Re: [R] Calling LAPACK functions directly from R
This was much more appropriate for R-devel, so please move any followup there. On Wed, 11 Apr 2007, William Constantine wrote: I am interested in tapping into LAPACK functionality directly from R. Using R-2.4.1 for Windows, I was able to do so ala: dyn.load(bin/Rlapack.dll) is.loaded(dstebz) # returns TRUE N - 100 NW - 4 n.tapers - 5 tpW - (2 * pi * NW)/N otNmo - 1:N D - as.double(cos(tpW) * ((N - 1 - 2 * (0:(N - 1)))/2)^2) E - as.double((otNmo * (N - otNmo))/2) z - .Fortran(dstebz, I, B, as.integer(N), double(1), double(1), as.integer(N - n.tapers + 1), as.integer(N), double(1), D, E, integer(1), integer(1), double(N), integer(N), integer(N), double(4 * N), integer(3 * N), integer(1))[13:15] I then extended this approach in developing an R package where I added the following to the appropriate .First.lib(): lapack.path - file.path(R.home(), ifelse(.Platform$OS.type == windows, file.path(bin, Rlapack), file.path(lib, libRlapack))) dyn.load(paste(lapack.path,.Platform$dynlib.ext, sep=)) which loads the LAPACK shared objects in lib/libRlapack.so for LINUX/UNIX and bin/Rlapack.dll for Windows. Q1: Is there a better or more robust way of loading LAPACK symbols into R, (e.g., one that is not platform dependent)? Why do you want to do that? All uses of LAPACK in R itself are via small C wrappers that make life a lot easier. - My R package does not currently contain any C or FORTRAN code. However, it has been suggested to me that creating a src/Makevars file containing the line: PKG_LIBS=$(LAPACK_LIBS) $(BLAS_LIBS) $(FLIBS) is a better means of loading LAPACK symbols as it would eliminate the need to use dyn.load() as shown in my .First.lib(). In Windows, however, this suggestion fails and results (for example) in the above code returning a missing dstebz symbol error. My understanding is that one need only create such a src/Makevar if they are interested in tapping into LAPACK functionality from their src/ C or FORTRAN code and so am doubtful of this suggestion. Q2: Given that I have no C/FORTRAN code in my package, am I correct to assume that creating a Makevars files in such a way does not eliminate the need to use dyn.load() as in the above? Q3: For future releases of R, should I expect the path of the LAPACK library to remain as they are noted above in my .First.lib() example? It may not work even now. Buiilds of R do not necessarily contain LAPACK code (you can link to an external library) even on Windows. The portable way to link to LAPACK is to use a C wrapper as described in 'Writing R Extensions'. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looping through series of names
Hallo [EMAIL PROTECTED] napsal dne 11.04.2007 22:45:19: Hi I am very new to R and have not able to find the answer to my question in the manual or any other post. I have a dataset that has various different sites names with data relating to each site. The data is in one dataset. I want to loop through the different site names and perform my test on each site. The sites are named not numbers for example YYC. How do I do this. I hope I was clear enough. Thanks for the help Instead of looping try to look at aggregate, tapply or by. And give a quick look to posting guide where you can find some general rules and hints what to do before posting and how to formulate your question to get helpful answer. Regards Petr [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] taskPR
Hello I tried to load the taskPR-package in R, but it doesn't work at all. Everytime I try /library(taskPR)/, I get the error-message: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so': /home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so: cannot open shared object file: No such file or directory Error in library(taskPR) : .First.lib failed for 'taskPR' Error in library.dynam.unload(taskPR, libpath) : shared library 'taskPR' was not loaded But the problem is, that the file /home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so exists. Furthermore the /library()/-Method shows, that the package is fully installed. Can anyone tell me, what is wrong in the dyn.load-function and how to correct my installation thanks Benno Willemsen __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] FFT output and phase
I'm doing the time-series data analysis using FFT. After fft(), I used spectrum() to calculate the power spectrum, which is used to define the main frequency. Now I want to get the phase information according to this frequency. For example, input 12 time-point data, d-c(1.2950487,0.8547907,0.5636902,0.3928039,0.4564737,0.2296953,-0.1319331,-0.3122781,-0.3773676,-0.5930794,-0.9185491,-1.3633293) fft(d) [1] 0.09596575+0.000i 0.84111080-4.3717147i 1.20570677-2.0131148i [4] 1.86094684-1.7742100i 1.39627707-0.8900564i 1.57888778-0.3601454i [7] 1.67875966+0.000i 1.57888778+0.3601454i 1.39627707+0.8900564i [10] 1.86094684+1.7742100i 1.20570677+2.0131148i 0.84111080+4.3717147i spectrum(fft(d))$freq [1] 0.0833 0.1667 0.2500 0. 0.4167 0.5000 spectrum(fft(d))$spec [1] 0.9772920 1.6402597 0.8764470 0.6785063 0.9855313 0.7042088 here, if i used the g-test to define the main component is the second frequency. I want to calculate the phase of it (Max Phase). Then which complex value in fft output is used for the atan function ? The first and seventh values are real, seems they are DC and Nyquist frequency. Your help should do me a big favor. Thanks, Guang'an Hu [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Assignment from list
I have a list of groups of xy positions I want to set to 0 in an array full of 1s. When the assignments are done directly from the list, they are incorrect, while if I use a temporary array derived from the list the assignments are correct. The following example will hopefully make my problem clearer. The matrices z and zz are initialised with 1. The z and zz values at coordinates in list x are then set to 0 using two different methods. Although either method should lead to the same results, z and zz are different. z is incorrect, because some values in its first row have been incorrectly set to 0. x - list( matrix( c( 9, 9, 74, 75 ), nrow=2 ), matrix( c( 11, 11, 34, 35 ), nrow=2 ), matrix( c( 14, 15, 58, 58 ), nrow=2 ), c( 16, 142 ), matrix( c( 19, 19, 94, 95 ), nrow=2 ), matrix( c( 19, 20, 127, 127 ), nrow=2 ), matrix( c( 22, 22, 112, 113 ), nrow=2 ), c( 23, 13 ), matrix( c( 26, 27, 81, 81 ), nrow=2 ), matrix( c( 31, 32, 153, 153 ), nrow=2 ) ) xx - do.call( rbind, x ) z - matrix( 1, ncol=249, nrow=240 ) zz - matrix( 1, ncol=249, nrow=240 ) for( k in x ) z[k] - 0 zz[xx] - 0 sum( z != zz ) c( sum( z[,1] != 1 ), sum( zz[,1] != 1 ) ) Somewhere, I must be doing something wrong, or assuming something incorrect. I would be very grateful if anybody could point me in the right direction. platform i386-apple-darwin8.8.1 arch i386 os darwin8.8.1 system i386, darwin8.8.1 status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) (same results with platform i386-pc-mingw32, R version 2.4.1 and platform x86_64-unknown-linux-gnu, R version 2.3.1) -- Dr Eric Blanc Lecturer in Bioinformatics MRC Centre for Developmental Neurobiology King's College London New Hunt's House Room 4.10B Guy's Hospital Campus London SE1 1UL E-mail: [EMAIL PROTECTED] Tel: +44 (0)20 7848 6532 Fax: +44 (0)20 7848 6550 [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] taskPR
It would have been helpful to have given the details the R posting guide asks for, such as your OS. That message probably does not mean that /home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so does not exist. It says 'cannot open shared object file', and this can have many causes. One is that there are dependent shared objects which are missing, so start from R CMD ldd /home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so and see if all are resolved. As far as I am aware package taskPR is based on private entry points in a version of R from long ago (it includes a copy of the private header file Defn.h from 2005) and is not really tested on CRAN (it uses --install=fake). On Thu, 12 Apr 2007, Benno Willemsen wrote: Hello I tried to load the taskPR-package in R, but it doesn't work at all. Everytime I try /library(taskPR)/, I get the error-message: Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so': /home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so: cannot open shared object file: No such file or directory Error in library(taskPR) : .First.lib failed for 'taskPR' Error in library.dynam.unload(taskPR, libpath) : shared library 'taskPR' was not loaded But the problem is, that the file /home/bw135690/R-2.4.1/library/taskPR/libs/taskPR.so exists. Furthermore the /library()/-Method shows, that the package is fully installed. Can anyone tell me, what is wrong in the dyn.load-function and how to correct my installation thanks Benno Willemsen __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Tweaking my plot of matrix as image
Greetings list, I have a rectangular 20 x 20 matrix with values in the range of [0 , 1]. I'd like to plot it as an image. To that end, I have used the image() function that seems to do what I want. Now, I just want to tweak it to look perfect. So here is my question: At the moment, the values of the axis range from [0, 1]. I want it to show the row and column of the matrix. How do I do that? Thanks in advance. Wee-Jin __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Assignment from list
try the following:
z - matrix(1, ncol = 249, nrow = 240)
zz - matrix(1, ncol = 249, nrow = 240)
for (k in seq_along(x)) {
z[rbind(x[[k]])] - 0
}
zz[xx] - 0
all.equal(z, zz)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Eric Blanc [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, April 12, 2007 10:36 AM
Subject: [R] Assignment from list
I have a list of groups of xy positions I want to set to 0 in an
array full of 1s. When the assignments are done directly from the
list, they are incorrect, while if I use a temporary array derived
from the list the assignments are correct. The following example
will
hopefully make my problem clearer.
The matrices z and zz are initialised with 1. The z and zz values at
coordinates in list x are then set to 0 using two different methods.
Although either method should lead to the same results, z and zz are
different. z is incorrect, because some values in its first row have
been incorrectly set to 0.
x - list( matrix( c( 9, 9, 74, 75 ), nrow=2 ),
matrix( c( 11, 11, 34, 35 ), nrow=2 ),
matrix( c( 14, 15, 58, 58 ), nrow=2 ),
c( 16, 142 ),
matrix( c( 19, 19, 94, 95 ), nrow=2 ),
matrix( c( 19, 20, 127, 127 ), nrow=2 ),
matrix( c( 22, 22, 112, 113 ), nrow=2 ),
c( 23, 13 ),
matrix( c( 26, 27, 81, 81 ), nrow=2 ),
matrix( c( 31, 32, 153, 153 ), nrow=2 ) )
xx - do.call( rbind, x )
z - matrix( 1, ncol=249, nrow=240 )
zz - matrix( 1, ncol=249, nrow=240 )
for( k in x ) z[k] - 0
zz[xx] - 0
sum( z != zz )
c( sum( z[,1] != 1 ), sum( zz[,1] != 1 ) )
Somewhere, I must be doing something wrong, or assuming something
incorrect. I would be very grateful if anybody could point me in the
right direction.
platform i386-apple-darwin8.8.1
arch i386
os darwin8.8.1
system i386, darwin8.8.1
status
major 2
minor 4.1
year 2006
month 12
day18
svn rev40228
language R
version.string R version 2.4.1 (2006-12-18)
(same results with platform i386-pc-mingw32, R version 2.4.1 and
platform x86_64-unknown-linux-gnu, R version 2.3.1)
--
Dr Eric Blanc
Lecturer in Bioinformatics
MRC Centre for Developmental Neurobiology
King's College London
New Hunt's House Room 4.10B
Guy's Hospital Campus
London SE1 1UL
E-mail: [EMAIL PROTECTED]
Tel: +44 (0)20 7848 6532
Fax: +44 (0)20 7848 6550
[[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.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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and provide commented, minimal, self-contained, reproducible code.
[R] LME: incompatible formulas for groups
Dear R-Users, I am currently working with LME to analyse repeated measures data. I encounter a problem when including both a random effect and a correlation structure with different grouping levels into the LME model. The error message is: Error in lme.formula(diameter ~ flowers*timef + competition*timef + population*timef, : Incompatible formulas for groups in random and correlation. The syntax for the model I want to calculate is as follows: model- lme(diameter ~ flowers*timef + competition*timef + population*timef, data= Timeseries, random= ~flowers|genotype, weights= varIdent(form= ~1|timef), correlation= corAR1(form= ~time|plantind) ) The model works when I specify random effects and correlation at the same level of grouping (either genotype or plantind). But this does not make too much sense in this analysis. The data structure (a subset of the whole data set is listed below): timef is a factor with three levels (time points 1, 2, 3; repeated measures of the same individuals); time is the same variable but in numerical format; genotype is a factor with ca. 300 levels in the whole data set; plantind is a factor with ca. 1200 levels in the whole data set; plantind is nested in genotype; flowers and competition are numerical variables and population is another factor (6 levels). I have read Pinheiro and Bates (2000) and searched the web and R-help archive. The above mentioned error message appears to occur occasionally, but I could not find an answer how to avoid it. Has anyone got an idea what the source of error might be in this case? I would be very thankful for any hints. Regards, Jan Thiele University of Copenhagen Faculty of Life Science Rolighedsvej 21 1958 Frederiksberg C Danmark Subset of the whole data set (csv format): Nr;diameter;flowers;competition;population;genotype;plantind;time;timef 14;1.05;0;-0.95;Baca;Baca13;14;1;1 1289;1.71;5;-1.29;Baca;Baca13;14;2;2 2546;1.16;0;-2.14;Baca;Baca13;14;3;3 21;0.55;0;-0.45;Baca;Baca12;21;1;1 1296;1.71;0;-1.79;Baca;Baca12;21;2;2 2553;-0.44;0;-2.04;Baca;Baca12;21;3;3 35;0.05;0;0.55;Baca;Baca22;35;1;1 1310;0.21;2;-0.29;Baca;Baca22;35;2;2 2567;0.86;0;0.86;Baca;Baca22;35;3;3 90;-0.95;10;1.05;Deru;Deru13;90;1;1 1364;-1.79;6;1.71;Deru;Deru13;90;2;2 106;0.05;1;0.55;Deru;Deru43;106;1;1 1379;-3.29;11;0.71;Deru;Deru43;106;2;2 120;-1.45;1;0.55;Deru;Deru80;120;1;1 1392;-3.79;0;1.71;Deru;Deru80;120;2;2 133;1.55;0;-0.45;Vejo;Vejo9;133;1;1 1405;2.21;0;-0.29;Vejo;Vejo9;133;2;2 2658;2.66;19;-0.34;Vejo;Vejo9;133;3;3 189;0.58;0;-3.42;Baca;Baca50;189;1;1 1461;1.44;0;-4.06;Baca;Baca50;189;2;2 2714;2.38;0;-5.23;Baca;Baca50;189;3;3 203;1.58;0;-0.42;Bile;Bile10;203;1;1 1475;0.94;0;-0.56;Bile;Bile10;203;2;2 2727;0.28;37;-4.43;Bile;Bile10;203;3;3 211;-0.92;0;-0.92;Bile;Bile2;211;1;1 1483;-1.06;0;-0.56;Bile;Bile2;211;2;2 2735;-0.02;23;3.08;Bile;Bile2;211;3;3 218;-0.42;0;0.58;Bile;Bile19;218;1;1 1489;-0.06;0;0.94;Bile;Bile19;218;2;2 2742;-1.53;1;-0.63;Bile;Bile19;218;3;3 221;-0.42;0;-0.42;Bile;Bile38;221;1;1 1492;0.44;0;0.44;Bile;Bile38;221;2;2 2745;-2.43;0;0.68;Bile;Bile38;221;3;3 240;-0.42;0;0.58;Bile;Bile43;240;1;1 1510;-0.56;0;-0.06;Bile;Bile43;240;2;2 2763;-3.43;2;-0.02;Bile;Bile43;240;3;3 261;0.08;0;2.08;Deru;Deru35;261;1;1 1531;-1.06;0;2.94;Deru;Deru35;261;2;2 2784;-5.03;0;-0.82;Deru;Deru35;261;3;3 264;-0.42;24;0.08;Deru;Deru25;264;1;1 1534;-3.06;3;0.44;Deru;Deru25;264;2;2 2787;-5.23;0;-0.02;Deru;Deru25;264;3;3 272;-2.42;1;0.58;Deru;Deru22;272;1;1 1542;-4.06;0;1.44;Deru;Deru22;272;2;2 2795;-5.23;0;2.78;Deru;Deru22;272;3;3 293;0.08;0;0.58;Vejo;Vejo9;293;1;1 1563;-0.56;0;1.94;Vejo;Vejo9;293;2;2 2815;-1.33;2;4.38;Vejo;Vejo9;293;3;3 308;-3.42;0;0.58;Vejo;Vejo54;308;1;1 1578;-4.06;0;2.44;Vejo;Vejo54;308;2;2 2829;-5.23;0;3.68;Vejo;Vejo54;308;3;3 313;-0.92;10;0.58;Vejo;Vejo27;313;1;1 1583;-1.06;8;1.44;Vejo;Vejo27;313;2;2 2834;-0.82;3;5.28;Vejo;Vejo27;313;3;3 338;-2.04;0;0.46;Baca;Baca27;338;1;1 1608;-3.73;0;1.27;Baca;Baca27;338;2;2 2858;-3.48;0;6.52;Baca;Baca27;338;3;3 340;-0.54;19;0.46;Baca;Baca33;340;1;1 1610;-0.23;8;-0.23;Baca;Baca33;340;2;2 2860;1.52;41;-0.48;Baca;Baca33;340;3;3 359;-0.04;0;2.96;Baca;Baca39;359;1;1 1628;-0.23;11;2.77;Baca;Baca39;359;2;2 2878;-1.48;4;2.52;Baca;Baca39;359;3;3 445;1.46;0;-0.04;Vejo;Vejo8;445;1;1 1711;1.27;2;-0.23;Vejo;Vejo8;445;2;2 2963;0.72;0;-0.98;Vejo;Vejo8;445;3;3 449;-1.04;5;0.96;Vejo;Vejo17;449;1;1 1715;-1.23;17;0.77;Vejo;Vejo17;449;2;2 2967;-2.38;10;-0.28;Vejo;Vejo17;449;3;3 476;-0.54;0;-0.04;Vejo;Vejo25;476;1;1 1742;-0.73;3;-0.23;Vejo;Vejo25;476;2;2 2994;-2.28;9;0.02;Vejo;Vejo25;476;3;3 480;-0.54;0;1.46;Vejo;Vejo21;480;1;1 1746;-0.23;1;1.77;Vejo;Vejo21;480;2;2 2998;-2.28;0;-0.48;Vejo;Vejo21;480;3;3 490;0.3;34;0.3;Baca;Baca10;490;1;1 1756;1.25;5;1.25;Baca;Baca10;490;2;2 3008;3.64;10;0.14;Baca;Baca10;490;3;3 505;-0.2;23;0.3;Baca;Baca32;505;1;1 1771;-0.25;23;0.25;Baca;Baca32;505;2;2 3022;-1.26;16;0.04;Baca;Baca32;505;3;3 518;0.8;29;0.3;Baca;Baca28;518;1;1
Re: [R] Tweaking my plot of matrix as image
Hi Did you by chance look at the help page of image? If you did, you could read x,y locations of grid lines at which the values in z are measured. These must be finite, non-missing and in (strictly) ascending order. By default, equally spaced values from 0 to 1 are used. If x is a list, its components x$x and x$y are used for x and y, respectively. If the list has component z this is used for z. z a matrix containing the values to be plotted (NAs are allowed). Note that x can be used instead of z for convenience. and therefore image(1:20, 1:20, your.matrix) shall do what you probably want. And if you have different matrix size then 1:dim(your.matrix)[1] can give you suitable sequence. Regards Petr Pikal [EMAIL PROTECTED] [EMAIL PROTECTED] napsal dne 12.04.2007 10:55:21: Greetings list, I have a rectangular 20 x 20 matrix with values in the range of [0 , 1]. I'd like to plot it as an image. To that end, I have used the image() function that seems to do what I want. Now, I just want to tweak it to look perfect. So here is my question: At the moment, the values of the axis range from [0, 1]. I want it to show the row and column of the matrix. How do I do that? Thanks in advance. Wee-Jin __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reasons to Use R
Charilaos Skiadas wrote: A new fortune candidate perhaps? On Apr 10, 2007, at 6:27 PM, Greg Snow wrote: Remember, everything is better than everything else given the right comparison. Only if we remove the grammatical blip that turns it into an infinite regress, i.e. Remember, anything is better than everything else given the right comparison Jim __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Tweaking my plot of matrix as image
Wee-Jin Goh wrote: Greetings list, I have a rectangular 20 x 20 matrix with values in the range of [0 , 1]. I'd like to plot it as an image. To that end, I have used the image() function that seems to do what I want. Now, I just want to tweak it to look perfect. So here is my question: At the moment, the values of the axis range from [0, 1]. I want it to show the row and column of the matrix. How do I do that? Perhaps color2D.matplot will suit you. It automatically shows the row and column indices on the plot in the same order as conventional matrix displays. Jim __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] reporting multinomial logistic regression results
I am hoping to obtain information regarding the reporting of multinomial logistic regression results - either recommendations of published studies employing R or suggestions about what output should be reported. I am used to SPSS which routinely produces a broader range of output than R (e.g wald statistic p values, exp(B) and associated confidence intervals). Below is the code I have been using for a series of analyses. mod.multacute - multinom(offence ~ in.acute.danger * violent.convictions, data = kc, na.action = na.omit) summary(mod.multacute, cor=F, Wald=T) Anova (mod.multacute) predictors - expand.grid(in.acute.danger = c(y,n), violent.convictions = c(y,n)) p.fita - predict(mod.multacute, predictors, type='probs') Any assistance is appreciated, regards Bob Green __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] using install.packages via proxy
Hello everybody, I'm trying to install rJava to use JRI in a Java program as JRI is a part of rJava. There should be the option to install this package via install.packages(rJava). Unfortunately I'm connected to the internet via a proxy. I have no idea how to tell R to connect via the proxy. The R-help tells me something about environment variables like http_proxy. But I have no idea how and where to set these variable as I'm new to R. I don't know whether this is the right way to solve my problem. Is there anyone out there who knows how to solve this problem? Any help is appreciated. Thank in advance Nicole Erbe Production Engineering Verigy Germany GmbH Herrenberger Str. 130 71034 Böblingen Tel: +49 7031 4357327 Mail: [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regression on two time series
Ron, you're right. It's not legitimate at all. I suggest you to take a look at the HUGE bibliography on cointegration, as a start up. Rogerio Dear all R user, Please forgive me if my question is too simple. My question is related to Statistics rather directly to R. Suppose I have two time series of spot prices of two commodities X and Y for two years. Now I want to see what percentage of spot price of X is explained by Y. Yes I can fit a regression equation of X on Y. But my question is, is it legitimate? Because both series are non-stationary with very high auto-correlation of order 1. And regression analysis is basically designed for cross-sectional data, not for time series data. Your help will be highly appreciable. Thanks and regards, Ron __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Positioning in xyplot
Wow! Thanks to Sundar and to Deepayan for this selfless effort.(I
rather suspect they had a good time figuring this out.) The outcome
is that I will be able to publish just the right figure for the job.
I will be sure to aknowledge your contribtution.
Michael
On Apr 12, 2007, at 12:48 AM, Deepayan Sarkar wrote:
On 4/11/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote:
Hi, Deepayan,
See the attached image for what your code produced. Not sure if
this is
what you intended.
Here's the correct version of callAfterMoving (I thought I had fixed
it, but I guess I pasted the wrong thing):
## this calls 'fun' after moving its viewport if panel.number() == 5
callAfterMoving -
function(fun, border = TRUE, move.x = 1, ...)
{
if (panel.number() != 5) { ## was == 5
fun(...)
if (border) grid.rect()
}
else {
cpl - current.limits()
pushViewport(viewport(x = move.x,
width = unit(1, npc),
xscale = cpl$xlim,
yscale = cpl$ylim,
clip = off))
fun(...)
if (border) grid.rect()
upViewport()
}
}
-Deepayan
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Re: [R] using install.packages via proxy
What operating system do you use? If you use Windows, than open the Control Panel, double click on the System icon, go to Advanced tab, press Environment variables button, then press New to create one more new variable, enter http_proxy as the Name of variable, and http://address.of.your.proxy:port; as its value. In unix you should consult with your shell manual for the proper command and proper startup script file (in bash, this would be export and ~/.bash_profile) You also could use Sys.putenv R function Erbe, Nicole wrote: I'm trying to install rJava to use JRI in a Java program as JRI is a part of rJava. There should be the option to install this package via install.packages(rJava). Unfortunately I'm connected to the internet via a proxy. I have no idea how to tell R to connect via the proxy. The R-help tells me something about environment variables like http_proxy. But I have no idea how and where to set these variable as I'm new to R. I don't know whether this is the right way to solve my problem. Is there anyone out there who knows how to solve this problem? Any help is appreciated. -- View this message in context: http://www.nabble.com/using-install.packages-via-proxy-tf3564991.html#a9958606 Sent from the R help mailing list archive at Nabble.com. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data file import - numbers and letters in a matrix(!)
Hello, I have a problem with the import of a date file. I seems verry tricky. I have a text file (end of the mail). Every file has a different number of measurments witch start with START OF HEIGHT DATA and ende with END OF HEIGHT DATA. I imported the file in a matrix but the letters before the numbers are my problem (S= ,S=,x=,y=). Because through the letters and the space after S= I got a different number of columns in my matrix and with letters in my matrix I can't count. My question. Is it possible to import the file to got 3 columns only with numbers and no letters like x=, y=? Thank's a lot Felix My R Code: -- # na.strings = S= Measure1 - matrix(scan(data.dat, n= 5063 * 4, skip = 20, what = character() ), 5063, 3, byrow = TRUE) Measure2 - matrix(scan(data.dat, n= 5063 * 4, skip = 5220, what = character() ), 5063, 3, byrow = TRUE) My data file: --- FILEDATE:02.02.2007 ... START OF HEIGHT DATA S= 0 y=0.0 x=0. S= 0 y=0.1 x=0.00055643 ... S= 9 y=4.9 x=1.67278117 S= 9 y=5.0 x=1.74873257 S=10 y=0.0 x=0. S=10 y=0.1 x=0.00075557 ... S=99 y=5.3 x=1.94719490 END OF HEIGHT DATA ... START OF HEIGHT DATA S= 0 y=0.0 x=0. S= 0 y=0.1 x=0.00055643 The imported matrix: [,1] [,2] [,3] [,4] [6,] S= 9y=4.9x=1.67278117 [7,] S= 9y=5.0x=1.74873257 [8,] S=10 y=0.0x=0. S=10 [9,] y=0.1x=0.00075557 S=10 y=0.2 [10,] x=0.00277444 S=10 y=0.3x=0.00605958 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using install.packages via proxy
I had that problme and ended up downloading the package and installing
locally.
there is probably a better way
Richard
Erbe, Nicole wrote:
Hello everybody,
I'm trying to install rJava to use JRI in a Java program as JRI is a part of
rJava. There should be the option to install this package via
install.packages(rJava). Unfortunately I'm connected to the internet via a
proxy. I have no idea how to tell R to connect via the proxy. The R-help tells
me something about environment variables like http_proxy. But I have no idea
how and where to set these variable as I'm new to R. I don't know whether this
is the right way to solve my problem.
Is there anyone out there who knows how to solve this problem? Any help is
appreciated.
Thank in advance
Nicole Erbe
Production Engineering
Verigy Germany GmbH
Herrenberger Str. 130
71034 Böblingen
Tel: +49 7031 4357327
Mail: [EMAIL PROTECTED]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Richard Gott
Professor of Science Education
School of Education
Durham University
[EMAIL PROTECTED]
0191 3348 354
DISCLAIMER:\ \ This e-mail is intended solely for the addres...{{dropped}}
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Re: [R] using install.packages via proxy
Hi, I'm (currently forced to) use Win XP. I actually found an easy way to solve the problem of installing the rJava package. To avoid the problems with the proxy I just downloaded the zip file and installed the package in R from a local source. Now I can use the JRI libraries in eclipse by adding the jar to the path. It works fine. Using the libraries is now another issue...I will see how well it works and if the results are what I need. Thanks a lot Nicole Erbe Production Engineering Verigy Germany GmbH Herrenberger Str. 130 71034 Böblingen Tel: +49 7031 4357327 Mail: [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Vladimir Eremeev Sent: Donnerstag, 12. April 2007 14:41 To: [EMAIL PROTECTED] Subject: Re: [R] using install.packages via proxy What operating system do you use? If you use Windows, than open the Control Panel, double click on the System icon, go to Advanced tab, press Environment variables button, then press New to create one more new variable, enter http_proxy as the Name of variable, and http://address.of.your.proxy:port; as its value. In unix you should consult with your shell manual for the proper command and proper startup script file (in bash, this would be export and ~/.bash_profile) You also could use Sys.putenv R function Erbe, Nicole wrote: I'm trying to install rJava to use JRI in a Java program as JRI is a part of rJava. There should be the option to install this package via install.packages(rJava). Unfortunately I'm connected to the internet via a proxy. I have no idea how to tell R to connect via the proxy. The R-help tells me something about environment variables like http_proxy. But I have no idea how and where to set these variable as I'm new to R. I don't know whether this is the right way to solve my problem. Is there anyone out there who knows how to solve this problem? Any help is appreciated. -- View this message in context: http://www.nabble.com/using-install.packages-via-proxy-tf3564991.html#a9958606 Sent from the R help mailing list archive at Nabble.com. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using install.packages via proxy
On Thu, 12 Apr 2007, Vladimir Eremeev wrote: What operating system do you use? If you use Windows, than open the Control Panel, double click on the System icon, go to Advanced tab, press Environment variables button, then press New to create one more new variable, enter http_proxy as the Name of variable, and http://address.of.your.proxy:port; as its value. For Windows, the rw-FAQ has two much more comprehensive (and accurate) answers with examples! [Please don't post partial answers to FAQs.] In unix you should consult with your shell manual for the proper command and proper startup script file (in bash, this would be export and ~/.bash_profile) You also could use Sys.putenv R function Or the command line or .Renviron (preferred, I think). Please do not set http_proxy globally: other programs may read it and have a different format. Erbe, Nicole wrote: I'm trying to install rJava to use JRI in a Java program as JRI is a part of rJava. There should be the option to install this package via install.packages(rJava). Unfortunately I'm connected to the internet via a proxy. I have no idea how to tell R to connect via the proxy. The R-help tells me something about environment variables like http_proxy. But I have no idea how and where to set these variable as I'm new to R. I don't know whether this is the right way to solve my problem. Is there anyone out there who knows how to solve this problem? Any help is appreciated. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reasons to Use R
Douglas Bates writes:
One
can do data analysis by using the computer as a blunt instrument with
which to bludgeon the problem to death but one can't do elegant data
analysis like that.
One nice thing about a blunt instrument like Stata is the ability to
hold an entire dataset in memory and interactively play with the model
and generate new variables all in one session. I figure out what I
want interactively and then separate the data management and analysis in
.do-files, then run them in batch mode.
However, when I first read of the approach of using Perl, sed or awk
to manage data and then only doing the analysis in R, I immediately
thought Wow, that is a really great idea, I never thought of it like
that before. It would really get me to think about the modelling and
the data management clearly. A little voice said Dude, you're not
using a PDP-11...(oh wait, that might be kinda cool) but the logic of
it immediately made sense. I consider it a big part of my
re-Unix-ization.
Joel
--
Joel J. Adamson
Biostatistician
Pediatric Psychopharmacology Research Unit
Massachusetts General Hospital
Boston, MA 02114
(617) 643-1432
(303) 880-3109
The information transmitted in this electronic communication...{{dropped}}
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Re: [R] Reasons to Use R
A re-interpretation of Zorn's lemma? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jim Lemon Sent: Thursday, April 12, 2007 5:14 AM To: [EMAIL PROTECTED] Subject: Re: [R] Reasons to Use R Charilaos Skiadas wrote: A new fortune candidate perhaps? On Apr 10, 2007, at 6:27 PM, Greg Snow wrote: Remember, everything is better than everything else given the right comparison. Only if we remove the grammatical blip that turns it into an infinite regress, i.e. Remember, anything is better than everything else given the right comparison Jim __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pattern
Dear R-experts,
I have the following function:
userInput - function() {
ANSWER - readline(saving place of the data (example
Z:/Software/test.mat)? )
x = c(.mat)
endingTest = x %in% ANSWER
print (endingTest)
if (endingTest == ??)
cat (saving place not accepted\n)
else
cat(Thank you! The current workspace will be stored in:
,ANSWER,\n\n)
print(ANSWER )
}
filename = userInput()
Before I enter the if loop I must test if the text stored in ANSWER has
the pattern .mat. If yes than endingTest = TRUE else endingTest =
FALSE.
Another problem is the last codeline. Later on in my program I need the
userinput to ad it to another variable. How can I manage this. I get the
following error message:
saving place of the data (example Z:/Software/test.mat)? Z:/data.mat
[1] FALSE
Thank you! The current workspace will be stored in: Z:/data.mat
[1] Z:/data.mat
Error in as.vector(x, mode) : cannot change into vector
Thanks, Corinna
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Re: [R] graphs superimposed on pictures?
At 14:39 11/04/2007, Robert Biddle wrote: Hi: I am doing some work that involves plotting points of interest superimposed on photographs and maps. I can produce the plots fine in R, but so far I have had to do the superimposition externally, which makes it tedious to do exploratory work. I have looked to see if there is some capability to put a background picture on a plot window, but I have not found anything. Advice, anyone? Although my situation was not exactly the same as yours you may find http://finzi.psych.upenn.edu/R/Rhelp02a/archive/42884.html a help Cheers Robert Biddle Michael Dewey http://www.aghmed.fsnet.co.uk __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Method dispatch for print() in package its
Dear all,
in the package its the print() method does not seem to correctly work in all
circumstances:
selectMethod(print, its)
Method Definition:
function (x, ...)
{
print([EMAIL PROTECTED] mailto:[EMAIL PROTECTED] , ...)
}
environment: namespace:its
Signatures:
x
target its
defined its
fundPME.lst[[1]]$irr
An object of class its
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
Slot dates:
[1] 2005-10-31 Westeuropäische Normalzeit
[EMAIL PROTECTED]
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
print(fundPME.lst[[1]]$irr)
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
So, is it necessary to define a print.its S3 method, or what should we do?
Many thanks and best regards,
Stefan
Dr. Stefan Albrecht, CFA
Allianz Private Equity Partners GmbH
Königinstr. 19 | 80539 Munich | Germany
Phone: +49.(0)89.3800.18317
Fax: +49.(0)89.3800.818317
EMail: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
Web: www.apep.com http://www.apep.com/
Allianz Private Equity Partners GmbH | Geschäftsführung: Wan Ching Ang, Karl
Ralf Jung
Sitz der Gesellschaft: München | Registergericht: München HRB 126221 |
Ust-ID-Nr.: DE 813 264 786
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Re: [R] data file import - numbers and letters in a matrix(!)
Try pasting this into an R session: Lines.raw - FILEDATE:02.02.2007 ... START OF HEIGHT DATA S= 0 y=0.0 x=0. S= 0 y=0.1 x=0.00055643 ... S= 9 y=4.9 x=1.67278117 S= 9 y=5.0 x=1.74873257 S=10 y=0.0 x=0. S=10 y=0.1 x=0.00075557 ... S=99 y=5.3 x=1.94719490 END OF HEIGHT DATA ... START OF HEIGHT DATA S= 0 y=0.0 x=0. S= 0 y=0.1 x=0.00055643 # next line would be replaced by # somthing like: Lines - readLines(myfile.dat) Lines - readLines(textConnection(Lines.raw)) # extract those lines that contain an = Lines - grep(=, Lines, value = TRUE) # get col names by removing all but letters spaces from line 1 cn - gsub([^a-zA-Z ], , Lines[1]) cn - scan(textConnection(cn), what = ) # remove anything that is not a number, dot or space and read in Lines - gsub([^ .0-9], , Lines) DF - read.table(textConnection(Lines), col.names = cn) closeAllConnections() DF On 4/12/07, Felix Wave [EMAIL PROTECTED] wrote: Hello, I have a problem with the import of a date file. I seems verry tricky. I have a text file (end of the mail). Every file has a different number of measurments witch start with START OF HEIGHT DATA and ende with END OF HEIGHT DATA. I imported the file in a matrix but the letters before the numbers are my problem (S= ,S=,x=,y=). Because through the letters and the space after S= I got a different number of columns in my matrix and with letters in my matrix I can't count. My question. Is it possible to import the file to got 3 columns only with numbers and no letters like x=, y=? Thank's a lot Felix My R Code: -- # na.strings = S= Measure1 - matrix(scan(data.dat, n= 5063 * 4, skip = 20, what = character() ), 5063, 3, byrow = TRUE) Measure2 - matrix(scan(data.dat, n= 5063 * 4, skip = 5220, what = character() ), 5063, 3, byrow = TRUE) My data file: --- FILEDATE:02.02.2007 ... START OF HEIGHT DATA S= 0 y=0.0 x=0. S= 0 y=0.1 x=0.00055643 ... S= 9 y=4.9 x=1.67278117 S= 9 y=5.0 x=1.74873257 S=10 y=0.0 x=0. S=10 y=0.1 x=0.00075557 ... S=99 y=5.3 x=1.94719490 END OF HEIGHT DATA ... START OF HEIGHT DATA S= 0 y=0.0 x=0. S= 0 y=0.1 x=0.00055643 The imported matrix: [,1] [,2] [,3] [,4] [6,] S= 9y=4.9x=1.67278117 [7,] S= 9y=5.0x=1.74873257 [8,] S=10 y=0.0x=0. S=10 [9,] y=0.1x=0.00075557 S=10 y=0.2 [10,] x=0.00277444 S=10 y=0.3x=0.00605958 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Method dispatch for print() in package its
Without a reproducible example I can't tell what the problem is
but as a workaround you could print it as a zoo object:
library(its)
library(zoo)
# create an its object
example(its)
ii - its(mat, times)
# print it without using its' print
print(as.zoo(ii))
# or from the command line just
as.zoo(ii)
On 4/12/07, Albrecht, Dr. Stefan (AZ Private Equity Partner)
[EMAIL PROTECTED] wrote:
Dear all,
in the package its the print() method does not seem to correctly work in all
circumstances:
selectMethod(print, its)
Method Definition:
function (x, ...)
{
print([EMAIL PROTECTED] mailto:[EMAIL PROTECTED] , ...)
}
environment: namespace:its
Signatures:
x
target its
defined its
fundPME.lst[[1]]$irr
An object of class its
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
Slot dates:
[1] 2005-10-31 Westeuropäische Normalzeit
[EMAIL PROTECTED]
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
print(fundPME.lst[[1]]$irr)
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
So, is it necessary to define a print.its S3 method, or what should we do?
Many thanks and best regards,
Stefan
Dr. Stefan Albrecht, CFA
Allianz Private Equity Partners GmbH
Königinstr. 19 | 80539 Munich | Germany
Phone: +49.(0)89.3800.18317
Fax: +49.(0)89.3800.818317
EMail: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
Web: www.apep.com http://www.apep.com/
Allianz Private Equity Partners GmbH | Geschäftsführung: Wan Ching Ang, Karl
Ralf Jung
Sitz der Gesellschaft: München | Registergericht: München HRB 126221 |
Ust-ID-Nr.: DE 813 264 786
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Re: [R] Method dispatch for print() in package its
Albrecht, Dr. Stefan (AZ Private Equity Partner) [EMAIL PROTECTED] writes: Dear all, in the package its the print() method does not seem to correctly work in all circumstances: My understanding is that one should define only a method for the show generic for S4 classes and leave print alone. Not sure if that helps you. + seth -- Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center http://bioconductor.org __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reasons to Use R
Lucke, Joseph F writes:
A re-interpretation of Zorn's lemma?
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Jim Lemon
Sent: Thursday, April 12, 2007 5:14 AM
To: [EMAIL PROTECTED]
Subject: Re: [R] Reasons to Use R
Charilaos Skiadas wrote:
A new fortune candidate perhaps?
On Apr 10, 2007, at 6:27 PM, Greg Snow wrote:
Remember, everything is better than everything else given the right
comparison.
Only if we remove the grammatical blip that turns it into an infinite
regress, i.e.
Remember, anything is better than everything else given the right
comparison
Jim
Anything is potentially better than any other thing given the right
comparison.
Joel
--
Joel J. Adamson
Biostatistician
Pediatric Psychopharmacology Research Unit
Massachusetts General Hospital
Boston, MA 02114
(617) 643-1432
(303) 880-3109
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Re: [R] Memory management
Okay thanks, I'm going through the docs now.. and I came through this..
The named field is set and accessed by the SET_NAMED and NAMED macros, and
take values 0, 1 and 2. R has a `call by value' illusion, so an assignment
like
b - a
appears to make a copy of a and refer to it as b. However, if neither a nor
b are subsequently altered there is no need to copy. What really happens is
that a new symbol b is bound to the same value as a and the named field on
the value object is set (in this case to 2). When an object is about to be
altered, the named field is consulted. A value of 2 means that the object
must be duplicated before being changed.
What does it mean the new symbol b is bound to the same value as a.
Does it mean b has a pointer pointing to a?
Thanks!!
- yooo
yoo wrote:
I guess I have more reading to do Are there any website that I can
read up on memory management, or specifically what happen when we 'pass
in' variables, which strategy is better at which situation?
Thanks~
- y
Prof Brian Ripley wrote:
On Tue, 10 Apr 2007, yoo wrote:
Hi all, I'm just curious how memory management works in R... I need to
run an
optimization that keeps calling the same function with a large set of
parameters... so then I start to wonder if it's better if I attach the
variables first vs passing them in (coz that involves a lot of copying..
)
Your paranethetical comment is wrong: no copying is needed to 'pass in' a
variable.
Thus, I do this
fn3 - function(x, y, z, a, b, c){ sum(x, y, z, a, b, c) }
fn4 - function(){ sum(x, y, z, a, b, c) }
rdn - rep(1.1, times=1e8)
r - proc.time()
for (i in 1:5)
fn3(rdn, rdn, rdn, rdn, rdn, rdn)
time1 - proc.time() - r
print(time1)
lt - list(x = rdn, y = rdn, z = rdn, a = rdn, b = rdn, c = rdn)
attach(lt)
r - proc.time()
for (i in 1:5)
fn4()
time2 - proc.time() - r
print(time2)
detach(lt)
The output is
[1] 25.691 0.003 25.735 0.000 0.000
[1] 25.822 0.005 25.860 0.000 0.000
Turns out attaching takes longer to run.. which is counter intuitive
(unless
the search to the pos=2 envir takes long time as well) Do you guys know
why
this is the case?
I would not trust timing differences of that nature: they often depend on
the state of the system, and in particular of the garbage collector.
You should be using system.time() for that reason: it calls the garbage
collector immediately before timing.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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View this message in context:
http://www.nabble.com/Memory-management-tf3556238.html#a9961010
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Method dispatch for print() in package its
Setting S4 methods for print() works correctly, as in your examples.
Your first example does not call print(), so it is not suprising that no
method dispatch takes place.
You seem to suppose that such a method should be used for top-level
auto-printing, and it will not be as your example shows. Auto-printing
calls show() on S4 objects, and in this case show() is handling the object
by its default method.
Note that if there is no S4 print method set, print(x) on an S4 object 'x'
will in fact call show(x). So had an S4 show() method been set, all three
examples would have given the same output (via show()).
On Thu, 12 Apr 2007, Albrecht, Dr. Stefan (AZ Private Equity Partner) wrote:
Dear all,
in the package its the print() method does not seem to correctly work in all
circumstances:
selectMethod(print, its)
Method Definition:
function (x, ...)
{
print([EMAIL PROTECTED] mailto:[EMAIL PROTECTED] , ...)
(sic!)
}
environment: namespace:its
Signatures:
x
target its
defined its
fundPME.lst[[1]]$irr
An object of class its
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
Slot dates:
[1] 2005-10-31 Westeuropäische Normalzeit
[EMAIL PROTECTED]
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
print(fundPME.lst[[1]]$irr)
IRR HSBC MEEM
2005-10-31 0.1926175 0.07802736
So, is it necessary to define a print.its S3 method, or what should we do?
Many thanks and best regards,
Stefan
Dr. Stefan Albrecht, CFA
Allianz Private Equity Partners GmbH
Königinstr. 19 | 80539 Munich | Germany
Phone: +49.(0)89.3800.18317
Fax: +49.(0)89.3800.818317
EMail: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
Web: www.apep.com http://www.apep.com/
Allianz Private Equity Partners GmbH | Geschäftsführung: Wan Ching Ang, Karl
Ralf Jung
Sitz der Gesellschaft: München | Registergericht: München HRB 126221 |
Ust-ID-Nr.: DE 813 264 786
[[alternative HTML version deleted]]
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
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[R] How to manipulate the pointer of a file?
Dear friends. With file( ) to obtain a pointer of a file, every time we use scan ( ) to read one row of it, the pointer will point to the next row of the file. In the following example, d1 and d2 are obtained the same way but they correspond to different rows of the same file because the pointer of the file moves down a row when a row of the file is read. The following is an example: a1 - list(name=Fred, wife=Mary, no.children=3) a2 - list(name=Tom, wife=Joy, no.children=9) a3 - list(name=Paul, wife=Alic, no.children=5) write.table(a1, file = tt.csv, sep=',',row.names=FALSE,col.name=TRUE) write.table(a2, file=tt.csv, sep=',', append=TRUE, row.names=FALSE, col.names=FALSE) write.table(a3, file=tt.csv, sep=',', append=TRUE, row.names=FALSE, col.names=FALSE) fp=file(tt.csv,r) c=scan(file=fp, sep=',', what=list(c1=, c2=, c3=), flush=TRUE, nlines=1) d1=scan(file=fp, sep=',', what=list(name=, wife=, no.kids=0), flush=TRUE, nlines=1) d1 R output: --- $name [1] Fred $wife [1] Mary $no.kids [1] 3 - d2=scan(file=fp, sep=',', what=list(name=, wife=, no.kids=0), flush=TRUE, nlines=1) d2 R Output: $name [1] Tom $wife [1] Joy $no.kids [1] 9 -- My question is, how to manipulate the pointer of the file further? For example, what if I need the pointer to go back to the previous row? Best Wishes Yuchen Luo [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data file import - numbers and letters in a matrix(!)
Here is the contents of my testdata.txt :
-
START OF HEIGHT DATA
S= 0y=0.0 x=0.
S= 0 y=0.1 x=0.00055643
S= 9 y=4.9 x=1.67278117
S= 9 y=5.0 x=1.74873257
S=10 y=0.0 x=0.
S=10y=0.1 x=0.00075557
S=99 y=5.3x=1.94719490
END OF HEIGHT DATA
-
If you have access to a shell command, you can try changing the input
file for read.delim using
cat testdata.txt | grep -v ^START | grep -v ^END | sed 's/ //g' |
sed 's/S=//' | sed 's/y=/\t/' | sed 's/x=/\t/'
or here is my ugly fix in R
my.read.file - function(file=file){
v1 - readLines( con=file, n=-1)
v2 - v1[ - grep( ^START|^END, v1 ) ]
v3 - gsub( , , v2)
v4 - gsub( S=|y=|x=, , v3 )
v5 - gsub(^ , , v4)
m - t( sapply( strsplit(v5, split= ), as.numeric ) )
colnames(m) - c(S, y, x )
return(m)
}
my.read.file( testdata.txt )
Regards, Adai
Felix Wave wrote:
Hello,
I have a problem with the import of a date file. I seems verry tricky.
I have a text file (end of the mail). Every file has a different number of
measurments
witch start with START OF HEIGHT DATA and ende with END OF HEIGHT DATA.
I imported the file in a matrix but the letters before the numbers are my
problem
(S= ,S=,x=,y=).
Because through the letters and the space after S= I got a different number
of columns in my matrix and with letters in my matrix I can't count.
My question. Is it possible to import the file to got 3 columns only with
numbers and
no letters like x=, y=?
Thank's a lot
Felix
My R Code:
--
# na.strings = S=
Measure1 - matrix(scan(data.dat, n= 5063 * 4, skip = 20, what =
character() ), 5063, 3, byrow = TRUE)
Measure2 - matrix(scan(data.dat, n= 5063 * 4, skip = 5220, what =
character() ), 5063, 3, byrow = TRUE)
My data file:
---
FILEDATE:02.02.2007
...
START OF HEIGHT DATA
S= 0 y=0.0 x=0.
S= 0 y=0.1 x=0.00055643
...
S= 9 y=4.9 x=1.67278117
S= 9 y=5.0 x=1.74873257
S=10 y=0.0 x=0.
S=10 y=0.1 x=0.00075557
...
S=99 y=5.3 x=1.94719490
END OF HEIGHT DATA
...
START OF HEIGHT DATA
S= 0 y=0.0 x=0.
S= 0 y=0.1 x=0.00055643
The imported matrix:
[,1] [,2] [,3] [,4]
[6,] S= 9y=4.9x=1.67278117
[7,] S= 9y=5.0x=1.74873257
[8,] S=10 y=0.0x=0. S=10
[9,] y=0.1x=0.00075557 S=10 y=0.2
[10,] x=0.00277444 S=10 y=0.3x=0.00605958
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Re: [R] how to reverse a list
Thanks all of you!
But my next question is, how to improve R programming skills? I never
have time in improving it but I feel I need to.
Regards,
W
On 4/11/07, Seth Falcon [EMAIL PROTECTED] wrote:
Weiwei Shi [EMAIL PROTECTED] writes:
I forgot to add my bad solution here:
reverseList - function(xlist){
blist - xlist[!is.na(xlist)]
x0 - unlist(blist)
l0 - length(blist)
d0 - as.data.frame(matrix(0, l0, 3))
d0[,1] - names(x0)
d0[,2] - x0
There is a helper function in Biobase that does this:
reverseSplit
reverseSplit
function (inList)
{
lens = sapply(inList, length)
nms = rep(names(inList), lens)
vals = unlist(inList)
split(nms, vals)
}
environment: namespace:Biobase
+ seth
--
Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center
http://bioconductor.org
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--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
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Re: [R] pattern
Schmitt, Corinna wrote:
Dear R-experts,
I have the following function:
userInput - function() {
ANSWER - readline(saving place of the data (example
Z:/Software/test.mat)? )
x = c(.mat)
endingTest = x %in% ANSWER
print (endingTest)
if (endingTest == ??)
cat (saving place not accepted\n)
else
cat(Thank you! The current workspace will be stored in:
,ANSWER,\n\n)
print(ANSWER )
}
filename = userInput()
Example:
userInput - function(){
ANSWER - readline(saving place of the data (example
Z:/Software/test.mat)?)
endingTest - length(grep(\\.mat$, ANSWER))
if(!endingTest)
cat(saving place not accepted\n)
else
cat(Thank you! The current workspace will be stored in:,
ANSWER, \n\n)
}
filename - userInput()
Uwe Ligges
Before I enter the if loop I must test if the text stored in ANSWER has
the pattern .mat. If yes than endingTest = TRUE else endingTest =
FALSE.
Another problem is the last codeline. Later on in my program I need the
userinput to ad it to another variable. How can I manage this. I get the
following error message:
saving place of the data (example Z:/Software/test.mat)? Z:/data.mat
[1] FALSE
Thank you! The current workspace will be stored in: Z:/data.mat
[1] Z:/data.mat
Error in as.vector(x, mode) : cannot change into vector
Thanks, Corinna
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[R] zoo merge() method
R users: I'd like to get some insight on an error I encounter when attempting to work with two moderately sized sets of time series data. FYI - I'm using the following versions of R and supporting packages on a Windows 2000 OS: - R version 2.4.1 (2006-12-18) - zoo version 1.2-2 - chron version 2.3-10 The two time series I'm working with are from the summer of 2004 and are: 1.) wet.bulb.air.temp: air temperatures recorded on an hourly basis, and 2.) creek.temperature: surface water body temperatures collected every 12 minutes. I would ultimately like to observe the difference in temperatures and attempted to get at this by merging the two time series (by union), interpolating the NAs, and finally, subtracting one vector from the other. The problem is that I can not combine the two zoo time series objects using the merge() or cbind() functions. I get the following error: Error in z[match0(index(a), indexes), ] - a[match0(indexes, index(a))] : number of items to replace is not a multiple of replacement length The input/output from a recent R Console session might help, so I've included it as follows: summary(creek.temperature) Index creek.temperature Min. :(07/21/04 00:03:00) Min. :12.82 1st Qu.:(08/11/04 03:00:00) 1st Qu.:16.28 Median :(09/01/04 03:45:00) Median :18.53 Mean :(09/01/04 04:45:13) Mean :18.87 3rd Qu.:(09/22/04 06:37:00) 3rd Qu.:21.48 Max. :(10/13/04 09:22:00) Max. :27.72 length(creek.temperature) [1] 10140 summary(wet.bulb.air.temp) Index wet.bulb.air.temp Min. :(07/01/04 00:00:00) Min. : 3.889 1st Qu.:(07/31/04 12:00:00) 1st Qu.:12.778 Median :(08/31/04 00:00:00) Median :14.444 Mean :(08/31/04 00:00:00) Mean :14.469 3rd Qu.:(09/30/04 12:00:00) 3rd Qu.:16.667 Max. :(10/31/04 00:00:00) Max. :22.222 length(wet.bulb.air.temp) [1] 2929 class(creek.temperature) [1] zoo class(wet.bulb.air.temp) [1] zoo merge(wet.bulb.air.temp, creek.temperature) Error in z[match0(index(a), indexes), ] - a[match0(indexes, index(a))] : number of items to replace is not a multiple of replacement length cbind(wet.bulb.air.temp, creek.temperature) Error in z[match0(index(a), indexes), ] - a[match0(indexes, index(a))] : number of items to replace is not a multiple of replacement length The really puzzling part about this error is that it does not occur when I pare down the data sets and only look at a 24 hour window of data (for brevity, the input/output from that exercise has not been included in this e-mail). My question to the R user community is as follows: What is this error? How do I get past it and get these two data sets to play nice with each other? Thanks, Matt Findley - Matthew C. Findley, CPSSc Environmental Scientist CH2M HILL 2300 NW Walnut Blvd Corvallis, OR 97330-3538 Tel: 541.768.3504 Fax: 541.752.0276 [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to manipulate the pointer of a file?
On Thu, 12 Apr 2007, Yuchen Luo wrote: With file( ) to obtain a pointer of a file, every time we use scan ( ) to read one row of it, the pointer will point to the next row of the file. In the following example, d1 and d2 are obtained the same way but they correspond to different rows of the same file because the pointer of the file moves down a row when a row of the file is read. Yes, as documented. Isn't that good? [...] My question is, how to manipulate the pointer of the file further? For example, what if I need the pointer to go back to the previous row? See ?seek, which is linked from the help page for file(). Using accurate terminology helps: file() gives you a connection. Part of a connection is two positions (one each for read and write). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to manipulate the pointer of a file?
?seek -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Yuchen Luo Sent: Thursday, April 12, 2007 9:33 AM To: [EMAIL PROTECTED] Subject: [R] How to manipulate the pointer of a file? Dear friends. With file( ) to obtain a pointer of a file, every time we use scan ( ) to read one row of it, the pointer will point to the next row of the file. In the following example, d1 and d2 are obtained the same way but they correspond to different rows of the same file because the pointer of the file moves down a row when a row of the file is read. The following is an example: a1 - list(name=Fred, wife=Mary, no.children=3) a2 - list(name=Tom, wife=Joy, no.children=9) a3 - list(name=Paul, wife=Alic, no.children=5) write.table(a1, file = tt.csv, sep=',',row.names=FALSE,col.name=TRUE) write.table(a2, file=tt.csv, sep=',', append=TRUE, row.names=FALSE, col.names=FALSE) write.table(a3, file=tt.csv, sep=',', append=TRUE, row.names=FALSE, col.names=FALSE) fp=file(tt.csv,r) c=scan(file=fp, sep=',', what=list(c1=, c2=, c3=), flush=TRUE, nlines=1) d1=scan(file=fp, sep=',', what=list(name=, wife=, no.kids=0), flush=TRUE, nlines=1) d1 R output: --- $name [1] Fred $wife [1] Mary $no.kids [1] 3 - d2=scan(file=fp, sep=',', what=list(name=, wife=, no.kids=0), flush=TRUE, nlines=1) d2 R Output: $name [1] Tom $wife [1] Joy $no.kids [1] 9 -- My question is, how to manipulate the pointer of the file further? For example, what if I need the pointer to go back to the previous row? Best Wishes Yuchen Luo [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Subsetting list of vectors with list of (boolean) vectors?
Dear Rologists, I'm stuck with this. How would you do this efficiently: aPGI [[1]] [1] 864 5576 aPGItest [[1]] [1] TRUE FALSE result - [magic box involving subset) result [[1]] [1] 864 Thanks for any hints, Joh __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] zoo merge() method
On Thu, 12 Apr 2007 [EMAIL PROTECTED] wrote: attempted to get at this by merging the two time series (by union), interpolating the NAs, and finally, subtracting one vector from the other. The problem is that I can not combine the two zoo time series objects using the merge() or cbind() functions. I get the following error: Error in z[match0(index(a), indexes), ] - a[match0(indexes, index(a))] : number of items to replace is not a multiple of replacement length Usually, problems like this occur when the time stamps in one of your time series are not unique. Maybe we should improve the error message by explicitely trying to catch this error. You can easily check this, e.g., via any(table(time(zoo_object)) 1) The input/output from a recent R Console session might help, Not really, the data itself would have been much more helpful...and even better some simplified artificial data set. hth, Z __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting list of vectors with list of (boolean) vectors?
On Thu, 2007-04-12 at 18:12 +0200, Johannes Graumann wrote: Dear Rologists, I'm stuck with this. How would you do this efficiently: aPGI [[1]] [1] 864 5576 aPGItest [[1]] [1] TRUE FALSE result - [magic box involving subset) result [[1]] [1] 864 Thanks for any hints, Joh lapply(seq(along = length(aPGI)), function(x) aPGI[[x]][aPGItest[[x]]]) [[1]] [1] 864 I think that this should be a generic solution for multiple (but common) levels in each list. HTH, Marc Schwartz __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting list of vectors with list of (boolean) vectors?
Johannes Graumann [EMAIL PROTECTED] writes: Dear Rologists, I'm stuck with this. How would you do this efficiently: aPGI [[1]] [1] 864 5576 aPGItest [[1]] [1] TRUE FALSE result - [magic box involving subset) result - aPGI[aPGItest[[1]]] result [[1]] [1] 864 + seth -- Seth Falcon | Computational Biology | Fred Hutchinson Cancer Research Center http://bioconductor.org __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting list of vectors with list of (boolean) vectors?
From: Marc Schwartz
On Thu, 2007-04-12 at 18:12 +0200, Johannes Graumann wrote:
Dear Rologists,
I'm stuck with this. How would you do this efficiently:
aPGI
[[1]]
[1] 864 5576
aPGItest
[[1]]
[1] TRUE FALSE
result - [magic box involving subset)
result
[[1]]
[1] 864
Thanks for any hints,
Joh
lapply(seq(along = length(aPGI)), function(x)
aPGI[[x]][aPGItest[[x]]])
[[1]]
[1] 864
Alternatively:
R mapply([, aPGI, aPGItest, SIMPLIFY=FALSE)
[[1]]
[1] 864
Cheers,
Andy
I think that this should be a generic solution for multiple
(but common) levels in each list.
HTH,
Marc Schwartz
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Re: [R] sas.get problem
--- Tim Churches [EMAIL PROTECTED] wrote: John Kane wrote: I have 3 SAS files all in the directory F:/sas, two data files and a format file : form.ea1.sas7bdat form.ea2.sas7bdat sas.fmts.sas7bdat F is a USB. I am trying import them to R using sas.get. I have not used SAS since I was downloading data from mainframe and having to write JCL. I had forgotten how bizarre SAS can be. I currently have not even figured out how to load the files into SAS but they look fine since I can import them with no problem into SPSS. I am using R2.4.1 under Windows XP SAS files were created with SAS 9.x They convert easily into SPSS 14 I n the example below I have tried various versions of the file names with with no luck. Can anyone suggest some approach(s) that I might take. Example. library(Hmisc) mydata - sas.get(library=F:/sas, mem=form.ea1, format.library=sas.fmts.sas7bdat, sasprog = 'C:Program Files/SAS/SAS 9.1/sas.exe') Error message (one of several that I have gotten while trying various things.) The filename, directory name, or volume label syntax is incorrect. Error in sas.get(library = F:/sas, mem = form.ea1, format.library = sas.fmts.sas7bdat, : SAS job failed with status 1 In addition: Warning messages: 1: sas.fmts.sas7bdat/formats.sc? or formats.sas7bcat not found. Formatting ignored. in: sas.get(library = F:/sas, mem = form.ea1, format.library = sas.fmts.sas7bdat, 2: 'cmd' execution failed with error code 1 in: shell(cmd, wait = TRUE, intern = output) The sas.get function in the Hmisc library is broken under Windows. Change line 127 from: status - sys(paste(shQuote(sasprog), shQuote(sasin), -log, shQuote(log.file)), output = FALSE) to: status - system(paste(shQuote(sasprog), shQuote(sasin), -log, shQuote(log.file))) Just how would I go about making this change in the code or where would I look? Thanks I found this fix in the R-help archives, sorry, don't have the original to hand so I can't give proper attribution, but the fix is not due to me. But it does work for me. I believe Frank Harrell has been notified of the problem and the fix. Once patched and working correctly, the sas.get function in the Hmisc library is fantastic - thanks Frank! Tim C __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Construct time series objects from raw data stored in csv files
Hi, I have time series data stored in csv files (see below for an example of the data). I understand that in order to analyze my data in R, I need to first transform it into a ts object. Howeve, I could not find an example on how exactly to do that. Is ts the only function I need? What are the steps that I need to go through to build a time series object from raw data like this? Thanks, Tom --- DATE,VALUE 1921-01-01,19.000 1921-02-01,18.400 1921-03-01,18.300 1921-04-01,18.100 1921-05-01,17.700 1921-06-01,17.600 1921-07-01,17.700 1921-08-01,17.700 1921-09-01,17.500 1921-10-01,17.500 1921-11-01,17.400 1921-12-01,17.300 1922-01-01,16.900 1922-02-01,16.900 1922-03-01,16.700 1922-04-01,16.700 1922-05-01,16.700 1922-06-01,16.700 1922-07-01,16.800 1922-08-01,16.600 1922-09-01,16.600 1922-10-01,16.700 1922-11-01,16.800 1922-12-01,16.900 [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Putting 2 breaks on Y axis
R plotting experts: I have a bivariate dataset composed of 300 (x,y) continuous datapoints. 297 of these points are located within the y range of [0,10], while 2 are located at 20 and one at 55. No coding errors, real outliers. When plotting these data with a scatterplot, I obviously have a problem. If I plot the full dataset with ylim = c(0,55), then I cannot see the structure in the data in the [0, 10] range. If I truncate the y axis with ylim = c(0,10), then I cannot see the 3 outliers. If I break the y axis from 10 to 20 (using plotrix functions), I still do not see the data optimally because of the white space from y=20 to y=55. What I would like to do is break the y axis at 2 points, roughly 10-20 and 20-55. Is there a function that can break an axis in 2 places? Thanks in advance for any suggestions. Brant __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sas.get problem
Cancel this! . I see that Hmisc is already changed. When I updated today it was changed. My thanks to Mr Dupont the maintainer. I still have a problem with sas.get but that's another post. --- John Kane [EMAIL PROTECTED] wrote: --- Tim Churches [EMAIL PROTECTED] wrote: John Kane wrote: I have 3 SAS files all in the directory F:/sas, two data files and a format file : form.ea1.sas7bdat form.ea2.sas7bdat sas.fmts.sas7bdat F is a USB. I am trying import them to R using sas.get. I have not used SAS since I was downloading data from mainframe and having to write JCL. I had forgotten how bizarre SAS can be. I currently have not even figured out how to load the files into SAS but they look fine since I can import them with no problem into SPSS. I am using R2.4.1 under Windows XP SAS files were created with SAS 9.x They convert easily into SPSS 14 I n the example below I have tried various versions of the file names with with no luck. Can anyone suggest some approach(s) that I might take. Example. library(Hmisc) mydata - sas.get(library=F:/sas, mem=form.ea1, format.library=sas.fmts.sas7bdat, sasprog = 'C:Program Files/SAS/SAS 9.1/sas.exe') Error message (one of several that I have gotten while trying various things.) The filename, directory name, or volume label syntax is incorrect. Error in sas.get(library = F:/sas, mem = form.ea1, format.library = sas.fmts.sas7bdat, : SAS job failed with status 1 In addition: Warning messages: 1: sas.fmts.sas7bdat/formats.sc? or formats.sas7bcat not found. Formatting ignored. in: sas.get(library = F:/sas, mem = form.ea1, format.library = sas.fmts.sas7bdat, 2: 'cmd' execution failed with error code 1 in: shell(cmd, wait = TRUE, intern = output) The sas.get function in the Hmisc library is broken under Windows. Change line 127 from: status - sys(paste(shQuote(sasprog), shQuote(sasin), -log, shQuote(log.file)), output = FALSE) to: status - system(paste(shQuote(sasprog), shQuote(sasin), -log, shQuote(log.file))) Just how would I go about making this change in the code or where would I look? Thanks I found this fix in the R-help archives, sorry, don't have the original to hand so I can't give proper attribution, but the fix is not due to me. But it does work for me. I believe Frank Harrell has been notified of the problem and the fix. Once patched and working correctly, the sas.get function in the Hmisc library is fantastic - thanks Frank! Tim C __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
On Thu, 12 Apr 2007, tom soyer wrote: Hi, I have time series data stored in csv files (see below for an example of the data). I understand that in order to analyze my data in R, I need to first transform it into a ts object. Howeve, I could not find an example on how exactly to do that. Is ts the only function I need? What are the steps that I need to go through to build a time series object from raw data like this? With the zoo package you can do library(zoo) z - read.zoo(yourdata.csv, sep = ,) plot(z) See vignette(zoo, package = zoo) for more information and also some more details about other time series classes. Z Thanks, Tom --- DATE,VALUE 1921-01-01,19.000 1921-02-01,18.400 1921-03-01,18.300 1921-04-01,18.100 1921-05-01,17.700 1921-06-01,17.600 1921-07-01,17.700 1921-08-01,17.700 1921-09-01,17.500 1921-10-01,17.500 1921-11-01,17.400 1921-12-01,17.300 1922-01-01,16.900 1922-02-01,16.900 1922-03-01,16.700 1922-04-01,16.700 1922-05-01,16.700 1922-06-01,16.700 1922-07-01,16.800 1922-08-01,16.600 1922-09-01,16.600 1922-10-01,16.700 1922-11-01,16.800 1922-12-01,16.900 [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
On 4/12/07, tom soyer [EMAIL PROTECTED] wrote: Hi, I have time series data stored in csv files (see below for an example of the data). I understand that in order to analyze my data in R, I need to first transform it into a ts object. Howeve, I could not find an example on how exactly to do that. Is ts the only function I need? What are the steps that I need to go through to build a time series object from raw data like this? Try pasting this into an R session: Lines.raw - DATE,VALUE 1921-01-01,19.000 1921-02-01,18.400 1921-03-01,18.300 1921-04-01,18.100 1921-05-01,17.700 1921-06-01,17.600 1921-07-01,17.700 1921-08-01,17.700 1921-09-01,17.500 1921-10-01,17.500 1921-11-01,17.400 1921-12-01,17.300 1922-01-01,16.900 1922-02-01,16.900 1922-03-01,16.700 1922-04-01,16.700 1922-05-01,16.700 1922-06-01,16.700 1922-07-01,16.800 1922-08-01,16.600 1922-09-01,16.600 1922-10-01,16.700 1922-11-01,16.800 1922-12-01,16.900 library(zoo) # replace next line with something like this: # z - read.zoo(myfile.dat, header = TRUE, sep = ,) z - read.zoo(textConnection(Lines.raw), header = TRUE, sep = ,) time(z) - as.yearmon(time(z)) as.ts(z) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
Thanks Gabor! I think your example works, but check this out: as.ts(z) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1921 19.0 18.4 18.3 18.1 17.7 17.6 17.7 17.7 17.5 17.5 17.4 17.3 1922 16.9 16.9 16.7 16.7 16.7 16.7 16.8 16.6 16.6 16.7 16.8 16.9 is.ts(z) [1] FALSE How come R does not recognize z as a ts object? It is a ts object, isn't it? On 4/12/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 4/12/07, tom soyer [EMAIL PROTECTED] wrote: Hi, I have time series data stored in csv files (see below for an example of the data). I understand that in order to analyze my data in R, I need to first transform it into a ts object. Howeve, I could not find an example on how exactly to do that. Is ts the only function I need? What are the steps that I need to go through to build a time series object from raw data like this? Try pasting this into an R session: Lines.raw - DATE,VALUE 1921-01-01,19.000 1921-02-01,18.400 1921-03-01,18.300 1921-04-01,18.100 1921-05-01,17.700 1921-06-01,17.600 1921-07-01,17.700 1921-08-01,17.700 1921-09-01,17.500 1921-10-01,17.500 1921-11-01,17.400 1921-12-01,17.300 1922-01-01,16.900 1922-02-01,16.900 1922-03-01,16.700 1922-04-01,16.700 1922-05-01,16.700 1922-06-01,16.700 1922-07-01,16.800 1922-08-01,16.600 1922-09-01,16.600 1922-10-01,16.700 1922-11-01,16.800 1922-12-01,16.900 library(zoo) # replace next line with something like this: # z - read.zoo(myfile.dat, header = TRUE, sep = ,) z - read.zoo(textConnection(Lines.raw), header = TRUE, sep = ,) time(z) - as.yearmon(time(z)) as.ts(z) -- Tom [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
On 4/12/07, tom soyer [EMAIL PROTECTED] wrote: Thanks Gabor! I think your example works, but check this out: as.ts(z) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1921 19.0 18.4 18.3 18.1 17.7 17.6 17.7 17.7 17.5 17.5 17.4 17.3 1922 16.9 16.9 16.7 16.7 16.7 16.7 16.8 16.6 16.6 16.7 16.8 16.9 is.ts(z) [1] FALSE The above outputs as.ts(z) on the console. If you want to assign it to a variable you need to do so: tz - as.ts(z) is.ts(tz) # TRUE __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
Oh yes, I forgot. Thanks!! On 4/12/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 4/12/07, tom soyer [EMAIL PROTECTED] wrote: Thanks Gabor! I think your example works, but check this out: as.ts(z) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1921 19.0 18.4 18.3 18.1 17.7 17.6 17.7 17.7 17.5 17.5 17.4 17.3 1922 16.9 16.9 16.7 16.7 16.7 16.7 16.8 16.6 16.6 16.7 16.8 16.9 is.ts(z) [1] FALSE The above outputs as.ts(z) on the console. If you want to assign it to a variable you need to do so: tz - as.ts(z) is.ts(tz) # TRUE -- Tom [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
Gabor, What should I do for weekly and daily data series? Are there functions similar to yearmon() for other time intervals? I see that there is a built-in yearqtr() function for quarterly data, but that's it. Thanks! On 4/12/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 4/12/07, tom soyer [EMAIL PROTECTED] wrote: Thanks Gabor! I think your example works, but check this out: as.ts(z) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1921 19.0 18.4 18.3 18.1 17.7 17.6 17.7 17.7 17.5 17.5 17.4 17.3 1922 16.9 16.9 16.7 16.7 16.7 16.7 16.8 16.6 16.6 16.7 16.8 16.9 is.ts(z) [1] FALSE The above outputs as.ts(z) on the console. If you want to assign it to a variable you need to do so: tz - as.ts(z) is.ts(tz) # TRUE -- Tom [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Putting 2 breaks on Y axis
On Thu, 2007-04-12 at 13:41 -0500, Inman, Brant A. M.D. wrote: R plotting experts: I have a bivariate dataset composed of 300 (x,y) continuous datapoints. 297 of these points are located within the y range of [0,10], while 2 are located at 20 and one at 55. No coding errors, real outliers. When plotting these data with a scatterplot, I obviously have a problem. If I plot the full dataset with ylim = c(0,55), then I cannot see the structure in the data in the [0, 10] range. If I truncate the y axis with ylim = c(0,10), then I cannot see the 3 outliers. If I break the y axis from 10 to 20 (using plotrix functions), I still do not see the data optimally because of the white space from y=20 to y=55. What I would like to do is break the y axis at 2 points, roughly 10-20 and 20-55. Is there a function that can break an axis in 2 places? Thanks in advance for any suggestions. Brant Brant, I am not a particular fan of broken axes (though others will disagree), much less two breaks. Presuming that your data might look something like this: http://www.itl.nist.gov/div898/handbook/eda/section3/scattera.htm A couple of thoughts: 1. Not being sure if your data range above actually includes 0, you may want to consider a log scaled axis, if not. 2. I might be tempted to use two plots: A. A first a plot of the entire data set, showing the 3 outliers B. A second plot of the 297 pairs with axes constrained to the appropriate ranges to enable better visualization of the data structure. If number 2 is more appropriate, you could also use par(mfcol) to set up side by side plots. See ?par. HTH, Marc Schwartz __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] making a counter of consecitive positive cases in time series
Hi all..RCounters!
I´m working with standarized time series, and i need make a counter of
consecutives positves numbers to make a cumulative experimental funtion. I
have x: the time series (0,1) and y: my counter, i have this for step. What is
wrong?.. any can help me please!
x-rbinom(15,1,.3)
y-NULL;s-0
for (i in 1: length (x))
{if (x[i]0)
{s-s+x[i]
s=0}
else
y-c(y,s)}
y
x
Thk u all!
José Bustos
-
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Re: [R] Putting 2 breaks on Y axis
On 4/12/07, Marc Schwartz [EMAIL PROTECTED] wrote: On Thu, 2007-04-12 at 13:41 -0500, Inman, Brant A. M.D. wrote: R plotting experts: I have a bivariate dataset composed of 300 (x,y) continuous datapoints. 297 of these points are located within the y range of [0,10], while 2 are located at 20 and one at 55. No coding errors, real outliers. When plotting these data with a scatterplot, I obviously have a problem. If I plot the full dataset with ylim = c(0,55), then I cannot see the structure in the data in the [0, 10] range. If I truncate the y axis with ylim = c(0,10), then I cannot see the 3 outliers. If I break the y axis from 10 to 20 (using plotrix functions), I still do not see the data optimally because of the white space from y=20 to y=55. What I would like to do is break the y axis at 2 points, roughly 10-20 and 20-55. Is there a function that can break an axis in 2 places? Thanks in advance for any suggestions. Brant Brant, I am not a particular fan of broken axes (though others will disagree), much less two breaks. Presuming that your data might look something like this: http://www.itl.nist.gov/div898/handbook/eda/section3/scattera.htm A couple of thoughts: 1. Not being sure if your data range above actually includes 0, you may want to consider a log scaled axis, if not. 2. I might be tempted to use two plots: A. A first a plot of the entire data set, showing the 3 outliers B. A second plot of the 297 pairs with axes constrained to the appropriate ranges to enable better visualization of the data structure. If number 2 is more appropriate, you could also use par(mfcol) to set up side by side plots. See ?par. HTH, Marc Schwartz I was thinking plot the data without the outliers and include a smaller inscribed plot in a corner showing all the data (the global view). But I couldn't figure out how to do this. (I think legend() works very hard to do this type of thing.) Stephen __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] LME: internal workings of QR factorization --repost
Hi: I've been reading Computational Methods for Multilevel Modeling by Pinheiro and Bates, the idea of embedding the technique in my own c-level code. The basic idea is to rewrite the joint density in a form to mimic a single least squares problem conditional upon the variance parameters. The paper is fairly clear except that some important level of detail is missing. For instance, when we first meet Q_(i): /\ / \ | Z_i X_i y_i | | R_11(i) R_10(i) c_1(i) | || = Q_(i) | | | Delta 0 0| | 0 R_00(i) c_0(i) | \/ \ / the text indicates that the Q-R factorization is limited to the first q columns of the augmented matrix on the left. If one plunks the first q columns of the augmented matrix on the left into a qr factorization, one obtains an orthogonal matrix Q that is (n_i + q) x q and a nonsingular upper triangular matrix R that is q x q. While the text describes R as a nonsingular upper triangular q x q, the matrix Q_(i) is described as a square (n_i + q) x (n_i + q) orthogonal matrix. The remaining columns in the matrix to the right are defined by applying transpose(Q_(i)) to both sides. The question is how to augment my Q which is orthogonal (n_i + q) x q with the missing (n_i + q) x n_i portion producing the orthogonal square matrix mentioned in the text? I tried appending the n_i x n_i identity matrix to the block diagonal, but this doesn't work as the resulting likelihood is insensitive to the variance parameters. Grant Izmirlian NCI __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Putting 2 breaks on Y axis
Try something like this (modify to how you like it): x - runif(100) y - rnorm(100, 5, 2) y[1:3] - c(19, 21, 50) layout(matrix( 3:1, ncol=1 ), heights=c(2,3,4)) par(mar=c(5,4,0,2)+0.1) plot(x,y, ylim=c(0,10), ylab='') par(mar=c(0.5,4,0,2)+0.1) plot(x,y, ylim=c(18,22), xlab='', xaxt='n' ) axis(1, labels=FALSE) par(mar=c(0.5,4,4,2)+0.1) plot(x,y, ylim=c(49,51),xlab='', main='my title', xaxt='n', ylab='' ) axis(1, labels=FALSE) -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Inman, Brant A. M.D. Sent: Thursday, April 12, 2007 12:41 PM To: [EMAIL PROTECTED] Subject: [R] Putting 2 breaks on Y axis R plotting experts: I have a bivariate dataset composed of 300 (x,y) continuous datapoints. 297 of these points are located within the y range of [0,10], while 2 are located at 20 and one at 55. No coding errors, real outliers. When plotting these data with a scatterplot, I obviously have a problem. If I plot the full dataset with ylim = c(0,55), then I cannot see the structure in the data in the [0, 10] range. If I truncate the y axis with ylim = c(0,10), then I cannot see the 3 outliers. If I break the y axis from 10 to 20 (using plotrix functions), I still do not see the data optimally because of the white space from y=20 to y=55. What I would like to do is break the y axis at 2 points, roughly 10-20 and 20-55. Is there a function that can break an axis in 2 places? Thanks in advance for any suggestions. Brant __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Putting 2 breaks on Y axis
Inman, Brant A. M.D. wrote: R plotting experts: I have a bivariate dataset composed of 300 (x,y) continuous datapoints. 297 of these points are located within the y range of [0,10], while 2 are located at 20 and one at 55. No coding errors, real outliers. When plotting these data with a scatterplot, I obviously have a problem. If I plot the full dataset with ylim = c(0,55), then I cannot see the structure in the data in the [0, 10] range. If I truncate the y axis with ylim = c(0,10), then I cannot see the 3 outliers. If I break the y axis from 10 to 20 (using plotrix functions), I still do not see the data optimally because of the white space from y=20 to y=55. What I would like to do is break the y axis at 2 points, roughly 10-20 and 20-55. Is there a function that can break an axis in 2 places? Hi Brant, gap.plot in the plotrix package can do one break, and it is possible to do two, as gap.boxplot does. It wouldn't be too difficult to recode gap.plot to get more than one break. I'll see what I can do today. Jim __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] LME: internal workings of QR factorization
Hi: I've been reading Computational Methods for Multilevel Modeling by Pinheiro and Bates, the idea of embedding the technique in my own c-level code. The basic idea is to rewrite the joint density in a form to mimic a single least squares problem conditional upon the variance parameters. The paper is fairly clear except that some important level of detail is missing. For instance, when we first meet Q_(i): /\ / \ | Z_i X_i y_i | | R_11(i) R_10(i) c_1(i) | || = Q_(i) | | | Delta 0 0| | 0 R_00(i) c_0(i) | \/ \ / the text indicates that the Q-R factorization is limited to the first q columns of the augmented matrix on the left. If one plunks the first q columns of the augmented matrix on the left into a qr factorization, one obtains an orthogonal matrix Q that is (n_i + q) x q and a nonsingular upper triangular matrix R that is q x q. While the text describes R as a nonsingular upper triangular q x q, the matrix Q_(i) is described as a square (n_i + q) x (n_i + q) orthogonal matrix. The remaining columns in the matrix to the right are defined by applying transpose(Q_(i)) to both sides. The question is how to augment my Q which is orthogonal (n_i + q) x q with the missing (n_i + q) x n_i portion producing the orthogonal square matrix mentioned in the text? I tried appending the n_i x n_i identity matrix to the block diagonal, but this doesn't work as the resulting likelihood is insensitive to the variance parameters. Grant Izmirlian NCI __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Random Sequence
Dear Friends, I'm trying to generate a sequence of 100 observations with either a 1 or -1. In other words the sequence should look something like this. y = 1 1 -1 1 -1 -1 -1 1 1 .. Can somebody please give me some direction on how I can do this in R. Thanks Anup Don't get soaked. Take a quick peak at the forecast __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random Sequence
Hi Anup, (runif(100).5)*1 #would give you 0's and 1's. sample(rep(c(-1,1),50),100) #A bit slower I think, gives you -1's and 1's Best, Matt On 4/12/07, Anup Nandialath [EMAIL PROTECTED] wrote: Dear Friends, I'm trying to generate a sequence of 100 observations with either a 1 or -1. In other words the sequence should look something like this. y = 1 1 -1 1 -1 -1 -1 1 1 .. Can somebody please give me some direction on how I can do this in R. Thanks Anup Don't get soaked. Take a quick peak at the forecast __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Matthew C Keller Postdoctoral Fellow Virginia Institute for Psychiatric and Behavioral Genetics __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random Sequence
Anup Nandialath wrote: Dear Friends, I'm trying to generate a sequence of 100 observations with either a 1 or -1. In other words the sequence should look something like this. y = 1 1 -1 1 -1 -1 -1 1 1 .. Can somebody please give me some direction on how I can do this in R. sample(c(-1, 1), 100, replace=TRUE) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random Sequence
There are many ways to do this. The first that comes to my mind is sample(c(1,-1),100,TRUE). Notice that sample also has a prob argument that may be useful for you. Francisco Anup Nandialath wrote: Dear Friends, I'm trying to generate a sequence of 100 observations with either a 1 or -1. In other words the sequence should look something like this. y = 1 1 -1 1 -1 -1 -1 1 1 .. Can somebody please give me some direction on how I can do this in R. Thanks Anup Don't get soaked. Take a quick peak at the forecast __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Putting 2 breaks on Y axis
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Stephen Weigand Sent: Thursday, April 12, 2007 2:32 PM To: Inman, Brant A. M.D. Cc: [EMAIL PROTECTED] Subject: Re: [R] Putting 2 breaks on Y axis [snip] I was thinking plot the data without the outliers and include a smaller inscribed plot in a corner showing all the data (the global view). But I couldn't figure out how to do this. (I think legend() works very hard to do this type of thing.) This is fairly easy to do with the subplot function in the TeachingDemos package: Stephen -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random Sequence
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Matthew Keller Sent: Thursday, April 12, 2007 2:32 PM To: Anup Nandialath Cc: [EMAIL PROTECTED] Subject: Re: [R] Random Sequence Hi Anup, (runif(100).5)*1 #would give you 0's and 1's. sample(rep(c(-1,1),50),100) #A bit slower I think, gives you -1's and 1's How about sample(c(-1,1), 100, replace=TRUE) Hope this is helpful, Dan Daniel J. Nordlund Research and Data Analysis Washington State Department of Social and Health Services Olympia, WA 98504-5204 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] GLM with random effects
Hi R-Users,
I have 3 replicates ('Replicate) of counts of parasites ('nor.tot.lep')
before and after an experiment ('In.Out'). I am trying to treat the
three replicates as a random effect in order to determine if the main
effect (In.Out) significantly influences my dependent variable
(nor.tot.lep) after the variance explained by the replicates is
accounted for. I have tried:
lmer(nor.tot.lep ~ In.Out + (In.Out|Replicate),data=coho, family=poisson)
Generalized linear mixed model fit using PQL
Formula: nor.tot.lep ~ In.Out + (In.Out | Replicate)
Data: coho
Family: Poisson
AIC BIC logLik deviance
849.2 867.4 -419.6 839.2
Random effects:
Groups Name Variance Std.Dev. Corr
Replicate (Intercept) 0.78861 0.88804
In.Out 0.67232 0.81995 -1.000
Residual 2.96308 1.72136
number of obs: 279, groups: Replicate, 3
Fixed effects:
Estimate Std. Error t value
(Intercept) -0.2431 0.6619 -0.3672
In.Out 1.6004 0.5645 2.8349
Correlation of Fixed Effects:
(Intr)
In.Out -0.975
There were 30 warnings (use warnings() to see them)
warnings()
Warning messages:
1: Estimated variance-covariance for factor ‘Replicate’ is singular
in: LMEopt(x = mer, value = cv)
2: nlminb returned message false convergence (8)
in: LMEopt(x = mer, value = cv)
but as Mr. Bates pointed out, this is inappropriate b/c I am trying to
use 3 distinct replicates to estimate 3 variance-covariance
parameters. It won't work. Notice that the estimated correlation is
-1.000. Your estimated variance-covariance matrix is singular
I have also tried:
glmmPQL(nor.tot.lep ~ In.Out, random = (In.Out|Replicate), family =
poisson, data = coho)
Error in glmmPQL(nor.tot.lep ~ In.Out, random = (In.Out | Replicate), :
object In.Out not found
and R cannot find In.Out
If anyone has any suggestions they would be extremely appreciated!
Cheers,
Brendan
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[R] problems in loading MASS
Hi, there: After I upgraded my R to 2.4.1, it is my first time of trying to use MASS and found the following error message: install.packages(MASS) --- Please select a CRAN mirror for use in this session --- trying URL 'http://cran.cnr.Berkeley.edu/bin/macosx/universal/contrib/2.4/VR_7.2-33.tgz' Content type 'application/x-gzip' length 995260 bytes opened URL == downloaded 971Kb The downloaded packages are in /tmp/RtmpmAzBwa/downloaded_packages library(MASS) Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/libs/i386/MASS.so': dlopen(/Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/libs/i386/MASS.so, 6): Library not loaded: /usr/local/gcc4.0/i686-apple-darwin8/lib/libgcc_s.1.0.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/libs/i386/MASS.so Reason: image not found Error: package/namespace load failed for 'MASS' sessionInfo() R version 2.4.1 (2006-12-18) i386-apple-darwin8.8.1 locale: en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: randomForestdprep 4.5-181.0 version _ platform i386-apple-darwin8.8.1 arch i386 os darwin8.8.1 system i386, darwin8.8.1 status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) Thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question on ridge regression with R
Hi, I am working on a project about hospital efficiency. Due to the high multicolinearlity of the data, I want to fit the model using ridge regression. However, I believe that the data from large hospital(indicated by the number of patients they treat a year) is more accurate than from small hosptials, and I want to put more weight on them. How do I do this with lm.ridge? I know I just need to put weights=*** in lm function, but for lm.ridge, i don't see that option in lm.ridge. Could somebody give me suggestions on that? thanks ben [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reasons to Use R
On 12-Apr-07 10:14:21, Jim Lemon wrote: Charilaos Skiadas wrote: A new fortune candidate perhaps? On Apr 10, 2007, at 6:27 PM, Greg Snow wrote: Remember, everything is better than everything else given the right comparison. Only if we remove the grammatical blip that turns it into an infinite regress, i.e. Remember, anything is better than everything else given the right comparison Jim Oh dear, I would be disappointed with that, Jim. I was rather enjoying the vision of a topological sort tree (ordered by better according to some comparison) in which every single thing had everything else hanging off it, and in turn was hanging off everything else! Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 12-Apr-07 Time: 11:45:05 -- XFMail -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] GLM with random effects
Hello Brendan,
do you have any particular objection to
lmer(nor.tot.lep ~ In.Out + (1|Replicate), data=coho, family=poisson)
?
Andrew
On Thu, Apr 12, 2007 at 02:58:07PM -0700, Brendan Connors wrote:
Hi R-Users,
I have 3 replicates ('Replicate) of counts of parasites ('nor.tot.lep')
before and after an experiment ('In.Out'). I am trying to treat the
three replicates as a random effect in order to determine if the main
effect (In.Out) significantly influences my dependent variable
(nor.tot.lep) after the variance explained by the replicates is
accounted for. I have tried:
lmer(nor.tot.lep ~ In.Out + (In.Out|Replicate),data=coho, family=poisson)
Generalized linear mixed model fit using PQL
Formula: nor.tot.lep ~ In.Out + (In.Out | Replicate)
Data: coho
Family: Poisson
AIC BIC logLik deviance
849.2 867.4 -419.6 839.2
Random effects:
Groups Name Variance Std.Dev. Corr
Replicate (Intercept) 0.78861 0.88804
In.Out 0.67232 0.81995 -1.000
Residual 2.96308 1.72136
number of obs: 279, groups: Replicate, 3
Fixed effects:
Estimate Std. Error t value
(Intercept) -0.2431 0.6619 -0.3672
In.Out 1.6004 0.5645 2.8349
Correlation of Fixed Effects:
(Intr)
In.Out -0.975
There were 30 warnings (use warnings() to see them)
warnings()
Warning messages:
1: Estimated variance-covariance for factor ?Replicate? is singular
in: LMEopt(x = mer, value = cv)
2: nlminb returned message false convergence (8)
in: LMEopt(x = mer, value = cv)
but as Mr. Bates pointed out, this is inappropriate b/c I am trying to
use 3 distinct replicates to estimate 3 variance-covariance
parameters. It won't work. Notice that the estimated correlation is
-1.000. Your estimated variance-covariance matrix is singular
I have also tried:
glmmPQL(nor.tot.lep ~ In.Out, random = (In.Out|Replicate), family =
poisson, data = coho)
Error in glmmPQL(nor.tot.lep ~ In.Out, random = (In.Out | Replicate), :
object In.Out not found
and R cannot find In.Out
If anyone has any suggestions they would be extremely appreciated!
Cheers,
Brendan
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Andrew Robinson
Department of Mathematics and StatisticsTel: +61-3-8344-9763
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] LME: internal workings of QR factorization
On 4/12/07, Izmirlian, Grant (NIH/NCI) [E] [EMAIL PROTECTED] wrote: Hi: I've been reading Computational Methods for Multilevel Modeling by Pinheiro and Bates, the idea of embedding the technique in my own c-level code. The basic idea is to rewrite the joint density in a form to mimic a single least squares problem conditional upon the variance parameters. The paper is fairly clear except that some important level of detail is missing. For instance, when we first meet Q_(i): /\ / \ | Z_i X_i y_i | | R_11(i) R_10(i) c_1(i) | || = Q_(i) | | | Delta 0 0| | 0 R_00(i) c_0(i) | \/ \ / the text indicates that the Q-R factorization is limited to the first q columns of the augmented matrix on the left. If one plunks the first q columns of the augmented matrix on the left into a qr factorization, one obtains an orthogonal matrix Q that is (n_i + q) x q and a nonsingular upper triangular matrix R that is q x q. While the text describes R as a nonsingular upper triangular q x q, the matrix Q_(i) is described as a square (n_i + q) x (n_i + q) orthogonal matrix. The remaining columns in the matrix to the right are defined by applying transpose(Q_(i)) to both sides. The question is how to augment my Q which is orthogonal (n_i + q) x q with the missing (n_i + q) x n_i portion producing the orthogonal square matrix mentioned in the text? I tried appending the n_i x n_i identity matrix to the block diagonal, but this doesn't work as the resulting likelihood is insensitive to the variance parameters. Grant Izmirlian The QR decomposition of an n by p matrix (n p) can be written as the product of an orthogonal n by n matrix Q and an n by p matrix R which is zero below the main diagonal. Because the rows of R beyond the pth are zero, there is no need to store them. For some purposes it is more convenient to write the decomposition as the product of Q1, an n by p matrix with orthonormal columns and R1 a p by p upper triangular matrix. If you are going to be incorporating calculations like this in your own code I would recommend looking at the Implementation vignette in the lme4 package. It describes the computational approach used in the latest version of lmer (currently called lmer2 but to become lmer at some point) which allows for multiple non-nested grouping factors. The techniques that Jose and I describe in the paper you mention only handles nested grouping factors cleanly. That vignette has been updated after the last release of the lme4 package. You can get the expanded version from the SVN repository or wait until after R-2.5.0 is released and we release new versions of the Matrix and lme4 packages for R-2.5.0. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] class and method
Hi, R-experts I defined a new class of object, called alpha, which is basically a data.frame. And (I think) I know how to create a method for alpha, such as summary.alpha, plot.alpha. The problem is, when I try to access alpha object by usual data.frame method, it won't. For example, suppose X is an object of class alpha. Then, the commands such as dim(X), X[1,1], none of these works. I want to know how to define a class so that when there is appropriate methods use it, but when there's not use the predetermined method for underlying object. Any comments would be appreciated. Thanks- -- View this message in context: http://www.nabble.com/class-and-method-tf3568696.html#a9969691 Sent from the R help mailing list archive at Nabble.com. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problems in loading MASS
Weiwei, I have never had much success installing packages from within R.app on MacOSX, because the location that it is supposed to save things, / Library/Frameworks/, needs elevated priviledges, which the app doesn't seem to try to get. So it at best ends up saving it in some temporary location, and it has to be downloaded again next time R is restarted. As a result, I have always downloaded the tgz file from my browser, then go to the terminal in that folder and do a sudo R CMD INSTALL packagename.tgz. But perhaps I am doing something wrong and one can do this properly from within R.app, I would love to be wrong on this one. sessionInfo() R version 2.4.1 (2006-12-18) powerpc-apple-darwin8.8.0 locale: en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base Haris Skiadas Department of Mathematics and Computer Science Hanover College On Apr 12, 2007, at 6:04 PM, Weiwei Shi wrote: Hi, there: After I upgraded my R to 2.4.1, it is my first time of trying to use MASS and found the following error message: install.packages(MASS) --- Please select a CRAN mirror for use in this session --- trying URL 'http://cran.cnr.Berkeley.edu/bin/macosx/universal/ contrib/2.4/VR_7.2-33.tgz' Content type 'application/x-gzip' length 995260 bytes opened URL == downloaded 971Kb The downloaded packages are in /tmp/RtmpmAzBwa/downloaded_packages library(MASS) Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/Library/Frameworks/R.framework/Versions/2.4/Resources/library/ MASS/libs/i386/MASS.so': dlopen(/Library/Frameworks/R.framework/Versions/2.4/Resources/ library/MASS/libs/i386/MASS.so, 6): Library not loaded: /usr/local/gcc4.0/i686-apple-darwin8/lib/libgcc_s.1.0.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/ libs/i386/MASS.so Reason: image not found Error: package/namespace load failed for 'MASS' sessionInfo() R version 2.4.1 (2006-12-18) i386-apple-darwin8.8.1 locale: en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: randomForestdprep 4.5-181.0 version _ platform i386-apple-darwin8.8.1 arch i386 os darwin8.8.1 system i386, darwin8.8.1 status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) Thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problems in loading MASS
Hi Charilaos: I think the installation has no problem as indicated by install.packages(). Anyway, I tried your way but unfortunately it does not help. Thanks though, -W On 4/12/07, Charilaos Skiadas [EMAIL PROTECTED] wrote: Weiwei, I have never had much success installing packages from within R.app on MacOSX, because the location that it is supposed to save things, / Library/Frameworks/, needs elevated priviledges, which the app doesn't seem to try to get. So it at best ends up saving it in some temporary location, and it has to be downloaded again next time R is restarted. As a result, I have always downloaded the tgz file from my browser, then go to the terminal in that folder and do a sudo R CMD INSTALL packagename.tgz. But perhaps I am doing something wrong and one can do this properly from within R.app, I would love to be wrong on this one. sessionInfo() R version 2.4.1 (2006-12-18) powerpc-apple-darwin8.8.0 locale: en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base Haris Skiadas Department of Mathematics and Computer Science Hanover College On Apr 12, 2007, at 6:04 PM, Weiwei Shi wrote: Hi, there: After I upgraded my R to 2.4.1, it is my first time of trying to use MASS and found the following error message: install.packages(MASS) --- Please select a CRAN mirror for use in this session --- trying URL 'http://cran.cnr.Berkeley.edu/bin/macosx/universal/ contrib/2.4/VR_7.2-33.tgz' Content type 'application/x-gzip' length 995260 bytes opened URL == downloaded 971Kb The downloaded packages are in /tmp/RtmpmAzBwa/downloaded_packages library(MASS) Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/Library/Frameworks/R.framework/Versions/2.4/Resources/library/ MASS/libs/i386/MASS.so': dlopen(/Library/Frameworks/R.framework/Versions/2.4/Resources/ library/MASS/libs/i386/MASS.so, 6): Library not loaded: /usr/local/gcc4.0/i686-apple-darwin8/lib/libgcc_s.1.0.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/ libs/i386/MASS.so Reason: image not found Error: package/namespace load failed for 'MASS' sessionInfo() R version 2.4.1 (2006-12-18) i386-apple-darwin8.8.1 locale: en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: randomForestdprep 4.5-181.0 version _ platform i386-apple-darwin8.8.1 arch i386 os darwin8.8.1 system i386, darwin8.8.1 status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) Thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Construct time series objects from raw data stored in csv files
On 4/12/07, tom soyer [EMAIL PROTECTED] wrote: What should I do for weekly and daily data series? Are there functions similar to yearmon() for other time intervals? I see that there is a built-in yearqtr() function for quarterly data, but that's it. ts series cannot directly represent daily and weekly series other than somehow deciding on a numeric representation for time. Here we will use the number of days and the number of weeks since the Epoch (1970-01-01) and assume we assume weeks start on Sunday. We will also do it over again using the number of weeks since the first point in the series and the number of days since the first point. Lines.raw is from my original post on this thread. z - read.zoo(textConnection(Lines.raw), header = TRUE, sep = ,) # ts series will represent days as no of days since Epoch zday - z frequency(zday) - 1 tsday - as.ts(zday) # ts series will represent weeks as no of weeks since Epoch zweek - zday offset - -3 # weeks start on Sun # offset - -4 # weeks start on Mon zweek - aggregate(z, (as.numeric(time(z)) + offset) %/% 7, mean) frequency(zweek) - 1 tsweek - as.ts(zweek) ## # alternately use number of days since first day in series # and number of weeks since first week in series zday0 - aggregate(zday, time(zday) - min(time(zday)), c) frequency(zday0) - 1 tsday0 - as.ts(zday0) zweek0 - aggregate(zweek, time(zweek) - min(time(zweek)), c) frequency(zweek0) - 1 tsweek0 - as.ts(zweek0) __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NYC Area R / S / S-Plus Users Group Forming, Thursday, April 26th
Hi gang. So I'm actually carrying through on the threat I made a couple of months back, and helping to start a local users group for the NYC Metropolitan area. Here's a copy of the announcement of our introductory / planning meeting, I also can send a flyer version of this to anyone who wants (Open Office document, should be viewable in MS-Office). Please feel free to attend, invite friends and colleagues, and spread the word around. Thanks! * !!!NYC Area R / S / S-Plus Users Group Forming!!! Yes, it's really happening! An organization for users of the popular R and S-Plus statistical / analytical / graphical software environments in the NYC Metropolitan Area to call their own. Swap ideas, share code, learn cool tricks and techniques! Introductory / Organizing Meeting WHEN: Thursday, April 26th, 2007, 6:30 8:00 PM WHERE: Pragma Financial Systems, 447 Broadway, 4th Floor, NYC, 10013 PFS is on the west side of Broadway, between Grand and Howard Streets, a block and a half north of the Canal / Lafayette Street subway stop (6,J,M,N,Q,R,W,Z trains) WHAT: Here's your opportunity to be a Founding MembeR give your input about what you'd like the group to be; we'll also have short presentations on Vectorization Efficiency, and on Robust Statistical procedures in R. Light refreshment will be served, so RSVP appreciated (not required). Even if you can't attend, you can still join, CONTACT: Talbot Katz, [EMAIL PROTECTED], 646.461.7840 If you don't already use R, you could (should!) start today download R for free from the official R Project website (http://www.r-project.org/). The R Project site contains a wealth of resources for users, including FAQ pages, a user Wiki, Help List / newsgroup, book suggestions, etc. There is already an active R / S / S-Plus user community online, but nothing compares with having live presentations, forums, workshops you'll see! * -- TMK -- __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] trouble getting bw plots using trellis graphics
I'm having trouble getting a black and white plot using the trellis device. I have color graphics, but I'd like to use bw for a paper. For example, trellis.device(color=F) d - read.table(textConnection( A,B 0,1 1,2), header=T, sep=,) xyplot(A ~ B, data=d, type=o, lty=c(1:2), col=2, pch=19:20) dev.off() gives me a red line even though I specify color=F. sessionInfo() R version 2.4.1 (2006-12-18) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods [7] base other attached packages: lattice 0.14-17 Thank you, Brian _ Need a break? Find your escape route with Live Search Maps. __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] trouble getting bw plots using trellis graphic
On 4/12/07, Brian Riordan [EMAIL PROTECTED] wrote: I'm having trouble getting a black and white plot using the trellis device. I have color graphics, but I'd like to use bw for a paper. For example, trellis.device(color=F) d - read.table(textConnection( A,B 0,1 1,2), header=T, sep=,) xyplot(A ~ B, data=d, type=o, lty=c(1:2), col=2, pch=19:20) dev.off() gives me a red line even though I specify color=F. Yes, the 'color' argument controls the default settings. If you don't want a non-default color, don't specify a color. You can't expect to say draw these points in red and then be surprised when the points are really drawn in red. If you expected that somehow col=2 would become a black and white color, then look at ?palette. For example, you can set the palette to shades of grey using palette(grey.colors(5)) -Deepayan __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] class and method
On Thu, 12 Apr 2007, jiho.han wrote: Hi, R-experts I defined a new class of object, called alpha, which is basically a data.frame. And (I think) I know how to create a method for alpha, such as summary.alpha, plot.alpha. The problem is, when I try to access alpha object by usual data.frame method, it won't. For example, suppose X is an object of class alpha. Then, the commands such as dim(X), X[1,1], none of these works. I want to know how to define a class so that when there is appropriate methods use it, but when there's not use the predetermined method for underlying object. Any comments would be appreciated. Thanks- class(X) - c(alpha, data.frame) is I think what you want. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] making a counter of consecitive positive cases in time series
On Thu, 2007-04-12 at 22:30 +0200, Jose Bustos Melo wrote:
Hi all..RCounters!
Im working with standarized time series, and i need make a counter
of consecutives positves numbers to make a cumulative experimental
funtion. I have x: the time series (0,1) and y: my counter, i have
this for step. What is wrong?.. any can help me please!
x-rbinom(15,1,.3)
y-NULL;s-0
for (i in 1: length (x))
{if (x[i]0)
{s-s+x[i]
s=0}
else
y-c(y,s)}
y
x
Thk u all!
Jos Bustos
I may be mis-understanding your desired result, but is this what you
want:
x - rbinom(15, 1, .3)
x
[1] 1 1 0 0 1 0 0 0 1 1 0 1 1 0 1
rle(x)
Run Length Encoding
lengths: int [1:9] 2 2 1 3 2 1 2 1 1
values : num [1:9] 1 0 1 0 1 0 1 0 1
See ?rle for more information.
HTH,
Marc Schwartz
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] problems in loading MASS
MASS is part of the standard R installation, as a recommended package. I suggest you try re-installing R and not then trying to download a different version. I would also suggest trying 2.5.0 beta at this stage in preference to 2.4.1 if you do need to re-install R. Beyond that, there is a list (r-sig-mac) for Mac-specific problems (and this is an OS-specific problem). On Thu, 12 Apr 2007, Weiwei Shi wrote: Hi, there: After I upgraded my R to 2.4.1, it is my first time of trying to use MASS and found the following error message: install.packages(MASS) --- Please select a CRAN mirror for use in this session --- trying URL 'http://cran.cnr.Berkeley.edu/bin/macosx/universal/contrib/2.4/VR_7.2-33.tgz' Content type 'application/x-gzip' length 995260 bytes opened URL == downloaded 971Kb The downloaded packages are in /tmp/RtmpmAzBwa/downloaded_packages library(MASS) Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library '/Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/libs/i386/MASS.so': dlopen(/Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/libs/i386/MASS.so, 6): Library not loaded: /usr/local/gcc4.0/i686-apple-darwin8/lib/libgcc_s.1.0.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.4/Resources/library/MASS/libs/i386/MASS.so Reason: image not found Error: package/namespace load failed for 'MASS' [...] -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
