has some fancier way of doing this. How to? Assuming randint(-2, 2) gives a value uniform on (-2, 2) a quick way to do what you want is vec - 100 + c(0, cumsum(runif(49, -2, 2))) Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Building a random walk vector
On 4 August 2006 at 01:17, Matthew Wilson wrote: | I'm studying R in my free time. I want to build a vector where each | element is equal to the element before it in the sequence plus some | random tweak. | | In python, I would write: | | vec = [100] * 50 # make a 50-element list with each element set to 100 | from random import randint | for i, v in enumerate(vec): | if i is not 0: # if we're not on the first element | vec[i] = vec[i-1] + randint(-2, 2) | | I suspect R has some fancier way of doing this. How to? Yup, cumsum() is your friend. You only need the first scalar of 100, vectorisation does the rest. Try set.seed(12345) Z - 100 + cumsum(runif(50,-2,2)) summary(Z) Min. 1st Qu. MedianMean 3rd Qu.Max. 98.54 100.80 102.30 102.10 103.40 105.70 plot(Z, type='l') Lastly, I think N(0, some_sd) is more customary that U(-2,2) but that is easy to change. Cheers, Dik -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot facet label font size
If you are willing to use grid then you could create only the sex factor in the left strips since its already in the desired position but when displaying it output a factor.level, i.e. label of A. (my.strip.left is modified from the prior post to do that.) Then after the plot is drawn, looping through all grobs looking for those with a label component of A producing a list of grob names, strip.left.names. We then mapply the real factor levels with those grobs editing them in reset.levels(), defined below. (I have used the fact, empirically determined that the stripts are produced in order of the factor levels.) Everything is the same as the last post except my.strip.left which has been modified and everything which comes after the call to histogram. Although this seems to work, maybe Deepayan or Paul can think of something slicker. library(ggplot) # data resides here library(lattice) library(grid) my.strip - function(which.given, which.panel, ...) if (which.given == 1 which.panel[2] == 2) strip.default(which.given, which.panel, ...) my.strip.left - function(which.given, which.panel, ..., factor.levels, horizontal) if (which.given == 1 which.panel[1] == 1) strip.default(which.given, which.panel, factor.levels = LETTERS, horizontal = FALSE, ...) histogram(~ tip/total_bill | sex + smoker, tips, strip = my.strip, strip.left = my.strip.left, par.settings = list(add.text = list(cex = 0.7))) is.strip.left - function(name) identical(grid.get(name)$label, A) strip.left.names - getNames()[sapply(getNames(), is.strip.left)] reset.levels - function(nam, lev) grid.edit(nam, label = lev) mapply(reset.levels , strip.left.names, levels(tips$smoker)) On 8/3/06, Walker, Sam [EMAIL PROTECTED] wrote: This works OK, but there is some extra spacing between the panels, the top axis and the strip on the top, and the left labels and panel. How can I remove these extra spaces? I've tried changing various layout.widths settings with no luck. It seems the spaces are calculated based on the number of conditioning variables, in this case 2 (sex+smoker). Thanks in advance... -Sam -Original Message- From: Gabor Grothendieck [mailto:[EMAIL PROTECTED] Sent: Wednesday, August 02, 2006 6:04 PM To: Walker, Sam Cc: r-help@stat.math.ethz.ch Subject: Re: [R] ggplot facet label font size On 8/2/06, Walker, Sam [EMAIL PROTECTED] wrote: How do I change the font size in the facet labels along the edges of the plot? For example (from the ggplot help file): p-ggplot(tips, sex ~ smoker, aesthetics=list(x=tip/total_bill)) gghistogram(p) In this plot, the facet labels are smoker: No, smoker: Yes, sex: Female, sex: Male. What command can I use to reduce the font size of these labels? In lattice terminology, cex is used to scale these strip labels. But I couldn't find the equivalent in ggplot. The reason I'm asking is I have a 9x7 array of plots which I've been plotting with lattice. I wanted to use ggplot because I like having the labels on the edge of the plots Note that lattice can do that by using custom strip functions: library(ggplot) # data resides here library(lattice) my.strip - function(which.given, which.panel, ...) if (which.given == 1 which.panel[2] == 2) strip.default(which.given, which.panel, ...) my.strip.left - function(which.given, which.panel, ..., horizontal) if (which.given == 2 which.panel[1] == 1) strip.default(which.given, which.panel, horizontal = FALSE, ...) histogram(~ tip/total_bill | sex + smoker, tips, strip = my.strip, strip.left = my.strip.left, par.settings = list(add.text = list(cex = 0.7))) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot facet label font size
Just sending this to you. One thing that might be easy to do yet give a lot of flexibility is to: 1. put meaningful names on the grobs. Even with just this it would be possible to do a getNames() in grid and then from inspection grid.edit the appropriate one(s). 2. create a routine that retrieves grobs so one does not have to use getNames with grep. trellis.focus and friends do this in lattice. Regards. On 8/3/06, hadley wickham [EMAIL PROTECTED] wrote: Hi Sam, How do I change the font size in the facet labels along the edges of the plot? Unfortunately, you can't currently change the size of those fonts. However, it is on my todo list (as well as completely custom strip functions) and should be available in the near future. One thing you could do is have a look at ggopt, where you can at least change the strip text, if not the size. Regards, Hadley On 8/2/06, Walker, Sam [EMAIL PROTECTED] wrote: For example (from the ggplot help file): p-ggplot(tips, sex ~ smoker, aesthetics=list(x=tip/total_bill)) gghistogram(p) In this plot, the facet labels are smoker: No, smoker: Yes, sex: Female, sex: Male. What command can I use to reduce the font size of these labels? In lattice terminology, cex is used to scale these strip labels. But I couldn't find the equivalent in ggplot. The reason I'm asking is I have a 9x7 array of plots which I've been plotting with lattice. I wanted to use ggplot because I like having the labels on the edge of the plots, but the label font size is too large and exceeding the size of the label box. Thanks in advance... -Sam __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
