[R] Replacing characters
Hello guys! May be I am lazy but I need to replace a character like \ or ' or to escape them in a character vector to write a SQL statement. How can I do that? Caveman [[alternative HTML version deleted]] __ R-help@r-project.org 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] Using R in combination with GRASS for image classification
Hello everybody! For my academic project work I intend to use only open-source software. My objective is to carry out classification of satellite digital images using Geostatistics. I found out GRASS is a good software for image processing. Also a good geostatistics software is R. It is also known that R can be used in combination with GRASS. But I am not able to confirm that satellite digital image classification can be carried out using geostatistical analysis with GRASS in combination with R. Can anybody confirm the capability of the software in question or give any alternative suggestions? Regards, Chethan S. -- Chethan S. PG Student | Remote Sensing and Geographic Information Systems Dept. of Applied Mechanics Hydraulics National Institute of Technology Karnataka | Surathkal Karnataka | India - 575 025 ***e-mail:* chethanuniver...@gmail.com *Blog:* http://gischethans.blogspot.com http://gischethans.wordpress.com *Twitter:* http://twitter.com/gischethans [[alternative HTML version deleted]] __ R-help@r-project.org 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] Why use numFold in evaluate_Weka_classifier of RWeka
s0300851 s0300851 at tp.edu.tw writes:
Hi everyone,
I have a question about using RWeka package，
we know that instruction make_Weka_classifier that can help
us to build a model,and evaluate_Weka_classifier instruction
can help us to evaluate the performance of the model using on new data.
But I have a question about how to using the parameter numFold in
evaluate_Weka_classifier.Cross-validation means that using some parts
to train our data,and some parts to do test,but it should be using in
the step of building our model not evaluation.
I try to think about the numFold=n in the evaluate_Weka_classifier may be
this:
randomly(but in proportion) to select data in the dataset then redo n times,
then to evaluate the performance.Is this correct?
No. It's preferable to learn about Weka right from the Weka manual.
About the number of folds ('numFold') it says:
A more elaborate method is cross-validation. Here, a number of
folds n is specified. The dataset is randomly reordered and then
split into n folds of equal size. In each iteration, one fold is
used for testing and the other n-1 folds are used for training the
classifier. The test results are collected and averaged over all
folds. This gives the cross-validation estimate of the accuracy.
Thanks.
Best regards ,
Hsiao
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[R] how to draw a spherical quadrant
Hi, Please, could you tell me if is there a R package that allow to draw spherical quadrants? The value of each point of the quadrant has to represent the power generated by a kite on that point tied to a line located in the center of the quadrant. Please, could you give me any clue? Thanks! Greetings, Ricardo -- Ricardo Rodríguez Your XEN ICT Team __ R-help@r-project.org 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 support for 64 bit integers
Are integers strictly a signed 32 bit number on R even if I am running a 64 bit version of R on a x86_64 bit machine? I ask because I have integers stored in a hdf5 file where some of the data is 64 bit integers. When I read that into R using the hdf5 library it seems any integer greater than 2**31 returns NA. Any solutions? Thanks, Theo [[alternative HTML version deleted]] __ R-help@r-project.org 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] PCA analysis, presence-absence of mammals in parks
Hi everyone, So I am trying to see which ecological parameter of different parks in nyc influence the most the diversity of the medium-sized mammals in those parks. I have a bunch of different parameters for each park I'm done studying and the presence (1) and absence (0) of each mammal. I wanted to do a principal component analysis of those data to know which ones of those parameters are the most important. Here is what I have: Parc PBP VCPPP CP surfacekm² 0.917 4.64 0.0213 3.4 woodenarea 0.514 0.6355 0.320.4952 scrubland 0.0905 0.0412 0.060.0498 lawn0.3654 0.2837 0.500.296 lake0.0028 0.0166 0.120.1488 wetareas0.0273 0.0230 0 0.0102 intertidalearea 0.0283 0 0 0 sportsfields0.0887 0.2579 0.120.1323 humanactivity 0.4196 0.7271 0.230.7912 #ofhighways 2 3 0 0 #ofroads4 22 7 11 distanceA 5.314.021.614.02 distanceB 6.124.513.7 4.02 distanceC 7.565.477.726.44 distanceD 9.669.6619.31 9.01 AvgPopN 11330 53806835 AvgPopS 0 702253804711 AvgPopE 0 359285364307 AvgPopW 3940322048409360 raccoon 1 1 1 1 skunk 1 1 1 1 opossum 1 1 1 0 rabbit 1 1 1 0 groundhog 0 0 0 0 muskrat 0 0 0 0 I get to open the .txt with R but then it gives me something pretty weird when I try to use the function dudi.pca(parcs)... Can anyone suggest me how I should to my analysis? I would really really be thankful for that! Geo. -- View this message in context: http://r.789695.n4.nabble.com/PCA-analysis-presence-absence-of-mammals-in-parks-tp2319051p2319051.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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 extract the conf.level out of t.test() data
Good afternoon everybody, I'm writing a little function to visualise hypothesis testing. Therefore I need to extract the confidence level of a t-test. Here a little example: x - str(t.test(1:10) gives List of 9 $ statistic : Named num 5.74 ..- attr(*, names)= chr t $ parameter : Named num 9 ..- attr(*, names)= chr df $ p.value: num 0.000278 $ conf.int : atomic [1:2] 3.33 7.67 ..- attr(*, conf.level)= num 0.95 $ estimate : Named num 5.5 ..- attr(*, names)= chr mean of x $ null.value : Named num 0 ..- attr(*, names)= chr mean $ alternative: chr two.sided $ method : chr One Sample t-test $ data.name : chr 1:10 - attr(*, class)= chr htest Now I can use x$conf.int what gives [1] 496.9141 499.6276 attr(,conf.level) [1] 0.95 In the example I try to extract the value 0.95 but I have no Idea how. I hope somebody can help me. Thanks in advance an greetings from Berlin Etienne __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coef(summary) and plyr
Upon reading the plyr documentation that was the distinct impression I got and
I´m glad that whatever expectations I had developed regarding plyr were
fulfilled. Thx for the input Hadley.
Maybe this is a cumbersome solution, but it works..
And Matthew, I will most definitively look into the datatable library.
mydf - data.frame(x1=rnorm(100), x2=rnorm(100), x3=rnorm(100))
mydf$fac-factor(sample((0:2),replace=T,100))
mydf$y- mydf$x1+0.01+mydf$x2*3-mydf$x3*19+rnorm(100)
dlply(mydf,.(fac),function(df) lm(y~x1+x2+x3,data=df))-dl
test-function(a){
coef(summary(a))-lo
a-colnames(lo)
b-rownames(lo)
c-length(a)
e-character(0)
r-NULL
for (x in (1:c)){
d-rep(paste(a[1:c],b[x],sep= ))
e-paste(c(e,d))
t-lo[x,]
r-c(r,t)
names(r)-e
}
return(r)
}
ldply(dl,function(x) test(x))-g
g
Regards,
Moleps
On 9. aug. 2010, at 19.55, Hadley Wickham wrote:
That's exactly what dlply does - so you should never have to do that
yourself.
I'm unclear what you are saying. Are you saying that the plyr function
_should_ have examined the objects in that list and determined that there
were 4 rows and properly labeled the rows to indicate which list they came
from?
Yes, exactly. It's the output from coef(summary(x)) that makes it
look like this isn't happening.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
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[R] how to interpolate multidimensional (3D) spatial data
Hello,
I am an archaeologist who wonders wether a
multidimensional (3D) spatial interpolation
is possible in R.
The data consists of measurements in 3D space
with coordinates X-Y-Z and a measured variable
U, that is, U is measured at points located
at X-Y-Z.
e.g. a data structure like the following:
X Y Z U
5.105.0216.03 6.674
5.414.3214.03 6.668
1.723.5912.28 7.521
. . . .
. . . .
I am used to interpolation in GIS with 2D-data
(x-y, u). The interpolation method should possible
be any I am familiar with like kriging, minimum
curvature (splines), inverse distance or something
like this.
Having done 3D kernel density estimation I think
of a visualization of the interpolated 3D field
of U like the density in 3D rgl-plots {function
'plot(kde(xyz,hpi(xyz)))', package=ks}, that is,
having the interpolated values shown as a kind of
semitransparent cloud in 3D with the colour and/or
the transparency representing the values of U.
Is there any function or package that does this
kind of computation and visualization?
Please excuse my bumpy English.
Thank You for any help.
And, thanks to the volunteer moderators allowing
mails for non-subscribers
Georg
*
Any technology sufficiently advanced *
is indistinguishable from magic. *
(Arthur C. Clarke)*
*
Georg Roth M.A. (prom.)
Historisches Seminar
Professur für Ur- und Frühgeschichte
Universität Leipzig
Ritterstraße 14
04109 Leipzig
Tel: 0049 - (0)341 - 97 37069
Fax: 0049 - (0)341 - 97 37046
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[R] Rcurl giving SSL grief
Do you use RGoogleDocs? If so can you please try run a script now and tell me if you get what I am getting? Error in curlPerform(curl = curl, .opts = opts, .encoding = .encoding) : SSL certificate problem, verify that the CA cert is OK. Details: error:14090086:SSL routines:SSL3_GET_SERVER_CERTIFICATE:certificate verify failed It came after I ran this: library(RGoogleDocs) ps -readline(prompt=get the password in ) sheets.con = getGoogleDocsConnection(getGoogleAuth(fx...@gmail.com, ps, service =wise)) ts2=getWorksheets(OnCall,sheets.con) I posted the question on StackOverflow at http://stackoverflow.com/questions/3442781/rgoogledocs-or-rcurl-giving-ssl-certificate-problem Is it a breach of etiquette to cross post? Farrel Buchinsky [[alternative HTML version deleted]] __ R-help@r-project.org 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 extract the conf.level out of t.test() data
Le 09/08/10 20:39, Etienne Stockhausen a écrit : Good afternoon everybody, I'm writing a little function to visualise hypothesis testing. Therefore I need to extract the confidence level of a t-test. Here a little example: x - str(t.test(1:10) gives List of 9 $ statistic : Named num 5.74 ..- attr(*, names)= chr t $ parameter : Named num 9 ..- attr(*, names)= chr df $ p.value : num 0.000278 $ conf.int : atomic [1:2] 3.33 7.67 ..- attr(*, conf.level)= num 0.95 $ estimate : Named num 5.5 ..- attr(*, names)= chr mean of x $ null.value : Named num 0 ..- attr(*, names)= chr mean $ alternative: chr two.sided $ method : chr One Sample t-test $ data.name : chr 1:10 - attr(*, class)= chr htest Now I can use x$conf.int what gives [1] 496.9141 499.6276 attr(,conf.level) [1] 0.95 In the example I try to extract the value 0.95 but I have no Idea how. I hope somebody can help me. Thanks in advance an greetings from Berlin Etienne You need the conf.level attribute, as in : x - t.test(1:10) attr( x$conf.int, conf.level ) Romain -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/aAyra4 : highlight 0.2-2 |- http://bit.ly/94EBKx : inline 0.3.6 `- http://bit.ly/aryfrk : useR! 2010 __ R-help@r-project.org 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] coef(summary) and plyr
correction...
Col and rows were mixed up and loop only worked when rows were less than or
equal to number of columns
//M
test-function(a){
coef(summary(a))-lo
a-colnames(lo)
b-rownames(lo)
c-length(a)
e-character(0)
r-NULL
for (x in (1:length(b))){
d-rep(paste(a[1:c],b[x],sep= ))
e-paste(c(e,d))
t-lo[x,]
r-c(r,t)
names(r)-e
}
return(r)
}
On 9. aug. 2010, at 19.55, Hadley Wickham wrote:
That's exactly what dlply does - so you should never have to do that
yourself.
I'm unclear what you are saying. Are you saying that the plyr function
_should_ have examined the objects in that list and determined that there
were 4 rows and properly labeled the rows to indicate which list they came
from?
Yes, exactly. It's the output from coef(summary(x)) that makes it
look like this isn't happening.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
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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] How to extract the conf.level out of t.test() data
try this: x - t.test(1:10) x$conf.int attr(x$conf.int, conf.level) I hope it helps. Best, Dimitris On 8/9/2010 8:39 PM, Etienne Stockhausen wrote: Good afternoon everybody, I'm writing a little function to visualise hypothesis testing. Therefore I need to extract the confidence level of a t-test. Here a little example: x - str(t.test(1:10) gives List of 9 $ statistic : Named num 5.74 ..- attr(*, names)= chr t $ parameter : Named num 9 ..- attr(*, names)= chr df $ p.value : num 0.000278 $ conf.int : atomic [1:2] 3.33 7.67 ..- attr(*, conf.level)= num 0.95 $ estimate : Named num 5.5 ..- attr(*, names)= chr mean of x $ null.value : Named num 0 ..- attr(*, names)= chr mean $ alternative: chr two.sided $ method : chr One Sample t-test $ data.name : chr 1:10 - attr(*, class)= chr htest Now I can use x$conf.int what gives [1] 496.9141 499.6276 attr(,conf.level) [1] 0.95 In the example I try to extract the value 0.95 but I have no Idea how. I hope somebody can help me. Thanks in advance an greetings from Berlin Etienne __ R-help@r-project.org 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. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org 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] Identifying integers (as opposed to real #s) in matrix
Is there a way to identify (for subsequent replacement) which rows in a matrix are comprised entirely of *integers*? I have a large set of *nx3 *matrices where each row either consists of a set of 3 integers or a set of 3 real numbers. A given matrix might looks something like this: [ ,1] [ ,2] [ ,3] [1, ] 121.-98. 276. [2, ] 10.1234 25.4573 -188.9204 [3, ] 121.-98. 276. [4, ] -214.4982 -99.1043-312.0495 . [n, ] 99. 1. -222. Ultimately, I'm going to replace the values in the integer-only rows with NAs. But first I need r to recognize the integer-only rows. I assume whatever function I write will be keyed off of the .s, but have no clue how to write that function. Any ideas? David Katz [[alternative HTML version deleted]] __ R-help@r-project.org 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] Identifying integers (as opposed to real #s) in matrix
Le 10/08/10 10:05, David Katz a écrit : Is there a way to identify (for subsequent replacement) which rows in a matrix are comprised entirely of *integers*? I have a large set of *nx3 *matrices where each row either consists of a set of 3 integers or a set of 3 real numbers. A given matrix might looks something like this: [ ,1] [ ,2] [ ,3] [1, ] 121.-98. 276. [2, ] 10.1234 25.4573 -188.9204 [3, ] 121.-98. 276. [4, ] -214.4982 -99.1043-312.0495 . [n, ] 99. 1. -222. Ultimately, I'm going to replace the values in the integer-only rows with NAs. But first I need r to recognize the integer-only rows. I assume whatever function I write will be keyed off of the .s, but have no clue how to write that function. Any ideas? David Katz Something like this perhaps: x - rbind( c(1, 2, 3), c(1.2,2.2, 3.2), c(1,2,3), c(1.2, 1.2, 1.3 ) ) x [,1] [,2] [,3] [1,] 1.0 2.0 3.0 [2,] 1.2 2.2 3.2 [3,] 1.0 2.0 3.0 [4,] 1.2 1.2 1.3 rowSums( x == round(x) ) == ncol(x) [1] TRUE FALSE TRUE FALSE x[ rowSums( x == round(x) ) == ncol(x) , ] - NA x [,1] [,2] [,3] [1,] NA NA NA [2,] 1.2 2.2 3.2 [3,] NA NA NA [4,] 1.2 1.2 1.3 Romain -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://bit.ly/aAyra4 : highlight 0.2-2 |- http://bit.ly/94EBKx : inline 0.3.6 `- http://bit.ly/aryfrk : useR! 2010 __ R-help@r-project.org 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] Results with name of dataset
Yes, this is exactly want I want. Thanks Michael -- View this message in context: http://r.789695.n4.nabble.com/Results-with-name-of-dataset-tp2318328p2319484.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Sweave with dev.new()
Dear list.
I am preparing a R package, and the last step is to write a package
vignette using Sweave. However, I am experiencing some trouble when trying
to include plots in my Sweave document. That is, in my package I have made
some plotting functions in which I start by calling 'new.dev()' to start a
graphics device of a certain width and height, and then proceed with
plot(). These functions work perfectly fine in R, and when I check and
build my package with R CMD build/check.
The problems start when I try to include the command and the output of my
plotting functions in the Sweave document, and I suspect it must have
something to do with the command 'dev.new()' in my code. Here is a basic
example of a Sweave file which can hopefully illustrate the problem:
\documentclass[a4paper]{article}
\usepackage{Sweave}
\RequirePackage{graphicx,fancyvrb}
\title{Sweavetest}
\begin{document}
\section{Plotting}
label=testfig,include=FALSE,echo=TRUE=
dev.new(width=7,height=6,record=TRUE)
plot(1:10,col=red)
@
\begin{figure}[ht]
\begin{center}
label=fig,fig=TRUE,echo=FALSE=
testfig
@
\end{center}
\caption{Testplot}
\label{fig:test}
\end{figure}
\end{document}
When I run Sweave on this file I get empty pdf and eps files named
Sweavetest-fig.pdf and Sweavetest-fig.eps, and the corresponding latex
file is unable to create a PDF-file.
Does anyone know what the problem is, and how I might solve it? This is
the first time that I have worked with Sweave, so I might be missing
something basic...
Best regards,
Gro Nilsen
PhD-student, University of Oslo
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Re: [R] Identifying integers (as opposed to real #s) in matrix
On Aug 10, 2010, at 10:13 AM, Romain Francois wrote: rowSums( x == round(x) ) == ncol(x) [1] TRUE FALSE TRUE FALSE For later readability, I think I'd prefer apply(x==round(x), 1, all) -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org 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] package for measurement error models
On Aug 10, 2010, at 3:52 AM, Carrie Li wrote: Thanks. I found the code in the link you gave me very helpful. But, I just have few questions regarding the code. It seems to me that in (from wikipdeia)Deming regression, it assumes that the ratios of the variances of two measurement errors are constant for all pairs of (x_i, y_i). However, if the ratios are not constant, (i.e. the variances of measurement are heterogeneous) , is it still appropriate to use Deming regression ? In a word, no. One way of looking at it is that as the ratio of variances varies from 0 to infinity, the analysis goes from regression of y on x to (inverse) regression of x on y, and those give different results, not just numerically but also asymptotically. I.e., getting the ratio wrong gives an inconsistent estimate; getting it wrong for some of the data, as is bound to happen if you assume it constant and it isn't, will also give a inconsistent estimate. Unless, that is, you can find a definition of average ratio that eliminates the bias, but I don't think it is worth the paperwork. Rather, I'd suggest direct minimization of the SSR (from the Wikipedia page), noting that you can plug in x_i^* as a function of beta also if the _individual_ ratios are known. (I get the feeling that someone must have been here before, so possibly others can fill in the gaps?) For modest sample sizes, it might also be possible to formulate the problem as a nonlinear model and use nls(). -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org 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] Turning a source into a Package
On 08/10/2010 02:51 AM, JH wrote:
I want to edit a function in the nlme package. I download the package source
nlme_3.1-96.tar.gz from the link below then edit it one of the scripts
inside of it. From this stage how can I turn it into a package that I can
use in R?
http://cran.r-project.org/web/packages/nlme/index.html
Hi JH,
The problem I have encountered with this is that installed packages
often do not have all the required directories and the source code for
checking and building. With a small package, you can get the source code
for all the functions by sinking the output produced by invoking each
function name without arguments to a file in the R subdirectory (copy
the whole installed package directory to another location first):
sink(anova.lme.R)
cat(anova.lme-)
anova.lme
sink()
but this is only practical for small packages. You would also have to
back translate the HTML help files to Rd format and place these in the
man directory, again a considerable task for a package like nlme. Your
best bet for something like this is to load the library:
library(nlme)
and then source the modified code:
source(anova.lme.R)
or whatever function you have modified. You can even write a function to
do this:
my.library(x,newfile) {
library(x)
source(newfile)
}
my.library(nlme,anova.lme.R)
Jim
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and provide commented, minimal, self-contained, reproducible code.
[R] one (small) sample wilcox.test confidence intervals
Dear R people, I notice that the confidence intervals of a very small sample (e.g. n=6) derived from the one-sample wilcox.test are just the maximum and minimum values of the sample. This only occurs when the required confidence level is higher than 0.93. Example: sample - c(1.22, 0.89, 1.14, 0.98, 1.37, 1.06) summary(sample) Min. 1st Qu. MedianMean 3rd Qu.Max. 0.891.001.101.111.201.37 wilcox.test(sample,conf.int=TRUE,conf.lev=0.95) Wilcoxon signed rank test data: sample V = 21, p-value = 0.03125 alternative hypothesis: true location is not equal to 0 95 percent confidence interval: 0.89 1.37 sample estimates: (pseudo)median 1.1 According to help, since my sample contains less than 50 values, an exact p-value is calculated that should enable the confidence interval to be obtained from Bauer (1972) (I have not read it): By default (if âexactâ is not specified), an exact p-value is computed if the samples contain less than 50 finite values and there are no ties. Otherwise, a normal approximation is used. ... If exact p-values are available, an exact confidence interval is obtained by the algorithm described in Bauer (1972), and the Hodges-Lehmann estimator is employed. Otherwise, the returned confidence interval and point estimate are based on normal approximations. These are continuity-corrected for the interval but _not_ the estimate (as the correction depends on the âalternativeâ). With small samples it may not be possible to achieve very high confidence interval coverages. If this happens a warning will be given and an interval with lower coverage will be substituted. The latter indeed happens if I ask for confidence level of 0.99: wilcox.test(sample,mu=0,conf.int=TRUE,conf.lev=0.99) Wilcoxon signed rank test data: sample V = 21, p-value = 0.03125 alternative hypothesis: true location is not equal to 0 96.9 percent confidence interval: 0.89 1.37 sample estimates: (pseudo)median 1.1 Warning message: In wilcox.test.default(sample, mu = 0, conf.int = TRUE, conf.lev = 0.99) : Requested conf.level not achievable My questions (finally!) are: 1. Why the above warning for conf.lev = 0.99 does not appear for 0.93 conf.lev 0.98 although it produces the same summary? 2. For conf.lev = 0.95, is there anything else I can do in order to obtain confidence intervals other than the max. and min. values of my sample or I am limited from my sample's size ? Thanks for your patience in reading this, Panos --- Dr Panos Hadjinicolaou Energy Environment Water Research Center (EEWRC) The Cyprus Institute --- [[alternative HTML version deleted]] __ R-help@r-project.org 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] Pie Chart in map
On 08/10/2010 03:04 AM, LCOG1 wrote: Hey R'rs, So im sick of dealing with ESRI products and am looking to stream line a process i now use GIS to do using R. I have made a lot of maps using R but have not yet seen a map that puts pie charts within the map to help represent data like the attachment. http://r.789695.n4.nabble.com/file/n2318816/template1.bmp I found Tanimura et al. work Proportional Symbol Mapping in R, but after a discussion with one of the authors i have learned that the function is out of date and would require a great deal of reworking. I also found hexbinpie on the R graphics gallery that i could incorporate. Before i go through the process of rewriting or out right creation of a process to replicate my GIS results i was hoping to make sure there wasnt something out there i could use to start or use altogether. Or perhaps a set of functions that already exists that i havent thought to use. Hi JR, You can also check out the floating.pie function in the plotrix package. Jim __ R-help@r-project.org 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] package for measurement error models
On Tue, 10 Aug 2010, peter dalgaard wrote: On Aug 10, 2010, at 3:52 AM, Carrie Li wrote: Thanks. I found the code in the link you gave me very helpful. But, I just have few questions regarding the code. It seems to me that in (from wikipdeia)Deming regression, it assumes that the ratios of the variances of two measurement errors are constant for all pairs of (x_i, y_i). However, if the ratios are not constant, (i.e. the variances of measurement are heterogeneous) , is it still appropriate to use Deming regression ? In a word, no. One way of looking at it is that as the ratio of variances varies from 0 to infinity, the analysis goes from regression of y on x to (inverse) regression of x on y, and those give different results, not just numerically but also asymptotically. I.e., getting the ratio wrong gives an inconsistent estimate; getting it wrong for some of the data, as is bound to happen if you assume it constant and it isn't, will also give a inconsistent estimate. Unless, that is, you can find a definition of average ratio that eliminates the bias, but I don't think it is worth the paperwork. Rather, I'd suggest direct minimization of the SSR (from the Wikipedia page), noting that you can plug in x_i^* as a function of beta also if the _individual_ ratios are known. (I get the feeling that someone must have been here before, so possibly others can fill in the gaps?) For modest sample sizes, it might also be possible to Yes, people have been there before. Mike Thompson and I published a now-much-cited-in-analytical-chemistry paper in The Analyst in 1987. A companion paper was rejected by a mainstream statistics journal as 'already known', but the journal editor was unable to get any prior publication out of the referee. formulate the problem as a nonlinear model and use nls(). Direct minimization is simple enough. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com -- Brian D. Ripley, rip...@stats.ox.ac.uk 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@r-project.org 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] Concatenate a mix of numbers and letters to create a vector name
I was not aware of the R-FAQ, it seems to have some very useful tips, thanks
for pointing there.
Regarding the 7.21 in the FAQ, I read it a few times but it did not lead me
anywhere. For the moment I am blaming my inexperience with some R basics, I
will come back after I do some more serious studying.
Thanks again,
Panos
Dr Panos Hadjinicolaou
Energy Environment Water Research Center (EEWRC)
The Cyprus Institute
-
_
From: Greg Snow [mailto:greg.s...@imail.org]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy],
r-help@r-project.org [mailto:r-h...@r-project.org]
Sent: Wed, 28 Jul 2010 00:39:41 +0300
Subject: RE: [R] Concatenate a mix of numbers and letters to create a vector
name
This is a frequently asked/answered question (7.21 in the FAQ). What searching
did you do and why did it not find this FAQ or previous discussion of it? How
could the documentation/search/etc. be improved so that you (and the next n
people with this question) will find the answer easier?
The most important of part of the FAQ answer is the last section where it
points out that using a list will be much simpler. You mentioned in the first
post that you were doing this in a loop, just start with an empty list, use
paste (or sprintf) to create the name, and then assign it as a new element of
the list with that name (e.g. mylist[[ sprintf('tmax.%d%d', var1, var2) ]] -
outputfromcomputations ).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Panos Hadjinicolaou
Sent: Monday, July 26, 2010 8:24 AM
To: r-help@r-project.org
Subject: Re: [R] Concatenate a mix of numbers and letters to create a
vector name
Thanks for the reply. Indeed the paste function results in
concatenation:
paste(c(tmax., 1950, 12), collapse=)
[1] tmax.195012
but I am looking for a way to subsequently get rid of the - - in
order to use tmax.195012 as an object (e.g. to define a vector with
that name). Any ideas?
Thanks,
Panos
_
From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy]
Cc: r-help@r-project.org
Sent: Mon, 26 Jul 2010 16:48:31 +0300
Subject: Re: [R] Concatenate a mix of numbers and letters to create a
vector name
have a look at function paste(), i.e., ?paste
I hope it helps.
Best,
Dimitris
On 7/26/2010 3:44 PM, Panos Hadjinicolaou wrote:
Dear all,
I am trying to create a vector name, for example tmax.195012 from
tmax., 1950 and 12. Obviously I don't wish to simply type it because
the 3 name components are changing in each iteration within a loop. Is
there any way of concatenating those 3 components (which are a mixture
of numbers and letters)?
Thanks for reading,
Panos
-
Dr Panos Hadjinicolaou
Energy Environment Water Research Center (EEWRC)
The Cyprus Institute
--
[[alternative HTML 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.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
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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] how to draw a spherical quadrant
On 08/10/2010 05:32 PM, [Ricardo Rodriguez] Your XEN ICT Team wrote: Hi, Please, could you tell me if is there a R package that allow to draw spherical quadrants? The value of each point of the quadrant has to represent the power generated by a kite on that point tied to a line located in the center of the quadrant. Hi Ricardo, You may find that the radial.plot function in the plotrix package will do what you want. I think you are looking at the polygon type of plot. Jim __ R-help@r-project.org 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] alternative in morantest
Dear list, I am learning moran.test in package spdep. There is a setting of alternative, defined as a character string specifying the alternative hypothesis, must be one of greater (default), less or two.sided. I would like to learn what a alternative hypothesis is in moran.test but found little in R-help. Please kindly share the references and thank you. Elaine [[alternative HTML version deleted]] __ R-help@r-project.org 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] Replacing characters
On Aug 10, 2010, at 2:02 AM, Orvalho Augusto wrote: Hello guys! May be I am lazy but I need to replace a character like \ or ' or to escape them in a character vector to write a SQL statement. ?sub ?regex How can I do that? Caveman -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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] Sweave with dev.new()
On Aug 10, 2010, at 3:51 AM, Gro Nilsen wrote:
Dear list.
I am preparing a R package, and the last step is to write a package
vignette using Sweave. However, I am experiencing some trouble when trying
to include plots in my Sweave document. That is, in my package I have made
some plotting functions in which I start by calling 'new.dev()' to start a
graphics device of a certain width and height, and then proceed with
plot(). These functions work perfectly fine in R, and when I check and
build my package with R CMD build/check.
The problems start when I try to include the command and the output of my
plotting functions in the Sweave document, and I suspect it must have
something to do with the command 'dev.new()' in my code. Here is a basic
example of a Sweave file which can hopefully illustrate the problem:
\documentclass[a4paper]{article}
\usepackage{Sweave}
\RequirePackage{graphicx,fancyvrb}
\title{Sweavetest}
\begin{document}
\section{Plotting}
label=testfig,include=FALSE,echo=TRUE=
dev.new(width=7,height=6,record=TRUE)
plot(1:10,col=red)
@
\begin{figure}[ht]
\begin{center}
label=fig,fig=TRUE,echo=FALSE=
testfig
@
\end{center}
\caption{Testplot}
\label{fig:test}
\end{figure}
\end{document}
When I run Sweave on this file I get empty pdf and eps files named
Sweavetest-fig.pdf and Sweavetest-fig.eps, and the corresponding latex
file is unable to create a PDF-file.
Does anyone know what the problem is, and how I might solve it? This is
the first time that I have worked with Sweave, so I might be missing
something basic...
Best regards,
Gro Nilsen
PhD-student, University of Oslo
It is not clear to me why you need the dev.new() call. By default using
Sweave, both PDF and EPS plot files will be created, for subsequent inclusion
in the document as appropriate. Just define the plot dimensions in the figure
chunk header:
label=fig,fig=TRUE,echo=FALSE,height=6,width=7=
Be aware however, that by default, the image size in the final document will be
set to \textwidth * 0.8, irrespective of the dimensions that you have
specified. You can alter that default behavior by using:
\usepackage[nogin]{Sweave}
in your document preamble. The figures will then be set to the sizes that you
define in the figure chunk header.
See ?RweaveLatex for more information.
HTH,
Marc Schwartz
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[R] FAQ package, anyone?
Dear list, Here's a reproducible example that fails, library(faq) Error in library(faq) : there is no package called 'faq' faq(lattice sweave) Error: could not find function faq As a fun challenge, I propose to remedy this by creating a fortune-like package dedicated to Frequently Asked Questions (FAQ) entries. The aim is multifold, 1- Provide a generic template for use in a variety of situations. I imagine that any package could then define its own FAQ entries and use the faq package to display them, search, etc. 2- Improve the visibility of the current FAQ: currently directing someone to a FAQ entry involves some web-browsing, which is not as convenient as it could be (compare with fortune()). 3- Once stored in a R-like data format, the FAQ could be straight-forwardly converted into a wide variety of output formats, including the original html page (thus made easier to maintain, perhaps). 4- Extend and generalise the fortune package to similar kinds of problems. In practical terms, I can imagine the following distinct steps, - check the FAQ license, and whether we can borrow its content - use some kind of html parsing package (XML?, i forget) - clean up the data and store it as a data.frame or higher class - modify / generalise the fortune functions to deal with the different fields and categories - write output formatting functions et voilà! Comments, contributions, etc. welcome Regards, Baptiste __ R-help@r-project.org 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] Replacing characters
Try this: library(RMySQL) #conn - dbConnect(...) mysqlEscapeStrings(conn, tes't) On Tue, Aug 10, 2010 at 3:02 AM, Orvalho Augusto orvaq...@gmail.com wrote: Hello guys! May be I am lazy but I need to replace a character like \ or ' or to escape them in a character vector to write a SQL statement. How can I do that? Caveman [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] one (small) sample wilcox.test confidence intervals
On Aug 10, 2010, at 1:30 PM, Panos Hadjinicolaou wrote:
My questions (finally!) are:
1. Why the above warning for conf.lev = 0.99 does not appear for 0.93
conf.lev 0.98 although it produces the same summary?
2. For conf.lev = 0.95, is there anything else I can do in order to obtain
confidence intervals other than the max. and min. values of my sample or I
am limited from my sample's size ?
Thanks for your patience in reading this,
Panos
It is as it should be.
I think it is instructive to look at explicitly shifted samples, e.g.
wilcox.test(sample-1.369)
Wilcoxon signed rank test
data: sample - 1.369
V = 1, p-value = 0.0625
alternative hypothesis: true location is not equal to 0
wilcox.test(sample-1.371)
Wilcoxon signed rank test
data: sample - 1.371
V = 0, p-value = 0.03125
alternative hypothesis: true location is not equal to 0
Notice how the p-value jumps as the shift crosses 1.37. You can shift the
distribution by 1.36 to the left and have nonsignificant test that the
center is at zero. However if you shift by more than 1.37, then you do get
significance. This is true for all significance levels between 0.03125 and
0.0625 (and 0.03125 == 1/32, the probability that all ranks have the same
sign).
The above explains almost everything if you think a little about the
definitions. The only slightly puzzling thing is why confidence levels larger
than 1-0.03125 are considered achievable. The actual code has
if (achieved.alpha - alpha alpha/2) {
warning(Requested conf.level not achievable)
conf.level - 1 - signif(achieved.alpha, 2)
}
so I have to assume that the author has considered this with some care.
--
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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Re: [R] R support for 64 bit integers
On Aug 9, 2010, at 2:45 PM, Theo Tannen wrote: Are integers strictly a signed 32 bit number on R even if I am running a 64 bit version of R on a x86_64 bit machine? I ask because I have integers stored in a hdf5 file where some of the data is 64 bit integers. When I read that into R using the hdf5 library it seems any integer greater than 2**31 returns NA. That's the limit. It's hard coded and not affected by the memory pointer size. Any solutions? I have heard of packages that handle big numbers. A bit of searching produces suggestions to look at gmp on CRAN and Rmpfr on R-Forge. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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] p-values with pvclust
Hi, if you look at the first image (Image1) you see that there are 2 main clusters 7 and 8 I wanted to use pvclust to calculate a p-value whether these clusters are due to chance or statistically significant. Unfortunately pvclust does not provide a p-value for the first brunch (7 and 8). So I added a row to my matrix which is very different to the rest of the data to create an additional brunch. See image here http://r.789695.n4.nabble.com/file/n2319732/Image2.png . I finally got my p-value for the brunch (7 and 8) which is 98 (9). I was happy to see that the p-value was significant until I realised that if I add an additional brunch which is not that different from the rest but still cluster in a sperate cluster (see image here http://r.789695.n4.nabble.com/file/n2319732/Image3.png ) the p-value is changing and not significant any longer (84). I was wondering why this happens because I thought that for each brunch the p-value is calculated independently? Does anybody know how to get a correct p-value for the first brunch (7 and 8) maybe without adding an additional brunch? Best regards syrvn -- View this message in context: http://r.789695.n4.nabble.com/p-values-with-pvclust-tp2319732p2319732.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] p-values with pvclust
Hi, if you look at the first image (Image1) you see that there are 2 main clusters 7 and 8 I wanted to use pvclust to calculate a p-value whether these clusters are due to chance or statistically significant. Unfortunately pvclust does not provide a p-value for the first brunch (7 and 8). So I added a row to my matrix which is very different to the rest of the data to create an additional brunch. See image 2. I finally got my p-value for the brunch (7 and 8) which is 98 (9). I was happy to see that the p-value was significant until I realised that if I add an additional brunch which is not that different from the rest but still cluster in a sperate cluster (see image 3) the p-value is changing and not significant any longer (84). http://r.789695.n4.nabble.com/file/n2319739/ImageNew2.png I was wondering why this happens because I thought that for each brunch the p-value is calculated independently? Does anybody know how to get a correct p-value for the first brunch (7 and 8) maybe without adding an additional brunch? Best regards syrvn -- View this message in context: http://r.789695.n4.nabble.com/p-values-with-pvclust-tp2319739p2319739.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Function to Define a Function
Gabor ... that worked perfectly. Thank you.
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Monday, August 09, 2010 10:20 PM
To: Derek Ogle
Cc: R (r-help@R-project.org)
Subject: Re: [R] Function to Define a Function
On Mon, Aug 9, 2010 at 9:31 PM, Derek Ogle do...@northland.edu wrote:
I am trying to define a general R function that has a function as the
output that depends on the user's input arguments (this may make more
sense by looking at the toy example below). My real use for this type
of code is to allow a user to choose from many parameterizations of the
same general model.
My issue is that when I compile a package with this type of code in
it I get a __warning__ that multiple local function definitions for
'm' with different formal arguments. While this is not a deadly
error I would like to avoid the warning if possible. Can someone
provide some guidance? Thank you in advance for any help you can
offer.
For what it is worth ... I am working on a Windows XP machine with R
2.11.1.
## A function that allows the user to create a new function that
depends on their
## choice in the type argument. As a simple example, if the user
chooses one
## then the output function is exponential growth, if the user
choses two then
## thhe output function is logistic growth.
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
switch(type,
one={ m - function(x,N0,r) N0*exp(x*r) },
two={ m - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) },
)
m
}
## define time steps
t - 0:10
## create a function -- junk1 -- that produces exponential growth
junk1 - mdlChooser(one)
junk1
res1 - junk1(t,500,0.2)
res1
## create a function -- junk2 -- that produces logistic growth
junk2 - mdlChooser(two)
junk2
res2 - junk2(t,500,0.2,1000)
res2
Try this:
mdlChooser - function(type = c(one, two)) {
one - function(x,N0,r) N0*exp(x*r)
two - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r))
type - match.arg(type)
get(type)
}
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Re: [R] optimization subject to constraints
Thanks, but I still cannot get to solve my problem: consider this simple
example:
f - function(x){
return(x[1]+x[2])
} # objective function
g - function(x){
return(x[1]^2+x[2]^2)
} # constraint
#
I wanna Maximize f(x) subject to g(x) = 1. By hand the solution is
(1/sqrt(2), 1/sqrt(2), sqrt(2)). This is to maximizing a linear function
subject to a nonlinear equality constraint. I didn't find any suitable
function in the packages I went through.
Thanks in advance,
Gildas
Spencer Graves a écrit :
To find every help page containing the term constrained
optimization, you can try the following:
library(sos)
co - findFn('constrained optimization')
Printing this co object opens a table in a web browser with
all matches sorted first by the package with the most matches and with
hot links to the documentation page.
writeFindFn2xls(co)
This writes an excel file, with the browser table as the second
tab and the first being a summary of the packages. This summary table
can be made more complete and useful using the installPackages
function, as noted in the sos vignette.
A shameless plug from the lead author of the sos package.
Spencer Graves
On 8/9/2010 10:01 AM, Ravi Varadhan wrote:
constrOptim can only handle linear inequality constraints. It cannot
handle
equality (linear or nonlinear) as well as nonlinear inequality
constraints.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On
Behalf Of Dwayne Blind
Sent: Monday, August 09, 2010 12:56 PM
To: Gildas Mazo
Cc: r-help@r-project.org
Subject: Re: [R] optimization subject to constraints
Hi !
Why not constrOptim ?
Dwayne
2010/8/9 Gildas Mazogildas.m...@curie.fr
Dear R users,
I'm looking for tools to perform optimization subject to constraints,
both linear and non-linear. I don't mind which algorithm may be
used, my
primary aim is to get something general and easy-to-use to study
simples
examples.
Thanks for helping,
Gildas
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[R] Fwd: List of lists ?
-- Forwarded message -- From: Carlos Petti carlos.pe...@gmail.com Date: 2010/8/10 Subject: Re: [R] List of lists ? To: David Winsemius dwinsem...@comcast.net Thanks for answer. I read the error messages but I did not find the solution :-( Your solution works. But, a new problem remains because I want to use the list of lists as follows : x - list(list()) x[[2]][[1]] - c(1, 2, 3) x[[2]][[2]] - c(3, 2, 1) Thanks in advance, Carlos 2010/8/9 David Winsemius dwinsem...@comcast.net: On Aug 9, 2010, at 12:57 PM, Carlos Petti wrote: Dear list, I have to use a list of lists containing vectors. For instance : [[1]] [[1]][[1]] [1] 1 2 3 [[1]][[2]] [1] 3 2 1 I want to attribute vectors to the main list without use of an intermediate list, but it does not work : More specifically it produces an error that has information in it. x[[1]][[1]] - c(1, 2, 3) Error in `*tmp*`[[1]] : subscript out of bounds x - list() x[[1]][[1]] - c(1, 2, 3) x[[1]][[2]] - c(3, 2, 1) So thinking perhaps we just needed another level of subscripting available I tried: x - list(list()) x[[1]][[1]] - c(1, 2, 3) x[[1]][[2]] - c(3, 2, 1) x [[1]] [[1]][[1]] [1] 1 2 3 [[1]][[2]] [1] 3 2 1 Success. Moral: Read the error messages for meaning or at least clues. (Further testing showed that almost anything inside the original list() call, even NULL, would have created enough structure for the interpreter to work with. David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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] Fwd: List of lists ?
-- Forwarded message -- From: Carlos Petti carlos.pe...@gmail.com Date: 2010/8/10 Subject: Re: [R] List of lists ? To: David Winsemius dwinsem...@comcast.net Perhaps a solution : x - list() x[[2]] - list() x[[2]][[1]] - c(1, 2, 3) x[[2]][[2]] - c(3, 2, 1) 2010/8/10 Carlos Petti carlos.pe...@gmail.com: Thanks for answer. I read the error messages but I did not find the solution :-( Your solution works. But, a new problem remains because I want to use the list of lists as follows : x - list(list()) x[[2]][[1]] - c(1, 2, 3) x[[2]][[2]] - c(3, 2, 1) Thanks in advance, Carlos 2010/8/9 David Winsemius dwinsem...@comcast.net: On Aug 9, 2010, at 12:57 PM, Carlos Petti wrote: Dear list, I have to use a list of lists containing vectors. For instance : [[1]] [[1]][[1]] [1] 1 2 3 [[1]][[2]] [1] 3 2 1 I want to attribute vectors to the main list without use of an intermediate list, but it does not work : More specifically it produces an error that has information in it. x[[1]][[1]] - c(1, 2, 3) Error in `*tmp*`[[1]] : subscript out of bounds x - list() x[[1]][[1]] - c(1, 2, 3) x[[1]][[2]] - c(3, 2, 1) So thinking perhaps we just needed another level of subscripting available I tried: x - list(list()) x[[1]][[1]] - c(1, 2, 3) x[[1]][[2]] - c(3, 2, 1) x [[1]] [[1]][[1]] [1] 1 2 3 [[1]][[2]] [1] 3 2 1 Success. Moral: Read the error messages for meaning or at least clues. (Further testing showed that almost anything inside the original list() call, even NULL, would have created enough structure for the interpreter to work with. David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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] optimization subject to constraints
try this (package Rsolnp)
library(Rsolnp)
g- function(x)
{
return(x[1]^2+x[2]^2)
} # constraint
f- function(x)
{
return(x[1]+x[2])
} # objective function
x0 = c(1, 1)
solnp(x0, fun=f, eqfun=g, eqB=c(1))
Am 10.08.2010 14:59, schrieb Gildas Mazo:
Thanks, but I still cannot get to solve my problem: consider this simple
example:
f- function(x){
return(x[1]+x[2])
} # objective function
g- function(x){
return(x[1]^2+x[2]^2)
} # constraint
#
I wanna Maximize f(x) subject to g(x) = 1. By hand the solution is
(1/sqrt(2), 1/sqrt(2), sqrt(2)). This is to maximizing a linear function
subject to a nonlinear equality constraint. I didn't find any suitable
function in the packages I went through.
Thanks in advance,
Gildas
Spencer Graves a écrit :
To find every help page containing the term constrained
optimization, you can try the following:
library(sos)
co- findFn('constrained optimization')
Printing this co object opens a table in a web browser with
all matches sorted first by the package with the most matches and with
hot links to the documentation page.
writeFindFn2xls(co)
This writes an excel file, with the browser table as the second
tab and the first being a summary of the packages. This summary table
can be made more complete and useful using the installPackages
function, as noted in the sos vignette.
A shameless plug from the lead author of the sos package.
Spencer Graves
On 8/9/2010 10:01 AM, Ravi Varadhan wrote:
constrOptim can only handle linear inequality constraints. It cannot
handle
equality (linear or nonlinear) as well as nonlinear inequality
constraints.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On
Behalf Of Dwayne Blind
Sent: Monday, August 09, 2010 12:56 PM
To: Gildas Mazo
Cc: r-help@r-project.org
Subject: Re: [R] optimization subject to constraints
Hi !
Why not constrOptim ?
Dwayne
2010/8/9 Gildas Mazogildas.m...@curie.fr
Dear R users,
I'm looking for tools to perform optimization subject to constraints,
both linear and non-linear. I don't mind which algorithm may be
used, my
primary aim is to get something general and easy-to-use to study
simples
examples.
Thanks for helping,
Gildas
__
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http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting
-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
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__
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[R] grep problem decimal points looping
Hi R Users,
I have been trying to work out how to rename column names using grep,
basically I have generated these column names using tapply:
[1] NAME X1.1 X2.1 X3.1 X4.1 X5.1 X6.1 X7.1 X8.1
[10] X1.2 X2.2 X3.2 X4.2 X5.2 X6.2 X7.2 X8.2 X1.3
[19] X2.3 X3.3 X4.3 X5.3 X6.3 X7.3 X8.3 X1.5 X2.5
[28] X3.5 X4.5 X5.5 X6.5 X7.5 X8.5 X1.6 X2.6 X3.6
[37] X4.6 X5.6 X6.6 X7.6 X8.6 X1.8 X2.8 X3.8 X4.8
[46] X5.8 X6.8 X7.8 X8.8 X1.9 X2.9 X3.9 X4.9 X5.9
[55] X6.9 X7.9 X8.9 X1.10 X2.10 X3.10 X4.10 X5.10
X6.10
[64] X7.10 X8.10 X1.12 X2.12 X3.12 X4.12 X5.12 X6.12
X7.12
[73] X8.12 X1.13 X2.13 X3.13 X4.13 X5.13 X6.13 X7.13
X8.13
[82] X1.14 X2.14 X3.14 X4.14 X5.14 X6.14 X7.14 X8.14
X1.15
[91] X2.15 X3.15 X4.15 X5.15 X6.15 X7.15 X8.15 X1.16
X2.16
[100] X3.16 X4.16 X5.16 X6.16 X7.16 X8.16 X1.17 X2.17
X3.17
[109] X4.17 X5.17 X6.17 X7.17 X8.17 X1.18 X2.18 X3.18
X4.18
[118] X5.18 X6.18 X7.18 X8.18 X1.19 X2.19 X3.19 X4.19
X5.19
[127] X6.19 X7.19 X8.19 X1.20 X2.20 X3.20 X4.20 X5.20
X6.20
[136] X7.20 X8.20 X1.21 X2.21 X3.21 X4.21 X5.21 X6.21
X7.21
[145] X8.21 X1.22 X2.22 X3.22 X4.22 X5.22 X6.22 X7.22
X8.22
[154] X1.23 X2.23 X3.23 X4.23 X5.23 X6.23 X7.23 X8.23
X1.24
[163] X2.24 X3.24 X4.24 X5.24 X6.24 X7.24 X8.24 X1.25
X2.25
[172] X3.25 X4.25 X5.25 X6.25 X7.25 X8.25 X1.26 X2.26
X3.26
[181] X4.26 X5.26 X6.26 X7.26 X8.26 X1.27 X2.27 X3.27
X4.27
[190] X5.27 X6.27 X7.27 X8.27 X1.28 X2.28 X3.28 X4.28
X5.28
[199] X6.28 X7.28 X8.28 X1.29 X2.29 X3.29 X4.29 X5.29
X6.29
[208] X7.29 X8.29 X1.30 X2.30 X3.30 X4.30 X5.30 X6.30
X7.30
[217] X8.30 X1.31 X2.31 X3.31 X4.31 X5.31 X6.31 X7.31
X8.31
[226] X1.32 X2.32 X3.32 X4.32 X5.32 X6.32 X7.32 X8.32
X1.33
[235] X2.33 X3.33 X4.33 X5.33 X6.33 X7.33 X8.33
What the names mean are behaviour.day the X is not important to the data, it
is the numbers I am trying to select on.
So I want to split the data by day i.e. selecting for the number after the
decimal.
I am using this code (where scananal is the data) with out looping so the
number following the decimal I change manually (NB the data have been
changed to character):
DAY - grep((X[[:digit:]]+).3,colnames(scananal))
However, this will select for day 3, 30, 31, 32, etc I have tried to use
fixed = TRUE, but that just returns integer(0). But if I use 30, it will
select only 30. Not sure what I'm doing wrong here, and I assumed that fixed
= T would fix this, but doesn't.
I have tried to loop this too, but with no luck, so if anyone can point me
in the right direction about how to loop using grep I would be most
grateful!
The main problem I have is where to put the loop, for example:
for(i in 1:33){
print(i)
DAY[[i]] - grep((X[[:digit:]]+).[[i]],colnames(scananal))
}
which doesn't work, and no doubt there are obvious reasons for this! Any
help would be much appreciated,
All the best,
Ross
--
View this message in context:
http://r.789695.n4.nabble.com/grep-problem-decimal-points-looping-tp2319773p2319773.html
Sent from the R help mailing list archive at Nabble.com.
__
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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] [Fwd: Re: optimization subject to constraints]
---BeginMessage---
Danke schön Matthias.
I had naively started with x0 = c(0,0) and I got a Redundant
constraints were found error. What's the problem with (0,0) ?
Matthias Gondan a écrit :
try this (package Rsolnp)
library(Rsolnp)
g- function(x)
{
return(x[1]^2+x[2]^2)
} # constraint
f- function(x)
{
return(x[1]+x[2])
} # objective function
x0 = c(1, 1)
solnp(x0, fun=f, eqfun=g, eqB=c(1))
Am 10.08.2010 14:59, schrieb Gildas Mazo:
Thanks, but I still cannot get to solve my problem: consider this simple
example:
f- function(x){
return(x[1]+x[2])
} # objective function
g- function(x){
return(x[1]^2+x[2]^2)
} # constraint
#
I wanna Maximize f(x) subject to g(x) = 1. By hand the solution is
(1/sqrt(2), 1/sqrt(2), sqrt(2)). This is to maximizing a linear function
subject to a nonlinear equality constraint. I didn't find any suitable
function in the packages I went through.
Thanks in advance,
Gildas
Spencer Graves a écrit :
To find every help page containing the term constrained
optimization, you can try the following:
library(sos)
co- findFn('constrained optimization')
Printing this co object opens a table in a web browser with
all matches sorted first by the package with the most matches and with
hot links to the documentation page.
writeFindFn2xls(co)
This writes an excel file, with the browser table as the second
tab and the first being a summary of the packages. This summary table
can be made more complete and useful using the installPackages
function, as noted in the sos vignette.
A shameless plug from the lead author of the sos package.
Spencer Graves
On 8/9/2010 10:01 AM, Ravi Varadhan wrote:
constrOptim can only handle linear inequality constraints. It cannot
handle
equality (linear or nonlinear) as well as nonlinear inequality
constraints.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On
Behalf Of Dwayne Blind
Sent: Monday, August 09, 2010 12:56 PM
To: Gildas Mazo
Cc: r-help@r-project.org
Subject: Re: [R] optimization subject to constraints
Hi !
Why not constrOptim ?
Dwayne
2010/8/9 Gildas Mazogildas.m...@curie.fr
Dear R users,
I'm looking for tools to perform optimization subject to constraints,
both linear and non-linear. I don't mind which algorithm may be
used, my
primary aim is to get something general and easy-to-use to study
simples
examples.
Thanks for helping,
Gildas
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting
-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org 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@r-project.org mailing list
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide
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---End Message---
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep problem decimal points looping
On Aug 10, 2010, at 9:51 AM, David Winsemius wrote:
On Aug 10, 2010, at 9:17 AM, RCulloch wrote:
Hi R Users,
I have been trying to work out how to rename column names using grep,
basically I have generated these column names using tapply:
[1] NAME X1.1 X2.1 X3.1 X4.1 X5.1 X6.1 X7.1
X8.1
[10] X1.2 X2.2 X3.2 X4.2 X5.2 X6.2 X7.2
X8.2 X1.3
[19] X2.3 X3.3 X4.3 X5.3 X6.3 X7.3 X8.3
X1.5 X2.5
[28] X3.5 X4.5 X5.5 X6.5 X7.5 X8.5 X1.6
X2.6 X3.6
[37] X4.6 X5.6 X6.6 X7.6 X8.6 X1.8 X2.8
X3.8 X4.8
[46] X5.8 X6.8 X7.8 X8.8 X1.9 X2.9 X3.9
X4.9 X5.9
[55] X6.9 X7.9 X8.9 X1.10 X2.10 X3.10 X4.10 X5.10
X6.10
[64] X7.10 X8.10 X1.12 X2.12 X3.12 X4.12 X5.12 X6.12
X7.12
[73] X8.12 X1.13 X2.13 X3.13 X4.13 X5.13 X6.13 X7.13
X8.13
[82] X1.14 X2.14 X3.14 X4.14 X5.14 X6.14 X7.14 X8.14
X1.15
[91] X2.15 X3.15 X4.15 X5.15 X6.15 X7.15 X8.15 X1.16
X2.16
[100] X3.16 X4.16 X5.16 X6.16 X7.16 X8.16 X1.17 X2.17
X3.17
[109] X4.17 X5.17 X6.17 X7.17 X8.17 X1.18 X2.18 X3.18
X4.18
[118] X5.18 X6.18 X7.18 X8.18 X1.19 X2.19 X3.19 X4.19
X5.19
[127] X6.19 X7.19 X8.19 X1.20 X2.20 X3.20 X4.20 X5.20
X6.20
[136] X7.20 X8.20 X1.21 X2.21 X3.21 X4.21 X5.21 X6.21
X7.21
[145] X8.21 X1.22 X2.22 X3.22 X4.22 X5.22 X6.22 X7.22
X8.22
[154] X1.23 X2.23 X3.23 X4.23 X5.23 X6.23 X7.23 X8.23
X1.24
[163] X2.24 X3.24 X4.24 X5.24 X6.24 X7.24 X8.24 X1.25
X2.25
[172] X3.25 X4.25 X5.25 X6.25 X7.25 X8.25 X1.26 X2.26
X3.26
[181] X4.26 X5.26 X6.26 X7.26 X8.26 X1.27 X2.27 X3.27
X4.27
[190] X5.27 X6.27 X7.27 X8.27 X1.28 X2.28 X3.28 X4.28
X5.28
[199] X6.28 X7.28 X8.28 X1.29 X2.29 X3.29 X4.29 X5.29
X6.29
[208] X7.29 X8.29 X1.30 X2.30 X3.30 X4.30 X5.30 X6.30
X7.30
[217] X8.30 X1.31 X2.31 X3.31 X4.31 X5.31 X6.31 X7.31
X8.31
[226] X1.32 X2.32 X3.32 X4.32 X5.32 X6.32 X7.32 X8.32
X1.33
[235] X2.33 X3.33 X4.33 X5.33 X6.33 X7.33 X8.33
What the names mean are behaviour.day the X is not important to the
data, it
is the numbers I am trying to select on.
So I want to split the data by day i.e. selecting for the number
after the
decimal.
I am using this code (where scananal is the data) with out looping
so the
number following the decimal I change manually (NB the data have been
changed to character):
You need to learn the special character$ which marks the no-
character end of string. After creating a replica of your column-
names with scan and grep:
inp - scan(what=character)
inX - inp[grep(X, inp)]
DAY - grep((X[[:digit:]]+).3$,inX)
inX[DAY]
[1] X1.3 X2.3 X3.3 X4.3 X5.3 X6.3 X7.3 X8.3
DAY - grep((X[[:digit:]]+).3,colnames(scananal))
However, this will select for day 3, 30, 31, 32, etc I have tried
to use
fixed = TRUE, but that just returns integer(0). But if I use 30, it
will
select only 30. Not sure what I'm doing wrong here, and I assumed
that fixed
= T would fix this, but doesn't.
I have tried to loop this too, but with no luck, so if anyone can
point me
in the right direction about how to loop using grep I would be most
grateful!
The main problem I have is where to put the loop, for example:
for(i in 1:33){
print(i)
DAY[[i]] - grep((X[[:digit:]]+).[[i]],colnames(scananal))
}
Hit the send button a bit prematurely. I have not figured out what
sort of process or result you hope to achieve but perhaps showing how
to improve the use of grep inside a loop will help:
for(i in 1:33){
patt - paste((X[[:digit:]]+)., i, $, sep=);
if (length(inX[grep(patt,inX)]) 0 ) { DAY[i] -
list( grep(patt,inX) ) }
}
DAY[1:5]
[[1]]
[1] 1 2 3 4 5 6 7 8
[[2]]
[1] 9 10 11 12 13 14 15 16
[[3]]
[1] 17 18 19 20 21 22 23 24
[[4]]
NULL
[[5]]
[1] 25 26 27 28 29 30 31 32
This first constructs a pattern. It also needs to test if there are
any results at each iteration because there are no days==4. Unless
you supply the result of grep() as a list it only records the first
day in a series, so it only gives you the starting locations. Maybe
if you clarified what you will be doing with this DAY construct, there
might be more of a target to shoot for. You could use lapply on those
column numbers at the moment.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] memory use without running gc()
Allan Engelhardt all...@cybaea.com writes: ### Method 2 ## Setup file - paste(/proc, Sys.getpid(), stat, sep = /) what - vector(list, 44); what[[23]] - integer(0) ## In your logging routine vsz - scan(file, what = what, quiet = TRUE)[[23]]/1024 cat(Virtual size: , vsz, \n, sep = ) [...] The scan method should be plenty fast. Change as appropriate if your definition of memory is different from virtual size. Thanks, that works well. I'm on an older kernel (2.6.18), so I only have 42 entries in /proc/pid/stat, but VSZ is still in position 23. Looking at the proc man page also shows /proc/[pid]/statm, which collects some of the memory information, but requires me to know what the page size is to make sense of it. Gegards, Johann __ R-help@r-project.org 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] grep problem decimal points looping
Try this also:
nm - scan('clipboard', what = '')
transform(structure(do.call(rbind, strsplit(nm[-1], \\.)), .Dimnames =
list(NULL, c('V1', 'V2'))), V1 = gsub(X, , V1))
On Tue, Aug 10, 2010 at 10:17 AM, RCulloch ross.cull...@dur.ac.uk wrote:
Hi R Users,
I have been trying to work out how to rename column names using grep,
basically I have generated these column names using tapply:
[1] NAME X1.1 X2.1 X3.1 X4.1 X5.1 X6.1 X7.1 X8.1
[10] X1.2 X2.2 X3.2 X4.2 X5.2 X6.2 X7.2 X8.2
X1.3
[19] X2.3 X3.3 X4.3 X5.3 X6.3 X7.3 X8.3 X1.5
X2.5
[28] X3.5 X4.5 X5.5 X6.5 X7.5 X8.5 X1.6 X2.6
X3.6
[37] X4.6 X5.6 X6.6 X7.6 X8.6 X1.8 X2.8 X3.8
X4.8
[46] X5.8 X6.8 X7.8 X8.8 X1.9 X2.9 X3.9 X4.9
X5.9
[55] X6.9 X7.9 X8.9 X1.10 X2.10 X3.10 X4.10 X5.10
X6.10
[64] X7.10 X8.10 X1.12 X2.12 X3.12 X4.12 X5.12 X6.12
X7.12
[73] X8.12 X1.13 X2.13 X3.13 X4.13 X5.13 X6.13 X7.13
X8.13
[82] X1.14 X2.14 X3.14 X4.14 X5.14 X6.14 X7.14 X8.14
X1.15
[91] X2.15 X3.15 X4.15 X5.15 X6.15 X7.15 X8.15 X1.16
X2.16
[100] X3.16 X4.16 X5.16 X6.16 X7.16 X8.16 X1.17 X2.17
X3.17
[109] X4.17 X5.17 X6.17 X7.17 X8.17 X1.18 X2.18 X3.18
X4.18
[118] X5.18 X6.18 X7.18 X8.18 X1.19 X2.19 X3.19 X4.19
X5.19
[127] X6.19 X7.19 X8.19 X1.20 X2.20 X3.20 X4.20 X5.20
X6.20
[136] X7.20 X8.20 X1.21 X2.21 X3.21 X4.21 X5.21 X6.21
X7.21
[145] X8.21 X1.22 X2.22 X3.22 X4.22 X5.22 X6.22 X7.22
X8.22
[154] X1.23 X2.23 X3.23 X4.23 X5.23 X6.23 X7.23 X8.23
X1.24
[163] X2.24 X3.24 X4.24 X5.24 X6.24 X7.24 X8.24 X1.25
X2.25
[172] X3.25 X4.25 X5.25 X6.25 X7.25 X8.25 X1.26 X2.26
X3.26
[181] X4.26 X5.26 X6.26 X7.26 X8.26 X1.27 X2.27 X3.27
X4.27
[190] X5.27 X6.27 X7.27 X8.27 X1.28 X2.28 X3.28 X4.28
X5.28
[199] X6.28 X7.28 X8.28 X1.29 X2.29 X3.29 X4.29 X5.29
X6.29
[208] X7.29 X8.29 X1.30 X2.30 X3.30 X4.30 X5.30 X6.30
X7.30
[217] X8.30 X1.31 X2.31 X3.31 X4.31 X5.31 X6.31 X7.31
X8.31
[226] X1.32 X2.32 X3.32 X4.32 X5.32 X6.32 X7.32 X8.32
X1.33
[235] X2.33 X3.33 X4.33 X5.33 X6.33 X7.33 X8.33
What the names mean are behaviour.day the X is not important to the data,
it
is the numbers I am trying to select on.
So I want to split the data by day i.e. selecting for the number after the
decimal.
I am using this code (where scananal is the data) with out looping so the
number following the decimal I change manually (NB the data have been
changed to character):
DAY - grep((X[[:digit:]]+).3,colnames(scananal))
However, this will select for day 3, 30, 31, 32, etc I have tried to use
fixed = TRUE, but that just returns integer(0). But if I use 30, it will
select only 30. Not sure what I'm doing wrong here, and I assumed that
fixed
= T would fix this, but doesn't.
I have tried to loop this too, but with no luck, so if anyone can point me
in the right direction about how to loop using grep I would be most
grateful!
The main problem I have is where to put the loop, for example:
for(i in 1:33){
print(i)
DAY[[i]] - grep((X[[:digit:]]+).[[i]],colnames(scananal))
}
which doesn't work, and no doubt there are obvious reasons for this! Any
help would be much appreciated,
All the best,
Ross
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[R] How to build R on Mac to keep packages running?
Hi there, I am trying to install RApache on my Mac (see http://worldofrcraft.blogspot.com/2010/08/installing-rapache-on-mac-os-x-snow.html). I needed to build R with --enable-R-shlib option, so very naively did the following ./configure --enable-R-shlib make sudo make install RApache runs like a charm now, but when I want to start my R.app, it does not start. Only my R64.app is starting, but not all packages are running with R64.app. Where can I find the magic configure-charms used to configure and build R for the Mac platform? Yours, Karsten === Karsten D. Wolf http://www.ifeb.uni-bremen.de/wolf __ R-help@r-project.org 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] Numerical Methods Course
I want to take this numerical methods course where the text is http://www.amazon.com/Numerical-Methods-J-Douglas-Faires/dp/0534407617 . The instructor recommends MATLAB, but states Fortran, C, Mathematica, or Maple will also do the job. Will R do the job as well? If not, where do you think it will be lacking in the context of this book/course. __ R-help@r-project.org 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] Plotting confidence bands around regression line
Dear R-helpers and graphics gurus, I have two problems with plotting confidence bands: 1. First is relatively simple. I am using the Passing-Bablok procedure to obtain unbiased regression coefficients. This procedure yields the a b coefficient values along with their confidence intervals. I then plot the raw data with the regression line, but I would like to add the confidence band for the line... and I can't figure out how to do it. In other words, given: Estimate 5%CI 95%CI Intercept -4.305562 -9.931152 -1.381792 Slope 1.257318 1.053025 1.678516 How to plot the regression line with confidence band? 2. Second problem is plotting confidence bands along fitted nls regression line. I tried predict(nls.object, int='c') - but doesn't work. Later I figured in the documentation that the 'int' parameter is currently ignored. I guess this means it's not a trivial thing to do. Does anyone have a suggestion on how to obtain confidence predictions for such a model? Please cc my email address when you reply. Thanks and best regards, -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 __ R-help@r-project.org 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: optimization subject to constraints]
I think the problem is because the the Hessian of the augmented Lagrangian iis
singular at c(0,0).
Try this:
require(alabama)
heq - function(x) {
x[1]^2+x[2]^2 - 1
}
constrOptim.nl(par=c(0,0), fn=f, heq=heq, control.outer=list(trace=FALSE))
$par
[1] -0.7071067 -0.7071067
$value
[1] -1.414213
$iterations
[1] 10
$lambda
[1] -0.7068717
$penalty
[1] -6.496021e-08
$counts
function gradient
100 30
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: Gildas Mazo gildas.m...@curie.fr
Date: Tuesday, August 10, 2010 10:11 am
Subject: [R] [Fwd: Re: optimization subject to constraints]
To: r-help@r-project.org
- Original Message -
From Gildas Mazo gildas.m...@curie.fr
Date Tue, 10 Aug 2010 15:49:19 +0200
To Matthias Gondan matthias-gon...@gmx.de
Subject Re: [R] optimization subject to constraints
Danke schön Matthias.
I had naively started with x0 = c(0,0) and I got a Redundant
constraints were found error. What's the problem with (0,0) ?
Matthias Gondan a écrit :
try this (package Rsolnp)
library(Rsolnp)
g- function(x)
{
return(x[1]^2+x[2]^2)
} # constraint
f- function(x)
{
return(x[1]+x[2])
} # objective function
x0 = c(1, 1)
solnp(x0, fun=f, eqfun=g, eqB=c(1))
Am 10.08.2010 14:59, schrieb Gildas Mazo:
Thanks, but I still cannot get to solve my problem: consider this
simple
example:
f- function(x){
return(x[1]+x[2])
} # objective function
g- function(x){
return(x[1]^2+x[2]^2)
} # constraint
#
I wanna Maximize f(x) subject to g(x) = 1. By hand the solution is
(1/sqrt(2), 1/sqrt(2), sqrt(2)). This is to maximizing a linear function
subject to a nonlinear equality constraint. I didn't find any suitable
function in the packages I went through.
Thanks in advance,
Gildas
Spencer Graves a écrit :
To find every help page containing the term constrained
optimization, you can try the following:
library(sos)
co- findFn('constrained optimization')
Printing this co object opens a table in a web browser
with
all matches sorted first by the package with the most matches and
with
hot links to the documentation page.
writeFindFn2xls(co)
This writes an excel file, with the browser table as the second
tab and the first being a summary of the packages. This summary
table
can be made more complete and useful using the installPackages
function, as noted in the sos vignette.
A shameless plug from the lead author of the sos package.
Spencer Graves
On 8/9/2010 10:01 AM, Ravi Varadhan wrote:
constrOptim can only handle linear inequality constraints. It cannot
handle
equality (linear or nonlinear) as well as nonlinear inequality
constraints.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org
[ On
Behalf Of Dwayne Blind
Sent: Monday, August 09, 2010 12:56 PM
To: Gildas Mazo
Cc: r-help@r-project.org
Subject: Re: [R] optimization subject to constraints
Hi !
Why not constrOptim ?
Dwayne
2010/8/9 Gildas Mazogildas.m...@curie.fr
Dear R users,
I'm looking for tools to perform optimization subject to constraints,
both linear and non-linear. I don't mind which algorithm may be
used, my
primary aim is to get something general and easy-to-use to study
simples
examples.
Thanks for helping,
Gildas
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-guide.html
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and provide commented, minimal, self-contained, reproducible code.
__
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide
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[R] influence measures for multivariate linear models
Barrett Ling, JASA, 1992, v.87(417), pp184-191 define general classes of influence measures for multivariate regression models, including analogs of Cook's D, Andrews Pregibon COVRATIO, etc. As in univariate response models, these are based on leverage and residuals based on omitting one (or more) observations at a time and refitting, although, in the univariate case, the computations can be optimized, as they are in stats::influence() and related methods. I'm interested in exploring the multivariate extension in R. I tried the following, and was surprised to find that R returned a result rather than an error -- presumably because mlm objects are not trapped before they get to lm.influence() # multivariate model data(Rohwer, package=heplots) rohwer.mod - lm(cbind(SAT, PPVT, Raven) ~ n + s + ns + na + ss, data=Rohwer) names(influence(rohwer.mod)) [1] hat coefficients sigmawt.res head(influence(rohwer.mod)$coefficients, 2) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2.25039 0.0254739 -0.025252 -0.06297 -0.121507 0.094355 [2,] 0.84649 -0.0062656 -0.077430 0.08345 -0.022579 -0.059480 Of course, the correct calculations would result from refitting, omitting each observation in turn, though doing this directly would be horribly inefficient. e.g, calculating B(i), deleting case i: coef(update(rohwer.mod, subset=1:69 !=1, data=Rohwer)) SAT PPVT Raven (Intercept) -2.466079 35.68664 11.510068 n1.888286 0.60949 0.075931 s -0.034524 -0.53040 0.160328 ns -2.739834 -0.67355 0.066392 na 2.219340 1.20481 -0.037272 ss 1.072300 0.99033 0.058509 coef(update(rohwer.mod, subset=1:69 !=2, data=Rohwer)) SAT PPVT Raven (Intercept) -1.062178 33.88199 10.8988006 n1.920026 0.59735 0.0713976 s0.017654 -0.47464 0.1774135 ns -2.886254 -0.67905 0.0673686 na 2.120411 1.29016 -0.0077484 ss 1.226135 0.96430 0.0471764 Is there anything existing for this case that I've missed, or does anyone have an interest in pursuing this topic? -Michael -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org 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] Plotting confidence bands around regression line
On Aug 10, 2010, at 10:56 AM, Michal Figurski wrote: Dear R-helpers and graphics gurus, I have two problems with plotting confidence bands: 1. First is relatively simple. I am using the Passing-Bablok procedure to obtain unbiased regression coefficients. This procedure yields the a b coefficient values along with their confidence intervals. I then plot the raw data with the regression line, but I would like to add the confidence band for the line... and I can't figure out how to do it. In other words, given: Estimate 5%CI 95%CI Intercept -4.305562 -9.931152 -1.381792 Slope 1.257318 1.053025 1.678516 How to plot the regression line with confidence band? Take a look at plotCI in either gplots or plotrix packages. Harrell's rms/Hmisc packages are nicely integrated with lattice and encourage you to create effective displays of models that remove simplistic linearity assumptions. -- David. 2. Second problem is plotting confidence bands along fitted nls regression line. I tried predict(nls.object, int='c') - but doesn't work. Later I figured in the documentation that the 'int' parameter is currently ignored. I guess this means it's not a trivial thing to do. Does anyone have a suggestion on how to obtain confidence predictions for such a model? Please cc my email address when you reply. Thanks and best regards, -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 __ R-help@r-project.org 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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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] grep problem decimal points looping
Hi David, Thanks very much for that reply! I might be a touch out of my comfort zone, but I can see how the loop script works and where I went wrong, but I'm not sure if I am asking the correct questions here, or perhaps more accurately I'm using the wrong command for the task in question - and as you say more info would be better! So. I want to split the data by day to look at the proportion of time an individual spent in each of the eight behaviours - there are 30 rows (i.e. individuals). So I'm going over old code trying to make it better (not that it could be worse!), especially trying to make it more efficient! So my old code did this (manually for each day): ##DAY1## DAY1 -cbind(scananal$X1.1, scananal$X2.1, scananal$X3.1, scananal$X4.1, scananal$X5.1, scananal$X6.1, scananal$X7.1, scananal$X8.1) head(DAY1) which would give, for example, head(DAY1) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 140212203 [2,] 230010000 [3,]00000000 [4,]00000000 [5,]00000000 [6,]00000000 I'd run the following script to get the proportions then bind that together with other data ###DAY1~~~### ## CALC NSCANS PER ID ## n - rowSums(DAY1) ## GIVE THE DAY NUMBER TO THE DATAFILE DAY - rep(1,30) ## CALC PROPORTION OF TIME IN EACH ACTIVITY ## scansprop - as.data.frame(prop.table(DAY1,1)) head(scansprop) ##CALC AS ARC_SINE_TRANSFORMED### transscan-asin(scansprop) head (transscan) ##gives column headings## names(transscan) ##CHECK IT ALL ADDS TO ONE!! ## rowSums(scansprop) ##MERGES ALL THE DATA FOR THE DAY DAY1_SUM - cbind(n,DAY,DAY1,scansprop,transscan) Then I would merge each of the days, so this script works, but I know it is rather a poor effort in R script to say the least! I'm trying to work through this myself, but hit a hurdle in the first instance! Not sure if this is any clearer? Cheers, Ross -- View this message in context: http://r.789695.n4.nabble.com/grep-problem-decimal-points-looping-tp2319773p2319941.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Numerical Methods Course
TGS, Given that you have to pay an outrageous $155.86 for that book, it seems reasonable to look for a free environment for numerical computing (like R!). If your instructor says that such a variety of programming languages would work, you could probably make a good argument to use R. But why not just ask your instructor? If your instructor insists on MATLAB, you could also consider using GNU Octave, a free MATLAB clone. -Matt On Tue, 2010-08-10 at 10:55 -0400, TGS wrote: I want to take this numerical methods course where the text is http://www.amazon.com/Numerical-Methods-J-Douglas-Faires/dp/0534407617 . The instructor recommends MATLAB, but states Fortran, C, Mathematica, or Maple will also do the job. Will R do the job as well? If not, where do you think it will be lacking in the context of this book/course. __ R-help@r-project.org 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. -- Matthew S. Shotwell Graduate Student Division of Biostatistics and Epidemiology Medical University of South Carolina __ R-help@r-project.org 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] grep problem decimal points looping
On Aug 10, 2010, at 9:17 AM, RCulloch wrote:
Hi R Users,
I have been trying to work out how to rename column names using grep,
basically I have generated these column names using tapply:
[1] NAME X1.1 X2.1 X3.1 X4.1 X5.1 X6.1 X7.1
X8.1
[10] X1.2 X2.2 X3.2 X4.2 X5.2 X6.2 X7.2 X8.2
X1.3
[19] X2.3 X3.3 X4.3 X5.3 X6.3 X7.3 X8.3 X1.5
X2.5
[28] X3.5 X4.5 X5.5 X6.5 X7.5 X8.5 X1.6 X2.6
X3.6
[37] X4.6 X5.6 X6.6 X7.6 X8.6 X1.8 X2.8 X3.8
X4.8
[46] X5.8 X6.8 X7.8 X8.8 X1.9 X2.9 X3.9 X4.9
X5.9
[55] X6.9 X7.9 X8.9 X1.10 X2.10 X3.10 X4.10 X5.10
X6.10
[64] X7.10 X8.10 X1.12 X2.12 X3.12 X4.12 X5.12 X6.12
X7.12
[73] X8.12 X1.13 X2.13 X3.13 X4.13 X5.13 X6.13 X7.13
X8.13
[82] X1.14 X2.14 X3.14 X4.14 X5.14 X6.14 X7.14 X8.14
X1.15
[91] X2.15 X3.15 X4.15 X5.15 X6.15 X7.15 X8.15 X1.16
X2.16
[100] X3.16 X4.16 X5.16 X6.16 X7.16 X8.16 X1.17 X2.17
X3.17
[109] X4.17 X5.17 X6.17 X7.17 X8.17 X1.18 X2.18 X3.18
X4.18
[118] X5.18 X6.18 X7.18 X8.18 X1.19 X2.19 X3.19 X4.19
X5.19
[127] X6.19 X7.19 X8.19 X1.20 X2.20 X3.20 X4.20 X5.20
X6.20
[136] X7.20 X8.20 X1.21 X2.21 X3.21 X4.21 X5.21 X6.21
X7.21
[145] X8.21 X1.22 X2.22 X3.22 X4.22 X5.22 X6.22 X7.22
X8.22
[154] X1.23 X2.23 X3.23 X4.23 X5.23 X6.23 X7.23 X8.23
X1.24
[163] X2.24 X3.24 X4.24 X5.24 X6.24 X7.24 X8.24 X1.25
X2.25
[172] X3.25 X4.25 X5.25 X6.25 X7.25 X8.25 X1.26 X2.26
X3.26
[181] X4.26 X5.26 X6.26 X7.26 X8.26 X1.27 X2.27 X3.27
X4.27
[190] X5.27 X6.27 X7.27 X8.27 X1.28 X2.28 X3.28 X4.28
X5.28
[199] X6.28 X7.28 X8.28 X1.29 X2.29 X3.29 X4.29 X5.29
X6.29
[208] X7.29 X8.29 X1.30 X2.30 X3.30 X4.30 X5.30 X6.30
X7.30
[217] X8.30 X1.31 X2.31 X3.31 X4.31 X5.31 X6.31 X7.31
X8.31
[226] X1.32 X2.32 X3.32 X4.32 X5.32 X6.32 X7.32 X8.32
X1.33
[235] X2.33 X3.33 X4.33 X5.33 X6.33 X7.33 X8.33
What the names mean are behaviour.day the X is not important to the
data, it
is the numbers I am trying to select on.
So I want to split the data by day i.e. selecting for the number
after the
decimal.
I am using this code (where scananal is the data) with out looping
so the
number following the decimal I change manually (NB the data have been
changed to character):
You need to learn the special character$ which marks the no-
character end of string. After creating a replica of your column-names
with scan and grep:
inp - scan(what=character)
inX - inp[grep(X, inp)]
DAY - grep((X[[:digit:]]+).3$,inX)
inX[DAY]
[1] X1.3 X2.3 X3.3 X4.3 X5.3 X6.3 X7.3 X8.3
DAY - grep((X[[:digit:]]+).3,colnames(scananal))
However, this will select for day 3, 30, 31, 32, etc I have tried to
use
fixed = TRUE, but that just returns integer(0). But if I use 30, it
will
select only 30. Not sure what I'm doing wrong here, and I assumed
that fixed
= T would fix this, but doesn't.
I have tried to loop this too, but with no luck, so if anyone can
point me
in the right direction about how to loop using grep I would be most
grateful!
The main problem I have is where to put the loop, for example:
for(i in 1:33){
print(i)
DAY[[i]] - grep((X[[:digit:]]+).[[i]],colnames(scananal))
}
which doesn't work, and no doubt there are obvious reasons for this!
Any
help would be much appreciated,
All the best,
Ross
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Re: [R] grep problem decimal points looping
On Aug 10, 2010, at 11:14 AM, RCulloch wrote:
Hi David,
Thanks very much for that reply! I might be a touch out of my
comfort zone,
but I can see how the loop script works and where I went wrong, but
I'm not
sure if I am asking the correct questions here, or perhaps more
accurately
I'm using the wrong command for the task in question - and as you
say more
info would be better! So.
I want to split the data by day to look at the proportion of time an
individual spent in each of the eight behaviours - there are 30 rows
(i.e.
individuals).
So I'm going over old code trying to make it better (not that it
could be
worse!), especially trying to make it more efficient!
So my old code did this (manually for each day):
##DAY1##
DAY1 -cbind($X1.1, scananal$X2.1, scananal$X3.1, scananal$X4.1,
scananal$X5.1, scananal$X6.1, scananal$X7.1, scananal$X8.1)
head(DAY1)
From our earlier efforts:
DAY[[1]]
[1] 1 2 3 4 5 6 7 8
If you were using the earlier results, then this could be made easier
and more generalizable to an iterative process by:
DAY1cols - scananal[ , DAY[[1]] ] or just ...
DAY1cols - scananal[ DAY[[1]] ]
You could replace the 1 by an i in a for loop, or create a list of
subsets with lapply.
which would give, for example,
head(DAY1)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 140212203
[2,] 230010000
[3,]00000000
[4,]00000000
[5,]00000000
[6,]00000000
I'd run the following script to get the proportions then bind that
together
with other data
###DAY1~~~###
## CALC NSCANS PER ID ##
n - rowSums(DAY1)
## GIVE THE DAY NUMBER TO THE DATAFILE
DAY - rep(1,30)
## CALC PROPORTION OF TIME IN EACH ACTIVITY ##
scansprop - as.data.frame(prop.table(DAY1,1))
head(scansprop)
##CALC AS ARC_SINE_TRANSFORMED###
transscan-asin(scansprop)
head (transscan)
##gives column headings##
names(transscan)
##CHECK IT ALL ADDS TO ONE!! ##
rowSums(scansprop)
##MERGES ALL THE DATA FOR THE DAY
DAY1_SUM - cbind(n,DAY,DAY1,scansprop,transscan)
Then I would merge each of the days, so this script works, but I
know it is
rather a poor effort in R script to say the least!
That may work perfectly well. I don't know. I didn't follow the logic
because your reference to proportions was not made with an
unambiguous definition of a denominator in several locations.
I'm trying to work through this myself, but hit a hurdle in the first
instance!
Not sure if this is any clearer?
Perhaps. Two options: Try to put that sequence into a function that
will return a structure. Then test it. Then lapply it.
Or put it into sequence that makes DAY1_SUM inside a for-loop, take
the results of for() {} on DAY[[i]] and have it assign values to a
structure like a list or a matrix.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] Fwd: List of lists ?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Carlos Petti
Sent: Tuesday, August 10, 2010 6:12 AM
To: r-help@r-project.org
Subject: [R] Fwd: List of lists ?
-- Forwarded message --
From: Carlos Petti carlos.pe...@gmail.com
Date: 2010/8/10
Subject: Re: [R] List of lists ?
To: David Winsemius dwinsem...@comcast.net
Thanks for answer.
I read the error messages but I did not find the solution :-(
Your solution works.
But, a new problem remains because I want
to use the list of lists as follows :
x - list(list())
x[[2]][[1]] - c(1, 2, 3)
x[[2]][[2]] - c(3, 2, 1)
You need to tell it that x[[i]] will be a list
for each i. You can do that with
x - list()
for(i in 1:3) {
x[[i]] - list()
for(j in seq_len(i)) {
x[[i]][[j]] - i*100 + seq_len(j)
}
}
It might be easier to read and perhaps faster to do
x - list()
for(i in 1:3) {
xi - list() # will become x[[i]]
for(j in seq_len(i)) {
xi[[j]] - i*100 + seq_len(j)
}
x[[i]] - xi
}
If you know the ultimate length of a list, and it will
be long, it may save time to allocate the whole thing
up front with
xlength - 3
x - vector(list, xlength)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Thanks in advance,
Carlos
2010/8/9 David Winsemius dwinsem...@comcast.net:
On Aug 9, 2010, at 12:57 PM, Carlos Petti wrote:
Dear list,
I have to use a list of lists containing vectors.
For instance :
[[1]]
[[1]][[1]]
[1] 1 2 3
[[1]][[2]]
[1] 3 2 1
I want to attribute vectors to the main list
without use of an intermediate list,
but it does not work :
More specifically it produces an error that has information in it.
x[[1]][[1]] - c(1, 2, 3)
Error in `*tmp*`[[1]] : subscript out of bounds
x - list()
x[[1]][[1]] - c(1, 2, 3)
x[[1]][[2]] - c(3, 2, 1)
So thinking perhaps we just needed another level of
subscripting available
I tried:
x - list(list())
x[[1]][[1]] - c(1, 2, 3)
x[[1]][[2]] - c(3, 2, 1)
x
[[1]]
[[1]][[1]]
[1] 1 2 3
[[1]][[2]]
[1] 3 2 1
Success. Moral: Read the error messages for meaning or at
least clues.
(Further testing showed that almost anything inside the
original list()
call, even NULL, would have created enough structure for
the interpreter to
work with.
David Winsemius, MD
West Hartford, CT
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Re: [R] Plotting confidence bands around regression line
David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. rms/Hmisc packages are very nice, but unfortunately they do not work with Passing-Bablok nor 'nls' models. -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 W dniu 2010-08-10 11:09, David Winsemius pisze: On Aug 10, 2010, at 10:56 AM, Michal Figurski wrote: Dear R-helpers and graphics gurus, I have two problems with plotting confidence bands: 1. First is relatively simple. I am using the Passing-Bablok procedure to obtain unbiased regression coefficients. This procedure yields the a b coefficient values along with their confidence intervals. I then plot the raw data with the regression line, but I would like to add the confidence band for the line... and I can't figure out how to do it. In other words, given: Estimate 5%CI 95%CI Intercept -4.305562 -9.931152 -1.381792 Slope 1.257318 1.053025 1.678516 How to plot the regression line with confidence band? Take a look at plotCI in either gplots or plotrix packages. Harrell's rms/Hmisc packages are nicely integrated with lattice and encourage you to create effective displays of models that remove simplistic linearity assumptions. __ R-help@r-project.org 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] useR! 2011: announcement and call for tutorials
We are pleased to announce that the R user conference useR! 2011 is scheduled for August 16-18, 2011, and will take place at the University of Warwick, Coventry, UK. As for the predecessor conferences, the program will consist of two parts: invited lectures and user-contributed sessions (abstract submission will be available online later this year). Prior to the conference, there will be tutorials on R (proposals for tutorials should be sent before October 29, 2010, see below). INVITED SPEAKERS Adrian Bowman, Lee Edlefsen, Ulrike Grömping, Wolfgang Huber, Brian Ripley, Jonathan Rougier, Simon Urbanek, and Brandon Whitcher. USER-CONTRIBUTED SESSIONS The conference will feature both talks and posters illustrating the use of R in practice. Contributions are welcome that introduce recent developments in the R Project (including CRAN packages), demonstrate applications of R in areas of current interest, or otherwise engage and inspire participants in their use of R. PRE-CONFERENCE TUTORIALS Before the official program, half-day tutorials will be offered on Monday, August 15. We invite R users to submit proposals for three hour tutorials on special topics regarding R. The proposals should give a brief description of the tutorial, including goals, detailed outline, justification of why the tutorial is important, background knowledge required and potential attendees. The proposals should be sent before October 29, 2010 to useR-2011_at_R-project.org. A web page offering more information on the `useR!' conference is available at http://www.R-project.org/useR-2011 We hope to see you in Coventry! The organizing committee: John Aston, Julia Brettschneider, David Firth, Ashley Ford, Ioannis Kosmidis, Tom Nichols, Elke Thönnes and Heather Turner ___ r-annou...@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ R-help@r-project.org 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] Plotting confidence bands around regression line
On Aug 10, 2010, at 11:23 AM, Michal Figurski wrote: David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. Wouldn't you also need to specify the range over which these estimates might be valid and to offer the means for the X values? What level of R knowledge are you at? You have provided no data or code. Many R methods offer predict methods that return CI's. -- david rms/Hmisc packages are very nice, but unfortunately they do not work with Passing-Bablok nor 'nls' models. -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 W dniu 2010-08-10 11:09, David Winsemius pisze: On Aug 10, 2010, at 10:56 AM, Michal Figurski wrote: Dear R-helpers and graphics gurus, I have two problems with plotting confidence bands: 1. First is relatively simple. I am using the Passing-Bablok procedure to obtain unbiased regression coefficients. This procedure yields the a b coefficient values along with their confidence intervals. I then plot the raw data with the regression line, but I would like to add the confidence band for the line... and I can't figure out how to do it. In other words, given: Estimate 5%CI 95%CI Intercept -4.305562 -9.931152 -1.381792 Slope 1.257318 1.053025 1.678516 How to plot the regression line with confidence band? Take a look at plotCI in either gplots or plotrix packages. Harrell's rms/Hmisc packages are nicely integrated with lattice and encourage you to create effective displays of models that remove simplistic linearity assumptions. David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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] Function to Define a Function
What if you change your function to:
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
switch(type,
one={ function(x,N0,r) N0*exp(x*r) },
two={ function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) },
)
}
Does that work for you?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Derek Ogle
Sent: Monday, August 09, 2010 7:31 PM
To: R (r-help@R-project.org)
Subject: [R] Function to Define a Function
I am trying to define a general R function that has a function as the
output that depends on the user's input arguments (this may make more
sense by looking at the toy example below). My real use for this type
of code is to allow a user to choose from many parameterizations of the
same general model.
My issue is that when I compile a package with this type of code in
it I get a __warning__ that multiple local function definitions for
'm' with different formal arguments. While this is not a deadly
error I would like to avoid the warning if possible. Can someone
provide some guidance? Thank you in advance for any help you can
offer.
For what it is worth ... I am working on a Windows XP machine with R
2.11.1.
## A function that allows the user to create a new function that
depends on their
## choice in the type argument. As a simple example, if the user
chooses one
## then the output function is exponential growth, if the user choses
two then
## thhe output function is logistic growth.
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
switch(type,
one={ m - function(x,N0,r) N0*exp(x*r) },
two={ m - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) },
)
m
}
## define time steps
t - 0:10
## create a function -- junk1 -- that produces exponential growth
junk1 - mdlChooser(one)
junk1
res1 - junk1(t,500,0.2)
res1
## create a function -- junk2 -- that produces logistic growth
junk2 - mdlChooser(two)
junk2
res2 - junk2(t,500,0.2,1000)
res2
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Concatenate a mix of numbers and letters to create a vector name
Others gave you some examples using assign, which is the same information as in
the FAQ. But I expect that you will be better served by using a list (you can
use the paste function to create the names for the list elements) rather than
assign and global variables.
Something like:
mylist - list()
for (i in c(1950, 1960, 1970) ) {
for (j in 1:12) {
mylist[[ paste( 'tmax.', i, j, sep='' ) ]] - somefunction(i, j)
}
}
Then you can access a single result like:
mylist[['tmax.15012']]
or you can do something with all the output using something like:
sapply( mylist, summary )
hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy]
Sent: Tuesday, August 10, 2010 6:07 AM
To: Greg Snow; r-help@r-project.org
Subject: RE: [R] Concatenate a mix of numbers and letters to create a
vector name
I was not aware of the R-FAQ, it seems to have some very useful tips,
thanks for pointing there.
Regarding the 7.21 in the FAQ, I read it a few times but it did not
lead me anywhere. For the moment I am blaming my inexperience with some
R basics, I will come back after I do some more serious studying.
Thanks again,
Panos
Dr Panos Hadjinicolaou
Energy Environment Water Research Center (EEWRC)
The Cyprus Institute
-
_
From: Greg Snow [mailto:greg.s...@imail.org]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy], r-h...@r-
project.org [mailto:r-h...@r-project.org]
Sent: Wed, 28 Jul 2010 00:39:41 +0300
Subject: RE: [R] Concatenate a mix of numbers and letters to create a
vector name
This is a frequently asked/answered question (7.21 in the FAQ). What
searching did you do and why did it not find this FAQ or previous
discussion of it? How could the documentation/search/etc. be improved
so that you (and the next n people with this question) will find the
answer easier?
The most important of part of the FAQ answer is the last section where
it points out that using a list will be much simpler. You mentioned in
the first post that you were doing this in a loop, just start with an
empty list, use paste (or sprintf) to create the name, and then assign
it as a new element of the list with that name (e.g. mylist[[
sprintf('tmax.%d%d', var1, var2) ]] - outputfromcomputations ).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Panos Hadjinicolaou
Sent: Monday, July 26, 2010 8:24 AM
To: r-help@r-project.org
Subject: Re: [R] Concatenate a mix of numbers and letters to create a
vector name
Thanks for the reply. Indeed the paste function results in
concatenation:
paste(c(tmax., 1950, 12), collapse=)
[1] tmax.195012
but I am looking for a way to subsequently get rid of the - - in
order to use tmax.195012 as an object (e.g. to define a vector with
that name). Any ideas?
Thanks,
Panos
_
From: Dimitris Rizopoulos [mailto:d.rizopou...@erasmusmc.nl]
To: Panos Hadjinicolaou [mailto:p.hadjinicol...@cyi.ac.cy]
Cc: r-help@r-project.org
Sent: Mon, 26 Jul 2010 16:48:31 +0300
Subject: Re: [R] Concatenate a mix of numbers and letters to create a
vector name
have a look at function paste(), i.e., ?paste
I hope it helps.
Best,
Dimitris
On 7/26/2010 3:44 PM, Panos Hadjinicolaou wrote:
Dear all,
I am trying to create a vector name, for example tmax.195012 from
tmax., 1950 and 12. Obviously I don't wish to simply type it because
the 3 name components are changing in each iteration within a loop.
Is
there any way of concatenating those 3 components (which are a
mixture
of numbers and letters)?
Thanks for reading,
Panos
-
Dr Panos Hadjinicolaou
Energy Environment Water Research Center (EEWRC)
The Cyprus Institute
-
-
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-
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guide.html
and provide commented, minimal, self-contained, reproducible
code.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Re: [R] Plotting confidence bands around regression line
David, I would consider myself intermediate in R, but a beginner in statistics. I need a formula that would allow me to calculate confidence boundaries of the regression line given the slope, intercept and their CIs (and *any* range). Passing-Bablok regression doesn't yet exist in R - I am developing it. Therefore I am sure there is no predict method for it ;) I believe I have provided sufficient data to address this problem, but if that would help anyone, here is more: # data frame a - structure(list(x = c(0.1, 1.43, 4.21, 3.67, 3.23, 7.72, 5.99, 9.16, 10.6, 9.84, 11.94, 12.03, 12.89, 11.26, 15.54, 15.58, 17.32, 17.65, 19.52, 20.48, 20.44, 20.51, 22.27, 23.58, 25.83, 26.04, 26.92, 28.44, 30.73, 28.78), y = c(1.08, 1.39, 1.84, 0.56, 7.23, 4.91, 3.35, 7.09, 3.16, 8.98, 16.37, 7.46, 15.46, 23.2, 4.63, 11.13, 15.68, 13.92, 26.44, 21.65, 21.01, 20.22, 22.69, 22.21, 23.6, 17.24, 45.24, 30.09, 40, 49.6)), .Names = c(x, y), row.names = c(NA, -30L), class = data.frame) Then I run the regression procedure (in development - now part of the 'MethComp' package): print(PBreg(a)) # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). I would appreciate if anyone could point me to a handbook or website where it is described. Regarding 2 - the predict method for 'nls' class currently *ignores* the interval parameter - as it is stated in documentation. Regards -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 11:38, David Winsemius wrote: On Aug 10, 2010, at 11:23 AM, Michal Figurski wrote: David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. Wouldn't you also need to specify the range over which these estimates might be valid and to offer the means for the X values? What level of R knowledge are you at? You have provided no data or code. Many R methods offer predict methods that return CI's. __ R-help@r-project.org 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] influence measures for multivariate linear models
Michael Friendly wrote: Barrett Ling, JASA, 1992, v.87(417), pp184-191 define general classes of influence measures for multivariate regression models, including analogs of Cook's D, Andrews Pregibon COVRATIO, etc. As in univariate response models, these are based on leverage and residuals based on omitting one (or more) observations at a time and refitting, although, in the univariate case, the computations can be optimized, as they are in stats::influence() and related methods. I'm interested in exploring the multivariate extension in R. I tried the following, and was surprised to find that R returned a result rather than an error -- presumably because mlm objects are not trapped before they get to lm.influence() # multivariate model data(Rohwer, package=heplots) rohwer.mod - lm(cbind(SAT, PPVT, Raven) ~ n + s + ns + na + ss, data=Rohwer) names(influence(rohwer.mod)) [1] hat coefficients sigmawt.res head(influence(rohwer.mod)$coefficients, 2) [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2.25039 0.0254739 -0.025252 -0.06297 -0.121507 0.094355 [2,] 0.84649 -0.0062656 -0.077430 0.08345 -0.022579 -0.059480 Of course, the correct calculations would result from refitting, omitting each observation in turn, though doing this directly would be horribly inefficient. e.g, calculating B(i), deleting case i: coef(update(rohwer.mod, subset=1:69 !=1, data=Rohwer)) SAT PPVT Raven (Intercept) -2.466079 35.68664 11.510068 n1.888286 0.60949 0.075931 s -0.034524 -0.53040 0.160328 ns -2.739834 -0.67355 0.066392 na 2.219340 1.20481 -0.037272 ss 1.072300 0.99033 0.058509 coef(update(rohwer.mod, subset=1:69 !=2, data=Rohwer)) SAT PPVT Raven (Intercept) -1.062178 33.88199 10.8988006 n1.920026 0.59735 0.0713976 s0.017654 -0.47464 0.1774135 ns -2.886254 -0.67905 0.0673686 na 2.120411 1.29016 -0.0077484 ss 1.226135 0.96430 0.0471764 Is there anything existing for this case that I've missed, or does anyone have an interest in pursuing this topic? Hmm, fitted coefficients in this sort multivariate models are the same as those in the univariate ones, so as long as you do whole-case deletions, I would think that you should be able to reuse the 1D code. I would conjecture that the main problem with what you currently get is that it only pertains to the 1st column -- looks like the differences between the two rows from lm.influence matches the differences between the first two colums from coef(update(...)). Since lm() only handles complete cases, casewise deletion diagnostics is probably the best you can get, otherwise it would be interesting to see the effect of deleting each coordinate separately. (As you know, these matters are within my general sphere of interest, but I'm afraid my time is too constrained at them moment for more than a sideline view.) -Michael -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org 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] Function to Define a Function
On Mon, 9 Aug 2010, Derek Ogle wrote:
I am trying to define a general R function that has a function as the output
that depends on the user's input arguments (this may make more sense by looking
at the toy example below). My real use for this type of code is to allow a
user to choose from many parameterizations of the same general model.
My issue is that when I compile a package with this type of code in it I get a __warning__ that
multiple local function definitions for 'm' with different formal arguments. While this is not a
deadly error I would like to avoid the warning if possible. Can someone provide some guidance?
Thank you in advance for any help you can offer.
snip
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
switch(type,
one={ m - function(x,N0,r) N0*exp(x*r) },
two={ m - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) },
)
m
}
One approach is to put the assignment outside the switch
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
m- switch(type,
one={ function(x,N0,r) N0*exp(x*r) },
two={ function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) },
)
m
}
or not even assign the result
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
switch(type,
one= function(x,N0,r) N0*exp(x*r) ,
two=function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) ,
)
}
-thomas
Thomas Lumley
Professor of Biostatistics
University of Washington, Seattle
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Re: [R] p-values with pvclust
I don't know much about pvclust itself, but you might consider the technique in this paper: Buja, A., Cook, D. Hofmann, H., Lawrence, M. Lee, E.-K., Swayne, D.F and Wickham, H. (2009) Statistical Inference for exploratory data analysis and model diagnostics Phil. Trans. R. Soc. A 2009 367, 4361-4383 doi: 10.1098/rsta.2009.0120 You just need to understand your null hypothesis enough to be able to simulate data from the null (possibly permuting original data, or generating data from a distribution similar to your data but without clusters). The vis.test function in the TeachingDemos package helps with an implementation of the test (you will still need to write some of your own code though). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of syrvn Sent: Tuesday, August 10, 2010 6:45 AM To: r-help@r-project.org Subject: [R] p-values with pvclust Hi, if you look at the first image (Image1) you see that there are 2 main clusters 7 and 8 I wanted to use pvclust to calculate a p-value whether these clusters are due to chance or statistically significant. Unfortunately pvclust does not provide a p-value for the first brunch (7 and 8). So I added a row to my matrix which is very different to the rest of the data to create an additional brunch. See image here http://r.789695.n4.nabble.com/file/n2319732/Image2.png . I finally got my p-value for the brunch (7 and 8) which is 98 (9). I was happy to see that the p-value was significant until I realised that if I add an additional brunch which is not that different from the rest but still cluster in a sperate cluster (see image here http://r.789695.n4.nabble.com/file/n2319732/Image3.png ) the p-value is changing and not significant any longer (84). I was wondering why this happens because I thought that for each brunch the p-value is calculated independently? Does anybody know how to get a correct p-value for the first brunch (7 and 8) maybe without adding an additional brunch? Best regards syrvn -- View this message in context: http://r.789695.n4.nabble.com/p-values- with-pvclust-tp2319732p2319732.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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@r-project.org 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] Function to Define a Function
Gabor Grothendieck ggrothendi...@gmail.com
on Mon, 9 Aug 2010 23:20:18 -0400 writes:
On Mon, Aug 9, 2010 at 9:31 PM, Derek Ogle do...@northland.edu wrote:
I am trying to define a general R function that has a
function as the output that depends on the user's input
arguments (this may make more sense by looking at the toy
example below). My real use for this type of code is to
allow a user to choose from many parameterizations of the
same general model.
My issue is that when I compile a package with this
type of code in it I get a __warning__ that multiple local
function definitions for 'm' with different formal
arguments. While this is not a deadly error I would like
to avoid the warning if possible. Can someone provide some
guidance?
Thank you in advance for any help you can offer.
For what it is worth ... I am working on a Windows XP machine with R
2.11.1.
## A function that allows the user to create a new function that depends
on their
## choice in the type argument. As a simple example, if the user
chooses one
## then the output function is exponential growth, if the user choses
two then
## thhe output function is logistic growth.
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
switch(type,
one={ m - function(x,N0,r) N0*exp(x*r) },
two={ m - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r)) },
)
m
}
## define time steps
t - 0:10
## create a function -- junk1 -- that produces exponential growth
junk1 - mdlChooser(one)
junk1
res1 - junk1(t,500,0.2)
res1
## create a function -- junk2 -- that produces logistic growth
junk2 - mdlChooser(two)
junk2
res2 - junk2(t,500,0.2,1000)
res2
Try this:
mdlChooser - function(type = c(one, two)) {
one - function(x,N0,r) N0*exp(x*r)
two - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r))
type - match.arg(type)
get(type)
}
or a bit more elegantly, I think,
mdlChooser - function(type=c(one,two)) {
switch(match.arg(type),
one= function(x,N0,r) N0*exp(x*r),
two= function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r))
)
}
which just leaves away some unnecessary code from Derek's
original code (and here you could even drop the last { .. }
pair).
Martin Maechler, ETH Zurich
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Re: [R] Plotting confidence bands around regression line
Please give the prescription. The article is not available on our extensive online library. I wonder if the method can compete with the bootstrap. Frank Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University On Tue, 10 Aug 2010, Michal Figurski wrote: David, I would consider myself intermediate in R, but a beginner in statistics. I need a formula that would allow me to calculate confidence boundaries of the regression line given the slope, intercept and their CIs (and *any* range). Passing-Bablok regression doesn't yet exist in R - I am developing it. Therefore I am sure there is no predict method for it ;) I believe I have provided sufficient data to address this problem, but if that would help anyone, here is more: # data frame a - structure(list(x = c(0.1, 1.43, 4.21, 3.67, 3.23, 7.72, 5.99, 9.16, 10.6, 9.84, 11.94, 12.03, 12.89, 11.26, 15.54, 15.58, 17.32, 17.65, 19.52, 20.48, 20.44, 20.51, 22.27, 23.58, 25.83, 26.04, 26.92, 28.44, 30.73, 28.78), y = c(1.08, 1.39, 1.84, 0.56, 7.23, 4.91, 3.35, 7.09, 3.16, 8.98, 16.37, 7.46, 15.46, 23.2, 4.63, 11.13, 15.68, 13.92, 26.44, 21.65, 21.01, 20.22, 22.69, 22.21, 23.6, 17.24, 45.24, 30.09, 40, 49.6)), .Names = c(x, y), row.names = c(NA, -30L), class = data.frame) Then I run the regression procedure (in development - now part of the 'MethComp' package): print(PBreg(a)) # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). I would appreciate if anyone could point me to a handbook or website where it is described. Regarding 2 - the predict method for 'nls' class currently *ignores* the interval parameter - as it is stated in documentation. Regards -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 11:38, David Winsemius wrote: On Aug 10, 2010, at 11:23 AM, Michal Figurski wrote: David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. Wouldn't you also need to specify the range over which these estimates might be valid and to offer the means for the X values? What level of R knowledge are you at? You have provided no data or code. Many R methods offer predict methods that return CI's. __ R-help@r-project.org 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@r-project.org 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] Plotting confidence bands around regression line
On Aug 10, 2010, at 12:12 PM, Michal Figurski wrote: David, I would consider myself intermediate in R, but a beginner in statistics. I need a formula that would allow me to calculate confidence boundaries of the regression line given the slope, intercept and their CIs (and *any* range). For ordinary regression the CI's for prediction intervals are going to be much wider than the CI's for parameter estimates. In both cases they are quadratic functions that depend on the mean_x_hat and on s_hat^2 (used as an estimate of sigma^2). These formulae should be available in any basic regression text. I am sufficiently aware of my non-statistician status to know that I could not comment on whether naively applying those functions to estimates from another method would have validity. -- David. Passing-Bablok regression doesn't yet exist in R - I am developing it. Therefore I am sure there is no predict method for it ;) I believe I have provided sufficient data to address this problem, but if that would help anyone, here is more: # data frame a - structure(list(x = c(0.1, 1.43, 4.21, 3.67, 3.23, 7.72, 5.99, 9.16, 10.6, 9.84, 11.94, 12.03, 12.89, 11.26, 15.54, 15.58, 17.32, 17.65, 19.52, 20.48, 20.44, 20.51, 22.27, 23.58, 25.83, 26.04, 26.92, 28.44, 30.73, 28.78), y = c(1.08, 1.39, 1.84, 0.56, 7.23, 4.91, 3.35, 7.09, 3.16, 8.98, 16.37, 7.46, 15.46, 23.2, 4.63, 11.13, 15.68, 13.92, 26.44, 21.65, 21.01, 20.22, 22.69, 22.21, 23.6, 17.24, 45.24, 30.09, 40, 49.6)), .Names = c(x, y), row.names = c(NA, -30L), class = data.frame) Then I run the regression procedure (in development - now part of the 'MethComp' package): print(PBreg(a)) # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). I would appreciate if anyone could point me to a handbook or website where it is described. Regarding 2 - the predict method for 'nls' class currently *ignores* the interval parameter - as it is stated in documentation. Regards -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 11:38, David Winsemius wrote: On Aug 10, 2010, at 11:23 AM, Michal Figurski wrote: David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. Wouldn't you also need to specify the range over which these estimates might be valid and to offer the means for the X values? What level of R knowledge are you at? You have provided no data or code. Many R methods offer predict methods that return CI's. David Winsemius, MD West Hartford, CT __ R-help@r-project.org 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 invert a list ?
Dear list, I have a list, as follows : a - 5 names(a) - a b - 9 names(b) - b c - 15 names(c) - c x - list(i = a, j = b, j = c) I want to invert the list, like this : $a i 5 $b j k 9 15 I do not find a clean solution. Could anyone give me elegant ideas ? Thanks in advance, Carlos __ R-help@r-project.org 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] Function to Define a Function
Neat. But why assign the functions to separate variables at all?
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
m - switch(type,
one=function(x,N0,r) N0*exp(x*r) ,
two=function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r))
)
m
}
also works without appearing to assign different functions to the same
variable.
(In this simple example, you wouldn't need the m- assignment either;
you could simply let the switch return its result. But I assume the real
intended use is more complicated than just returning the function)
Derek Ogle do...@northland.edu 10/08/2010 13:48:13
Gabor ... that worked perfectly. Thank you.
-Original Message-
Try this:
mdlChooser - function(type = c(one, two)) {
one - function(x,N0,r) N0*exp(x*r)
two - function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r))
type - match.arg(type)
get(type)
}
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Re: [R] Plotting confidence bands around regression line
Michal Figurski wrote: # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). The answer is that you can't. You can't even do it with ordinary linear regression without knowing the correlation between slope and intercept. However, if you can get a CI for the intercept then you could subtract x0 from all the x and get a CI for the value at x0. (This brings echos from a distant past. My master's thesis was about some similar median-type estimators. I can't remember whether I looked at the Passing-Bablok paper at the time (1985!!) but my general recollection is that this group of methods is littered with unstated assumptions.) -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org 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 invert a list ?
Here is one way: xst - stack(x) let - letters[cumsum(duplicated(match(xst$ind, letters))) + match(xst$ind, letters)] with(xst, structure(split(structure(values, names = let), ind), .Names = row.names(xst)[1:length(unique(ind))])) On Tue, Aug 10, 2010 at 1:58 PM, Carlos Petti carlos.pe...@gmail.comwrote: Dear list, I have a list, as follows : a - 5 names(a) - a b - 9 names(b) - b c - 15 names(c) - c x - list(i = a, j = b, j = c) I want to invert the list, like this : $a i 5 $b j k 9 15 I do not find a clean solution. Could anyone give me elegant ideas ? Thanks in advance, Carlos __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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 in R2Bugs
Hello,
I am running a GLMM using R2Bugs, but am getting the below error message. I am
including the entire output, although the 2nd and 3rd lines seem to indicate
the problem. Note that I do define N (it is an integer) and send it to a
datalist (see the R commands and model below). Interestingly, when I put all
of this into OpenBugs directly, the model runs. I do have reasons, however,
that I would like to do this through R.
model is syntactically correct
expected variable name error pos 341553 (error on line 1)
variable N is not defined
Initializing chain 1: model must be compiled before initial values loaded
Initializing chain 2: model must be compiled before initial values loaded
Initializing chain 3: model must be compiled before initial values loaded
model must be compiled before generating initial values
Sampling has been started...
model must be initialized before updating
model must be initialized before DIC an be monitored
Error in BRugs::samplesSet(parametersToSave) :
model must be initialized before monitors used
Thanks for your help!
John
R commands
datalist-list(r, N,n,Lev, Light, Soil1, Soil2, Soil3,
Consp, Plot, DConsp,
Spp, n.plot, n.sp)
bugs.data(datalist)
parmlist-c(b.lev, b.light, b.con, b.dcon, b.s1, b.s2, b.s3,
sigma.ind, sigma.plt, sigma.sp, sigma.sppl)
initlist-list(list(b.lev=0.01, b.light=0.01, b.con=0.01, b.s1=0.01,
b.s2=0.01, b.s3=0.01, b.dcon=0.01, tau.ind=rep(0,N),
tau.plt=rep(0,n.plot), tau.sp=rep(0,n.sp), tau=array(0, c(n.sp,
n.plot))),
list(b.lev=0.01, b.light=0.01, b.con=0.01, b.s1=0.01,
b.s2=0.01, b.s3=0.01, b.dcon=0.01, tau.ind=rep(0,N),
tau.plt=rep(0,n.plot), tau.sp=rep(0,n.sp), tau=array(0, c(n.sp,
n.plot))),
list(b.lev=0.01, b.light=0.01, b.con=0.01, b.s1=0.01,
b.s2=0.01, b.s3=0.01, b.dcon=0.01, tau.ind=rep(0,N),
tau.plt=rep(0,n.plot), tau.sp=rep(0,n.sp), tau=array(0, c(n.sp,
n.plot
modfile-(GLMM Model.txt)
GLMMcom1-bugs(data=datalist, inits=initlist, parameters=parmlist,
model.file=modfile,
n.chains=3, n.iter=2000, DIC=TRUE, n.burnin=500,
bugs.directory=/Program Files/OpenBUGS, program=openbugs,
debug=TRUE)
model {
for (i in 1:N){ # No. of sect x dist plots
r[i]~dbin(p[i], n[i])
eps.ind[i]~dnorm(0, tau.ind)
logit(p[i]) - b.lev*Lev[i] + b.light*Light[i] + b.con*Consp[i] +
b.dcon*DConsp[i] +
b.s1*Soil1[i] + b.s2*Soil2[i] + b.s3*Soil3[i] +
eps.sp[Sp[i]] + eps.plt[Plot[i]] + eps[Sp[i], Plot[i]] +
eps.ind[i]
}
for(k in 1:n.sp){for(s in 1:n.plot){eps[k, s] ~ dnorm(0.0,tau)}}
for(s in 1:n.plot){eps.plt[s] ~ dnorm(0.0, tau.plt)}
for(k in 1:n.sp){eps.sp[k] ~ dnorm(0.0, tau.sp)}
b.lev~dnorm(0, 0.01)
b.light~dnorm(0, 0.01)
b.con~dnorm(0,0.01)
b.s1~dnorm(0, 0.01)
b.s2~dnorm(0, 0.01)
b.s3~dnorm(0, 0.01)
b.dcon~dnorm(0, 0.01)
tau.ind~dunif(0,100)
sigma.ind-1/sqrt(tau.ind)
tau.plt ~ dunif(0,100)
sigma.plt-1/sqrt(tau.plt)
tau.sp ~ dunif(0,100)
sigma.sp-1/sqrt(tau.sp)
tau ~ dunif(0,100)
sigma.sppl-1/sqrt(tau) #sd of species x plot random effect
}
-
John Poulsen, PhD
Assistant Scientist
Woods Hole Research Center
149 Woods Hole Road
Falmouth, MA 02540-1644 USA
Tel: 508.540.9900 x168
www.whrc.org
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[R] matrix problem
Hi, I have a file like this: 1 2 0.1 2 3 0.2 3 1 0.3 And I want to read it to create a matrix like this: [,1] [,2][,3] [1,]0 0.1 0 [2,]0 00.2 [3,]0.300 How can I do it efficiently? Thanks. -- Best, Zhenjiang [[alternative HTML version deleted]] __ R-help@r-project.org 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] matrix problem
Try this: Lines - '1 2 0.1 2 3 0.2 3 1 0.3' DF - read.table(textConnection(Lines)) m - matrix(0, ncol = nrow(DF), nrow = nrow(DF)) m[as.matrix(DF[1:2])] - DF[[3]] On Tue, Aug 10, 2010 at 3:03 PM, zhenjiang xu zhenjiang...@gmail.comwrote: Hi, I have a file like this: 1 2 0.1 2 3 0.2 3 1 0.3 And I want to read it to create a matrix like this: [,1] [,2][,3] [1,]0 0.1 0 [2,]0 00.2 [3,]0.300 How can I do it efficiently? Thanks. -- Best, Zhenjiang [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] Plotting confidence bands around regression line
Frank, I had to order this article through Inter-Library Loan and wait for it for a week! I'll try to make it short. In Passing-Bablok the principle is to calculate slopes between all possible pairs of points in the dataset, and then to take a shifted median of those slopes, where the offset is the number of slopes of value (-1). Because of this, bootstrap is out of question - it would take too much time. Let n be the number of data points, N - the number of slopes and K - the offset. The locations of CI boundaries in the set of N slopes are then calculated with formulas: M1 - N - qnorm(1 - conf.level/2) * sqrt((n*(n-1)*(2*n+5))/18))/2 M2 - N - M1 + 1 CIs for intercept are calculated as medians of y(i) - slopes[M1,M2]*x(i) I hope I don't confuse anyone. The article has a mathematical derivations section and justification for these formulas. It looks solid to me, although after 25 years the flaws may be apparent to some of you. -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 12:29, Frank Harrell wrote: Please give the prescription. The article is not available on our extensive online library. I wonder if the method can compete with the bootstrap. Frank Frank E Harrell Jr Professor and Chairman School of Medicine Department of Biostatistics Vanderbilt University On Tue, 10 Aug 2010, Michal Figurski wrote: David, I would consider myself intermediate in R, but a beginner in statistics. I need a formula that would allow me to calculate confidence boundaries of the regression line given the slope, intercept and their CIs (and *any* range). Passing-Bablok regression doesn't yet exist in R - I am developing it. Therefore I am sure there is no predict method for it ;) I believe I have provided sufficient data to address this problem, but if that would help anyone, here is more: # data frame a - structure(list(x = c(0.1, 1.43, 4.21, 3.67, 3.23, 7.72, 5.99, 9.16, 10.6, 9.84, 11.94, 12.03, 12.89, 11.26, 15.54, 15.58, 17.32, 17.65, 19.52, 20.48, 20.44, 20.51, 22.27, 23.58, 25.83, 26.04, 26.92, 28.44, 30.73, 28.78), y = c(1.08, 1.39, 1.84, 0.56, 7.23, 4.91, 3.35, 7.09, 3.16, 8.98, 16.37, 7.46, 15.46, 23.2, 4.63, 11.13, 15.68, 13.92, 26.44, 21.65, 21.01, 20.22, 22.69, 22.21, 23.6, 17.24, 45.24, 30.09, 40, 49.6)), .Names = c(x, y), row.names = c(NA, -30L), class = data.frame) Then I run the regression procedure (in development - now part of the 'MethComp' package): print(PBreg(a)) # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). I would appreciate if anyone could point me to a handbook or website where it is described. Regarding 2 - the predict method for 'nls' class currently *ignores* the interval parameter - as it is stated in documentation. Regards -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 11:38, David Winsemius wrote: On Aug 10, 2010, at 11:23 AM, Michal Figurski wrote: David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. Wouldn't you also need to specify the range over which these estimates might be valid and to offer the means for the X values? What level of R knowledge are you at? You have provided no data or code. Many R methods offer predict methods that return CI's. __ R-help@r-project.org 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@r-project.org 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] Plotting confidence bands around regression line
Peter, Since in PB the procedure is to calculate a whole list of slopes and intercepts, wouldn't it be a solution to determine the correlation and go from there? How do I do it? -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 13:12, Peter Dalgaard wrote: Michal Figurski wrote: # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). The answer is that you can't. You can't even do it with ordinary linear regression without knowing the correlation between slope and intercept. However, if you can get a CI for the intercept then you could subtract x0 from all the x and get a CI for the value at x0. (This brings echos from a distant past. My master's thesis was about some similar median-type estimators. I can't remember whether I looked at the Passing-Bablok paper at the time (1985!!) but my general recollection is that this group of methods is littered with unstated assumptions.) __ R-help@r-project.org 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] matrix problem
Let me give you a not that efficient one... assume you have read the matrix (named as x) into R: n=dim(x)[1] y=matrix(0,n,n) for (i in 1:n) y[x[i,1],x[i,2]]=x[i,3] -- View this message in context: http://r.789695.n4.nabble.com/matrix-problem-tp2320193p2320219.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] matrix problem
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of zhenjiang xu Sent: Tuesday, August 10, 2010 11:03 AM To: R-help@r-project.org Subject: [R] matrix problem Hi, I have a file like this: 1 2 0.1 2 3 0.2 3 1 0.3 And I want to read it to create a matrix like this: [,1] [,2][,3] [1,]0 0.1 0 [2,]0 00.2 [3,]0.300 How can I do it efficiently? Thanks. Use a k-column matrix as a subscript into your k-dimensional output array. (k is 2 in your case.) E.g., 'input' is your matrix in a form that one can paste into an R session: input - cbind(c(1,2,3), c(2,3,1), c(.1,.2,.3)) size - max(input[,1:2]) # you may want something else here output - matrix(0.0, size, size) output[input[,1:2]] - input[,3] output [,1] [,2] [,3] [1,] 0.0 0.1 0.0 [2,] 0.0 0.0 0.2 [3,] 0.3 0.0 0.0 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -- Best, Zhenjiang [[alternative HTML version deleted]] __ R-help@r-project.org 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@r-project.org 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] extracting information from an object
I was working on a project involving a linear model, and wanted to extract the standard error of a predictor. I am able to do so, but not in the way I would expect. I would have expected that if a created a model such as Model1 - lm(y~x,z,d), the object Model1 would contain that information even though it does not print it out when I simply type Model1. I would also have (wrongly) suspected that if I type summary(Model1) R would simply look at the object Model1 and find whatever it needs. But it doesn't work that way. If I want that standard error I have to first create a summary of Model1 and then extract the standard error from the summary with something like summary(Model1)$coefficients or, more specifically, summary(Model)$coefficients[2,2]. [I know that I can cram all of that into one line if I want to.] But doesn't that mean that when I ask for a summary R has to recreate the linear model all over again before pulling out the standard error. (Venables and Ripley, p. 77) suggest that this could happen if the method is not written correctly, but how is it not happening anyway?) And if so, if Model1 doesn't contain the raw data, how does summary produce an answer even if I delete one of the variables before calling it? As you can see, I have figured out how to get what I want, but I don't understand the process of building objects, which is the important thing to understand. Perhaps I don't understand methods well enough. Below is sample code: #Sample for linear model x - c(3,7,9,15,18) y - c(5,4,8,6,9) reg - lm(y~x) reg #Produces only the regression coefficients and using str(reg) indicates that # that is all that it has. regsummary - summary(reg) #Produces what I need and str(regsummary) shows that st. errors are part of the object. regsummary$coefficients[1:2, 1:4] rm(y) out - summary(reg) # works just fine although y is no longer available and reg doesn't look like it # could supply it. Thanks, Dave Howell __ R-help@r-project.org 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] Plotting confidence bands around regression line
Thanks Michael, That's the method that Dana Quade taught me in his intro nonparametrics course at UNC in the mid 1970s, at least for a single predictor. His method did not incorporate the shift you mentioned though. The method looks robust. Not sure about efficiency. Frank Frank E Harrell Jr Professor and ChairmanSchool of Medicine Department of Biostatistics Vanderbilt University On Tue, 10 Aug 2010, Michal Figurski wrote: Frank, I had to order this article through Inter-Library Loan and wait for it for a week! I'll try to make it short. In Passing-Bablok the principle is to calculate slopes between all possible pairs of points in the dataset, and then to take a shifted median of those slopes, where the offset is the number of slopes of value (-1). Because of this, bootstrap is out of question - it would take too much time. Let n be the number of data points, N - the number of slopes and K - the offset. The locations of CI boundaries in the set of N slopes are then calculated with formulas: M1 - N - qnorm(1 - conf.level/2) * sqrt((n*(n-1)*(2*n+5))/18))/2 M2 - N - M1 + 1 CIs for intercept are calculated as medians of y(i) - slopes[M1,M2]*x(i) I hope I don't confuse anyone. The article has a mathematical derivations section and justification for these formulas. It looks solid to me, although after 25 years the flaws may be apparent to some of you. -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 12:29, Frank Harrell wrote: Please give the prescription. The article is not available on our extensive online library. I wonder if the method can compete with the bootstrap. Frank Frank E Harrell Jr Professor and Chairman School of Medicine Department of Biostatistics Vanderbilt University On Tue, 10 Aug 2010, Michal Figurski wrote: David, I would consider myself intermediate in R, but a beginner in statistics. I need a formula that would allow me to calculate confidence boundaries of the regression line given the slope, intercept and their CIs (and *any* range). Passing-Bablok regression doesn't yet exist in R - I am developing it. Therefore I am sure there is no predict method for it ;) I believe I have provided sufficient data to address this problem, but if that would help anyone, here is more: # data frame a - structure(list(x = c(0.1, 1.43, 4.21, 3.67, 3.23, 7.72, 5.99, 9.16, 10.6, 9.84, 11.94, 12.03, 12.89, 11.26, 15.54, 15.58, 17.32, 17.65, 19.52, 20.48, 20.44, 20.51, 22.27, 23.58, 25.83, 26.04, 26.92, 28.44, 30.73, 28.78), y = c(1.08, 1.39, 1.84, 0.56, 7.23, 4.91, 3.35, 7.09, 3.16, 8.98, 16.37, 7.46, 15.46, 23.2, 4.63, 11.13, 15.68, 13.92, 26.44, 21.65, 21.01, 20.22, 22.69, 22.21, 23.6, 17.24, 45.24, 30.09, 40, 49.6)), .Names = c(x, y), row.names = c(NA, -30L), class = data.frame) Then I run the regression procedure (in development - now part of the 'MethComp' package): print(PBreg(a)) # And the result of the Passing-Bablok regression on this data frame: Estimate 5%CI 95%CI Intercept -4.306197 -9.948438 -1.374663 Slope 1.257584 1.052696 1.679290 The original Passing Bablok article on this method has an easy prescription for CIs on coefficients, so I implemented that. Now I need a way to calculate CI boundaries for individual points - this may be a basic handbook stuff - I just don't know it (I'm not a statistician). I would appreciate if anyone could point me to a handbook or website where it is described. Regarding 2 - the predict method for 'nls' class currently *ignores* the interval parameter - as it is stated in documentation. Regards -- Michal J. Figurski, PhD HUP, Pathology Laboratory Medicine Biomarker Research Laboratory 3400 Spruce St. 7 Maloney Philadelphia, PA 19104 tel. (215) 662-3413 On 2010-08-10 11:38, David Winsemius wrote: On Aug 10, 2010, at 11:23 AM, Michal Figurski wrote: David, I may have stated my problem incorrectly - my problem is to *obtain the coordinates* for confidence boundary lines. As input data I have only CIs for slope and intercept. Wouldn't you also need to specify the range over which these estimates might be valid and to offer the means for the X values? What level of R knowledge are you at? You have provided no data or code. Many R methods offer predict methods that return CI's. __ R-help@r-project.org 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@r-project.org 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] partial match of one column in data frame to another character vector
Here is some data (dput output below) myData id group 1 D599 A 2 002-0004 B 3 F01932 A 18 F16 B 19 F28 A 20 A94 B and a vector of IDs (the full label). fullID [1] F16-284 ACC-A94-AB ADAD599 002-0004BCC CDCF01932.AB F28DDB NOMATCH-EX For each id in myData, there could be a partial match in fullID. For example D599 in myData matches ADAD599. I would like to add a column to myData that contains the corresponding fullID or NA if a match was not found. Thanks for your help. Juliet # #Data # myData - structure(list(id = structure(c(6L, 5L, 1L, 2L, 3L, 4L), .Label = c( F01932, F16 ,F28 , A94, 002-0004, D599), class = factor), group = structure(c(5L, 4L, 1L, 3L, 2L, 3L), .Label = c( A, A,B, B, A), class = factor)), .Names = c(id, group ), class = data.frame, row.names = c(1, 2, 3, 18, 19 , 20 )) fullID - c(F16-284, ACC-A94-AB, ADAD599, 002-0004BCC, CDCF01932.AB, F28DDB, NOMATCH-EX) __ R-help@r-project.org 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] partial match of one column in data frame to another character vector
Try this: myData$fullID - sapply(gsub(^ +| +$, , myData$id), grep, x = fullID, value = TRUE) On Tue, Aug 10, 2010 at 3:39 PM, Juliet Hannah juliet.han...@gmail.comwrote: Here is some data (dput output below) myData id group 1 D599 A 2 002-0004 B 3 F01932 A 18 F16 B 19 F28 A 20 A94 B and a vector of IDs (the full label). fullID [1] F16-284 ACC-A94-AB ADAD599 002-0004BCC CDCF01932.AB F28DDB NOMATCH-EX For each id in myData, there could be a partial match in fullID. For example D599 in myData matches ADAD599. I would like to add a column to myData that contains the corresponding fullID or NA if a match was not found. Thanks for your help. Juliet # #Data # myData - structure(list(id = structure(c(6L, 5L, 1L, 2L, 3L, 4L), .Label = c( F01932, F16 ,F28 , A94, 002-0004, D599), class = factor), group = structure(c(5L, 4L, 1L, 3L, 2L, 3L), .Label = c( A, A,B, B, A), class = factor)), .Names = c(id, group ), class = data.frame, row.names = c(1, 2, 3, 18, 19 , 20 )) fullID - c(F16-284, ACC-A94-AB, ADAD599, 002-0004BCC, CDCF01932.AB, F28DDB, NOMATCH-EX) __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] extracting information from an object
David, I was working on a project involving a linear model, and wanted to extract the standard error of a predictor. I am able to do so, but not in the way I would expect. I would have expected that if a created a model such as Model1 - lm(y~x,z,d), the object Model1 would contain that information even though it does not print it out when I simply type Model1. You can always see what information objects contain by using the ?str function on them. In this instance, str(Model1) will show the components of the Model1 object. I would also have (wrongly) suspected that if I type summary(Model1) R would simply look at the object Model1 and find whatever it needs. But it doesn't work that way. Correct. You can always see what a function does by printing its definition at the R prompt. In this case, typing summary.lm will show you what the summary function does when passed an 'lm' object. If I want that standard error I have to first create a summary of Model1 and then extract the standard error from the summary with something like summary(Model1)$coefficients or, more specifically, summary(Model)$coefficients[2,2]. Use the ?coef function for this purpose, which works with most modelling functions, including 'lm'. [I know that I can cram all of that into one line if I want to.] But doesn't that mean that when I ask for a summary R has to recreate the linear model all over again before pulling out the standard error. (Venables and Ripley, p. 77) Which book? I don't think MASS 4th edition. suggest that this could happen if the method is not written correctly, but how is it not happening anyway?) The Model1 object contains the necessary components to calculate the quantity. Look at how it's done in summary.lm. And if so, if Model1 doesn't contain the raw data, how does summary produce an answer even if I delete one of the variables before calling it? By default, an 'lm' object will contain the model.matrix. See the ?lm value section for other components. In general, it would be bad practice to have a function like summary.lm depend not only on a supplied argument, but on some other object that it is not a function of being present in the workspace. As you can see, I have figured out how to get what I want, but I don't understand the process of building objects, which is the important thing to understand. Perhaps I don't understand methods well enough. I think a combination of the ?str function, looking at the actual lm and summary.lm functions, and a careful reading of the help pages will help. Below is sample code: #Sample for linear model x - c(3,7,9,15,18) y - c(5,4,8,6,9) reg - lm(y~x) reg #Produces only the regression coefficients and using str(reg) indicates that # that is all that it has. regsummary - summary(reg) #Produces what I need and str(regsummary) shows that st. errors are part of the object. regsummary$coefficients[1:2, 1:4] rm(y) out - summary(reg) # works just fine although y is no longer available and reg doesn't look like it # could supply it. reg$model reg$qr __ R-help@r-project.org 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] sapply/lapply instead of loop
Using the input below, can I do something more elegant (and more efficient)
than the loop also listed below to pad strings to a width of 5? The true
matrix is about 300K rows and 31 columns.
###
#INPUT
###
temp
DX1 DX2 DX3
1 13761 8125 49178
2 63371 v75 22237
3 51745 77703 93500
4 64081 32826 v72
5 78477 43828 87645
###
#CODE
###
ssize - c(nrow(temp), ncol(temp))
aa - c(1:ssize[2])
aa - paste(DX, aa, sep = )
record - matrix(?, nrow = ssize, ncol = ssize[2])
colnames(record) - aa
mm - 0
#for (j in 1:1) {
for (j in 1:ssize[1]) {
mm - j
a - as.character(as.matrix(as.data.frame(temp[j,])))
len2 - sum(a != ?)
mi - 0
for (k in 1:len2) {
aa - a[k]
a0 - 5 - nchar(aa)
if (a0 0) {
for (st in 1:a0) {
aa - paste(aa, , sep = )
}
}
record[j, k] - aa
}
}
###
#OUTPUT
###
DX1 DX2 DX3
1 13761 8125 49178
2 63371 v75 22237
3 51745 77703 93500
4 64081 32826 v72
5 78477 43828 87645
--
View this message in context:
http://r.789695.n4.nabble.com/sapply-lapply-instead-of-loop-tp2320265p2320265.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] problem installing boot package on Mac
Dear all, I cannot seem to get boot running. Can anybody help? I've tried downloading and compiling the binary from http://cran.r-project.org/web/packages/boot/index.html, as well as the below. (Please see the scary-sounding error message, and the sessionInfo.) Neither works. Any help would be much appreciated. - Malcolm install.packages(boot, dependencies=T) trying URL 'http://cran.uk.r-project.org/bin/macosx/leopard/contrib/2.11/boot_1.2-42.tgz' Content type 'application/x-gzip' length 448826 bytes (438 Kb) opened URL == downloaded 438 Kb The downloaded packages are in /var/folders/8O/8OQqiVUgEBSa2PwHYc1b1E+++TI/-Tmp-//RtmpXuArKA/downloaded_packages library(boot) Error in FUN(melanoma[[1L]], ...) : internal error -3 in R_decompress1 boot Error: object 'boot' not found sessionInfo() R version 2.11.1 (2010-05-31) x86_64-apple-darwin9.8.0 locale: [1] en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] coda_0.13-5deldir_0.0-12 maptools_0.7-34nlme_3.1-96 [5] MASS_7.3-7 Matrix_0.999375-42 lattice_0.18-8 sp_0.9-66 [9] foreign_0.8-40 loaded via a namespace (and not attached): [1] boot_1.2-42 grid_2.11.1 spdep_0.5-17 tools_2.11.1 __ R-help@r-project.org 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] sapply/lapply instead of loop
will this do what you want:
newTemp[] - lapply(newTemp, function(.col){
+ # convert to character and pad to 5 space
+ sprintf(%5s, as.character(.col))
+ })
str(newTemp)
'data.frame': 5 obs. of 3 variables:
$ DX1: chr 13761 63371 51745 64081 ...
$ DX2: chr 8125 v75 77703 32826 ...
$ DX3: chr 49178 22237 93500 v72 ...
On Tue, Aug 10, 2010 at 2:55 PM, GL pfl...@shands.ufl.edu wrote:
Using the input below, can I do something more elegant (and more efficient)
than the loop also listed below to pad strings to a width of 5? The true
matrix is about 300K rows and 31 columns.
###
#INPUT
###
temp
DX1 DX2 DX3
1 13761 8125 49178
2 63371 v75 22237
3 51745 77703 93500
4 64081 32826 v72
5 78477 43828 87645
###
#CODE
###
ssize - c(nrow(temp), ncol(temp))
aa - c(1:ssize[2])
aa - paste(DX, aa, sep = )
record - matrix(?, nrow = ssize, ncol = ssize[2])
colnames(record) - aa
mm - 0
#for (j in 1:1) {
for (j in 1:ssize[1]) {
mm - j
a - as.character(as.matrix(as.data.frame(temp[j,])))
len2 - sum(a != ?)
mi - 0
for (k in 1:len2) {
aa - a[k]
a0 - 5 - nchar(aa)
if (a0 0) {
for (st in 1:a0) {
aa - paste(aa, , sep = )
}
}
record[j, k] - aa
}
}
###
#OUTPUT
###
DX1 DX2 DX3
1 13761 8125 49178
2 63371 v75 22237
3 51745 77703 93500
4 64081 32826 v72
5 78477 43828 87645
--
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+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] matrix problem
Hi, I guess you just want to reshape your data to wide format. strs - Index Time Value 1 2 0.1 2 3 0.2 3 1 0.3 DF - read.table(textConnection(strs),header=T) rDF - reshape(DF, idvar=Index, timevar=Time, direction=wide) rDF[is.na(rDF)] - 0 - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/matrix-problem-tp2320193p2320287.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] sapply/lapply instead of loop
Try this:
formatC(as.matrix(temp))
On Tue, Aug 10, 2010 at 3:55 PM, GL pfl...@shands.ufl.edu wrote:
Using the input below, can I do something more elegant (and more efficient)
than the loop also listed below to pad strings to a width of 5? The true
matrix is about 300K rows and 31 columns.
###
#INPUT
###
temp
DX1 DX2 DX3
1 13761 8125 49178
2 63371 v75 22237
3 51745 77703 93500
4 64081 32826 v72
5 78477 43828 87645
###
#CODE
###
ssize - c(nrow(temp), ncol(temp))
aa - c(1:ssize[2])
aa - paste(DX, aa, sep = )
record - matrix(?, nrow = ssize, ncol = ssize[2])
colnames(record) - aa
mm - 0
#for (j in 1:1) {
for (j in 1:ssize[1]) {
mm - j
a - as.character(as.matrix(as.data.frame(temp[j,])))
len2 - sum(a != ?)
mi - 0
for (k in 1:len2) {
aa - a[k]
a0 - 5 - nchar(aa)
if (a0 0) {
for (st in 1:a0) {
aa - paste(aa, , sep = )
}
}
record[j, k] - aa
}
}
###
#OUTPUT
###
DX1 DX2 DX3
1 13761 8125 49178
2 63371 v75 22237
3 51745 77703 93500
4 64081 32826 v72
5 78477 43828 87645
--
View this message in context:
http://r.789695.n4.nabble.com/sapply-lapply-instead-of-loop-tp2320265p2320265.html
Sent from the R help mailing list archive at Nabble.com.
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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Re: [R] sapply/lapply instead of loop
Both of those approaches seem to return ( v75) instead of (v75 ). -- View this message in context: http://r.789695.n4.nabble.com/sapply-lapply-instead-of-loop-tp2320265p2320305.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] sapply/lapply instead of loop
So try: format(as.matrix(temp)) On Tue, Aug 10, 2010 at 4:13 PM, GL pfl...@shands.ufl.edu wrote: Both of those approaches seem to return ( v75) instead of (v75 ). -- View this message in context: http://r.789695.n4.nabble.com/sapply-lapply-instead-of-loop-tp2320265p2320305.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org 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] tikzDevice 0.5.0 released to CRAN
# tikzDevice
---
## Description
The tikzDevice package new graphics device for R which enables direct
output of graphics in a LaTeX-friendly way. Plotting commands issued
by R functions are transformed into LaTeX code blocks. These blocks
are interpreted with the help of TikZ-- a graphics library for TeX and
friends written by Till Tantau.
The tikzDevice supports three main modes of output:
- Figure chunks: placed in .tex files and suitable for inclusion in
LaTeX documents via the \input{} command.
- Stand alone figures: Complete LaTeX documents containing figure
code that can be compiled into stand-alone images. Pages are
cropped to the size of the figure using the preview package.
- Console output: TikZ code is returned directly to the R console
as a character vector for further manipulation.
## Beta Notice
The tikzDevice is currently flagged as a beta work. The package is
reasonably stable and has been used by the authors to produce graphics
for academic publications for over a year. The reason for beta status
is that there are several open design issues- two of which are:
- Providing support for UTF8 text.
- Supporting TeX variants other than LaTeX.
Resolving these issues may require changes to the tikzDevice that
break backwards compatibility with previous versions. The beta flag
is a reminder that such changes may occur- although we will strive to
avoid them if possible.
The beta flag will be removed upon release of version 1.0. At this
time maintaining backwards compatibility will become a primary concern.
## Obtaining the Package
Stable versions of the tikzDevice may be downloaded from CRAN:
install.packages( 'tikzDevice' )
Development versions may be obtained from R-Forge:
install.packages( 'tikzDevice',
repos='http://r-forge.r-project.net' )
## Reporting Bugs and Getting Help
The tikzDevice has a dedicated mailing list courtesy of R-Forge. The
mailing list is the easiest way to get answers for questions related
to usage:
tikzdevice-bugs @at@ lists.r-forge.r-project.org
Primary development takes place on GitHub. Bugs and feature requests
may be made by opening issues at the primary repository:
http://github.com/Sharpie/RTikZDevice/issues
Adventurous users are encouraged to fork the repository and contribute
to the development of the device!
## Latest Changes
*See the CHANGELOG for changes that occurred in previous releases*
---
### Version 0.5.0 Beta
---
Contributors
The following people contributed to this release of the tikzDevice:
- Lorenzo Isella contributed bug reports and examples that led to the
discovery of a bug in fontsize calculations that appeared when
certain LaTeX commands were used to change the active font.
- Vivianne Vilar for spotting spelling and grammar errors in the
vignette.
- Gabor Grothendieck for the idea for sending output to the screen
for use with sink() (i.e. the console option)
New Features
- console option for directing tikz() output back into the R console
instead of to a file.
- Preliminary support for a sanitize option which allows automatic
escaping of characters that have special meaning to TeX like $ and
%.
- tikzAnnotate() and tikzCoord() functions. tikzAnnotate() allows
arbitrary LaTeX code to be injected into the output stream of an
active tikz() graphics device. tikzCoord() is a wrapper for
tikzAnnotate() that inserts named locations into the graphics code.
These locations may be referenced by other TikZ drawing commands.
Bug Fixes
- Removed bad colon in the DESCRIPTION file.
- Proper fontsize calculations now include ps from par() and fontsize
from gpar(). This fixes issues with lattice-based graphics such as
ggplot2.
- Metrics are now calculated properly when commands like
\renewcommand\rmdefault are used to adjust the active font.
- Sanitization of % signs in labels.
- The package no longer overwrites user customizations set in places like
.Rprofile with default values when loaded.
- Attempting to use new graphics functions such as rasterImage() now
produces error messages instead of fatal crashes in R 2.11.0 and
above.
[[alternative HTML version deleted]]
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Re: [R] sapply/lapply instead of loop
That works great, and is ever so much simpler. Thanks much! -- View this message in context: http://r.789695.n4.nabble.com/sapply-lapply-instead-of-loop-tp2320265p2320317.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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 draw a spherical quadrant
Thanks, Jim! Jim Lemon wrote: You may find that the radial.plot function in the plotrix package will do what you want. I think you are looking at the polygon type of plot. It seems to me that I've a lot of things to learn only from plotrix package! With the functions included in this package seems relatively easy to get a planar projection of the spherical quarter holding all power values. But, please, do you know if is there any package that could be able to add some 3D effect to the representation? For instance, look at this MATLAB example: http://www.matrixlab-examples.com/3D-plot-part3.html Thanks! Greetings, Ricardo -- Ricardo Rodríguez Your XEN ICT Team __ R-help@r-project.org 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 draw a spherical quadrant
On 10/08/2010 3:34 PM, [Ricardo Rodriguez] Your XEN ICT Team wrote: Thanks, Jim! Jim Lemon wrote: You may find that the radial.plot function in the plotrix package will do what you want. I think you are looking at the polygon type of plot. It seems to me that I've a lot of things to learn only from plotrix package! With the functions included in this package seems relatively easy to get a planar projection of the spherical quarter holding all power values. But, please, do you know if is there any package that could be able to add some 3D effect to the representation? For instance, look at this MATLAB example: http://www.matrixlab-examples.com/3D-plot-part3.html The rgl package can do general 3D plotting, but I don't know of anyone who has put together what you're looking for, so you'll need to compute the individual line segments or polygons yourself. Duncan Murdoch __ R-help@r-project.org 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] Importing arguments for use by functions in a script
Thank you for your response. Re: I guess my point with all that is that you might reconsider problem and what your goal is You are quite right, I just got caught up with the not being able to do something - and now see the error of my ways. But I am glad to see your solution. -- View this message in context: http://r.789695.n4.nabble.com/Importing-arguments-for-use-by-functions-in-a-script-tp2317758p2320383.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Intersecting list vs rows in matrix
Know that if I have List_1 and List_2 that I can check to see if the intersect via the code below: List _1: a, b, c, d, e, f, g List_2: z, y, x, w, v, u, b length(intersect(List_1, List_2)) 0 return = true If instead I wanted to check a dataframe that is a list of lists, how would I do that by record without looping? List _1: a, b, c, d, e, f, g List_2: z, y, x, w, v, u, b y, z, w, v, v, u, m z, y, x, a, b, c . . . return true false true , , , -- View this message in context: http://r.789695.n4.nabble.com/Intersecting-list-vs-rows-in-matrix-tp2320427p2320427.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] regression line of 2 independent variables
If appropriate for your data, you could try errors-in-variables (package leiv) or use the first component from principal components. HTH, -- David -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of array chip Sent: Monday, August 09, 2010 11:34 AM To: r-help@r-project.org Subject: [R] regression line of 2 independent variables Hi, I would like to a draw a scatterplot of x1 and x2 (plot (x1, x2)), and also want to draw a sort of regression line across the data points. But x1 and x2 are just 2 independent variables, so in this case a regression of x1 over x2, or vice versa, is not appropriate per se. What would be an appropriate way to do this? Thanks for any suggestions. John __ R-help@r-project.org 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. This e-mail and any materials attached hereto, including, without limitation, all content hereof and thereof (collectively, XR Content) are confidential and proprietary to XR Trading, LLC (XR) and/or its affiliates, and are protected by intellectual property laws. Without the prior written consent of XR, the XR Content may not (i) be disclosed to any third party or (ii) be reproduced or otherwise used by anyone other than current employees of XR or its affiliates, on behalf of XR or its affiliates. THE XR CONTENT IS PROVIDED AS IS, WITHOUT REPRESENTATIONS OR WARRANTIES OF ANY KIND. TO THE MAXIMUM EXTENT PERMISSIBLE UNDER APPLICABLE LAW, XR HEREBY DISCLAIMS ANY AND ALL WARRANTIES, EXPRESS AND IMPLIED, RELATING TO THE XR CONTENT, AND NEITHER XR NOR ANY OF ITS AFFILIATES SHALL IN ANY EVENT BE LIABLE FOR ANY DAMAGES OF ANY NATURE WHATSOEVER, INCLUDING, BUT NOT LIMITED TO, DIRECT, INDIRECT, CONSEQUENTIAL, SPECIAL AND PUNITIVE DAMAGES, LOSS OF PROFITS AND TRADING LOSSES, RESULTING FROM ANY PERSON'S USE OR RELIANCE UPON, OR INABILITY TO USE, ANY XR CONTENT, EVEN IF XR IS ADVISED OF THE POSSIBILITY OF SUCH DAMAGES OR IF SUCH DAMAGES WERE FORESEEABLE. __ R-help@r-project.org 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 invert a list ?
Hi Carlos, I give a handmade code, hope it helps. y - list() y$a - a y$b - c(b,c) names(y$a) - i names(y$b) - c(j,k) Carlos Petti wrote: a - 5 names(a) - a b - 9 names(b) - b c - 15 names(c) - c x - list(i = a, j = b, j = c) - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-invert-a-list-tp2320108p2320433.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Importing arguments for use by functions in a script
I just got caught up with the not being able to do something I have done that so frequently I think I should include at the beginning of each of my R sessions: When you find the path hard and steep, it's probably because you missed the trail. It always seems the more frustrating the problem, the simpler the solution. Cheers, Josh On Tue, Aug 10, 2010 at 1:11 PM, EvansA annaevan...@hotmail.com wrote: Thank you for your response. Re: I guess my point with all that is that you might reconsider problem and what your goal is You are quite right, I just got caught up with the not being able to do something - and now see the error of my ways. But I am glad to see your solution. -- View this message in context: http://r.789695.n4.nabble.com/Importing-arguments-for-use-by-functions-in-a-script-tp2317758p2320383.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org 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] Intersecting list vs rows in matrix
GL wrote: Know that if I have List_1 and List_2 that I can check to see if the intersect via the code below: List _1: a, b, c, d, e, f, g List_2: z, y, x, w, v, u, b length(intersect(List_1, List_2)) 0 return = true If instead I wanted to check a dataframe that is a list of lists, how would I do that by record without looping? List _1: a, b, c, d, e, f, g List_2: z, y, x, w, v, u, b y, z, w, v, v, u, m z, y, x, a, b, c . . . return true false true *Please* use actual R code and objects to represent your examples, not pseudo-code! I believe you may have some confusion about the difference between lists and vectors in R. A data.frame is a list, but not a list of lists. Since you haven't given a reproducible example, it's hard for me to know exactly what you want. Assume a data.frame df1 composed of all character data, and a character vector cv to match against. apply(df1, 1, function(x) length(intersect(x, cv)) 0) may do what you want. __ R-help@r-project.org 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.
