[R] Adding Grid lines
Dear all R users, Can anyone please tell me how to add grid lines in any plot in R including in Histogram, QQ plot etc? Thanks and regards, Arun [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [R-pkgs] New version of glmmML
On 8/21/06, Martin Maechler [EMAIL PROTECTED] wrote: Hi Göran, GB == Göran Broström [EMAIL PROTECTED] on Mon, 21 Aug 2006 11:12:49 +0200 writes: GB A new version, 0.65-1, of glmmML is now on CRAN. It is a major rewrite GB of the inner structures, so frequent updates (bug fixes) may be GB expected for some time. GB News: [] Sorry for my lazy question : What does the package do that lmer() does not? Hi Martin, most of the things I mentioned in the announcement. To appreciate it I think you have to study 'lmer' as well (defeat your laziness!;). One thing I didn't mention is the possibility to fit a model at a fixed value of the standard deviation of the random effects. Utilising that, it is easy to create a profile likelihood (as a function of sigma), which can be very helpful in certain cases. HTH, Göran __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] return tree from .Call
Please see the posting guide: low-level programming questions to R-devel.
On Mon, 21 Aug 2006, Sender wrote:
Hello:
I was hoping to get some advice about how to return a tree (basically a
linked list -- with each node containing a parent, left, and right node
pointers) from a C routine back into R. Each node itself contains several
attributes (a double, a char *, an int, and a void * )
Initially I was thinking I could just return to R a SEXP containing a
pointer to the Root Node, but then realized that the pointer would be
useless since C probably frees the memory after I leave the routine.
That's not true in general: it depends how you allocated them. With
malloc, you need to do a free. External pointer types and finalizers are
available for just this, and you can also use the memory of R structures.
Since trees vary in length (or height) I need to be able to loop thru the
tree until all Nodes have been visited and saved in a SEXP. I'm really new
to handling R objects from within C, and converting and returning a large
tree structure is daunting. Here is some rough code I was thinking about
using to do this. Any suggestions or help will be greatly appreciated!
Since elements of a protected list are protected, you can simplify the use
of PROTECTs (and may have to in order to avoid overflows) by linking into
a list at allocation time. Do node first, then its elements.
SEXP list ;
PROTECT( list = allocVector(VECSXP, tree-size) ) ;
for( i = 0; i tree-size; ++i ){
SEXP node, tree_double, tree_char, tree_int, tree_voidPTR ;
PROTECT( tree_double = NEW_NUMERIC( tree-weight ) ) ;
PROTECT( tree_char = NEW_CHARACTER( tree-name ) ) ;
PROTECT( tree_int = NEW_INTEGER( tree-exons ) ) ;
PROTECT( tree_voidPTR = NEW_CHARACTER( tree-data ) ) ; ??? not sure
about this...
PROTECT( node = allocVector( VECSXP, 4 ) ) ;
SET_VECTOR_ELT( node, 0, tree_double) ;
SET_VECTOR_ELT( node, 1, tree_char) ;
SET_VECTOR_ELT( node, 2, tree_int) ;
SET_VECTOR_ELT( node, 3, tree_voidPTR ) ;
SET_VECTOR_ELT( list, i, node ) ;
}
UNPROTECT( tree-size * 4 ) ; ??? Tricky..
return( list ) ;
Look reasonable ?
THANKS !
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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)
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Re: [R] Adding Grid lines
Arun Kumar Saha wrote: Dear all R users, Can anyone please tell me how to add grid lines in any plot in R including in Histogram, QQ plot etc? Have you ever typed ?grid before posting? Uwe Ligges Thanks and regards, Arun [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding Grid lines
Perhaps ?grid will help you. -- Bjørn-Helge Mevik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] ANN: 'weaver' package, caching for Sweave
Hi all, I've added a new package 'weaver' to the BioC repository: http://www.bioconductor.org/packages/1.9/bioc/html/weaver.html The weaver package provides extensions to the Sweave utilities included in R's base package. The focus of the extensions is on caching computationally expensive (time consuming) code chunks in Sweave documents. Why would you want to cache code chunks? If your Sweave document includes one or more code chunks that take a long time to compute, you may find it frustrating to make small changes to the document. Each run requires recomputing the expensive code chunks. If these chunks aren't changing, you can benefit from the caching provided by weaver. To install: source(http://bioconductor.org/biocLite.R;) biocLite(weaver) If you give it a try and have any feedback, drop me a note. Best Wishes, + seth ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Successive subsets from a vector?
I'd like to pick every imbricated five character long subsets from a
vector. I guess there is some efficient way to do this without loops...
Here is a for-loop-version and a model for output:
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6);
ADDRESSES=c();
for(i in 1:(length(VECTOR)-4)){
ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse=)
}
ADDRESSES
[1] 14265 42650 265011 6501110 5011104 0111043
1110436 104368
[9] 43686
Atte Tenkanen
University of Turku, Finland
[[alternative text/enriched version deleted]]
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Re: [R] How to share variables
Hi Thanks again. I hope not to waste to much of your time. I delete some lines of your answer Each time myfun is run a new environment is created to hold its local variables. The parent of that environment is e in this example by construction. So e and the environment that is temporarily created to hold myfun's variables are distinct. This means that the enviroment is duplicated, ie it is present twince in memory? I must keep some big variables and it will be a waste of memory; moreover if I update a value it will be lost. If I can use inside myfun the variable as e$dat (without changing the enviroment (no environment(myfun) - e statement)) than it will be ok. Yes you can. You can either make sure that e is visible to myfun via normal scoping rules or pass it explicitly: e - new.env() e$dat - 1:3 myfun - function(x) sum(x + e$dat) myfun(10) Hit!!! It solves the problem. A small drawback is that I need to modify the name of each occurrence of the variable. # or passing e explicitly myfun2 - function(x, e) sum(x + e$dat) myfun2(10, e) Any overhead in passing the environment? Is it a pointer? Sergio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fwd: Re: Finney's fiducial confidence intervals of LD50
Finney's method for finding the confidence interval for a ratio of parameters is quite simple and is probably described in his book 'Probit analysis'. It is also known as Fieller's method so an RSiteSearch on 'Fieller' might show something useful. On Mon, Aug 21, 2006 at 08:46:24AM -0700, carlos riveira wrote: thanks a lot Renaud. but i was interested in Finney's fiducial confidence intervals of LD50 so to obtain comparable results with SPSS. But your reply leads me to the next question: does anybody know what is the best method (asymptotic, bootstrap etc.) for calculating confidence intervals of LD50? i could get rid of Finney's fiducial confidence intervals but only if there was a better method.. any idea? Renaud Lancelot [EMAIL PROTECTED] wrote: Date: Mon, 21 Aug 2006 16:35:49 +0200 From: Renaud Lancelot [EMAIL PROTECTED] To: carlos riveira [EMAIL PROTECTED] Subject: Re: [R] Finney's fiducial confidence intervals of LD50 CC: r-help@stat.math.ethz.ch Sorry there was a typo in my previous reply: D50 - 10^c(logD50 + c(0, -1.96, 1.96) * attr(logD50, SE)) names(D50) - c(D50, lower, upper) D50 D50 lower upper 140.8353 103.3171 191.9777 Best, Renaud 2006/8/21, Renaud Lancelot : I don't know what Finney's fiducial confidence interval is but if your problem is to handle the output of dose.p (from MASS), you can do as follows: library(MASS) Response - c(0, 7, 26, 27, 0, 5, 13, 29, 0, 4, 11, 25) Tot - rep(30.5, 12) Dose - rep(c(10, 40, 160, 640), 3) fm - glm(Response/Tot ~ log10(Dose), family = quasibinomial(link = probit)) logD50 - dose.p(fm, cf = 1:2, p = 0.5) D50 - 10^c(logD50 + c(1, -1.96, 1.96) * attr(logD50, SE)) names(D50) - c(D50, lower, upper) D50 D50 lower upper 164.9506 103.3171 191.9777 Best, Renaud 2006/8/21, carlos riveira : I am working with Probit regression (I cannot switch to logit) can anybody help me in finding out how to obtain with R Finney's fiducial confidence intervals for the levels of the predictor (Dose) needed to produce a proportion of 50% of responses(LD50, ED50 etc.)? If the Pearson chi-square goodness-of-fit test is significant (by default), a heterogeneity factor should be used to calculate the limits. Response-c(0,7,26,27,0,5,13,29,0,4,11,25) Tot-rep(30.5,12) Dose-rep(c(10,40,160,640),3) probit-glm(formula = Response/Tot~ log10(Dose), family=quasibinomial (link=probit)) D50- round(10^(dose.p(probit,cf=1:2,p=0.5))) #This is what SPSS calculates. I would like to reproduce these results with R: #SPSS RESULTS: #PRNT50= 140,83525 #CI = 98,37857;205,34483 #Regr.coeff= 1,91676 (S.E.=0,16765) #Intercept=-4,11856 (S.E.=0,36355) Thanks a lot for your help. Carlos __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Renaud LANCELOT D?partement Elevage et M?decine V?t?rinaire (EMVT) du CIRAD Directeur adjoint charg? des affaires scientifiques CIRAD, Animal Production and Veterinary Medicine Department Deputy director for scientific affairs Campus international de Baillarguet TA 30 / B (B?t. B, Bur. 214) 34398 Montpellier Cedex 5 - France T?l +33 (0)4 67 59 37 17 Secr. +33 (0)4 67 59 39 04 Fax +33 (0)4 67 59 37 95 -- Renaud LANCELOT D?partement Elevage et M?decine V?t?rinaire (EMVT) du CIRAD Directeur adjoint charg? des affaires scientifiques CIRAD, Animal Production and Veterinary Medicine Department Deputy director for scientific affairs Campus international de Baillarguet TA 30 / B (B?t. B, Bur. 214) 34398 Montpellier Cedex 5 - France T?l +33 (0)4 67 59 37 17 Secr. +33 (0)4 67 59 39 04 Fax +33 (0)4 67 59 37 95 - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- *I.White * *University of Edinburgh * *Ashworth Laboratories, West Mains Road* *Edinburgh EH9 3JT * *Fax: 0131 650 6564 Tel: 0131 650 5490 * *E-mail: [EMAIL PROTECTED] * __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the
Re: [R] Successive subsets from a vector?
embed(VECTOR, 5)[, 5:1]
gives the subsets, so something like
apply(embed(VECTOR, 5)[, 5:1], 1, paste, collapse=)
does the job.
The following is a bit more efficient
ind - 1:(length(VECTOR)-4)
do.call(paste, c(lapply(0:4, function(j) VECTOR[ind+j]), sep=))
but by looking at how embed() works it could be made as efficient.
Larger example:
VECTOR - sample(1:10, 1e5, replace=TRUE)
system.time(apply(embed(VECTOR, 5)[, 5:1], 1, paste, collapse=))
[1] 5.73 0.05 5.81 NA NA
system.time({ind - 1:(length(VECTOR)-4)
+ do.call(paste, c(lapply(0:4, function(j) VECTOR[ind+j]), sep=))
+ })
[1] 1.00 0.01 1.01 NA NA
The loop method took 195 secs. Just assigning to an answer of the correct
length reduced this to 5 secs. e.g. use
ADDRESSES - character(length(VECTOR)-4)
Moral: don't grow vectors repeatedly.
On Tue, 22 Aug 2006, kone wrote:
I'd like to pick every imbricated five character long subsets from a
vector. I guess there is some efficient way to do this without loops...
Here is a for-loop-version and a model for output:
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6);
ADDRESSES=c();
You do not need the semicolons, and they just confuse readers.
for(i in 1:(length(VECTOR)-4)){
ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse=)
}
ADDRESSES
[1] 14265 42650 265011 6501110 5011104 0111043
1110436 104368
[9] 43686
Atte Tenkanen
University of Turku, Finland
[[alternative text/enriched version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Successive subsets from a vector?
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6)
x - lapply(seq(length(VECTOR)-4),function(z)paste(VECTOR[z:(z+4)],
collapse=''))
unlist(x)
[1] 14265 42650 265011 6501110 5011104 0111043 1110436
104368 43686
On 8/22/06, kone [EMAIL PROTECTED] wrote:
I'd like to pick every imbricated five character long subsets from a
vector. I guess there is some efficient way to do this without loops...
Here is a for-loop-version and a model for output:
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6);
ADDRESSES=c();
for(i in 1:(length(VECTOR)-4)){
ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse=)
}
ADDRESSES
[1] 14265 42650 265011 6501110 5011104 0111043
1110436 104368
[9] 43686
Atte Tenkanen
University of Turku, Finland
[[alternative text/enriched version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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[R] Rgraphviz installation Problem
Dear Robert, Thanks for your time. I have downloaded Rgraphviz (windows binary) from www.bioconductor.org and put inside R2.3.0 library then i installed from the local zip its says package 'graph' couldnot be loaded. Am i doing the installation correctly? Still the new user. Can you guide me sir? JJ -- Lecturer J. Joshua Thomas KDU College Penang Campus Research Student, University Sains Malaysia [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] big numbers
Hi
Can I get R to handle really big numbers?I am not interested
in more than (say) 10 sig figs, but I would like to deal with numbers
up to, say, 10^1.
If
a - 10^1
b - pi* a
I would like a+b to return 3.1415926e1.
Toy example, illustrating why I can't deal with log(a) and log(b),
follows.
f - function(a,n=100){
out - rep(0,n)
out[1] - a
for(i in 2:n){
out[i] - sum(exp(out[1:i])) + rexp(1)
}
return(log(out))
}
then f(1,10) has infinities in it, even though the values should be
moderate size.
What are my options here?
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] Rgraphviz installation Problem
You also need to install the 'graph' package, i think it is also available from bioconductor.org. Other packages might be needed as well Gabor On Tue, Aug 22, 2006 at 06:42:31PM +0800, j.joshua thomas wrote: Dear Robert, Thanks for your time. I have downloaded Rgraphviz (windows binary) from www.bioconductor.org and put inside R2.3.0 library then i installed from the local zip its says package 'graph' couldnot be loaded. Am i doing the installation correctly? Still the new user. Can you guide me sir? JJ -- Lecturer J. Joshua Thomas KDU College Penang Campus Research Student, University Sains Malaysia [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Successive subsets from a vector?
Thanks!
I have used tons of for- and while-loops (I'm ashamed to reveal these scripts,
but I'm primarily a musician;-)
http://users.utu.fi/attenka/SetTheoryScripts.r), taken some or more cup of
cocoa and mostly been happy ;-) Now I got so many new ways to do these things,
that it takes a while to ruminate all the ideas here.
Atte
embed(VECTOR, 5)[, 5:1]
gives the subsets, so something like
apply(embed(VECTOR, 5)[, 5:1], 1, paste, collapse=)
does the job.
The following is a bit more efficient
ind - 1:(length(VECTOR)-4)
do.call(paste, c(lapply(0:4, function(j) VECTOR[ind+j]), sep=))
but by looking at how embed() works it could be made as efficient.
Larger example:
VECTOR - sample(1:10, 1e5, replace=TRUE)
system.time(apply(embed(VECTOR, 5)[, 5:1], 1, paste, collapse=))
[1] 5.73 0.05 5.81 NA NA
system.time({ind - 1:(length(VECTOR)-4)
+ do.call(paste, c(lapply(0:4, function(j) VECTOR[ind+j]), sep=))
+ })
[1] 1.00 0.01 1.01 NA NA
The loop method took 195 secs. Just assigning to an answer of the
correct
length reduced this to 5 secs. e.g. use
ADDRESSES - character(length(VECTOR)-4)
Moral: don't grow vectors repeatedly.
On Tue, 22 Aug 2006, kone wrote:
I'd like to pick every imbricated five character long subsets
from a
vector. I guess there is some efficient way to do this without
loops... Here is a for-loop-version and a model for output:
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6);
ADDRESSES=c();
You do not need the semicolons, and they just confuse readers.
for(i in 1:(length(VECTOR)-4)){
ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse=)
}
ADDRESSES
[1] 14265 42650 265011 6501110 5011104 0111043
1110436 104368
[9] 43686
Atte Tenkanen
University of Turku, Finland
[[alternative text/enriched version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html and provide commented, minimal, self-contained,
reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] listing a sequence of vectors in a matrix
Hi,
I'm having trouble applying the matrix function. I'd like to be able to
create a matrix of vectors filled in by rows, which are not all the same
length, and so I need it to fill in NAs where applicable.
It's easiest to explain with a simple example:
Suppose vec = c(3,4,5). How can I form a matrix of the vectors 1:vec[j]
for j=1:3?
i.e. 1 2 3 NA NA
1 2 3 4 NA
1 2 3 45
I've tried matrix(c(1:vec[j]),nrow=max(j),ncol=max(vec)) but it will
only give me a matrix with repeated values for j=1, like 1 2 3 1
2
3 1 2 3 1
2 3 1 2 3
Also using the list function hasn't got me anywhere either..
Any help/ideas would be greatly appreciated!
Many thanks,
Sara-Jane Dunn
--
This message (and any attachments) is for the recipient only...{{dropped}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate example : where is the state.region variable?
--- MARK LEEDS [EMAIL PROTECTED] wrote: these people/experts provide all these packages and documentation as a FAVOR and for the fact that they enjoy spreading knowledge/statistical computing abilities etc. It's not their job so I think criticism of the docs and the fact that they use a variable from another place is kind of harsh. Mark I am very appeciative of the time, expertise and great helpfulness that I have seen in the R community. If there is no criticism of R then how do we find out about problems that may exist? - Original Message - From: John Kane [EMAIL PROTECTED] To: Gabor Grothendieck [EMAIL PROTECTED] Cc: R R-help r-help@stat.math.ethz.ch Sent: Monday, August 21, 2006 6:59 PM Subject: Re: [R] aggregate example : where is the state.region variable? --- Gabor Grothendieck [EMAIL PROTECTED] wrote: Its not part of state.x77. Its a completely separate variable. Try ls(package:datasets) and notice its in the list or try ?state.region and note that its a variable in datasets. Thanks. I was wondering if it was going something like that. However, it is a bloody stupid example, at least to a newbie. A call to another data.set in what is supposed to be a simple example is very confusing. When someone is apparently illustrating a function with a simple one line command I don't expect them to call another data set, apparently create a new variable (Region), and use that new variable as the grouping variable without a word of explanation of what the example is doing. If I sound a bit annoyed it is because I am. It might be nice to have an example illlustate the funtion,not do a couple of other undocumented things as well. On 8/21/06, John Kane [EMAIL PROTECTED] wrote: I was looking ?aggregate and ran the first example aggregate(state.x77, list(Region = state.region), mean) The variables in state.x77 appear to be : state.x77 Population Income Illiteracy Life Exp Murder HS Grad Frost Area Where is the state.region variable coming from? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] boxplot order of the levels
hello i drew a boxplot with: boxplot(voxit$AGE ~ voxit$A02X) and the boxes are from left to right: Ja Leer Nein wn k.A. levels(voxit$A02X) [1] Ja Leer Nein wn k.A. 98 there are no entries with 98 but i want: Ja Nein Leer wn k.A. how can i change the order of the boxes bye thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to interrupt running computation?
If I start a computation in R, how can I interrupt it? I' using R 2.1.0. Thanks for your help, Lothar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to share variables
On 8/22/06, Sergio Martino [EMAIL PROTECTED] wrote: Hi Thanks again. I hope not to waste to much of your time. I delete some lines of your answer Each time myfun is run a new environment is created to hold its local variables. The parent of that environment is e in this example by construction. So e and the environment that is temporarily created to hold myfun's variables are distinct. This means that the enviroment is duplicated, ie it is present twince in memory? Each time myfun starts up a new environment comes into being that contains x and each time it completes that environment is destroyed. I must keep some big variables and it will be a waste of memory; moreover if I update a value it will be lost. If you update a local variable then its lost upon exit (of course you could return the variable or return the environment inside the function) but if you update it in e then its not lost. If I can use inside myfun the variable as e$dat (without changing the enviroment (no environment(myfun) - e statement)) than it will be ok. Yes you can. You can either make sure that e is visible to myfun via normal scoping rules or pass it explicitly: e - new.env() e$dat - 1:3 myfun - function(x) sum(x + e$dat) myfun(10) Hit!!! It solves the problem. A small drawback is that I need to modify the name of each occurrence of the variable. That's why in an earlier example we set the environment of myfun to e. # or passing e explicitly myfun2 - function(x, e) sum(x + e$dat) myfun2(10, e) Any overhead in passing the environment? Is it a pointer? ?system.time to experiment with timings. Sergio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Successive subsets from a vector?
The loop method took 195 secs. Just assigning to an answer of the correct length reduced this to 5 secs. e.g. use ADDRESSES - character(length(VECTOR)-4) Moral: don't grow vectors repeatedly. Other languages (eg. Java) grow the size of the vector independently of the number of observations in it (I think Java doubles the size whenever the vector is filled), thus changing O(n) behaviour to O(log n). I've always wondered why R doesn't do this. Hadley __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to interrupt running computation?
On 8/22/06, Lothar Schmid [EMAIL PROTECTED] wrote: If I start a computation in R, how can I interrupt it? I' using R 2.1.0. Hi Lothar, What do you mean by interrupt - cancel completely and return to the prompt? Or pause, then resume? Your operating system would be helpful too. If the former, try Esc on Windows or Ctrl-C on Linux. If the latter, Ctrl-Z will send your R session to the background on Linux, and fg will bring it back. I don't think there's a Windows equivalent. Sarah -- Sarah Goslee http://www.stringpage.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to interrupt running computation?
Lothar, Which system do you use? Windows, Linux, Mac? Stefan Lothar Schmid schrieb: If I start a computation in R, how can I interrupt it? I' using R 2.1.0. Thanks for your help, Lothar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] boxplot order of the levels
On Tue, 2006-08-22 at 14:27 +0200, Thomas Kuster wrote: hello i drew a boxplot with: boxplot(voxit$AGE ~ voxit$A02X) and the boxes are from left to right: Ja Leer Nein wn k.A. levels(voxit$A02X) [1] Ja Leer Nein wn k.A. 98 there are no entries with 98 but i want: Ja Nein Leer wn k.A. how can i change the order of the boxes bye thomas See the following post from last week: https://stat.ethz.ch/pipermail/r-help/2006-August/111289.html HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] listing a sequence of vectors in a matrix
Here are two solutions. seq(length = ...) instead of
just seq(...) is so that v can possibly contain zeros.
# data
v - 3:5
# solution 1 - rbind/lapply
f - function(n) {
s = seq(length = n)
replace(rep(NA, max(v)), s, s)
}
do.call(rbind, lapply(v, f))
# solution 2 - loop
mat - matrix(NA, length(v), max(v))
for(i in seq(v)) {
s - seq(length = v[i])
mat[i, s] - s
}
On 8/22/06, Sara-Jane Dunn [EMAIL PROTECTED] wrote:
Hi,
I'm having trouble applying the matrix function. I'd like to be able to
create a matrix of vectors filled in by rows, which are not all the same
length, and so I need it to fill in NAs where applicable.
It's easiest to explain with a simple example:
Suppose vec = c(3,4,5). How can I form a matrix of the vectors 1:vec[j]
for j=1:3?
i.e. 1 2 3 NA NA
1 2 3 4 NA
1 2 3 45
I've tried matrix(c(1:vec[j]),nrow=max(j),ncol=max(vec)) but it will
only give me a matrix with repeated values for j=1, like 1 2 3 1
2
3 1 2 3 1
2 3 1 2 3
Also using the list function hasn't got me anywhere either..
Any help/ideas would be greatly appreciated!
Many thanks,
Sara-Jane Dunn
--
This message (and any attachments) is for the recipient only...{{dropped}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] listing a sequence of vectors in a matrix
Hi
f - function(a,n){(1:a)[1:n]}
t(sapply(c(2,3,4,4,4,5,6),f,n=5))
[,1] [,2] [,3] [,4] [,5]
[1,]12 NA NA NA
[2,]123 NA NA
[3,]1234 NA
[4,]1234 NA
[5,]1234 NA
[6,]12345
[7,]12345
HTH
rksh
On 22 Aug 2006, at 12:29, Sara-Jane Dunn wrote:
Hi,
I'm having trouble applying the matrix function. I'd like to be
able to
create a matrix of vectors filled in by rows, which are not all the
same
length, and so I need it to fill in NAs where applicable.
It's easiest to explain with a simple example:
Suppose vec = c(3,4,5). How can I form a matrix of the vectors 1:vec
[j]
for j=1:3?
i.e. 1 2 3 NA NA
1 2 3 4 NA
1 2 3 45
I've tried matrix(c(1:vec[j]),nrow=max(j),ncol=max(vec)) but it will
only give me a matrix with repeated values for j=1, like 1 2 3 1
2
3 1 2 3 1
2 3 1 2 3
Also using the list function hasn't got me anywhere either..
Any help/ideas would be greatly appreciated!
Many thanks,
Sara-Jane Dunn
--
This message (and any attachments) is for the recipient on...{{dropped}}
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] listing a sequence of vectors in a matrix
Gabor makes a good point about seq() vs a:b [a common gotcha
for me].
I'll revise my original function to:
f - function(a,n){(seq(length=a))[1:n]}
t(sapply(c(2,3,4,4,4,5,6,0),f,n=5))
[,1] [,2] [,3] [,4] [,5]
[1,]12 NA NA NA
[2,]123 NA NA
[3,]1234 NA
[4,]1234 NA
[5,]1234 NA
[6,]12345
[7,]12345
[8,] NA NA NA NA NA
rksh
On 22 Aug 2006, at 13:55, Gabor Grothendieck wrote:
Here are two solutions. seq(length = ...) instead of
just seq(...) is so that v can possibly contain zeros.
# data
v - 3:5
# solution 1 - rbind/lapply
f - function(n) {
s = seq(length = n)
replace(rep(NA, max(v)), s, s)
}
do.call(rbind, lapply(v, f))
# solution 2 - loop
mat - matrix(NA, length(v), max(v))
for(i in seq(v)) {
s - seq(length = v[i])
mat[i, s] - s
}
On 8/22/06, Sara-Jane Dunn [EMAIL PROTECTED] wrote:
Hi,
I'm having trouble applying the matrix function. I'd like to be
able to
create a matrix of vectors filled in by rows, which are not all
the same
length, and so I need it to fill in NAs where applicable.
It's easiest to explain with a simple example:
Suppose vec = c(3,4,5). How can I form a matrix of the vectors
1:vec[j]
for j=1:3?
i.e. 1 2 3 NA NA
1 2 3 4 NA
1 2 3 45
I've tried matrix(c(1:vec[j]),nrow=max(j),ncol=max(vec)) but it will
only give me a matrix with repeated values for j=1, like 1 2 3 1
2
3 1 2 3 1
2 3 1 2 3
Also using the list function hasn't got me anywhere either..
Any help/ideas would be greatly appreciated!
Many thanks,
Sara-Jane Dunn
--
This message (and any attachments) is for the recipient only...
{{dropped}}
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate example : where is the state.region variable?
Gabor == Gabor Grothendieck [EMAIL PROTECTED] on Mon, 21 Aug 2006 21:03:49 -0400 writes: Gabor It is worthwhile to note that what is being Gabor illustrated here is aggregating a numeric matrix by a Gabor factor using the aggregate.default method and, of Gabor course, a factor can't be part of a numeric matrix. Gabor Of course, that is not say that the examples could Gabor not be improved in terms of clarity, simplicity and Gabor comprehensiveness (there is no example of Gabor aggregate.data.frame). yes, thank you, Gabor . and we (the R developers) have accepted and incorporated quite a few constructive proposals for improvement. Just offending the original authors (bloody ..) without adding any constructive proposal for improvement doesn't really help. You can always get the money back you paid for R. You can also decide to leave this mailing list and get the money back you paid for that service. Unfortunately, we can't get the time and energy back we've lost when dealing with such postings... Martin Maechler, ETH Zurich Gabor On 8/21/06, John Kane [EMAIL PROTECTED] wrote: --- Gabor Grothendieck [EMAIL PROTECTED] wrote: Its not part of state.x77. Its a completely separate variable. Try ls(package:datasets) and notice its in the list or try ?state.region and note that its a variable in datasets. Thanks. I was wondering if it was going something like that. However, it is a bloody stupid example, at least to a newbie. A call to another data.set in what is supposed to be a simple example is very confusing. When someone is apparently illustrating a function with a simple one line command I don't expect them to call another data set, apparently create a new variable (Region), and use that new variable as the grouping variable without a word of explanation of what the example is doing. If I sound a bit annoyed it is because I am. It might be nice to have an example illlustate the funtion,not do a couple of other undocumented things as well. On 8/21/06, John Kane [EMAIL PROTECTED] wrote: I was looking ?aggregate and ran the first example aggregate(state.x77, list(Region = state.region), mean) The variables in state.x77 appear to be : state.x77 Population Income Illiteracy Life Exp Murder HS Grad Frost Area Where is the state.region variable coming from? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com Gabor __ Gabor R-help@stat.math.ethz.ch mailing list Gabor https://stat.ethz.ch/mailman/listinfo/r-help PLEASE Gabor do read the posting guide Gabor http://www.R-project.org/posting-guide.html and Gabor provide commented, minimal, self-contained, Gabor reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R2WinBugs
We had set R2WinBugs to use the burnin as the adaptive phase. I think this was changed very slightly in the latest version of R2WinBUGS so that the adaptive phase would equal burnin minus 1. This was to allow DIC to be calculated. So I think it should work OK now. If you send an example I can take a look. Andrew Uwe Ligges wrote: Bowden, J.M. wrote: Hi all, I am having problems using the R2Winbugs function When I perform an analysis directly in Winbugs I can specify that the first 'n' iterations are to be done using an 'adaptive' phase. After this phase the markov chain seems to mix a lot better. I don't seem to be able to specify R2winbugs to carry out this adaptive phase, I can just specify the burnin length but this (to my knowledge) is not the same thing. As a result my models are not fitting as well as I would like, has anyone had a similar experience? I think this is a question for Andrew Gelman (CCing), who wrote the underlying code for automatically setting the adaptive phase. Uwe Ligges Jack __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Andrew Gelman Professor, Department of Statistics Professor, Department of Political Science [EMAIL PROTECTED] www.stat.columbia.edu/~gelman Statistics department office: Social Work Bldg (Amsterdam Ave at 122 St), Room 1016 212-851-2142 Political Science department office: International Affairs Bldg (Amsterdam Ave at 118 St), Room 731 212-854-7075 Mailing address: 1255 Amsterdam Ave, Room 1016 Columbia University New York, NY 10027-5904 212-851-2142 (fax) 212-851-2164 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to interrupt running computation?
I'm using Linux. And I'd like just to cancel a running computation, not the entire R prompt. Lothar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Finney's fiducial confidence intervals of LD50
Thanks for the tip!
unfortunately though conf limits calculated with Fieller and delta methods do
not seem to be in agreement with (and seem to be worse than..) my SPSS
results.. Am i doing something wrong?
thanks a lot in advance for your help!!
An RSiteSearch on 'Fieller' gave me the following 2 posts
one from S.B.Cox using fieller's conf limits, and one from Varadhan using
delta method:
1-Fieller's Conf Limits and EC50's
From: Stephen B. Cox stephen.cox
Date: Wed, 13 Jul 2005 11:42:52 -0500
ec.calc-function(obj,conf.level=.95,p=.5) {
call - match.call()
coef = coef(obj)
vcov = summary.glm(obj)$cov.unscaled
b0-coef[1]
b1-coef[2]
var.b0-vcov[1,1]
var.b1-vcov[2,2]
cov.b0.b1-vcov[1,2]
alpha-1-conf.level
zalpha.2 - -qnorm(alpha/2)
gamma - zalpha.2^2 * var.b1 / (b1^2)
eta = family(obj)$linkfun(p) #based on calcs in VR's dose.p
EC50 - (eta-b0)/b1
const1 - (gamma/(1-gamma))*(EC50 + cov.b0.b1/var.b1)
const2a - var.b0 + 2*cov.b0.b1*EC50 + var.b1*EC50^2 -
gamma*(var.b0 - cov.b0.b1^2/var.b1)
const2 - zalpha.2/( (1-gamma)*abs(b1) )*sqrt(const2a)
LCL - EC50 + const1 - const2
UCL - EC50 + const1 + const2
conf.pts - c(LCL,EC50,UCL)
names(conf.pts) - c(Lower,EC50,Upper)
return(conf.pts,conf.level,call=call)
}
#when i apply it to my data (see below)
# i get with Fieller's method by Cox
# Lower EC50Upper
#11.47628 140.8351 8423.78
# while SPSS gives me:
# Lower EC50Upper
#98,37857 140,83525 205,34483
library(MASS)
Response-c(0,7,26,27,0,5,13,29,0,4,11,25)
Tot-rep(30.5,12)
Dose-rep(c(10,40,160,640),3)
probit-glm(formula = Response/Tot~ log10(Dose), family=quasibinomial
(link=probit))
ec.calc(probit,conf.level=.95,p=.5)
$conf.pts
Lower EC50Upper
1.059801 2.148711 3.925507
$conf.level
[1] 0.95
$call
ec.calc(obj = probit, conf.level = 0.95, p = 0.5)
Warning message:
multi-argument returns are deprecated in: return(conf.pts, conf.level, call =
call)
2-Fieller's Conf Limits and EC50's
From: Ravi Varadhan rvaradha
Date: Thu, 14 Jul 2005 09:44:47 -0400
varEC50 - 1/b1^2 * (var.b0 + EC50^2*var.b1 + 2*EC50*cov.b0.b1)
LCL - EC50 - zalpha.2 * sqrt(varEC50)
UCL - EC50 + zalpha.2 * sqrt(varEC50)
#when i apply it to my data
# i get with delta method by varadhan
# Lower EC50Upper
# 46.15749 140.8351 429.7151
# while SPSS gives me:
# Lower EC50Upper
#98,37857 140,83525 205,34483
i.m.s.white [EMAIL PROTECTED] wrote:
Finney's method for finding the confidence interval for a ratio of
parameters is quite simple and is probably described in his book
'Probit analysis'. It is also known as Fieller's method so an
RSiteSearch on 'Fieller' might show something useful.
On Mon, Aug 21, 2006 at 08:46:24AM -0700, carlos riveira wrote:
thanks a lot Renaud.
but i was interested in Finney's fiducial confidence intervals of LD50 so to
obtain comparable results with SPSS.
But your reply leads me to the next question: does anybody know what is the
best method (asymptotic, bootstrap etc.) for calculating confidence intervals
of LD50?
i could get rid of Finney's fiducial confidence intervals but only if there
was a better method..
any idea?
Renaud Lancelot wrote:
Date: Mon, 21 Aug 2006 16:35:49 +0200
From: Renaud Lancelot
To: carlos riveira
Subject: Re: [R] Finney's fiducial confidence intervals of LD50
CC: r-help@stat.math.ethz.ch
Sorry there was a typo in my previous reply:
D50 - 10^c(logD50 + c(0, -1.96, 1.96) * attr(logD50, SE))
names(D50) - c(D50, lower, upper)
D50
D50 lower upper
140.8353 103.3171 191.9777
Best,
Renaud
2006/8/21, Renaud Lancelot :
I don't know what Finney's fiducial confidence interval is but if your
problem is to handle the output of dose.p (from MASS), you can do as
follows:
library(MASS)
Response - c(0, 7, 26, 27, 0, 5, 13, 29, 0, 4, 11, 25)
Tot - rep(30.5, 12)
Dose - rep(c(10, 40, 160, 640), 3)
fm - glm(Response/Tot ~ log10(Dose), family = quasibinomial(link =
probit))
logD50 - dose.p(fm, cf = 1:2, p = 0.5)
D50 - 10^c(logD50 + c(1, -1.96, 1.96) * attr(logD50, SE))
names(D50) - c(D50, lower, upper)
D50
D50 lower upper
164.9506 103.3171 191.9777
Best,
Renaud
2006/8/21, carlos riveira :
I am working with Probit regression (I cannot switch to logit) can
anybody help me in finding out how to obtain with R Finney's fiducial
confidence intervals for the levels of the predictor (Dose) needed to
produce a proportion of 50% of responses(LD50, ED50 etc.)?
If the Pearson chi-square goodness-of-fit test is significant (by
default), a heterogeneity factor should be
[R] :Circular-Linear Correlation
Dear All, I'm looking for code that does circular-linear correlations as proposed by Mardia (1976). Thank you, Agnes __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to interrupt running computation?
Then Ctrl-C should do the trick. Sarah On 8/22/06, Lothar Schmid [EMAIL PROTECTED] wrote: I'm using Linux. And I'd like just to cancel a running computation, not the entire R prompt. Lothar -- Sarah Goslee http://www.stringpage.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] summary(lm ... conrasts=...)
Hi Folks, I've encountered something I hadn't been consciously aware of previously, and I'm wondering what the explanation might be. In (on another list) using R to demonstrate the difference between different contrasts in 'lm' I set up an example where Y is sampled from three different normal distributions according to the levels (A,B,C) of a factor X: Y-c(rnorm(mean=0,n=12),rnorm(mean=2,n=12),rnorm(mean=4,n=12)) X-factor(c(rep(A,12),rep(B,12),rep(C,12))) Then I do a summary(lm(Y~X)...) using first Treatment contrasts and then Helmert contrasts. Here are the coefficient parts of the results in each case: summary(lm(Y~X,contrasts=list(X=contr.treatment))) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 0.2303 0.3220 0.715 0.47944 XB1.3057 0.4554 2.867 0.00716 ** XC3.4204 0.4554 7.511 1.23e-08 *** summary(lm(Y~X,contrasts=list(X=contr.helmert))) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 1.8057 0.1859 9.713 3.34e-11 *** X10.6529 0.2277 2.867 0.00716 ** X20.9225 0.1315 7.017 5.00e-08 *** What I'm wondering is why the effect names are X.B and X.C for Treatment, and X1, X2 for Helmert. Why not X.B and X.C in both cases? Just as XB contrasts B with the overall mean and XC contrasts C with the overall mean, XA being implicit, in the Treatment contrasts, so X1 contrasts B with A and X2 contrasts C with (A+B) in Helmert, so there is to my mind just as definite an association of B with the first contrast, and C with the second, in the Helmert case as in the Treatment case! I know it's just a matter of notation, but in the Helmert case the association with the names of the factor levels has been lost, and it could be useful to have it explicit. (Or is it intended simply as a reminder that one is using a particular system of contrasts?) Thanks, and best wishes to all, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 22-Aug-06 Time: 14:45:17 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] big numbers
The 'gmp' package may be of use here, but I'm not sure.
-roger
Robin Hankin wrote:
Hi
Can I get R to handle really big numbers?I am not interested
in more than (say) 10 sig figs, but I would like to deal with numbers
up to, say, 10^1.
If
a - 10^1
b - pi* a
I would like a+b to return 3.1415926e1.
Toy example, illustrating why I can't deal with log(a) and log(b),
follows.
f - function(a,n=100){
out - rep(0,n)
out[1] - a
for(i in 2:n){
out[i] - sum(exp(out[1:i])) + rexp(1)
}
return(log(out))
}
then f(1,10) has infinities in it, even though the values should be
moderate size.
What are my options here?
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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[R] new version of The R Guide available on CRAN
Hello, Version 2.2 of The R Guide is available for download in the Contributed Documents section on CRAN. The R Guide is written for the beginning R user. I use the guide in my undergraduate probability and math stat sequence, but anyone with a basic understanding of statistics (who wants to learn R) should find it useful. This updated version includes sections on multiple comparisons, optimization, along with some improvements suggested by fellow R users from around the world. The entire document is under 60 pages in length. Jason -- Assistant Professor of Statistics Mathematics and Computer Science Department University of Richmond, Virginia 23173 (804) 289-8081 fax:(804) 287-6664 http://www.mathcs.richmond.edu/~wowen This is R. There is no if. Only how. Simon (Yoda) Blomberg __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Link
Its a very nice document. Here is the link: http://cran.r-project.org/doc/contrib/Owen-TheRGuide.pdf -- View this message in context: http://www.nabble.com/new-version-of-%22The-R-Guide%22-available-on-CRAN-tf2146496.html#a5926389 Sent from the R help forum at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate example : where is the state.region variable?
there is no factor in the dataset but why there is not one and why a call to another dataset is totally opaque. The reason is purely historical. The state dataset is about 10 years older than the data.frame concept. At the time the state.* variables were constructed it was not possible to put numeric data and factor data into the same rectangular structure. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] listing a sequence of vectors in a matrix
Here is another variation using Robin's idea of t(sapply(...))
v - c(3:5, 0)
t(sapply(lapply(v, function(n) seq(length = n)), length-, max(v) ))
# which can be shortened even further for the case where
# there are no zeros in v
v - 3:5
t(sapply(lapply(v, seq), length-, max(v) ))
On 8/22/06, Robin Hankin [EMAIL PROTECTED] wrote:
Gabor makes a good point about seq() vs a:b [a common gotcha
for me].
I'll revise my original function to:
f - function(a,n){(seq(length=a))[1:n]}
t(sapply(c(2,3,4,4,4,5,6,0),f,n=5))
[,1] [,2] [,3] [,4] [,5]
[1,]12 NA NA NA
[2,]123 NA NA
[3,]1234 NA
[4,]1234 NA
[5,]1234 NA
[6,]12345
[7,]12345
[8,] NA NA NA NA NA
rksh
On 22 Aug 2006, at 13:55, Gabor Grothendieck wrote:
Here are two solutions. seq(length = ...) instead of
just seq(...) is so that v can possibly contain zeros.
# data
v - 3:5
# solution 1 - rbind/lapply
f - function(n) {
s = seq(length = n)
replace(rep(NA, max(v)), s, s)
}
do.call(rbind, lapply(v, f))
# solution 2 - loop
mat - matrix(NA, length(v), max(v))
for(i in seq(v)) {
s - seq(length = v[i])
mat[i, s] - s
}
On 8/22/06, Sara-Jane Dunn [EMAIL PROTECTED] wrote:
Hi,
I'm having trouble applying the matrix function. I'd like to be
able to
create a matrix of vectors filled in by rows, which are not all
the same
length, and so I need it to fill in NAs where applicable.
It's easiest to explain with a simple example:
Suppose vec = c(3,4,5). How can I form a matrix of the vectors
1:vec[j]
for j=1:3?
i.e. 1 2 3 NA NA
1 2 3 4 NA
1 2 3 45
I've tried matrix(c(1:vec[j]),nrow=max(j),ncol=max(vec)) but it will
only give me a matrix with repeated values for j=1, like 1 2 3 1
2
3 1 2 3 1
2 3 1 2 3
Also using the list function hasn't got me anywhere either..
Any help/ideas would be greatly appreciated!
Many thanks,
Sara-Jane Dunn
--
This message (and any attachments) is for the recipient only...
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guide.html
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guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] lean and mean lm/glm?
On Mon, 21 Aug 2006, Damien Moore wrote: For very large regression problems there is the biglm package (put you data into a database, read in 500,000 rows at a time, and keep updating the fit). thanks. I took a look at biglm and it seems pretty easy to use and, looking at the source, avoids much of the redundancy of lm. Correct me if i'm wrong, but I think it would be virtually impossible to extend to glm, because of the non-linearity in glm models. No, it is quite straightforward if you are willing to make multiple passes through the data. It is hard with a single pass and may not be possible unless the data are in random order. Fisher scoring for glms is just an iterative weighted least squares calculation using a set of 'working' weights and 'working' response. These can be defined chunk by chunk and fed to biglm. Three iterations should be sufficient. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Escaping ' character
Dear marc: thank you for your tip.
cat(x.new, \n)
3\',5\'-cyclic-nucleotide phosphodiesterase activity
here cat is printing on screen.
how can I direct the output to an object.
I cannot do:
y - cat(x.new, \n)
is there any other way.
thanks
srini
Try this:
x - 3',5'-cyclic-nucleotide phosphodiesterase
activity
x
[1] 3',5'-cyclic-nucleotide phosphodiesterase
activity
gsub(', ', x)
[1] 3\\',5\\'-cyclic-nucleotide phosphodiesterase
activity
Note that I use gsub() to replace both instances of
the ', whereas sub()
will only replace the first.
The escape character itself needs to be escaped when
used within the
replacement regex, since the \ is a metacharacter.
You end up with
four \s since R also treats the \ specially.
When you cat() the output, you get:
x.new
[1] 3\\',5\\'-cyclic-nucleotide phosphodiesterase
activity
cat(x.new, \n)
3\',5\'-cyclic-nucleotide phosphodiesterase activity
HTH,
Marc Schwartz
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[R] HH and Rcmdr.HH packages available
The software for my recent book Statistical Analysis and Data Display Richard M. Heiberger and Burt Holland http://springeronline.com/0-387-40270-5 is now available as an R package from the book's website. The package has been submitted to CRAN. Description: Support software for Statistical Analysis and Data Display (Springer, ISBN 0-387-40270-5). This contemporary presentation of statistical methods features extensive use of graphical displays for exploring data and for displaying the analysis. The authors demonstrate how to analyze data---showing code, graphics, and accompanying computer listings---for all the methods they cover. They emphasize how to construct and interpret graphs, discuss principles of graphical design, and show how accompanying traditional tabular results are used to confirm the visual impressions derived directly from the graphs. Many of the graphical formats are novel and appear here for the first time in print. All chapters have exercises. Software that adds additional functions to Rcmdr 1.7 is now available from my website and has been submitted to CRAN. http://astro.ocis.temple.edu/~rmh/Rcmdr.HH/ Description: Our introductory course spends time on several topics that are not yet in the R Commander. Therefore we wrote the menu items and make them available. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Comparing pre-post effect sizes and areas under the curve, resp.
Hello! Does anybody know how to compare pre-post-effect sizes of different variables from the same sample with statistical tests? I have the same problem with areas under the curve (AUC) from ROC-Analysis. Any recommendations on references or programms are welcome! Thank you, Will __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [R-pkgs] ANN: 'weaver' package, caching for Sweave
Hi again, Sorry for the noise, but I need to make a correction: Seth Falcon [EMAIL PROTECTED] writes: To install: source(http://bioconductor.org/biocLite.R;) biocLite(weaver) At present, the above install sequence will _only_ work if you are using a development version of R. If you are using the current R release, you will have to work a bit harder to install (put weaver works there too): First install weaver's dependencies: digest (on CRAN) codetools: http://bioconductor.org/packages/1.9/omegahat/html/codetools.html Then install weaver: http://www.bioconductor.org/packages/1.9/bioc/html/weaver.html Finally, I will try to make Windows binaries available by the end of the week. + seth ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Successive subsets from a vector?
On Tue, 22 Aug 2006, hadley wickham wrote: The loop method took 195 secs. Just assigning to an answer of the correct length reduced this to 5 secs. e.g. use ADDRESSES - character(length(VECTOR)-4) Moral: don't grow vectors repeatedly. Other languages (eg. Java) grow the size of the vector independently of the number of observations in it (I think Java doubles the size whenever the vector is filled), thus changing O(n) behaviour to O(log n). I've always wondered why R doesn't do this. At one point at least that was too expensive on memory/address space (and it may still be for 32-bit OSes). There is even a 'truelength' field in the vector header to allow for such a strategy, and the strategy is used in scan() and elsewhere. In my experience it is relatively rare not to know the vector length in advance in R code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 3 September Courses: (1) Regression Modeling Strategies in R/Splus, (2) R/Splus Advanced Programming, (3) R/Splus Fundamentals
XLSolutions Corporation (www.xlsolutions-corp.com) is proud to announce our September courses: (1) Regression Modeling Strategies in R/Splus --- by Prof Frank Harrell http://xlsolutions-corp.com/Rstats2.htm *** Washington DC, September 28-29, 2006 *** (2) R/Splus Advanced Programming --- by the R Development Core Team Guru! *** New York City / September 11-12,2006*** http://www.xlsolutions-corp.com/Radv.htm (3) R/Splus Fundamentals and Programming Techniques *** Raleigh / September 12-13, 2006 *** http://www.xlsolutions-corp.com/Rfund.htm Ask for group discount and reserve your seat Now - Earlybird Rates. Payment due after the class! Email Sue Turner: [EMAIL PROTECTED] (1) Regression Modeling Strategies at http://xlsolutions-corp.com/Rstats2.htm (2) R/Splus Advanced Programming Course Outline: - Overview of R/S fundamentals: Syntax and Semantics - Class and Inheritance in R/S-Plus - Concepts, Construction and good use of language objects - Coercion and efficiency - Object-oriented programming in R and S-Plus - Advanced manipulation tools: Parse, Deparse, Substitute, etc. - How to fully take advantage of Vectorization - Generic and Method Functions - Search path, databases and frames Visibility - Working with large objects - Handling Properly Recursion and iterative calculations - Managing loops; For (S-Plus) and for() loops - Consequences of Lazy Evaluation - Efficient Code practices for large computations - Memory management and Resource monitoring - Writing R/S-Plus functions to call compiled code - Writing and debugging compiled code for R/S-Plus system - Connecting R/S-Plus to External Data Sources - Understanding the structure of model fitting functions in R/S-Plus - Designing and Packaging efficiently a new model function It'll also deal with lots of S-Plus efficiency issues and any special topics from participants is welcome. (3) R/Splus Fundamentals and Programming Techniques Course outline. - An Overview of R and S - Data Manipulation and Graphics - Using Lattice Graphics - A Comparison of R and S-Plus - How can R Complement SAS? - Writing Functions - Avoiding Loops - Vectorization - Statistical Modeling - Project Management - Techniques for Effective use of R and S - Enhancing Plots - Using High-level Plotting Functions - Building and Distributing Packages (libraries) - Connecting; ODBC, Rweb, Orca via sockets and via Rjava Email us for group discounts. Email Sue Turner: [EMAIL PROTECTED] Phone: 206-686-1578 Visit us: www.xlsolutions-corp.com/training.htm Please let us know if you and your colleagues are interested in this class to take advantage of group discount. Register now to secure your seat! Cheers, Elvis Miller, PhD Manager Training. XLSolutions Corporation 206 686 1578 www.xlsolutions-corp.com [EMAIL PROTECTED] 3 Courses - (1) Regression Modeling Strategies in R/Splus (2) R/Splus Advanced Programming (3) R/Splus Fundamentals __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] HH and Rcmdr.HH packages available
Dear R-helpers, For those of you who wish to use the useful and interesting HH package on the Mac: I successfully downloaded the package from http:// astro.temple.edu/~rmh/HH/HH_1.4.tar.gz Safari routinely upacks archives, so after downloading I had to (in the Terminal) gzip Desktop/H_1.4.tar and then installed it with the command sudo R CMD INSTALL Desktop/HH_1.4.tar.gz _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NonLinearLeastSquares Trust-Region
Hello! I am running R-2.3.1-i386-1 on Slackware Linux 10.2. I am a former matlab user, moving to R. In matlab, via the cftool, I performed nonlinear curve fitting using the method nonlinear least squares with the Trust-Region algorithm and not using robust fitting. Is it possible to perform the same analysis in R? I read quite a lot of R documentation, but I could not find an alternative solution. If there is such, please forgive my ignorance (I am a newbie in R) and tell me which function from which package is capable of performing the same analysis. If the same analysis is not possible to carry out in R, I would be grateful if you suggest to me some alternative procedure. I found that the nls function performs nonlinear least squares. The problem is that I do not want to implement the Gauss-Newton algorithm. In the worst case I would be contented with the Levenberg-Marquardt algorithm, if it is implemented in R. R nls's documentation mentions the port package and the ‘nl2sol’ algorithm, but I could not find that package in the CRAN repository, so that I could read and judge whether that algorithm would be appropriate. Thank you very much in advance. I am looking forward to your answer. Regards, Martin - http://ide.li/ - портал за българите по света. Статии, новини, форуми, снимки, информация. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Escaping ' character
On Tue, 2006-08-22 at 07:32 -0700, Srinivas Iyyer wrote:
Dear marc: thank you for your tip.
cat(x.new, \n)
3\',5\'-cyclic-nucleotide phosphodiesterase activity
here cat is printing on screen.
how can I direct the output to an object.
I cannot do:
y - cat(x.new, \n)
is there any other way.
thanks
srini
Srini,
Going back to your initial post, try something like the following using
paste():
x - 3',5'-cyclic-nucleotide phosphodiesterase activity
x.new - gsub(', ', x)
#Note the escaping of the single quotes here:
sql.cmd - paste(fetch_count_fterm_sql(\', (x.new), \');,
sep = )
# Beware any line wrapping here
sql.cmd
[1] fetch_count_fterm_sql('3\\',5\\'-cyclic-nucleotide phosphodiesterase
activity');
This way the character vector 'sql.cmd' has the full sql query, which
you can then pass to your statement processing code.
I'm not sure how you are passing the code, but if in a text file as
input, you can do something like:
sink(sqlfile.txt)
cat(sql.cmd, \n)
sink()
Where the text file 'sqlfile.txt' will contain the single line:
fetch_count_fterm_sql('3\',5\'-cyclic-nucleotide phosphodiesterase activity');
See ?sink for more information.
HTH,
Marc
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[R] Authoring a book
Me and some colleagues are planning to write a textbook together (Statistics using R) where the target audience for the book is psychologists and students of psychology. We thought that it might be a good idea to use a Wiki when writing the text. Is that a good idea? Does anybody have any experience in that direction? What alternatives are there? The tool (Wiki) would have to be able to handle tables and mathematical formulas in some manner, and of course, some mechanism to export the contents to a word processor in the final stages. I have my own server, Windows, based on Apache, PhP, and MySQL. Tom __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] summary(lm ... conrasts=...)
On Tue, 22 Aug 2006, [EMAIL PROTECTED] wrote: Hi Folks, I've encountered something I hadn't been consciously aware of previously, and I'm wondering what the explanation might be. Try contr.helmert(letters[1:3]) [,1] [,2] a -1 -1 b1 -1 c02 contr.treatment(letters[1:3]) b c a 0 0 b 1 0 c 0 1 and note the difference in column names. Those who made the decision to use those column names determined this. I agreed with them that labelling the second Helmert contrast here as 'c' would be confusing, especially easy to confuse with treatment contrasts. However, I thought the treatment contrasts should be labelled b-a and c-a. We also had arguments about xc vs x.c vs x:c. AFAIR brevity won. Once you know how it is done, it is easy to change the behaviour, of course: just roll your own contrasts function with the colnames you want. In (on another list) using R to demonstrate the difference between different contrasts in 'lm' I set up an example where Y is sampled from three different normal distributions according to the levels (A,B,C) of a factor X: Y-c(rnorm(mean=0,n=12),rnorm(mean=2,n=12),rnorm(mean=4,n=12)) X-factor(c(rep(A,12),rep(B,12),rep(C,12))) Then I do a summary(lm(Y~X)...) using first Treatment contrasts and then Helmert contrasts. Here are the coefficient parts of the results in each case: Just coef() or print() gives you the coefficient names: this is not done by summary(). summary(lm(Y~X,contrasts=list(X=contr.treatment))) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 0.2303 0.3220 0.715 0.47944 XB1.3057 0.4554 2.867 0.00716 ** XC3.4204 0.4554 7.511 1.23e-08 *** summary(lm(Y~X,contrasts=list(X=contr.helmert))) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 1.8057 0.1859 9.713 3.34e-11 *** X10.6529 0.2277 2.867 0.00716 ** X20.9225 0.1315 7.017 5.00e-08 *** What I'm wondering is why the effect names are X.B and X.C for Treatment, and X1, X2 for Helmert. Why not X.B and X.C in both cases? Just as XB contrasts B with the overall mean and XC contrasts C with the overall mean, XA being implicit, in the Treatment contrasts, so X1 contrasts B with A and X2 contrasts C with (A+B) in Helmert, so there is to my mind just as definite an association of B with the first contrast, and C with the second, in the Helmert case as in the Treatment case! I know it's just a matter of notation, but in the Helmert case the association with the names of the factor levels has been lost, and it could be useful to have it explicit. (Or is it intended simply as a reminder that one is using a particular system of contrasts?) Thanks, and best wishes to all, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 22-Aug-06 Time: 14:45:17 -- XFMail -- -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lean and mean lm/glm?
On Mon, 21 Aug 2006, Damien Moore wrote: For very large regression problems there is the biglm package (put you data into a database, read in 500,000 rows at a time, and keep updating the fit). thanks. I took a look at biglm and it seems pretty easy to use and, looking at the source, avoids much of the redundancy of lm. Correct me if i'm wrong, but I think it would be virtually impossible to extend to glm, because of the non-linearity in glm models. I might hack around at the source code for glm.fit -- I think I can avoid some of the redundancy involved in that routine pretty easily, but it will mean rewriting the summary output code... Damien, If you know what is 'under the hood' of glm, you can use the biglm approach to perform a one-step update of the coefficients of a glm model. There is plenty of theory for one-step estimators that use consistent estimates as starting values. You can probably get a good starting value by averaging all of the results returned by slicing the data set into smaller pieces and running glm.fit on each of them. Chuck cheers Damien --- On Mon 08/21, Greg Snow [EMAIL PROTECTED] wrote:From: Greg Snow [mailto: [EMAIL PROTECTED]: [EMAIL PROTECTED], [EMAIL PROTECTED]: Mon, 21 Aug 2006 12:01:06 -0600Subject: RE: [R] lean and mean lm/glm? For very large regression problems there is the biglm package (put you data into a database, read in 500,000 rows at a time, and keep updating the fit). This has not been extended to glm yet. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Damien Moore Sent: Monday, August 21, 2006 11:49 AM To: r-help@stat.math.ethz.ch Subject: [R] lean and mean lm/glm? Hi All: I'm new to R and have a few questions about getting R to run efficiently with large datasets. I'm running R on Windows XP with 1Gb ram (so about 600mb-700mb after the usual windows overhead). I have a dataset that has 4 million observations and about 20 variables. I want to run probit regressions on this data, but can't do this with more than about 500,000 observations before I start running out of ram (you could argue that I'm getting sufficient precision with 500,000 obs but lets pretend otherwise). Loading 500,000 observations into a data frame only takes about 100Mb of ram, so that isn't the problem. Instead it seems R uses huge amount of memory when running the glm methods. I called the Fortran routines that lm and glm use directly but even they create a large number of extraneous variables in the output (e.g. the Xs, ys, residuals etc) and during processing. For instance (sample code) x=runif(100) y=3*x+rnorm(100) #I notice this step chews up a lot more than the 7mb of ram required to store y during processing, but cleans up ok afterwards with a gc() call X=cbind(x) p=ncol(X) n=NROW(y) ny=NCOL(y) tol=1e-7 #this is the fortran routine called by lm - regressing y on X here z - .Fortran(dqrls, qr = X, n = n, p = p, y = y, ny = ny, tol = as.double(tol), coefficients = mat.or.vec(p, ny), residuals = y, effects = y, rank = integer(1), pivot = 1:p, qraux = double(p), work = double(2 * p), PACKAGE = base) This code runs very quickly - suggesting that in principle R should have no problem at all handling very large data sets, but uses 100mb during processing and z is about a 20mb object. Scaling this up to a much larger dataset with many variables its easy to see i'm going to run into problems My questions: 1. are there any memory efficient alternatives to lm/glm in R? 2. is there any way to prevent the Fortran routine dqrls from producing so much output? (I suspect not since its output has to be compatible with the summary method, which seems to rely on having a copy of all variables instead of just references to the relevant variables - correct me if i'm wrong on this) 3. failing 1 2 how easy would it be to create new versions of lm and glm that don't use so much memory? (Not that I'm volunteering or anything ;) ). There is no need to hold individual residuals in memory or make copies of the variables (at least for my purposes). How well documented is the source code? cheers Damien Moore __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [ Part 3.53: Included Message ] Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717
Re: [R] Rgraphviz installation Problem
j.joshua thomas [EMAIL PROTECTED] writes: Dear Robert, Thanks for your time. I have downloaded Rgraphviz (windows binary) from www.bioconductor.org and put inside R2.3.0 library then i installed from the local zip its says package 'graph' couldnot be loaded. Am i doing the installation correctly? Still the new user. Can you guide me sir? Questions about BioC packages are best directed to the bioconductor mailing list. I would recommend trying: source(http://bioconductor.org/biocLite.R;) biocLite(Rgraphviz) Best, + seth __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Successive subsets from a vector?
Like this:
do.call( paste, c( list(sep=), lapply(1:5,function(x)
VECTOR[x:(length(VECTOR)-5+x)]) ))
[1] 14265 42650 265011 6501110 5011104 0111043 1110436
104368 43686
HTH,
Chuck
On Tue, 22 Aug 2006, kone wrote:
I'd like to pick every imbricated five character long subsets from a
vector. I guess there is some efficient way to do this without loops...
Here is a for-loop-version and a model for output:
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6);
ADDRESSES=c();
for(i in 1:(length(VECTOR)-4)){
ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse=)
}
ADDRESSES
[1] 14265 42650 265011 6501110 5011104 0111043
1110436 104368
[9] 43686
Atte Tenkanen
University of Turku, Finland
[[alternative text/enriched version deleted]]
[ Part 3.64: Included Message ]
Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED] UC San Diego
http://biostat.ucsd.edu/~cberry/ La Jolla, San Diego 92093-0717
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[R] Selection on dataframe based on order of rows
I have a dataframe with the following structure iddate value - 122/08/2006 48 124/08/2006 50 128/08/2006 150 130/08/2006 100 101/09/2006 30 211/08/2006 30 222/08/2006 100 228/08/2006 11 202/09/2006 5 301/07/2006 3 301/08/2006 100 301/09/2006 100 422/08/2006 48 424/08/2006 50 428/08/2006 150 430/08/2006 100 401/09/2006 30 403/09/2006 100 406/09/2006 100 N.B.: dates in european format; ordered dataframe For each ID I need to select the first occurrence of all the rows which are the first of at least two with value = 50. Rather convoluted explication. I mean that for each id I have to select the first row in which value is 50 only if at least the following row has value 50 too. If this is not true I repeat the test for all the following rows in which value 50 untill I find a record that respects the condition this means that with my example dataframe the result is : iddate value - 128/08/2006 150 301/08/2006 100 428/08/2006 150 It's clear that a for loop would work but I think that that is a better way. I tried by and could obtain the first row for wich value is 50. I thought of an iterative process (delete the first row 50, find the second row 50, examine if there are rows in the middle) but it is quite inelegant as if the first value is not the good one I have to repeat the process for a a priori unknown number of times. Thanks in advance for Your help Sandro Bonfigli __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Marginal Predicitions from nlme and lme4
Is there a way (simple or not) to get the marginal prediction from lme (in nlme) and/or lmer (in lme4)? Rick B. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Successive subsets from a vector?
Here is a solution that uses gsub with a negative lookahead perl-style
regexp to do it:
VECTOR - c(1,4,2,6,5,0,11,10,4,3,6,8,6)
e -
([[:digit:]]+),(?=([[:digit:]]+),([[:digit:]]+),([[:digit:]]+),([[:digit:]]+))
out - gsub(e, \\1\\2\\3\\4\\5 , paste(VECTOR, collapse = ,), perl = TRUE)
head(strsplit(out, )[[1]], -1) # uses head from R 2.4.0
On 8/22/06, kone [EMAIL PROTECTED] wrote:
I'd like to pick every imbricated five character long subsets from a
vector. I guess there is some efficient way to do this without loops...
Here is a for-loop-version and a model for output:
VECTOR=c(1,4,2,6,5,0,11,10,4,3,6,8,6);
ADDRESSES=c();
for(i in 1:(length(VECTOR)-4)){
ADDRESSES[i]=paste(VECTOR[i:(i+4)],collapse=)
}
ADDRESSES
[1] 14265 42650 265011 6501110 5011104 0111043
1110436 104368
[9] 43686
Atte Tenkanen
University of Turku, Finland
[[alternative text/enriched version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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[R] Mac os
Dear Sir / Madam Hi, I have written some code For R that uses for loops to do 2-dimensional grid searches for maximum likelhood combined with iterated GLS estimation. As can be expected, depending on the szie of the grid, estimation can take quite some time. However, I have noticed that the same code run on a windows operating system is much faster than when run on a Mac (I basically paste the code into the console and then run things from there). I was wondering if anyone knew if this is typical, or if there is some good reason for this? (I am not that familiar with the mac operating system, so might be missing something obvious). I am writing a user manual for the code and would like to have some explanation (or possible improvement) for mac users, so any info on this would be much appreciated Sincerely Jason Pienaar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] new version of The R Guide available on CRAN
Hi, thanks for this. I'll keep it in mind next time in teaching/referring someone to R. BTW, before the R-core guys get you ;) Just replace all places where you use library to refer to a package (see all comments on the definition of these on r-help/r-devel), e.g. Page 17: FYI, .GlobalEnv is your workspace and the package quantities are libraries that contain (among other things) the functions and datasets that we are learning about in this manual. Cheers Henrik On 8/22/06, Owen, Jason [EMAIL PROTECTED] wrote: Hello, Version 2.2 of The R Guide is available for download in the Contributed Documents section on CRAN. The R Guide is written for the beginning R user. I use the guide in my undergraduate probability and math stat sequence, but anyone with a basic understanding of statistics (who wants to learn R) should find it useful. This updated version includes sections on multiple comparisons, optimization, along with some improvements suggested by fellow R users from around the world. The entire document is under 60 pages in length. Jason -- Assistant Professor of Statistics Mathematics and Computer Science Department University of Richmond, Virginia 23173 (804) 289-8081 fax:(804) 287-6664 http://www.mathcs.richmond.edu/~wowen This is R. There is no if. Only how. Simon (Yoda) Blomberg __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] a generic Adaptive Gauss Quadrature function in R?
Hi there, I am using SAS Proc NLMIXED to maximize a likelihood with multivariate normal random effects. An example is the two part random effects model for repeated measures semi-continous data with a cluster at 0. I use the model y ~ general(loglike) statement in Proc NLMIXED, so I can specify a general log likelihood function constructed by SAS programming statements. Then the likelihood can be maximized by AGQ. Is there a similar generic AGQ function in R to let me write explicitly the log likelihood and then maximize it accordingly? Can nlme do the work? Thanks! Lei Liu Assistant Professor Division of Biostatistics and Epidemiology Department of Public Health Sciences School of Medicine University of Virginia 3181 Hospital West Complex Charlottesville, VA 22908-0717 1-434-982-3364 (o) 1-434-806-8086 (c) 1-434-243-5787 (f) [EMAIL PROTECTED] [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate example : where is the state.region variable?
--- Richard M. Heiberger [EMAIL PROTECTED] wrote: there is no factor in the dataset but why there is not one and why a call to another dataset is totally opaque. The reason is purely historical. The state dataset is about 10 years older than the data.frame concept. At the time the state.* variables were constructed it was not possible to put numeric data and factor data into the same rectangular structure. I see. So originally the example would have been more obvious. Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question on R Training
Dear R users, The R Project website doesn't seem to have any links devoted to R training. Are there any R trainers out there? Thank you for your help! Jeffrey V. Ludlow - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Marginal Predicitions from nlme and lme4
Rick, if by marginal prediction, you mean the prediction without random effects, then use the level argument. See ?predict.lme or ?fitted.lme If not then I don't know :) Cheers Andrew On Tue, Aug 22, 2006 at 02:27:06PM -0400, Rick Bilonick wrote: Is there a way (simple or not) to get the marginal prediction from lme (in nlme) and/or lmer (in lme4)? Rick B. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to run ANCOVA?
Dear all, I would like to know how to run an analysis of covariance in R. For example, I have a data frame (data) consisting of two second-degree categorical variables (diagnosis and gender), one continous independent variable (age) and one continous dependent variable (response). I ran a simple anova to see the effects of diagnosis and gender (and interaction): aov.out - aov(response~diagnosis*gender,data) anova(aov.out) Now I would like to covary for age, how can I add age as a covariate to this equation? Thanks, Sasha __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] error message from lm.ridge() in MASS library
Dear all, I got a wierd problem when using lm.ridge() in MASS library. When my X matrix has few columns, there is no problem. But when my X matrix gets larger (over 1000 columns), I got the following error: Error in Xs$v %*% a : non-conformable arguments In addition: Warning messages: 1: longer object length is not a multiple of shorter object length in: d^2 + rep(lambda, rep(p, k)) 2: longer object length is not a multiple of shorter object length in: drop(d * rhs)/div The R code I use for the calculation is lm.ridge( y ~ x,lambda=seq(1,15,1)). Please advice. Thanks a lot! Jeny __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R is wonderful
I have always find heaveyweight solutions while use R. If I were rich, I would have made a great donation to R fundation ... [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Marginal Predicitions from nlme and lme4
On Wed, 2006-08-23 at 06:43 +1000, Andrew Robinson wrote: Rick, if by marginal prediction, you mean the prediction without random effects, then use the level argument. See ?predict.lme or ?fitted.lme If not then I don't know :) Cheers Andrew Thanks. I'm familiar with level in predict and fitted for lme. These allow you to select the fixed effects and/or the random effects. The marginal prediction integrates out the random effects and is what a GEE marginal model produces. From what I've read, the marginal effects seem to be less desirable than the fixed effects from an lme or a generalized lme. But I would still like to compute them for comparison. Rick B. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rpart output: rule extraction beyond path.rpart()
Greetings -
Is there a way to automatically perform what I believe is called rule
extraction (by Quinlan and the machine learning community at least) for
the leaves of trees generated by rpart? I can use path.rpart() to
automatically extract the paths to the leaves, but these can be
needlessly cumbersome. For example, one path returned by path.rpart()
might be:
[1] root y=-0.1905 y 0.1495 z=-0.19 z 0.1785
[6] y=-0.1385 z=-0.153 x 0.37x=-0.363
But the y = -0.1905 and z=-.19 are both redundant, given restrictions
placed further down the tree. Simplifying the paths by hand is feasible
for small trees but quite cumbersome when dimensionality increases. I
can think of ways to write code to do this automatically, but would
prefer not to if it's already implemented. I have done extensive
searching and turned up nothing, but I fear I might just be lacking the
right terminology. Any thoughts?
Much appreciated,
-Ben
Ben Bryant
Doctoral Fellow
Pardee RAND Graduate School
[EMAIL PROTECTED]
This email message is for the sole use of the intended recip...{{dropped}}
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Re: [R] Marginal Predicitions from nlme and lme4
On Tue, 22 Aug 2006, Rick Bilonick wrote: On Wed, 2006-08-23 at 06:43 +1000, Andrew Robinson wrote: Rick, if by marginal prediction, you mean the prediction without random effects, then use the level argument. See ?predict.lme or ?fitted.lme If not then I don't know :) Cheers Andrew Thanks. I'm familiar with level in predict and fitted for lme. These allow you to select the fixed effects and/or the random effects. The marginal prediction integrates out the random effects and is what a GEE marginal model produces. From what I've read, the marginal effects seem to be less desirable than the fixed effects from an lme or a generalized lme. But I would still like to compute them for comparison. I don't agree that they are less useful, but they are not in general easy to obtain from a GLMM. For any linear link model or a log link model that has only random intercepts the marginal and conditional effects are the same. For the probit model there is a conversion formula, but for other models they typically require high-dimensional integration to compute. It's easy just to fit a marginal glm if you want marginal coefficients and a mixed model if you want conditional coefficients. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selection on dataframe based on order of rows
Try this:
# data
DF - structure(list(id = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4,
4, 4, 4, 4, 4, 4), date = structure(c(8, 9, 10, 11, 3, 7, 8,
10, 4, 1, 2, 3, 8, 9, 10, 11, 3, 5, 6), .Label = c(01/07/2006,
01/08/2006, 01/09/2006, 02/09/2006, 03/09/2006, 06/09/2006,
11/08/2006, 22/08/2006, 24/08/2006, 28/08/2006, 30/08/2006
), class = factor), value = c(48, 50, 150, 100, 30, 30, 100,
11, 5, 3, 100, 100, 48, 50, 150, 100, 30, 100, 100)), .Names = c(id,
date, value), class = data.frame, row.names = c(1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19))
f - function(x) {
idx - which(x$value 50 c(x$value[-1], 0) 50)
if (length(idx) 0) x[idx[1],]
}
do.call(rbind, by(DF, DF$id, f))
On 8/22/06, Bonfigli Sandro [EMAIL PROTECTED] wrote:
I have a dataframe with the following structure
iddate value
-
122/08/2006 48
124/08/2006 50
128/08/2006 150
130/08/2006 100
101/09/2006 30
211/08/2006 30
222/08/2006 100
228/08/2006 11
202/09/2006 5
301/07/2006 3
301/08/2006 100
301/09/2006 100
422/08/2006 48
424/08/2006 50
428/08/2006 150
430/08/2006 100
401/09/2006 30
403/09/2006 100
406/09/2006 100
N.B.: dates in european format; ordered dataframe
For each ID I need to select the first occurrence of
all the rows which are the first of at least two with
value = 50.
Rather convoluted explication. I mean that for each id I have to select
the first row in which value is 50 only if at least the following row
has value 50 too. If this is not true I repeat the test for all the
following rows in which value 50 untill I find a record that respects
the condition
this means that with my example dataframe the result is :
iddate value
-
128/08/2006 150
301/08/2006 100
428/08/2006 150
It's clear that a for loop would work but I think that that is a better
way.
I tried by and could obtain the first row for wich value is 50.
I thought of an iterative process (delete the first row 50, find the
second row 50, examine if there are rows in the middle) but it
is quite inelegant as if the first value is not the good one I have to
repeat the process for a a priori unknown number of times.
Thanks in advance for Your help
Sandro Bonfigli
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[R] Rtangle in R-2.3.1
Simply using Rtangle(file.Rnw) gives the following error message in R-2.3.1. It used to work fine with the earlier versions of R. Rtangle(qtlbim.Rnw) Error in Rtangle(qtlbim.Rnw) : unused argument(s) ( ...) RtangleSetup(file=qtlbim.Rnw,output=qtlbim.R) Writing to file qtlbim.R Error in RtangleSetup(file = qtlbim.Rnw, output = qtlbim.R) : argument syntax is missing, with no default - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on R Training
Check out with XLSolutions Corp [EMAIL PROTECTED] --- jeffrey ludlow [EMAIL PROTECTED] wrote: Dear R users, The R Project website doesn't seem to have any links devoted to R training. Are there any R trainers out there? Thank you for your help! Jeffrey V. Ludlow - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to complete this task on data management
Dear friends, When i clean my dataset , i met a difficulty suppose my data set is : * data-data.frame(x=c(1:5,1,2,3)) data x 1 1 2 2 3 3 4 4 5 5* 6 1 7 2 8 3 Now i need to add the data which are less than 3.5 at the bottom, not including the top data, so the results should be : x 1 1 2 2 3 3 4 4 5 5 *6 6* I tried to use data[data$x3.5,] to do it , but it also delete the first several numbers,* How to finish it ?* Thanks very much. -- Kind Regards, Zhi Jie,Zhang [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] glm inside one self-defined function
Hi list,
I've searched in R-help and found some related discussions but still could
not understand this type of error. My own function is pretty complex, so I
would not put it here, but the basic algorithm is like this:
myfun-function(k){
mydata-...#by someway I create a data frame
mymodel-glm(y~.,family=binomial(),data=mydata)
...#some other stuff
}
as I execute this function, it gives error like this
Error in inherits(x, data.frame) : object mydata not found
So I guess glm here tries to find mydata in the parent environment. Why
doesn't it take mydata inside the function? How to let glm correctly
locate it? Is this (scope/environment) mentioned in R manual? Thanks,
Mike
[[alternative HTML version deleted]]
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Re: [R] glm inside one self-defined function
Mike Wolfgang asks:
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Mike Wolfgang
Sent: Wednesday, 23 August 2006 1:31 PM
To: R-help list
Subject: [R] glm inside one self-defined function
Hi list,
I've searched in R-help and found some related discussions but still
could
not understand this type of error. My own function is pretty complex,
so I
would not put it here, but the basic algorithm is like this:
myfun-function(k){
mydata-...#by someway I create a data frame
mymodel-glm(y~.,family=binomial(),data=mydata)
...#some other stuff
}
I think you are leaving out something. Here is a test of what you
claim gives a problem (R 2.3.1, Windows):
myfun - function(n) {
+ z - rnorm(n)
+ mydata - data.frame(x = z,
+ y = rbinom(n, size = 1, prob = exp(z)/(1+exp(z
+ fm - glm(y ~ x, binomial, mydata)
+ fm
+ }
myfun(100)
Call: glm(formula = y ~ x, family = binomial, data = mydata)
Coefficients:
(Intercept)x
0.1587 1.0223
Degrees of Freedom: 99 Total (i.e. Null); 98 Residual
Null Deviance: 137.6
Residual Deviance: 118.3AIC: 122.3
Not even a murmur of complaint. (This also works in S-PLUS 7.0 but
earlier versions of S-PLUS gave a problem rather like the one you note,
curiously.)
Look again at your code and see if the abstract version you give
really matches what you did, may I suggest?
as I execute this function, it gives error like this
Error in inherits(x, data.frame) : object mydata not found
So I guess glm here tries to find mydata in the parent environment.
Why
doesn't it take mydata inside the function? How to let glm correctly
locate it? Is this (scope/environment) mentioned in R manual? Thanks,
Mike
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Re: [R] R is wonderful
Hi All,
I have also similar feelings for R. I really thank each and every one in
the R community for joining together to Create R and Spread R.
thanks
Sayonara With Smile With Warm Regards :-)
G a u r a v Y a d a v
Senior Executive Officer,
Economic Research Surveillance Department,
Clearing Corporation Of India Limited.
Address: 5th, 6th, 7th Floor, Trade Wing 'C', Kamala City, S.B. Marg,
Mumbai - 400 013
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Re: [R] how to run ANCOVA?
aov.out - aov(response~diagnosis*gender,data) Just add it where you think it belongs in the sequential sum of squares To adjust the factors for the covariate use aov.out - aov(response ~ age + diagnosis*gender, data) To adjust the covariate for the factors aov.out - aov(response ~ diagnosis*gender + age, data) If you want to check for interaction of the factors with the covariate, then use * instead of + in the formula. Please note that I added spaces to your statement to improve human legibility. Rich __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
