[R] Weird problem with median on a factor
Hi all, I hope this isn't a naive newbie question again. Here you can see column 264 of a data frame containing data of the same interview in May and September. Column 264 contains the answers of 49 persons to a question in May. fbhint.spss1[,264] [1] teils/teils sehr wichtig NA NA sehr wichtig [6] sehr wichtig sehr wichtig sehr wichtig NA NA [11] NA NA wichtig NA NA [16] sehr wichtig NA NA NA NA [21] NA NA NA wichtig NA [26] NA NA NA NA NA [31] NA NA NA NA teils/teils [36] sehr wichtig NA NA NA NA [41] wichtig NA sehr wichtig NA NA [46] sehr wichtig wichtig NA NA Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig Column 566 contains the answers from the same persons to the same question in September. fbhint.spss1[,566] [1] NA NA NA wichtig wichtig [6] sehr wichtig sehr wichtig wichtig wichtig NA [11] NA NA sehr wichtig sehr wichtig sehr wichtig [16] sehr wichtig NA unwichtigwichtig wichtig [21] NA NA teils/teils teils/teils NA [26] unwichtigNA NA NA NA [31] wichtig sehr wichtig sehr wichtig NA unwichtig [36] sehr wichtig NA NA teils/teils wichtig [41] wichtig wichtig NA NA wichtig [46] NA sehr wichtig teils/teils NA Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig The following works: median(fbhint.spss1[,264], na.rm=T) [1] sehr wichtig Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig ... but here it doesn't: median(fbhint.spss1[,566], na.rm=T) Error in Summary.factor(..., na.rm = na.rm) : sum not meaningful for factors I don't have any ideas why! Can somebody give me a hint? TIA Best regards, Christoph __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] inverse function for positive definite matrix
Hi, there. Can anyone tell me inverse function for positive definite matrix in R? In GAUSS, this function is called invpd and is known to be faster than the normal inverse function. Thanks! Yulei $$$ Yulei He 1586 Murfin Ave. Apt 37 Ann Arbor, MI 48105-3135 [EMAIL PROTECTED] 734-647-0305(H) 734-763-0421(O) 734-763-0427(O) 734-764-8263(fax) $$ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] RMySQL and '_' character in column names
Prof Brian Ripley [EMAIL PROTECTED] writes: ?make.names may help. Or backtick quoting in 1.8.0 (which this is clearly not since _ now gives a syntax error). On Thu, 30 Oct 2003, Xavier Fernández i Marín wrote: Hi, I'm using RMySQL in order to obtain data from my MySQL server. In my databases sometimes I have columns with names that contain '_' character (ex: 'gdp_capita', 'population_total', etc...). When these names appear as the names of the vectors in the data frame that I get, sometimes I have problems as: cor(gdp_capita, population_total) Error: object _capita not found use of _ is soon to be removed. Is there an automatic way to transform the '_' characters in MySQL to '.' in R using RMysql? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] long algo
Alessandro Semeria [EMAIL PROTECTED] wrote: Is well know that R is inefficent on loops. This is a dangerous half-truth. R is an interpreted language. The interpreter uses techniques similar to those used in Scheme interpreters. As interpreters go, it's pretty good. For comparison, in processing XML documents, I've had interpreted Scheme running rings around compiled Java (by doing the task a different way, of course). Also for comparison, years ago I had a Prolog program for median polish that made a published Fortran program for median polish look sick (by using a much better data structure). With Luke Tierney's byte-code compiler, I expect R loops will become close to as efficient as Python ones, and people run entire web sites with Python. It is more accurate to say that R code qua R code is not as efficient as the large body of primitives that operate on entire arrays. When you have to perform heavy loop is better to use a call to fortran or c code (.Fortran() , .C() functions) Even if the premiss were literally and exactly true, the conclusion would not follow. When you have a speed problem with R code, (1) Find out where the problem is, exactly. People's intuition about performance bottlenecks is notoriously bad. Do what the experts do: *measure*. (2) Try to restructure the code *entirely in R* to be as clear and high level as possible. If there have to be subscripts, at least let them be vector subscripts. (3) Measure again. Chances are that making the code clear and high level has fixed the performance problem. (4) If that fails, try restructuring the code a couple of ways, *entirely in R*. The two basic techniques for optimising a calculation are (a) eliminate it entirely and (b) if you can't eliminate the first evaluation of an expression, eliminate the second by saving the result. As a special case of (b), try moving things out of loops; try splitting a calculation into a part that changes a lot and a part that changes very little, and update the small-change part only when you have to. Perhaps apply the idea of program differentiation. (NOT the idea of taking a function that computes a value and automatically computing a function that computes the derivative of the first, but the idea of saying if I have z-f(x,y) and I make a small change to x, do I have to recompute z completely or can I came a small change to z?) Try to use built in operations as much as possible on data structures that are as large as appropriate. (5) Measure again. This will probably have fixed the performance problem. (6) If all else fails, now it's time to try Fortran or C. It's too bad there isn't an existing Fortran or C module you can just call, if there had been you'd have used that before writing the original R code. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] cross-classified random factors in lme without blocking
On page 165 of Mixed-Effects Models in S and S-Plus by Pinheiro and Bates there is an example of using lme() in the nlme package to fit a model with crossed random factors. The example assumes though that the data is grouped. Is it possible to use lme() to fit crossed random factors when the data is not grouped? E.g., y - rnorm(12); a=gl(4,1,12); b=gl(3,4,12). Can I fit an additive model like y~a+b but with a and b random? Everything I've tried gives an error: lme(y~1,random=~1|(a+b)) Error in switch(mode(object), name = , numeric = , call = object, character = as.name(object), : [[ cannot be of mode ( lme(y~1,random=~a+b) Error in getGroups.data.frame(dataMix, groups) : Invalid formula for groups Thanks Gordon __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] which argument is missing from my boot...
Hi,
I just cannot figure out what is missing from my boot.. This is what I have in
my R codes:
-
payment.data-read.csv(maintenance payment.csv.,header=TRUE)
fee-lm(log(payment.data[,2])~payment.data[,1])
g-fee$coefficients[2]
g
phat.data-read.csv(phat.csv.,header=TRUE)
phat1-phat.data[,2]+0.01
PreOLS-lm(log(phat1)~phat.data[,1])
resid-PreOLS$residuals
esqu-resid^2
sigma-(1-phat1)/(phat1*nrow(phat.data))
cnsOLS-lm((esqu-sigma)~1)
f-cnsOLS$fitted.values-sigma
window-phat.data[,1]
phat2-phat.data[,2]
data-data.frame(window,phat2)
maxpay-max(payment.data[2])
minpay-min(payment.data[2])
delta.fun-function(data)
{
wOLS-lm(log(data[,2]+0.01)~data[,1],weights=f^(-0.5))
g -1*wOLS$coefficients[2]*log(maxpay/minpay)/wOLS$coefficients[1]
}
boot(data,delta.fun,100)
Then, I get this message:
Error in statistic(data, original, ...) : unused argument(s) ( ...)
I tried so many combinations by defining variables in the statistic...I could
not figure out what arguments are unused...I checked the delta.fun, and it
seems to be working...
Could anyone help me, PLEASE???
Thank you
soyoko
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Ms. Soyoko Umeno
Graduate Research Assitant for the Illinois-Missouri Biotechnology Alliance (IMBA) at
http://www.imba.missouri.edu/
Ph.D. Student at the Department of Agricultural and Consumer Economics
at the University of Illinois at Urbana-Champaign
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RE: [R] Weird problem with median on a factor
What do you expect the median of a factor to be? Are
you sure you don't want the *mode* (most common
value)?
If you want the median numeric code of ordered
factors, maybe use `as.numeric` first. Consider what you
think should happen if this median is not an integer.
HTH
-Original Message-
From: Christoph Bier [mailto:[EMAIL PROTECTED]
Sent: 30 October 2003 21:35
To: [EMAIL PROTECTED]
Subject: [R] Weird problem with median on a factor
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Hi all,
I hope this isn't a naive newbie question again. Here you can
see column 264 of
a data frame containing data of the same interview in May and
September. Column
264 contains the answers of 49 persons to a question in May.
fbhint.spss1[,264]
[1] teils/teils sehr wichtig NA NA sehr wichtig
[6] sehr wichtig sehr wichtig sehr wichtig NA NA
[11] NA NA wichtig NA NA
[16] sehr wichtig NA NA NA NA
[21] NA NA NA wichtig NA
[26] NA NA NA NA NA
[31] NA NA NA NA teils/teils
[36] sehr wichtig NA NA NA NA
[41] wichtig NA sehr wichtig NA NA
[46] sehr wichtig wichtig NA NA
Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig
Column 566 contains the answers from the same persons to the
same question in
September.
fbhint.spss1[,566]
[1] NA NA NA wichtig wichtig
[6] sehr wichtig sehr wichtig wichtig wichtig NA
[11] NA NA sehr wichtig sehr wichtig sehr wichtig
[16] sehr wichtig NA unwichtigwichtig wichtig
[21] NA NA teils/teils teils/teils NA
[26] unwichtigNA NA NA NA
[31] wichtig sehr wichtig sehr wichtig NA unwichtig
[36] sehr wichtig NA NA teils/teils wichtig
[41] wichtig wichtig NA NA wichtig
[46] NA sehr wichtig teils/teils NA
Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig
The following works:
median(fbhint.spss1[,264], na.rm=T)
[1] sehr wichtig
Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig
... but here it doesn't:
median(fbhint.spss1[,566], na.rm=T)
Error in Summary.factor(..., na.rm = na.rm) :
sum not meaningful for factors
I don't have any ideas why! Can somebody give me a hint?
TIA
Best regards,
Christoph
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Simon Fear
Senior Statistician
Syne qua non Ltd
Tel: +44 (0) 1379 69
Fax: +44 (0) 1379 65
email: [EMAIL PROTECTED]
web: http://www.synequanon.com
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[R] weighted rank correlation test?
Hi R-gurus! Is there a package that implements corr.test with weights, or will I have to deal with this myself? I found corr() from the boot package which will calculate r for me and I could certainly do the ranking myself but it does not give me a p-value... Thanks for any hints. cu Philipp -- Dr. Philipp PagelTel. +49-89-3187-3675 Institute for Bioinformatics / MIPS Fax. +49-89-3187-3585 GSF - National Research Center for Environment and Health Ingolstaedter Landstrasse 1 85764 Neuherberg, Germany __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Weird problem with median on a factor
Christoph Bier [EMAIL PROTECTED] writes: The following works: median(fbhint.spss1[,264], na.rm=T) [1] sehr wichtig Levels: sehr wichtig wichtig teils/teils unwichtig ganz unwichtig ... but here it doesn't: median(fbhint.spss1[,566], na.rm=T) Error in Summary.factor(..., na.rm = na.rm) : sum not meaningful for factors I don't have any ideas why! Can somebody give me a hint? Offhand, I'd guess that the median is inbetween two factor levels in one case and not in the other. However, both cases should give an error, especially for unordered factors, but it is not well-defined for ordered factors either. If you want to interpret your factor as a numeric scale, use as.numeric first. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] cross-classified random factors in lme without blocking
Gordon Smyth [EMAIL PROTECTED] writes: On page 165 of Mixed-Effects Models in S and S-Plus by Pinheiro and Bates there is an example of using lme() in the nlme package to fit a model with crossed random factors. The example assumes though that the data is grouped. Is it possible to use lme() to fit crossed random factors when the data is not grouped? E.g., y - rnorm(12); a=gl(4,1,12); b=gl(3,4,12). Can I fit an additive model like y~a+b but with a and b random? Everything I've tried gives an error: lme(y~1,random=~1|(a+b)) Error in switch(mode(object), name = , numeric = , call = object, character = as.name(object), : [[ cannot be of mode ( lme(y~1,random=~a+b) Error in getGroups.data.frame(dataMix, groups) : Invalid formula for groups A standard trick is to define a grouping with one level: one - rep(1,length(y) lme(, random=~pdBlocked(.)|one) (Sorry, I'm a little rusty on the syntax, but just follow the example in PB) AFAIR, it also works with random=list(a=~1,one=~b) and vice versa. (The model is the same but you get different DF calculations, none of which are correct in the completely balanced case...) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Weird problem with median on a factor
Peter Dalgaard wrote: Offhand, I'd guess that the median is inbetween two factor levels in one case and not in the other. Hm, maybe. A problem, that would not occur, if I used the median on the numeric data of this factor. However, both cases should give an error, especially for unordered factors, but it is not well-defined That's what I expected. for ordered factors either. If you want to interpret your factor as a numeric scale, use as.numeric first. Yes, I already understood this :-) (At last we received your book or rather three of your books and the MASS-book). Thanks for your answer. Regards, Christoph -- Christoph Bier, Dipl.Oecotroph., Email: [EMAIL PROTECTED] Universitaet Kassel, FG Oekologische Lebensmittelqualitaet und Ernaehrungskultur \\ Postfach 12 52 \\ 37202 Witzenhausen Tel.: +49 (0) 55 42 / 98 -17 21, Fax: -17 13 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Weird problem with median on a factor
Simon Fear schrieb: The last part of my message is what I thought might be the cause - maybe your median is not an integer, so what category should it be mapped to? What do you get if you slip in an `as.numeric` before calculating the median? [if still an error, then there is definitely something else going wrong to report] Then everything is ok. as.numeric(fbhint.spss1$V15.SP1) - tmp.data2 median(tmp.data2, na.rm=T) [1] 2 Christoph -- Christoph Bier, Dipl.Oecotroph., Email: [EMAIL PROTECTED] Universitaet Kassel, FG Oekologische Lebensmittelqualitaet und Ernaehrungskultur \\ Postfach 12 52 \\ 37202 Witzenhausen Tel.: +49 (0) 55 42 / 98 -17 21, Fax: -17 13 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] too long
william ritchie wrote:
Hi everyone,
I ve been using R for months and find it really
practical and straight forward.
However (the inevitable however), I am finding it very
slow for one of my operations:
it s basically an itertation over i and j in a pretty
big table (4* 4608). It takes 30 minutes
Thanks
You got at least 3 replies on your first message. Why do you post it
again???
Uwe Ligges
Ps:if it can help here is the source:
median1-matrix(nrow=4608,ncol=1)
median2-matrix(nrow=4608,ncol=1)
median3-matrix(nrow=4608,ncol=1)
median4-matrix(nrow=4608,ncol=1)
v-c(18,19,20,21,23)
for (i in 0:11)
{
for (j in 1:384)
{
median1[j+(i*384),]-puce[j+(i*384),5]+median(puce[v+384*i,2]-puce[v+384*i,5])
median2[j+(i*384),]-puce[j+(i*384),19]+median(puce[v+384*i,16]-puce[v+384*i,19])
median3[j+(i*384),]-puce[j+(i*384),12]+median(puce[v+384*i,9]-puce[v+384*i,12])
median4[j+(i*384),]-puce[j+(i*384),26]+median(puce[v+384*i,23]-puce[v+384*i,26])
puce[,5]-median1
puce[,19]-median2
puce[,12]-median3
puce[,26]-median4
}
}
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RE: [R] Weird problem with median on a factor
Final guess as to observed behaviour: in the first case after
removal of NAs there were an odd number of observations
(so that sum was not called within the code for median).
In your second call I suspect that even though you got
an integer answer, it was found as sum(2,2)/2.
It seems to me the best way to deal with this bug would
be to make calling median with a factor argument be an
immediate error. Or just trust users never to attempt such
a thing ...
Simon Fear
Senior Statistician
Syne qua non Ltd
Tel: +44 (0) 1379 69
Fax: +44 (0) 1379 65
email: [EMAIL PROTECTED]
web: http://www.synequanon.com
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[R] strange logLik results in gls (nlme)
I am trying to analyse a data with gls/lm using the following set of models prcn.0.lm - lm( log10(Y)~(cond-1)+(cond-1):t ,prcn) prcn.1.gls - gls( log10(Y)~(cond-1)+(cond-1):t ,prcn,cor=corAR1()) prcn.0.gls - gls( log10(Y)~(cond-1)+(cond-1):t ,prcn) prcn.1m.gls - gls( log10(Y)~(cond-1)+(cond-1):t ,prcn,cor=corAR1(),method=ML) I get the following AICs for these models: AIC(prcn.1m.gls) [1] -78.3 AIC(prcn.1.gls) [1] -46.3 AIC(prcn.0.gls) [1] -24.7 AIC(prcn.0.lm) [1] -59.8 It is the difference between the last two, which puzzles me. They are the same models. So I can't compare the AICs of prcn.0.lm and prcn.1.gls directly. When using anova() for the comparison, I get a sensible result: anova(prcn.1.gls,prcn.0.lm) Model df AICBIC logLik Test L.Ratio p-value prcn.1.gls 1 6 -46.3 -28.62 29.1 prcn.0.lm 2 5 -24.7 -9.97 17.3 1 vs 223.6 .0001 Multiple arguments in AIC() give: AIC(prcn.1.gls,prcn.0.lm) df AIC prcn.1.gls 6 -46.3 prcn.0.lm 5 -59.8 How can I be sure to make it right? __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] How to grow an R object from C (.Call Interface)
What is the best way to grow an R return object in writing a C function using the Rdefines.h macros. In my application, the final size of the return object is not known during construction. My understanding is that each time I grow an R object I have to use PROTECT() again, probably before UNPROTECTing the smaller version. However, due to the stack character of the PROTECT mechanism, UNPROTECT would not work to remove the smaller one, after the bigger has been protected. Is this an indication to use UNPROTECT_PTR ? Or is another approach recommended? May be the solution to this is worth a sentence in Wrtiting R Extensions. Thanks for any help Jens Oehlschlägel -- NEU FÜR ALLE - GMX MediaCenter - für Fotos, Musik, Dateien... Fotoalbum, File Sharing, MMS, Multimedia-Gruß, GMX FotoService Jetzt kostenlos anmelden unter http://www.gmx.net +++ GMX - die erste Adresse für Mail, Message, More! +++ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] How to grow an R object from C (.Call Interface)
I think the solution is PROTECT_WITH_INDEX. From the Rinternals.h file /* We sometimes need to coerce a protected value and place the new coerced value under protection. For these cases PROTECT_WITH_INDEX saves an index of the protection location that can be used to replace the protected value using REPROTECT. */ typedef int PROTECT_INDEX; #define PROTECT_WITH_INDEX(x,i) R_ProtectWithIndex(x,i) #define REPROTECT(x,i) R_Reprotect(x,i) You can see examples in dataentry.c, optim.c and elsewhere. I agree that should be mentioned in R-exts. On Fri, 31 Oct 2003, Jens Oehlschlägel wrote: What is the best way to grow an R return object in writing a C function using the Rdefines.h macros. In my application, the final size of the return object is not known during construction. My understanding is that each time I grow an R object I have to use PROTECT() again, probably before UNPROTECTing the smaller version. However, due to the stack character of the PROTECT mechanism, UNPROTECT would not work to remove the smaller one, after the bigger has been protected. Is this an indication to use UNPROTECT_PTR ? Or is another approach recommended? May be the solution to this is worth a sentence in Wrtiting R Extensions. Thanks for any help Jens Oehlschlägel -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] How to grow an R object from C (.Call Interface)
Jens Oehlschlägel [EMAIL PROTECTED] writes: What is the best way to grow an R return object in writing a C function using the Rdefines.h macros. In my application, the final size of the return object is not known during construction. My understanding is that each time I grow an R object I have to use PROTECT() again, probably before UNPROTECTing the smaller version. However, due to the stack character of the PROTECT mechanism, UNPROTECT would not work to remove the smaller one, after the bigger has been protected. Is this an indication to use UNPROTECT_PTR ? Or is another approach recommended? May be the solution to this is worth a sentence in Wrtiting R Extensions. Thanks for any help It depends on the kind of object, I'll assume we're talking vector objects here. If you're *extending* an existing object (the new object becomes part of the old object) matters are quite different. I think you are looking for the PROTECT_WITH_INDEX and REPROTECT mechanism that Luke added (I almost said recently, but it was in fact three years ago!). If that is not in the manual, it should be. UNPROTECT_PTR is older and solves a similar problem, but where the routine that unprotects is not the same as the one that protects (this happens a lot in the parser code) and you cannot be sure that it is the top item that needs to be removed. UNPROTECT_PTR works by searching the stack for the pointer, pulling the record out of the stack, and dropping every stack element on top of it. In contrast REPROTECT knows which record to extract and just changes the pointer value in it. I.e. UNPROTECT_PTR is potentially much less efficient, although I don't think there are practical cases where more than an handful of items have been pushed on the stack (the usual cases are zero or one). Also remember not to call any routine that could trigger a garbage collection while you need access to both the old and the new extended object. So, allocate, copy, reprotect, and *then* start computing new entries. -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] cross-classified random factors in lme without blocking
Peter Dalgaard [EMAIL PROTECTED] writes: Gordon Smyth [EMAIL PROTECTED] writes: On page 165 of Mixed-Effects Models in S and S-Plus by Pinheiro and Bates there is an example of using lme() in the nlme package to fit a model with crossed random factors. The example assumes though that the data is grouped. Is it possible to use lme() to fit crossed random factors when the data is not grouped? E.g., y - rnorm(12); a=gl(4,1,12); b=gl(3,4,12). Can I fit an additive model like y~a+b but with a and b random? Everything I've tried gives an error: lme(y~1,random=~1|(a+b)) Error in switch(mode(object), name = , numeric = , call = object, character = as.name(object), : [[ cannot be of mode ( lme(y~1,random=~a+b) Error in getGroups.data.frame(dataMix, groups) : Invalid formula for groups A standard trick is to define a grouping with one level: one - rep(1,length(y) lme(, random=~pdBlocked(.)|one) (Sorry, I'm a little rusty on the syntax, but just follow the example in PB) AFAIR, it also works with random=list(a=~1,one=~b) and vice versa. Not sure about that. (The model is the same but you get different DF calculations, none of which are correct in the completely balanced case...) I realize that it is awkward to use lme to fit models with crossed random effects. As Saikat DebRoy and I described in a recent preprint http://www.stat.wisc.edu/~bates/reports/MultiComp.pdf we now have a good handle on the computational methods for mixed-effects models with nested or crossed or partially crossed random effects. Both the nlme and the lme4 packages are based on structures that are tuned to nested random effects and do not easily accomodate crossed random effects. I have a draft of the contents of classes and methods for fitting linear mixed-effects models with nested or crossed or ... but it is a long way from the draft to working, tested code. Although it will take some time to get all the pieces in place I do offer some encouragement that this awkward phrasing of crossed random effects will some day be behind us. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] cross-classified random factors in lme without blocking
Douglas Bates [EMAIL PROTECTED] writes: (Sorry, I'm a little rusty on the syntax, but just follow the example in PB) AFAIR, it also works with random=list(a=~1,one=~b) and vice versa. Not sure about that. Sorry. It's certainly not correct as written. It has to be something like list(a=1,one=pdIdent(form=~b-1)) otherwise you get a general symmetric covariance for the effect of b. (The model is the same but you get different DF calculations, none of which are correct in the completely balanced case...) I realize that it is awkward to use lme to fit models with crossed random effects. As Saikat DebRoy and I described in a recent preprint http://www.stat.wisc.edu/~bates/reports/MultiComp.pdf .../MixedComp.pdf, right? we now have a good handle on the computational methods for mixed-effects models with nested or crossed or partially crossed random effects. Both the nlme and the lme4 packages are based on structures that are tuned to nested random effects and do not easily accomodate crossed random effects. I have a draft of the contents of classes and methods for fitting linear mixed-effects models with nested or crossed or ... but it is a long way from the draft to working, tested code. Although it will take some time to get all the pieces in place I do offer some encouragement that this awkward phrasing of crossed random effects will some day be behind us. Looking forward to it... :-) -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] cross-classified random factors in lme without blocking
Douglas Bates [EMAIL PROTECTED] writes: ... I realize that it is awkward to use lme to fit models with crossed random effects. As Saikat DebRoy and I described in a recent preprint http://www.stat.wisc.edu/~bates/reports/MultiComp.pdf we now have a good handle on the computational methods for mixed-effects models with nested or crossed or partially crossed random effects. That URL should have been http://www.stat.wisc.edu/~bates/reports/MixedComp.pdf __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Re: packaging a package addon
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED] On Thu, 30 Oct 2003, Ross Boylan wrote: [...] Finally, a comment on R CMD check: perhaps it could produce some more of the output when things fail? I found that to diagnose the loading problems as I developed this, I had to attempt to load the package myself in R to see what the actual problem was. There wasn't enough info in the R CMD check to tell what exactly the problem was. I think there usually is, but you have to look in the log file or one of the example files. But I would not be doing R CMD check until I had both installed and loaded the package and run a few examples. I often do what Ross does, just because it's easy to do... Is it possible to get R to at least print the trackback on error when checking the package, so there's a bit more info in the log file? Best, Andy -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Creating packages in 1.8
Hi,
I decided to upgrade to 1.8 today... :-)
Anyway, we are writing our own package that is dependent on a
bioconductor library - 'affy'.
I've checked and when I fire up R, library(affy) behaves as expected...
so it all seems to be installed and OK...
In the DESCRIPTION file in my package source I have the line:
Depends: affy
When I run R CMD check simpleaffy
I get to:
...
* checking package dependencies ... ERROR
Packages required but not available:
WARNING: ignoring environment value of R_HOME
Prompt
Any ideas what is going on - as far as I can see the only dependency is
to affy which is there and OK... I get no list of packages that are
missing :-(
I'm assuming the warning comes because I have R_ENVIRON pointing to a
.Renviron file in my home directory...
Any help would be much appreciated!
Cheers,
Crispin
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[R] print(), cat() and simple I/O in R
I am trying to produce rather mundane output of the form e.g. pi, e = 3.14 2.718 The closest result I achieved so far with print() is: print (c(pi, exp(1)), digits = 3) [1] 3.14 2.72 print(c(pi, e =, pi, exp(1)), digits = 3) [1] pi, e = 3.14159265358979 2.71828182845905 I understand that c() promotes floats to strings and this is why I get what I get. and with cat() (it apparently does not have equivalent of digits parameter) cat (pi, e =, pi, exp(1), \n) pi, e = 3.141593 2.718282 Any pointers with respect how can I print what I want to print would be greatly appreciated. Ryszard [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] print(), cat() and simple I/O in R
Have you considered round and paste? hope this helps. spencer graves [EMAIL PROTECTED] wrote: I am trying to produce rather mundane output of the form e.g. pi, e = 3.14 2.718 The closest result I achieved so far with print() is: print (c(pi, exp(1)), digits = 3) [1] 3.14 2.72 print(c(pi, e =, pi, exp(1)), digits = 3) [1] pi, e = 3.14159265358979 2.71828182845905 I understand that c() promotes floats to strings and this is why I get what I get. and with cat() (it apparently does not have equivalent of digits parameter) cat (pi, e =, pi, exp(1), \n) pi, e = 3.141593 2.718282 Any pointers with respect how can I print what I want to print would be greatly appreciated. Ryszard [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] print(), cat() and simple I/O in R
See format(), formatC() and sprintf(). /Henrik -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: den 31 oktober 2003 16:48 To: [EMAIL PROTECTED] Subject: [R] print(), cat() and simple I/O in R I am trying to produce rather mundane output of the form e.g. pi, e = 3.14 2.718 The closest result I achieved so far with print() is: print (c(pi, exp(1)), digits = 3) [1] 3.14 2.72 print(c(pi, e =, pi, exp(1)), digits = 3) [1] pi, e = 3.14159265358979 2.71828182845905 I understand that c() promotes floats to strings and this is why I get what I get. and with cat() (it apparently does not have equivalent of digits parameter) cat (pi, e =, pi, exp(1), \n) pi, e = 3.141593 2.718282 Any pointers with respect how can I print what I want to print would be greatly appreciated. Ryszard [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailma n/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] print(), cat() and simple I/O in R
?format ?formatC Date: Fri, 31 Oct 2003 10:47:36 -0500 From: [EMAIL PROTECTED] Sender: [EMAIL PROTECTED] Precedence: list I am trying to produce rather mundane output of the form e.g. pi, e = 3.14 2.718 The closest result I achieved so far with print() is: print (c(pi, exp(1)), digits = 3) [1] 3.14 2.72 print(c(pi, e =, pi, exp(1)), digits = 3) [1] pi, e = 3.14159265358979 2.71828182845905 I understand that c() promotes floats to strings and this is why I get what I get. and with cat() (it apparently does not have equivalent of digits parameter) cat (pi, e =, pi, exp(1), \n) pi, e = 3.141593 2.718282 Any pointers with respect how can I print what I want to print would be greatly appreciated. Ryszard [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- __ [ ] [ Giovanni Petris [EMAIL PROTECTED] ] [ Department of Mathematical Sciences ] [ University of Arkansas - Fayetteville, AR 72701 ] [ Ph: (479) 575-6324, 575-8630 (fax) ] [ http://definetti.uark.edu/~gpetris/ ] [__] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] constrained nonlinear optimisation in R?
Hello. I have searched the archives but have not found anything. I need to solve a constrained optimisation problem for a nonlinear function (maximum entropy formalism). Specifically, Optimise: -1*SUM(p_ilog(p_i)) for a vector p_i of probabilities, conditional on a series of constraints of the form: SUM(T_i*p_i)=k_i for given values of T_i and k_i (these are constraints on expectations). Can this be done in R? Bill Shipley Associate Editor, Ecology North American Editor, Annals of Botany Département de biologie, Université de Sherbrooke, Sherbrooke (Québec) J1K 2R1 CANADA [EMAIL PROTECTED] http://callisto.si.usherb.ca:8080/bshipley/ http://callisto.si.usherb.ca:8080/bshipley/ [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Change in 'solve' for r-patched
Dear R help Thanks to Professor Bates for his information about how to calculate least square estimates [not using solve] in ``the right way''. This is very useful indead, I am clearly one of of the package maintainers who is not using using solve in a proper way at the moment. However, the calculations in my code look more like GLS than LS. ## GLS could in principlpe be implemented like this : betahat - solve(t(X) %*% solve(Omega)%*% X) %*% t(X)%*%solve(Omega)%*% y ## where Omega is a strictly p.d. symmetric matrix Does someone have a recommendation on how to do this in ``the right way'' ? My first attempt (trying to imitate the LS solution recommended by Prof. Bates) is : temp - backsolve(chol(Omega),cbind(X,y)) betahat - qr.coef(qr(temp[,1:ncol(X)]), temp[,ncol(X)+1]) Thank you in advance for any help Cheers Ole -- Ole F. Christensen Center for Bioinformatik Datalogisk Institut Aarhus Universitet Ny Munkegade, Bygning 540 8000 Aarhus C Denmark __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] constrained nonlinear optimisation in R?
Hello. I have searched the archives but have not found anything. I need to solve a constrained optimisation problem for a nonlinear function (maximum entropy formalism). Specifically, Optimise: -1*SUM(p_ilog(p_i)) for a vector p_i of probabilities, conditional on a series of constraints of the form: SUM(T_i*p_i)=k_i for given values of T_i and k_i (these are constraints on expectations). A better answer may exist to this question, but here goes anyway Could you use sequential quaratic programming here (i.e. just constrain the QP problem generated at each iterate of Newton's method)? There's an R library for quadratic programming Simon _ Simon Wood [EMAIL PROTECTED]www.stats.gla.ac.uk/~simon/ Department of Statistics, University of Glasgow, Glasgow, G12 8QQ Direct telephone: (0)141 330 4530 Fax: (0)141 330 4814 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] constrained nonlinear optimisation in R?
Other alternatives to the R library for quadratic programming: 1. What are the nature of your constraints? optim will optimize a function with optional box constraints. constrOptim will optimize a function subject to linear inequality constraints. 2. If you want to estimate the p[i]'s, i = 1, ..., k, I would recommend a multivariate logistic transformation to (k-1) unconstrained variables. I have had serious difficulties with constrained optimizers testing values outside the constraints and then stopping because the objective function misbehaved. I don't know if optim does this, but I don't even try constrained optimization if I can find a sensible, unconstrained parameterization. Often, confidence regions, etc., are better behaved in the unconstrained space as well. hope this helps. spencer graves Simon Wood wrote: Hello. I have searched the archives but have not found anything. I need to solve a constrained optimisation problem for a nonlinear function (?maximum entropy formalism?). Specifically, Optimise: -1*SUM(p_ilog(p_i)) for a vector p_i of probabilities, conditional on a series of constraints of the form: SUM(T_i*p_i)=k_i for given values of T_i and k_i (these are constraints on expectations). A better answer may exist to this question, but here goes anyway Could you use sequential quaratic programming here (i.e. just constrain the QP problem generated at each iterate of Newton's method)? There's an R library for quadratic programming Simon _ Simon Wood [EMAIL PROTECTED]www.stats.gla.ac.uk/~simon/ Department of Statistics, University of Glasgow, Glasgow, G12 8QQ Direct telephone: (0)141 330 4530 Fax: (0)141 330 4814 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Creating packages in 1.8
Hi,
Firstly, sorry to reply to my own posting... Also for not having done
more trawling before I sent the original message...
The problems we are having were down to R being installed in a different
directory to the default one (for various sysadmin reasons)...
We've successfully got R runnning on a single machine in the default
directories where it likes to be - it is happy, and 'R CMD check' works
fine... :-)
Alas, this is not a good long-term solution for us can anyone help us
work out the easiest way to get R installed in another directory
structure instead of the default ones?
I appreciate that very similar questions have been asked before, but I
suspect that 1.8 is slightly different (since R CMD check ... worked
fine for us in 1.7.1).
Crispin
-Original Message-
From: Crispin Miller
Sent: 31 October 2003 14:51
To: [EMAIL PROTECTED]
Subject: [R] Creating packages in 1.8
Hi,
I decided to upgrade to 1.8 today... :-)
Anyway, we are writing our own package that is dependent on a
bioconductor library - 'affy'. I've checked and when I fire
up R, library(affy) behaves as expected... so it all seems to
be installed and OK...
In the DESCRIPTION file in my package source I have the line:
Depends: affy
When I run R CMD check simpleaffy
I get to:
...
* checking package dependencies ... ERROR
Packages required but not available:
WARNING: ignoring environment value of R_HOME
Prompt
Any ideas what is going on - as far as I can see the only
dependency is to affy which is there and OK... I get no list
of packages that are missing :-(
I'm assuming the warning comes because I have R_ENVIRON
pointing to a .Renviron file in my home directory...
Any help would be much appreciated!
Cheers,
Crispin
This email is confidential and intended solely for the use
o...{{dropped}}
__
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https://www.stat.math.ethz.ch/mailman/listinfo /r-help
This email is confidential and intended solely for the use o...{{dropped}}
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Re: [R] Change in 'solve' for r-patched
GLS is usually solved by taking a matrix square root and converting to least squares (and BTW you might want to use lm.fit not solve for least squares to allow for aliased columns). So your idea is right (and is the one given in my 1981 Spatial Statistics book). These days I would usually use an eigendecomposition instead of Cholesky as it will enable you to cope better with nearly non-positive-definite Omega. See lm.gls in package MASS for an outline implementation. On Fri, 31 Oct 2003, Ole F. Christensen wrote: Dear R help Thanks to Professor Bates for his information about how to calculate least square estimates [not using solve] in ``the right way''. This is very useful indead, I am clearly one of of the package maintainers who is not using using solve in a proper way at the moment. However, the calculations in my code look more like GLS than LS. ## GLS could in principlpe be implemented like this : betahat - solve(t(X) %*% solve(Omega)%*% X) %*% t(X)%*%solve(Omega)%*% y ## where Omega is a strictly p.d. symmetric matrix Does someone have a recommendation on how to do this in ``the right way'' ? My first attempt (trying to imitate the LS solution recommended by Prof. Bates) is : temp - backsolve(chol(Omega),cbind(X,y)) betahat - qr.coef(qr(temp[,1:ncol(X)]), temp[,ncol(X)+1]) Thank you in advance for any help Cheers Ole -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] constrained nonlinear optimisation in R?
Simon Wood [EMAIL PROTECTED] writes:
Hello. I have searched the archives but have not found anything. I
need to solve a constrained optimisation problem for a nonlinear
function (maximum entropy formalism). Specifically,
Optimise: -1*SUM(p_ilog(p_i)) for a vector p_i of probabilities,
conditional on a series of constraints of the form:
SUM(T_i*p_i)=k_i for given values of T_i and k_i (these are
constraints on expectations).
A better answer may exist to this question, but here goes anyway
Could you use sequential quaratic programming here (i.e. just constrain
the QP problem generated at each iterate of Newton's method)? There's an R
library for quadratic programming
Simon
_
Simon Wood [EMAIL PROTECTED]www.stats.gla.ac.uk/~simon/
Department of Statistics, University of Glasgow, Glasgow, G12 8QQ
Direct telephone: (0)141 330 4530 Fax: (0)141 330 4814
__
[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
help.search(constrained) suggests:
constrOptim(base) Linearly constrained optimisation
which might do the trick.
--
[EMAIL PROTECTED]http://www.analytics.washington.edu/
Biomedical and Health Informatics University of Washington
Biostatistics, SCHARP/HVTN Fred Hutchinson Cancer Research Center
UW (Tu/Th/F): 206-616-7630 FAX=206-543-3461 | Voicemail is unreliable
FHCRC (M/W): 206-667-7025 FAX=206-667-4812 | use Email
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RE: [R] Creating packages in 1.8
On what OS?
It's easy on Windows: choose the directory in the installer.
It's easy for a source build on Unix: set --prefix at configure time.
It's easy for rpms if you know how.
I suspect it is easy for .debs.
And there is no R 1.8: it is R 1.8.0.
On Fri, 31 Oct 2003, Crispin Miller wrote:
Hi,
Firstly, sorry to reply to my own posting... Also for not having done
more trawling before I sent the original message...
The problems we are having were down to R being installed in a different
directory to the default one (for various sysadmin reasons)...
We've successfully got R runnning on a single machine in the default
directories where it likes to be - it is happy, and 'R CMD check' works
fine... :-)
Alas, this is not a good long-term solution for us can anyone help us
work out the easiest way to get R installed in another directory
structure instead of the default ones?
I appreciate that very similar questions have been asked before, but I
suspect that 1.8 is slightly different (since R CMD check ... worked
fine for us in 1.7.1).
Crispin
-Original Message-
From: Crispin Miller
Sent: 31 October 2003 14:51
To: [EMAIL PROTECTED]
Subject: [R] Creating packages in 1.8
Hi,
I decided to upgrade to 1.8 today... :-)
Anyway, we are writing our own package that is dependent on a
bioconductor library - 'affy'. I've checked and when I fire
up R, library(affy) behaves as expected... so it all seems to
be installed and OK...
In the DESCRIPTION file in my package source I have the line:
Depends: affy
When I run R CMD check simpleaffy
I get to:
...
* checking package dependencies ... ERROR
Packages required but not available:
WARNING: ignoring environment value of R_HOME
Prompt
Any ideas what is going on - as far as I can see the only
dependency is to affy which is there and OK... I get no list
of packages that are missing :-(
I'm assuming the warning comes because I have R_ENVIRON
pointing to a .Renviron file in my home directory...
Any help would be much appreciated!
Cheers,
Crispin
This email is confidential and intended solely for the use
o...{{dropped}}
__
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] Fatal error in SJava.
Hi can you please send me your compiled SJava package with the modified REmbed.c because in Windows i'm not able to recompile the package because of errors Thanks very much Markus Helbig [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] dnorm() lead to a probability 1
Howdee, One of my student spotted something I can't explain: a probability 1 vs a normal probability density function. dnorm(x=1, mean=1, sd=0.4) [1] 0.9973557 dnorm(x=1, mean=1, sd=0.39) [1] 1.022929 dnorm(x=1, mean=1, sd=0.3) [1] 1.329808 dnorm(x=1, mean=1, sd=0.1) [1] 3.989423 dnorm(x=1, mean=1, sd=0.01) [1] 39.89423 dnorm(x=1, mean=1, sd=0.001) [1] 398.9423 Is there a bug with the algorithm? Thanks, Marc Marc Bélisle Professeur adjoint Département de biologie Université de Sherbrooke 2500 boul. de l'Université Sherbrooke, Québec J1K 2R1 CANADA Tél: +1-819-821-8000 poste 1313 Fax: +1-819-821-8049 Courriél: [EMAIL PROTECTED] Site Web: www.usherbrooke.ca/biologie/recherche/ecologie/Belisle/belisle.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] question about optim
Hello, When we use optim and run into errors, such as generates NA/NaN/Inf, and the routine stops because of this error, is there a way to print the values of the argument of the functions where the error occurs? For example, I am doing minimizing negative log likelihood and when the optim stops because of those errors, I'd like to know what was the parameter values at that time. I heard nlm has something called print.level and I am wondering there is a way for optim as well. Thank you. Mikyoung Jun __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] dnorm() lead to a probability 1
On Fri, Oct 31, 2003 at 02:40:15PM -0500, Marc Belisle wrote: Howdee, One of my student spotted something I can't explain: a probability 1 vs a normal probability density function. The integral has to be 1 --- but dnorm doesn't compute that. You were probably looking for pnorm(), and it will give you 0.5 for all those cases (where x==mean) as you'd expect. Hth, Dirk dnorm(x=1, mean=1, sd=0.4) [1] 0.9973557 dnorm(x=1, mean=1, sd=0.39) [1] 1.022929 dnorm(x=1, mean=1, sd=0.3) [1] 1.329808 dnorm(x=1, mean=1, sd=0.1) [1] 3.989423 dnorm(x=1, mean=1, sd=0.01) [1] 39.89423 dnorm(x=1, mean=1, sd=0.001) [1] 398.9423 Is there a bug with the algorithm? Thanks, Marc Marc B?lisle Professeur adjoint D?partement de biologie Universit? de Sherbrooke 2500 boul. de l'Universit? Sherbrooke, Qu?bec J1K 2R1 CANADA T?l: +1-819-821-8000 poste 1313 Fax: +1-819-821-8049 Courri?l: [EMAIL PROTECTED] Site Web: www.usherbrooke.ca/biologie/recherche/ecologie/Belisle/belisle.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Those are my principles, and if you don't like them... well, I have others. -- Groucho Marx __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] strange sprintf() behaviour ?
This is quite strange behaviour - at least for R-novice as myself
Consider this:
testf - function() { x -2; sprintf(%s %f, x =, x); return(x) }
result - testf()
testf - function() { x -2; sprintf(%s %f, x =, x) }
result - testf()
testf()
[1] x = 2.00
Apparently adding return() statement and invoking function like this
result - testf()
suppresses output from sprintf()
Output from print() is NOT suppressed:
testf - function() { x -2; print(c(x =, x)) }
result - testf()
[1] x = 2
testf - function() { x -2; print(c(x =, x)); return(x) }
result - testf()
[1] x = 2
Is there a way to use sprintf() inside a function ?
I guess I can say: print(sprintf()) - is it the only solution for this ?
R
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Re: [R] question about optim
You might want to try using options(error = recover) or perhaps options(error = browser). -roger Mikyoung Jun wrote: Hello, When we use optim and run into errors, such as generates NA/NaN/Inf, and the routine stops because of this error, is there a way to print the values of the argument of the functions where the error occurs? For example, I am doing minimizing negative log likelihood and when the optim stops because of those errors, I'd like to know what was the parameter values at that time. I heard nlm has something called print.level and I am wondering there is a way for optim as well. Thank you. Mikyoung Jun __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] dnorm() lead to a probability 1
On Friday 31 October 2003 20:40, Marc Belisle wrote: Howdee, One of my student spotted something I can't explain: a probability 1 vs a normal probability density function. dnorm(x=1, mean=1, sd=0.4) [1] 0.9973557 dnorm(x=1, mean=1, sd=0.39) [1] 1.022929 dnorm(x=1, mean=1, sd=0.3) [1] 1.329808 dnorm(x=1, mean=1, sd=0.1) [1] 3.989423 dnorm(x=1, mean=1, sd=0.01) [1] 39.89423 dnorm(x=1, mean=1, sd=0.001) [1] 398.9423 Is there a bug with the algorithm? The *area* under the density curve corresponds to the probability in the corresponding interval...as you might have learned in a statistics course. So it's perfeclty alright for a density function to exceed 1 if the area under the whole curve still equals one. Immediately obvious for curve(dunif(x, min = 0, max = 0.5)) hth, Z Thanks, Marc Marc Bélisle Professeur adjoint Département de biologie Université de Sherbrooke 2500 boul. de l'Université Sherbrooke, Québec J1K 2R1 CANADA Tél: +1-819-821-8000 poste 1313 Fax: +1-819-821-8049 Courriél: [EMAIL PROTECTED] Site Web: www.usherbrooke.ca/biologie/recherche/ecologie/Belisle/belisle.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] dnorm() lead to a probability 1
Dnorm isn't the probability - it's the y-value on the density function. Try plotting it - it makes a nice normal plot. See ?dnorm for definition. Bob -Original Message- From: Marc Belisle [mailto:[EMAIL PROTECTED] Sent: Friday, October 31, 2003 2:40 PM To: R-Help Subject: [R] dnorm() lead to a probability 1 Howdee, One of my student spotted something I can't explain: a probability 1 vs a normal probability density function. dnorm(x=1, mean=1, sd=0.4) [1] 0.9973557 dnorm(x=1, mean=1, sd=0.39) [1] 1.022929 dnorm(x=1, mean=1, sd=0.3) [1] 1.329808 dnorm(x=1, mean=1, sd=0.1) [1] 3.989423 dnorm(x=1, mean=1, sd=0.01) [1] 39.89423 dnorm(x=1, mean=1, sd=0.001) [1] 398.9423 Is there a bug with the algorithm? Thanks, Marc Marc Bélisle Professeur adjoint Département de biologie Université de Sherbrooke 2500 boul. de l'Université Sherbrooke, Québec J1K 2R1 CANADA Tél: +1-819-821-8000 poste 1313 Fax: +1-819-821-8049 Courriél: [EMAIL PROTECTED] Site Web: www.usherbrooke.ca/biologie/recherche/ecologie/Belisle/belisle.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Change in 'solve' for r-patched
Hi all, Thanks Douglas for bringing up numerical stability and OLS regression coefficients. I'm often worried about the speed of regression runs, as I sometimes run large simulations or resamplings. Inspired, and prone to run many regressions, I did a little simulation as follows. XP results are on a 2.4Ghz Pentium 4 in Windows from the binaries (without BLAS?) and Linux results are on a 1.5Ghz Pentium M laptop with a Pentium Atlas BLAS. Setting up the data: x is rnormals with a vector of ones prepended... x - cbind(rep(1,100), matrix(rnorm(1000), nc=10)) beta0 - runif(11) y - x %*% beta0 + rnorm(100) Estimating Slope Coefficients: system.time(for (i in 1:1000) lm( y~ x -1)) XP: [1] 5.91 0.00 5.90 NA NA Linux: [1] 5.27 0.01 5.28 0.00 0.00 system.time(for (i in 1:1000) solve(t(x) %*% x) %*% t(x) %*% y) XP: [1] 0.64 0.01 0.65 NA NA Linux: [1] 0.51 0.01 0.57 0.00 0.00 system.time(for (i in 1:1000) qr.coef(qr(x), y) ) XP: [1] 0.75 0.00 0.75 NA NA Linux: [1] 0.76 0.00 0.77 0.00 0.00 system.time(for (i in 1:1000) solve(crossprod(x), crossprod(x,y))) XP: [1] 0.45 0.00 0.53 NA NA Linux: [1] 0.35 0.00 0.36 0.00 0.00 On both platforms, BLAS or not, the solve(crossprod()) method works the fastest, with the naïve solve(t(x)%*%x) second-fastest. Calculating (t(x)%*%x)^-1: system.time(for (i in 1:1000) chol2inv(qr.R(qr(x XP: [1] 0.44 0.00 0.44 NA NA Linux: [1] 0.40 0.00 0.40 0.00 0.00 system.time(for (i in 1:1000) solve(crossprod(x)) ) XP: [1] 0.58 0.01 0.59 NA NA Linux: [1] 0.34 0.00 0.34 0.00 0.00 Where the chol2inv() method was faster in Windows but slower in Linux, which I'm guessing is due to differential uses of BLAS functions. None of these address the problem of numerical accuracy, which was the thrust of Doug's comments. Has anyone done a quick simulation to investigate the stability of the solutions? Is it small coefficients or near-collinearity (or both) that one has to worry about? Are the solve(crossprod()) methods obviously unstable? They surely do work quickly! Thanks again R-universe. I've needed no other statistical software for the last 2 years. I hope to get around to contributing a package or two soon. Elliot. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Summing elements in a list
The following seems to give what you want.
tmp - rep(w, each=nrow1*ncol1+nrow2*ncol2) * unlist(matlist)
tmp - rowSums(matrix(tmp,nr=nrow1*ncol1+nrow2*ncol2))
mat1 - tmp[1:(nrow1*ncol1)]; dim(mat1) - c(nrow1,ncol1)
mat2 - tmp[nrow1*ncol1+1:(nrow2*ncol2)]; dim(mat2) - c(nrow2,ncol2)
Giovanni
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[ Department of Mathematical Sciences ]
[ University of Arkansas - Fayetteville, AR 72701 ]
[ Ph: (479) 575-6324, 575-8630 (fax) ]
[ http://definetti.uark.edu/~gpetris/ ]
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Date: Fri, 31 Oct 2003 14:16:31 -0500
From: Angelo Canty [EMAIL PROTECTED]
Sender: [EMAIL PROTECTED]
Organization: McMaster University
Precedence: list
Hi,
Suppose that I have a list where each component is a list of two
matrices. I also have a vector of weights. How can I collapse my
list of lists into a single list of two matrices where each matrix
in the result is the weighted sum of the corresponding matrices.
I could use a loop but this is a nested calculation so I was hoping
there is a more efficient way to do this. To help clarify, here is
the code I would use with a for loop
result - list(mat1=matrix(0,nrow1,ncol1),
mat2=matrix(0,nrow2,ncol2))
for (i in seq(along=matlist)) {
result$mat1 - result$mat1+w[i]*matlist[[i]]$mat1
result$mat2 - result$mat2+w[i]*matlist[[i]]$mat2
}
I apologise if this is a trivial question. Unfortunately I don't have
my copy of VR S Programming to hand.
Thanks for your help,
Angelo
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| Mathematics and Statistics Phone: (905) 525-9140 x 27079 |
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Re: [R] Summing elements in a list
Angelo Canty wrote:
Hi,
Suppose that I have a list where each component is a list of two
matrices. I also have a vector of weights. How can I collapse my
list of lists into a single list of two matrices where each matrix
in the result is the weighted sum of the corresponding matrices.
I could use a loop but this is a nested calculation so I was hoping
there is a more efficient way to do this. To help clarify, here is
the code I would use with a for loop
result - list(mat1=matrix(0,nrow1,ncol1),
mat2=matrix(0,nrow2,ncol2))
for (i in seq(along=matlist)) {
result$mat1 - result$mat1+w[i]*matlist[[i]]$mat1
result$mat2 - result$mat2+w[i]*matlist[[i]]$mat2
}
I apologise if this is a trivial question. Unfortunately I don't have
my copy of VR S Programming to hand.
Here is one possibility:
result - list(
mat1 = matrix(rowSums(sapply(matlist, function(x)x$mat1) %*% diag(w)), nrow1, ncol1)
mat2 = matrix(rowSums(sapply(matlist, function(x)x$mat2) %*% diag(w)), nrow2, ncol2)
)
Warning: It doesn't have the readability that the original code has though.
Paul.
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[R] problem with tune.svm
rng - list(gamma = 2^(-1:1), cost = 2^(2:4)) rng $gamma [1] 0.5 1.0 2.0 $cost [1] 4 8 16 obj - tune.svm(pIC50 ~ ., data = data, ranges = rng) Error in tune(svm, train.x = x, data = data, ranges = ranges, ...) : formal argument ranges matched by multiple actual arguments Ay idea why ??? Ryszard [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Weird problem with median on a factor
Beating a dead horse...
I am an R beginner trying to understand this factor business. While the
entire business of finding the median of factor may be silly from a
practical point of view, this email chain has helped me understand
something.
I have looked at the median function and it tests to see if what is passed
to it is numeric. If I were building a function, if I tested for mode
numeric, and if something told me it was numeric then like the median
function I would naively assume that I could do arithmetic on it:
saywhut-as.factor(c(NA,1,1,1,1,2,10))
mode(saywhut)
[1] numeric
It appears to me that the when the median function tests for numeric it
doesn't have the desired result with an object of class factor (and maybe
other classes?) as was shown by the example.
I have a suspicion that something of class factor has at least two pieces,
one of which is the levels which can possibly be character or something else
and the other piece is the ordering of the levels which is of storage.mode
integer. Is it this ordering that determines the mode of the factor??
But if the mode of factor is truly numeric, why doesn't the median function
use the numeric piece for finding the median (like it did with odd n - not
that anyone would ever really want the median of a factor:)?? I think that
Simon Fear hit on the right idea because of the definition of median that is
used for an even number of observations takes the sum of the ordered middle
two observations. It is the sum (called by the median function) that chokes
on a factor.
sum(saywhut,na.rm=T)
Error in Summary.factor(..., na.rm = na.rm) :
sum not meaningful for factors
It appears that whoever built the sum function built in a test for factor
(Simon Fear's first suggestion for median)
On the other hand:
sd(saywhut,na.rm=T)
[1] 3.614784
(Simon Fear's second suggestion for median)
Bytheway, mean treats factor in different way:
mean(saywhut)
[1] NA
Warning message:
argument is not numeric or logical: returning NA in: mean.default(saywhut).
There is an R-FAQ that tells one how to convert a factor to 'numeric' but if
I had tested for something being numeric to begin with I never would have
guessed that I needed to convert it to numeric. I think what this
conversion is really doing is getting rid of the machinery associated with
the class factor:
#from the R-FAQ
test-as.numeric(as.character(saywhut))
mode(test)
[1] numeric
median(test,na.rm=T)
[1] 1
and bytheway:
not.a.factor-c(NA,1,1,2,10)
mode(not.a.factor)
[1] character
median(not.a.factor,na.rm=T)
Error in median(not.a.factor, na.rm = T) :
need numeric data
Simon Fear: It seems to me the best way to deal with this bug would
be to make calling median with a factor argument be an immediate error.
Do you think that all base functions (sum, sd, mean, median,...) should deal
with this in a consistent way (This might be much more work.)? Another
thing that would make things consistent would be to take the stop-work
behavior out of sum:)
I don't think there is any real problem in the current behavior of factor as
long as the interaction between functions and classes produces this
stop-work behavior - preferably with a warning - and not unexpected side
effects. I am curious if there are other classes of mode numeric which
median-mean-sum-sd-etc might choke on.
tongue-in-cheek on
Of course, R would produce a median for factors by using the correct
defintion of a median of samples i.e., one that agrees with the definition
of median on a CDF, even though this concept gives most people apoplexy.
off
Thanks
Bob
Usual disclaimers
-Original Message-
From: Simon Fear [mailto:[EMAIL PROTECTED]
Sent: Friday, October 31, 2003 6:18 AM
To: Christoph Bier
Cc: [EMAIL PROTECTED]
Subject: RE: [R] Weird problem with median on a factor
Final guess as to observed behaviour: in the first case after
removal of NAs there were an odd number of observations
(so that sum was not called within the code for median).
In your second call I suspect that even though you got
an integer answer, it was found as sum(2,2)/2.
It seems to me the best way to deal with this bug would
be to make calling median with a factor argument be an
immediate error. Or just trust users never to attempt such
a thing ...
Simon Fear
Senior Statistician
Syne qua non Ltd
Tel: +44 (0) 1379 69
Fax: +44 (0) 1379 65
email: [EMAIL PROTECTED]
web: http://www.synequanon.com
Number of attachments included with this message: 0
This message (and any associated files) is confidential and\...{{dropped}}
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[R] Array Dimension Names
I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, workers, sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks for your help. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Weird problem with median on a factor
median() has this test in it:
if (mode(x) != numeric)
stop(need numeric data)
Note the following:
is.numeric(factor(letters))
[1] FALSE
mode(factor(letters))
[1] numeric
It seems as though median() is using the wrong test.
-- Tony Plate
At Friday 03:37 PM 10/31/2003 -0500, [EMAIL PROTECTED] wrote:
Beating a dead horse...
I am an R beginner trying to understand this factor business. While the
entire business of finding the median of factor may be silly from a
practical point of view, this email chain has helped me understand
something.
I have looked at the median function and it tests to see if what is passed
to it is numeric. If I were building a function, if I tested for mode
numeric, and if something told me it was numeric then like the median
function I would naively assume that I could do arithmetic on it:
saywhut-as.factor(c(NA,1,1,1,1,2,10))
mode(saywhut)
[1] numeric
It appears to me that the when the median function tests for numeric it
doesn't have the desired result with an object of class factor (and maybe
other classes?) as was shown by the example.
I have a suspicion that something of class factor has at least two pieces,
one of which is the levels which can possibly be character or something else
and the other piece is the ordering of the levels which is of storage.mode
integer. Is it this ordering that determines the mode of the factor??
But if the mode of factor is truly numeric, why doesn't the median function
use the numeric piece for finding the median (like it did with odd n - not
that anyone would ever really want the median of a factor:)?? I think that
Simon Fear hit on the right idea because of the definition of median that is
used for an even number of observations takes the sum of the ordered middle
two observations. It is the sum (called by the median function) that chokes
on a factor.
sum(saywhut,na.rm=T)
Error in Summary.factor(..., na.rm = na.rm) :
sum not meaningful for factors
It appears that whoever built the sum function built in a test for factor
(Simon Fear's first suggestion for median)
On the other hand:
sd(saywhut,na.rm=T)
[1] 3.614784
(Simon Fear's second suggestion for median)
Bytheway, mean treats factor in different way:
mean(saywhut)
[1] NA
Warning message:
argument is not numeric or logical: returning NA in: mean.default(saywhut).
There is an R-FAQ that tells one how to convert a factor to 'numeric' but if
I had tested for something being numeric to begin with I never would have
guessed that I needed to convert it to numeric. I think what this
conversion is really doing is getting rid of the machinery associated with
the class factor:
#from the R-FAQ
test-as.numeric(as.character(saywhut))
mode(test)
[1] numeric
median(test,na.rm=T)
[1] 1
and bytheway:
not.a.factor-c(NA,1,1,2,10)
mode(not.a.factor)
[1] character
median(not.a.factor,na.rm=T)
Error in median(not.a.factor, na.rm = T) :
need numeric data
Simon Fear: It seems to me the best way to deal with this bug would
be to make calling median with a factor argument be an immediate error.
Do you think that all base functions (sum, sd, mean, median,...) should deal
with this in a consistent way (This might be much more work.)? Another
thing that would make things consistent would be to take the stop-work
behavior out of sum:)
I don't think there is any real problem in the current behavior of factor as
long as the interaction between functions and classes produces this
stop-work behavior - preferably with a warning - and not unexpected side
effects. I am curious if there are other classes of mode numeric which
median-mean-sum-sd-etc might choke on.
tongue-in-cheek on
Of course, R would produce a median for factors by using the correct
defintion of a median of samples i.e., one that agrees with the definition
of median on a CDF, even though this concept gives most people apoplexy.
off
Thanks
Bob
Usual disclaimers
-Original Message-
From: Simon Fear [mailto:[EMAIL PROTECTED]
Sent: Friday, October 31, 2003 6:18 AM
To: Christoph Bier
Cc: [EMAIL PROTECTED]
Subject: RE: [R] Weird problem with median on a factor
Final guess as to observed behaviour: in the first case after
removal of NAs there were an odd number of observations
(so that sum was not called within the code for median).
In your second call I suspect that even though you got
an integer answer, it was found as sum(2,2)/2.
It seems to me the best way to deal with this bug would
be to make calling median with a factor argument be an
immediate error. Or just trust users never to attempt such
a thing ...
Simon Fear
Senior Statistician
Syne qua non Ltd
Tel: +44 (0) 1379 69
Fax: +44 (0) 1379 65
email: [EMAIL PROTECTED]
web: http://www.synequanon.com
Number of attachments included with this message: 0
This message (and any associated files) is confidential and\...{{dropped}}
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Re: [R] Array Dimension Names
Not that I know of. BUT dimnames can themselves have names attributes,
so a very small hack to apply() will do what you want.
I did
dump(apply,file=apply.R)
and added the following lines after dn - dimnames(X) (line 14) [this is
in R 1.7.1].
if (is.character(MARGIN)) {
if (is.null(dn) stop(dimnames(X) must have names)
MARGIN - match(MARGIN,names(dn))
}
and then did
source(apply.R)
x = array(1,dim=c(2,2,2))
dimnames(x) = list(a=1:2,b=1:2,c=1:2)
apply(x,a,sum)
apply(x,c(a,b),sum)
On Fri, 31 Oct 2003 [EMAIL PROTECTED] wrote:
I would like to reference array dimensions by name in an apply and a summary
function. For example:
apply(x, workers, sum)
Is there a better way to do this than creating a new attribute for the array
and then creating new methods for apply and summary? I don't want to name
the individual elements of each dimension (such as with dimnames) but rather
name the dimensions. Thanks for your help.
Benjamin Stabler
Transportation Planning Analysis Unit
Oregon Department of Transportation
555 13th Street NE, Suite 2
Salem, OR 97301 Ph: 503-986-4104
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[R] Problems with help.start()
Hello there, I've just installed R for Windows 1.8.0 and I'm experiencing problems with help.start(). It opens the help page as it supposed to do, then I go to Search Engine Keywords, but when I click anything there, it returns me a (aparently) JavaScript error, something like The object does not support this method or property. The Search Engine doesn't work as well. I tried to open with Internet Explorer 6 and Mozilla (the last one). Have anybody already seen this problem? Thanks, Henrique. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Re: packaging a package addon
On Fri, 2003-10-31 at 03:14, Prof Brian Ripley wrote: But I would not be doing R CMD check until I had both installed and loaded the package and run a few examples. That's interesting; I thought R CMD check was supposed to be done before hand. So are you saying the proper development sequence is 1. Do as much as you can with the basic R and C (Fortran, whatever) code to check it's OK. 2. R CMD build test, revise R CMD build etc 3. then, when everything looks OK R CMD check ? __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Re: packaging a package addon
On Fri, 2003-10-31 at 06:41, A.J. Rossini wrote: Ross Boylan [EMAIL PROTECTED] writes: I also added library(survival) to my .First.lib. Is library, rather than require, the right choice here? I want it to fail if survival doesn't load. test the results from require, something like: if (!require(survival)) stop(can't load survival) Doesn't using library do about the same thing? What's the advantage of this, clearer diagnostics? __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Array Dimension Names
Oh yeah, thanks. I thought I might write a function such as getNames() that returns the dimension number of the names of the dimnames of an object. That way I don't have to rewrite apply, sweep, and aperm. But even better would be for R to allow character names in addition to index numbers for the MARGIN argument to apply and sweep, and the perm argument to aperm. Thanks again, Ben Stabler -Original Message- From: Tony Plate [mailto:[EMAIL PROTECTED] Sent: Friday, October 31, 2003 1:07 PM To: STABLER Benjamin; [EMAIL PROTECTED] Subject: Re: [R] Array Dimension Names You can already name dimensions using standard arrays (but you can't use these names for the MARGIN argument of apply) e.g.: x - array(1:6, 3:2, dimnames=list(rows=letters[1:3],cols=LETTERS[24:25])) x cols rows X Y a 1 4 b 2 5 c 3 6 apply(x, 2, sum) X Y 6 15 apply(x, cols, sum) Error in -MARGIN : Invalid argument to unary operator You could pretty easily create your own version of apply() that checked if MARGIN was character, and if it were, matched it against names(dimnames(X)) hope this helps, Tony Plate At Friday 12:50 PM 10/31/2003 -0800, [EMAIL PROTECTED] wrote: I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, workers, sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks for your help. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problems with help.start()
No, so looks like a problem with your browser/OS. Are you sure you have working Java and JavaScript? Both a fully patched IE6 and Netscape 7.1 work for me (as do Opera 6 and Mozilla 1.2). On Fri, 31 Oct 2003, Henrique Patrício Sant'Anna Branco wrote: Hello there, I've just installed R for Windows 1.8.0 and I'm experiencing problems with help.start(). It opens the help page as it supposed to do, then I go to Search Engine Keywords, but when I click anything there, it returns me a (aparently) JavaScript error, something like The object does not support this method or property. The Search Engine doesn't work as well. I tried to open with Internet Explorer 6 and Mozilla (the last one). Have anybody already seen this problem? Thanks, Henrique. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] help with constrOptim function
On Fri, 31 Oct 2003, Bill Shipley wrote: Hello. I had previously posted a question concerning the optimization of a nonlinear function conditional on equality constraints. I was pointed towards the contrOptim function. Perhaps mistakenly, then. constrOptim() does linear *inequality* constraints. Equality constraints are probably best handled by Lagrange multipliers or reparametrisation. -thomas __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Re: packaging a package addon
On Fri, 2003-10-31 at 13:29, Prof Brian Ripley wrote: I would do 1 3 2 as R CMD check works on a source dir, not a .tar.gz. Your mileage may vary and all that. I'm not following something. 1 3 2 was what I was doing. I thought you said (below on Fri) that you'd do 2, then 3. I may be misunderstanding what the phrase installed and loaded the package means. I thought installing and loading it referred to doing an R CMD build to make the package, and then R CMD INSTALL on the result. I'd say R CMD check was a final check before distribution via R CMD build. On Fri, 31 Oct 2003, Ross Boylan wrote: On Fri, 2003-10-31 at 03:14, Prof Brian Ripley wrote: But I would not be doing R CMD check until I had both installed and loaded the package and run a few examples. That's interesting; I thought R CMD check was supposed to be done before hand. So are you saying the proper development sequence is 1. Do as much as you can with the basic R and C (Fortran, whatever) code to check it's OK. 2. R CMD build test, revise R CMD build etc 3. then, when everything looks OK R CMD check ? -- Ross Boylan wk: (415) 502-4031 530 Parnassus Avenue (Library) rm 115-4 [EMAIL PROTECTED] Dept of Epidemiology and Biostatistics fax: (415) 476-9856 University of California, San Francisco San Francisco, CA 94143-0840 hm: (415) 550-1062 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Re: packaging a package addon
On Fri, 31 Oct 2003, Ross Boylan wrote:
On Fri, 2003-10-31 at 06:41, A.J. Rossini wrote:
Ross Boylan [EMAIL PROTECTED] writes:
I also added library(survival) to my .First.lib. Is library, rather
than require, the right choice here? I want it to fail if survival
doesn't load.
test the results from require, something like:
if (!require(survival)) stop(can't load survival)
Doesn't using library do about the same thing? What's the advantage of
this, clearer diagnostics?
If you are going to fail when survival isn't found you should probably
just use library, though Tony's suggestion is effectively equivalent, and
I have also seen the Perly
require(tcltk) || stop(error message)
The main point of require() is when failure isn't completely fatal: eg
hypothetically
if (!require(boot)) {
warning(No `boot' package -- you're not getting confidence intervals)
conf.int-FALSE
}
-thomas
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Re: [R] Problems with help.start()
On Fri, 2003-10-31 at 15:05, Henrique Patrcio Sant'Anna Branco wrote: Hello there, I've just installed R for Windows 1.8.0 and I'm experiencing problems with help.start(). It opens the help page as it supposed to do, then I go to Search Engine Keywords, but when I click anything there, it returns me a (aparently) JavaScript error, something like The object does not support this method or property. The Search Engine doesn't work as well. I tried to open with Internet Explorer 6 and Mozilla (the last one). Have anybody already seen this problem? Thanks, Henrique. This has been discussed previously. It is tied to having both Java and JavaScript functioning in your browser. If Java is installed on your computer, be sure that both are enabled in your browser. Both must be functional in order to use the help.start() search engine which is a java applet. The following link provides information for Mozilla 1.5, which is the latest version, regarding installing Java: http://www.mozilla.org/releases/mozilla1.5/installation-extras.html There is also a FAQ at Sun here: http://www.java.com/en/download/help/enable_browser.jsp If you do not have Java installed on your computer yet, you can go here: http://www.java.com/en/download/help/auto_install.jsp for instructions on how to download and install it. HTH, Marc Schwartz __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Re: packaging a package addon
On Fri, 31 Oct 2003, Ross Boylan wrote: On Fri, 2003-10-31 at 13:29, Prof Brian Ripley wrote: I would do 1 3 2 as R CMD check works on a source dir, not a .tar.gz. Your mileage may vary and all that. I'm not following something. 1 3 2 was what I was doing. I thought you said (below on Fri) that you'd do 2, then 3. I may be misunderstanding what the phrase installed and loaded the package means. I thought installing and loading it referred to doing an R CMD build to make the package, and then R CMD INSTALL on the result. NO, do R CMD INSTALL on the sources, and the R run some tests. I'd say R CMD check was a final check before distribution via R CMD build. On Fri, 31 Oct 2003, Ross Boylan wrote: On Fri, 2003-10-31 at 03:14, Prof Brian Ripley wrote: But I would not be doing R CMD check until I had both installed and loaded the package and run a few examples. That's interesting; I thought R CMD check was supposed to be done before hand. So are you saying the proper development sequence is 1. Do as much as you can with the basic R and C (Fortran, whatever) code to check it's OK. 2. R CMD build test, revise R CMD build etc 3. then, when everything looks OK R CMD check ? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] problem with tune.svm
On Fri, 31 Oct 2003 [EMAIL PROTECTED] wrote: rng - list(gamma = 2^(-1:1), cost = 2^(2:4)) rng $gamma [1] 0.5 1.0 2.0 $cost [1] 4 8 16 obj - tune.svm(pIC50 ~ ., data = data, ranges = rng) Error in tune(svm, train.x = x, data = data, ranges = ranges, ...) : formal argument ranges matched by multiple actual arguments The function `tune.svm' has no `range' argument, use `gamma' and `cost' separately. The idea is to make `tune.foo' a `vectorized' function of `foo' in the parameters. If you want to preconstruct a list, use tune(svmobj, ranges = ...) instead. g., -d Ay idea why ??? Ryszard [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Array Dimension Names
You could add attributes to your array when creating it and then retrieve them: x - matrix(1:8,2,4) attr(x,workers) - 1 attr(x,variables) - 2 apply(x,attr(x,variables),sum) or perhaps: y - matrix(1:8,2,4) attr(y,margins) - list(workers = 1, variables = 2) apply(y,attr(y,margins)$variables,sum) These are simple enough that you might not need to develop your own apply but you could pretty them up even more if you did: my.apply - function(x,dim,fn) apply(x,attr(x,dim),fn) my.apply(x,variables,sum) --- From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: [R] Array Dimension Names I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, workers, sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks for your help. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 ___ No banners. No pop-ups. No kidding. Introducing My Way - http://www.myway.com __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Change in 'solve' for r-patched
On Fri, 31 Oct 2003, Williams, Elliot - BLS wrote: None of these address the problem of numerical accuracy, which was the thrust of Doug's comments. Has anyone done a quick simulation to investigate the stability of the solutions? Is it small coefficients or near-collinearity (or both) that one has to worry about? Are the solve(crossprod()) methods obviously unstable? They surely do work quickly! They are usually not *very* unstable. That is, solve(crossprod()) potentially loses about twice as many digits to rounding error as Doug's proposals. In many cases this is not that serious -- if you have 15 digits accuracy then losing 2 instead of 1 (or even 8 instead of 4) of them may not be a big deal. Back in the days of single precision this was more obviously important and you would get visible differences between statistical packages based on their linear algebra routines. (There was a notorious econometric data set that caused problems in single precision with the solve(crossprod()) approach but was fine with QR decomposition). On the other hand, even with double precision in extreme situations the harm can be noticeable, and the cost of working with QR or Cholesky decompositions instead is fairly small (and often negative -- with larger matrices you would probably see the cholesky-based method being faster) In linear regression the problem happens with collinearity, or when one variable has much smaller variance than another. In both of these cases the eigenvalues of the information matrix are of very different sizes. The ratio of the smallest to the largest eigenvalue is called the condition number, and the tolerance in solve() is based on condition numbers. So how bad is the error? For any matrix M there exists some vector b and error e in b so that the relative error in solve(M,b+e) compared to solve(M,b) is the condition number of the matrix M times the relative error in b. This means that condition numbers greater than 10^7 (and finding these in package examples is what prompted Doug's message) can amplify numerical error ten million times. This is starting to get undesirable even when you have 15 digits to work with. Working with very ill-conditioned matrices is possible, but it requires care in tracking and bounding errors, and if I wanted to do that I'd be in applied math, not biostatistics. That's why a simple fix that reduces condition numbers from, say, 10^10 to 10^5 really is worthwhile, especially for software that will be used by other people who don't understand its internals. -thomas __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Partial least squares.
Dear R-helpers, I am looking, quite unsuccesfully, for a number of functions/packages. Firstly, I am interested in a package for partial least squares. I have found that there seemed to exist a package called pls, but which seems not to run any more with modern versions of R. I have not been able to find certain chemometrics package I found some people discussing about in this list some time ago and that, it seems, included these kind of functions. Secondly, I have not been able to find a function equivalent to the SAS procedure STEPDISC which performs a step process (only available for lm and glm on R) on linear discriminant analysis (lda on R). Does anybody know of a top-the-shelf answer to these questions? Carlos J. Gil Bellosta Sigma Consultores Estadísticos http://www.consultoresestadisticos.com __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Partial least squares.
The package `pls.pcr' has NIPALS and SIMPLS. There's also gpls in BioConductor 1.3. Andy -Original Message- From: Carlos J. Gil Bellosta [mailto:[EMAIL PROTECTED] Sent: Friday, October 31, 2003 8:02 PM To: [EMAIL PROTECTED] Subject: [R] Partial least squares. Dear R-helpers, I am looking, quite unsuccesfully, for a number of functions/packages. Firstly, I am interested in a package for partial least squares. I have found that there seemed to exist a package called pls, but which seems not to run any more with modern versions of R. I have not been able to find certain chemometrics package I found some people discussing about in this list some time ago and that, it seems, included these kind of functions. Secondly, I have not been able to find a function equivalent to the SAS procedure STEPDISC which performs a step process (only available for lm and glm on R) on linear discriminant analysis (lda on R). Does anybody know of a top-the-shelf answer to these questions? Carlos J. Gil Bellosta Sigma Consultores Estadísticos http://www.consultoresestadisticos.com __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo /r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] html glitches with help?
Looking at the html generated help pages for a package I'm working on, I
noticed a couple of things that looked a little funny. I suspect they
are general features of the html for R (I don't usually look at it).
First is a problem of vertical alignment in tables. The first column
consistently aligned vertically *below* the alignment line of the bottom
line of the second column. This was a problem even when both columns
were a single line; it was worse when they were multiple lines.
In slightly exagerated form, the output looked like this:
long discussion of what paramater pp does
and its wonderful features
pp
Likely at least two separate issues: why is it aligning with the last,
rather than the first, line of the second column, and why is it below
that?
It may be relevant that the first column was in \code{a} and the whole
thing was in an \arguments{\item{foo}{\tabular{ll} section.
Second, the items marked with \code{} appeared fainter than the other
text, and were a little hard to read. I'd expect them to be bolder.
Perhaps the R.css style sheet could be tweaked for this?
R 1.7.1-1 on Debian GNU/Linux
Viewed with Mozilla 1.4
Konqueror from KDE 3.1.3 looked very similar, except the typeface for
\code was identical to that of the rest.
Sample excerpt from the .Rd file:
\value{returns a list:
\item{singles}{data frame, one row per simulation, with the following
columns:
\tabular{ll}{
coefficients \tab one column per variable\cr
\code{pi1} \tab conditional probability of sampling cases \cr
Similar behavior in \arguments section. It is the pi1 above, for
example, that is aligned poorly. Alignment of the outer level \item's
is better (in fact, the opposite problem, their baseline is above the
baseline of the first line of the second column, though their tops are
not much higher).
--
Ross Boylan wk: (415) 502-4031
530 Parnassus Avenue (Library) rm 115-4 [EMAIL PROTECTED]
Dept of Epidemiology and Biostatistics fax: (415) 476-9856
University of California, San Francisco
San Francisco, CA 94143-0840 hm: (415) 550-1062
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[R] Question about the high dimensional density estimation
Hi, I found that the R package KernSmooth can deal with only 1D and 2D data. But now I have a collection of 4-dimensional data (x1,x2,x3,x4) and would like to estimate the mode of the underlying density. What can I do about it ? Thanks a lot. -- Ying-Chao Hung Assistant Professor Graduate Institute of Statistics National Central University Chung-Li, Taiwan TEL: 886-3-426-7219 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Therotical basis of Kriging
hello
I want to know about therotical basis of Kriging in elemantary level.
I will appreciate if anyone sends me address,link,e-documents, etc..
kind regards
--
Ahmet Temiz
General Directory of Disaster Affairs
Ankara TURKEY
__
Inflex - installed on mailserver for domain @deprem.gov.tr
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[R] Re: [STATSGRASS] Therotical basis of Kriging
Hello, here is a link about kriging, I think is clear enough to understand it. http://www.geomatics.ucalgary.ca/~nel-shei/DTMLectureNotes/ENGO%20573%20-%20Chapter%203%20Kriging%20and%20variograms.doc Bye temiz wrote: hello I want to know about therotical basis of Kriging in elemantary level. I will appreciate if anyone sends me address,link,e-documents, etc.. kind regards -- Ahmet Temiz General Directory of Disaster Affairs Ankara TURKEY __ Inflex - installed on mailserver for domain @deprem.gov.tr Queries to: [EMAIL PROTECTED] __ The views and opinions expressed in this e-mail message are the sender's own and do not necessarily represent the views and the opinions of Earthquake Research Dept. of General Directorate of Disaster Affairs. Bu e-postadaki fikir ve gorusler gonderenin sahsina ait olup, yasal olarak T.C. B.I.B. Afet Isleri Gn.Mud. Deprem Arastirma Dairesi'ni baglayici nitelikte degildir. ___ statsgrass mailing list [EMAIL PROTECTED] http://grass.itc.it/mailman/listinfo/statsgrass -- - note: change my e-mail reference to [EMAIL PROTECTED] because the old one will be deleted soon. - Ing. Massimiliano Cannata Istituto di Scienze della Terra - SUPSI C.P. 72 - CH-6952 Canobbio (Ticino, Switzerland) Tel +41 91 /935 12 25 - Fax +41 91 /935 12 09 eMail: [EMAIL PROTECTED] Internet: http://www.ist.supsi.ch __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
