Re: [R] re ading tokens
j daniel wrote: I am not familiar with processing text in R. Can someone tell me how to read each line of words as separate elements in a list? FE, I would like to turn: word1 word2 word3 word2 word4 into a list of length two with three character elements in the first list and two elements in the second. I know that this should be easy, but I am a little confused by the text functions. You could use scan. Have a look at package gsubfn, where there is a demo, that show additional features you are going to use library(gsubfn) demo(gsubfn-gries) The example code is a bit overnested, but to better understand what is going on, unwrap it: So tail(sort(table(unlist(strapply(Lines1, \\w+, perl = TRUE) is: x1 = strapply(Lines1, \\w+, perl = TRUE) x1 x2 = ulist(x2) x2 x3 = table(x2) x3 x4 = sort(x3) x4 tail(x4) Dieter -- View this message in context: http://old.nabble.com/reading-tokens-tp26159931p26160018.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill map with gradient: package?
Hi, I'd like to fill an existing svg (or png) map with gradient colors. In detail: The file http://commons.wikimedia.org/wiki/File:Karte_%C3%96sterreich_Bundesl%C3%A4nder.svg should be filled with the population density data from this table: http://de.wikipedia.org/wiki/%C3%96sterreich#Verwaltungsgliederung choosing the right color saturation (or whatever). The final result should be something like this map: http://commons.wikimedia.org/wiki/Image:Bevoelkerungsdichte_-_Oesterreich.svg Is there a package or so for these two tasks (filling and color density ploting)? Thanks for your help, Thomas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Graph to show very few classes has maximum best performers
Hello I need help for the following: my data is class no. best performers 101 10 102 21 103 1 104 3 105 5 106 190 ... ... ... ... ... ... ... ... ... ... I want to plot a graph using R, to show that very few classes has maximum best performers. can someone guide me for this? I tried using cumulative frequencies and then plotting a graph but it does not serve the purpose. Thank you in advance Regards, Sunita -- View this message in context: http://old.nabble.com/Graph-to-show-very-few-classes-has-maximum-best-performers-tp26160026p26160026.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R2WinBUGS and R-2.10.0: The school example not running
Dear R users I have been encountering an error (in the Winbugs window: Blackbox - index out of range . ). So to me it seems R2WinBUGS is not linking up well with the new R version since the same model runs well under r-2.9.2. So is my assumption correct or am I doing something wrong. The data I used and the code are in the help of R2WinBUGS (?bugs). Regards, -- Luwis Diya, Leuven Biostatistics and Statistical Bioinformatics Centre (L-BioStat), Kapucijnenvoer 35 blok d - bus 7001, 3000 Leuven, Belgium Tel: +32 16 336886 or +32 16 336892 Fax: +32 16 337015 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] why is adnonis function called adonis {packageveg an}
stvienna wiener stvienna at gmail.com writes: I used google and looked at the documentation to find out why the ADONIS function is called adonis (in the vegan package). Because it is ANOSIM resurrected? The background is that vegan already had function anosim() which should come from ANOSIM = Analysis of Similarities. However, like all R stuff, it actually used dissimilarities and so should be called anodis() to be linguistically correct. When Hank Stevens started to write current adonis(), the first working versions were called anodis(), but that was that was confusing when anosim() alreaded handled dissimilarities. So there was a nifty swap of letters to give us adonis(), and that stuck. By the way, Peter Solymos wrote a metafunction using adonis(), and he called that function persephone(). We haven't released that (yet). Cheap thrills, but thrills anyway (and I can't afford expensive thrills). I am writing a document and would like to include a abbreviation list (similar to ANOSIM = Analysis of Similarities). Just go on. However, the name of the function is not necessarily the acronym of the method. Cheers, Jari Oksanen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plots with k-means
Hello Iuri, Code sent is a sample with basic functionality from the package I´m due to send to CRAN. Final package allows inteactivity (dragging and selecting points, zooming in/out fisheye effect, ...) and full exploration of the hyperbolic-alike space simulated. Example sent just give you a taste of the visual methaphor implemented, as described in http://www.antsearch.univ-tours.fr/publi/DacVenADMA2006.pdf Plot generated with sample code allows you to analize the way kmeans have worked. You are plotting every point in the sample and six groups appear clearly (groups in the sample are so intragroup homogeneus that in most cases you will see six points plotted). Regards, Eduardo San Miguel Martin BI Consultant 2009/11/2 Iuri Gavronski i...@ufrgs.br: David, Eduardo, Thanks for the code. I have run it and I'm not sure what to do with the graph when it comes up. Can I interact with it, such as an RGL graph? I've tried clicking or dragging with the mouse and nothing happens. My system is a Windows Vista and R2.9. Best, Iuri. On Mon, Nov 2, 2009 at 7:35 PM, David Winsemius dwinsem...@comcast.net wrote: The attached file did not come through to the list. I think you have some non-standard characters (or at least non-standard in my locale). I was able to get the code to run after using the Zap Gremlins function in TextWrangler. Prior to that treatment pretty much every line threw an error of this sort: setClass(Class = 'POI', + representation(matrizSim = 'matrix',cos.query.docs = 'vector', Error: unexpected input in: setClass(Class = 'POI', ¬ wordsInQuery = 'ANY',docs = 'matrix', objeto = 'matrix', objetoC Error: unexpected input in ¬ = 'matrix', Error: unexpected '=' in = Pcoords = 'matrix', PcoordsFI = 'matrix', newPcoords = 'matrix', Error: unexpected input in ¬ newcoords = 'numeric' , Error: unexpected ',' in newcoords = 'numeric' , newcoords_1 = 'numeric', M = 'numeric', poisTextCol = Error: unexpected input in ¬ I also needed to remove a couple of spaces between function names and parentheses when these occurred at ends-of-lines. Attached is a working version as a .txt file (which should make it through the list-serv: -- David. sessionInfo() R version 2.10.0 Patched (2009-10-29 r50258) x86_64-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] rms_2.1-0 Hmisc_3.7-0 survival_2.35-7 loaded via a namespace (and not attached): [1] cluster_1.12.1 grid_2.10.0 lattice_0.17-26 On Nov 2, 2009, at 3:43 PM, eduardo san miguel wrote: I send r-code in an attached file. 2009/11/2 Iuri Gavronski i...@proxima.adm.br: Eduardo, Would you mind sending me the R code in an attached file. Your code didn't work here and I am not sure it is because of line breaks from the email program. Iuri. On Mon, Nov 2, 2009 at 10:53 AM, eduardo san miguel eduardosa...@gmail.com wrote: Hello all, I have almost finished the development of a new package where ideas from Tamara Munzner, George Furnas and Costa and Venturini are implemented. 1.- Da Costa, David Venturini, Gilles (2006). An Interactive Visualization Environment for Data Exploration Using Points of Interest. adma 2006: 416-423 2.- Furnas, George (1986). Generalized Fisheye Views. Human Factors in computing systems, CHI '86 conference proceedings, ACM, New York, pp. 16-23. 3.- Heidi Lam, Ronald A. Rensink, and Tamara Munzner (2006). Effects of 2D Geometric Transformations on Visual Memory. Proc. Applied Perception in Graphics and Visualization (APGV 2006), 119-126, 2006. 4.- Keith Lau, Ron Rensink, and Tamara Munzner (2004). Perceptual Invariance of Nonlinear Focus+Context Transformations. Proc. First Symposium on Applied Perception in Graphics and Visualization (APGV 04) 2004, pp 65-72. This is a sample with some basic functionality and a VERY BASIC example with kmeans plotting. Comments will be greatly appreciated. Regards -- R CODE require(methods) setClass(Class = 'POI', representation(matrizSim = 'matrix',cos.query.docs = 'vector', wordsInQuery = 'ANY',docs = 'matrix', objeto = 'matrix', objetoC = 'matrix', Pcoords = 'matrix', PcoordsFI = 'matrix', newPcoords = 'matrix', newcoords = 'numeric' , newcoords_1 = 'numeric', M = 'numeric', poisTextCol = 'character' , colores = 'vector' , poisCircleCol = 'character' , linesCol = 'character', itemsCol = 'character', LABELS = 'logical', vscale = 'numeric', hscale = 'numeric', circleCol = 'character', plotCol = 'character', itemsFamily = 'character', lenteDefault = 'numeric', zoomDefault = 'numeric' , rateDefault = 'numeric' , topKDefault = 'numeric' , pal = 'character', selected = 'numeric' , circRadio = 'numeric' , IncVscale = 'numeric',
Re: [R] RMySQL and Stored procedures
Orvalho Augusto wrote: Can someone provide me a good way to circumvent the lack of calling Stored Procedures from RMySQL? I can not rewrite those stored procedures on R because there a lot more folks here that only understands SQL. The stored procedure returns a resultset. Use RODBC instead. Assuming two stored procedures : CREATE PROCEDURE getusers (in userid INT) BEGIN SELECT name from jos_users where id userid; END; CREATE PROCEDURE getusers0() BEGIN SELECT name from jos_users; END; The following works # library(RODBC) con = odbcConnect(mysql,uid=xxx,pwd=yyy) users0 = sqlQuery(con,CALL getusers0) users0 users = sqlQuery(con,CALL getusers(20)) users close(con) # Dieter -- View this message in context: http://old.nabble.com/RMySQL-and-Stored-procedures-tp26159085p26160033.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] bargraph.ci - CI and color question.
Hello, When using bargraph.CI in package sciplot can the bars for each group be different colors? How do I select the color for each group? When I use this instead of the default (SD vs SE): bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) Am I getting 95% CI bars? Thank you kindly, Michael Just __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to exclude certain columns by column names?
Linlin Yan wrote:
Try this:
x[, colnames(x) != 'a']
And more general:
x[, !colnames(x) %in% c('a','b')]
Paul
[1] 3 4
On Tue, Nov 3, 2009 at 9:31 AM, Peng Yu pengyu...@gmail.com wrote:
I can exclude columns by column number using '-'. But I wondering if
there is an easy way to exclude some columns by column names.
x=cbind(c(1,2),c(3,4))
x
[,1] [,2]
[1,]13
[2,]24
colnames(x)=c('a','b')
x
a b
[1,] 1 3
[2,] 2 4
x[,-'a']
Error in -a : invalid argument to unary operator
x[,-1]
[1] 3 4
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Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
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P.O. Box 80.115
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Phone: +3130 274 3113 Mon-Tue
Phone: +3130 253 5773 Wed-Fri
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[R] D'agostino-Pearson K2 omnibus test.
Hi, Is there an implementation of the D'agostino K2 omnibus test in an R package. I've found the D test in the fBasics package, but this doesn't give the same answer as my algorithm. I've already coded a version of the K2 and need to check the results. I have a set of spectra with gaussian distributed noise and I want a statistical test to separate those with no signal (so the data should be distributed normally) from those spectra with a line. Has anyone any experience of this? Any help would be greatly appreciated. James __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lmer and estimable
Hi everyone,
I'm using lmer and estimable (from packages lme4 and gmodels respectively) and
have the disconcerting happening that when I run exactly the same code, I get
different results! In checking this out by running the code 50x, it seems to
be that answers may be randomly deviating around those which I get from another
stats package (GenStat, using the linear mixed models functionality in this).
As such, my guess is that perhaps there is a random seed involved somewhere in
the estimable function, and that the answer is simply converging to a slightly
different place each time, based on a slightly different starting place. But
this is only a guess - if anyone could either confirm or otherwise explain this
unexpected-to-me happening, that'd be fantastic!!
The code that I'm running is:
design.1 = rbind(c(0,0,0,1,0,0,0,0,0,0,0,0))
lme1 - lmer(var_x~AGE*GENDER*DAM_DIET + (1|CHIP) + (1|LITTER),x_temp)
est - estimable(lme1,design.1)
print(est)
est - estimable(lme1,design.1) # Repeat of code
print(est)
est - estimable(lme1,design.1) # Another repeat of code
print(est)
And this is the output - you can see in this that the results vary each time
the code is run...
print(est)
Estimate Std. Error p value
(0 0 0 1 0 0 0 0 0 0 0 0) 0.02241752 0.2185279 0.902
est - estimable(lme1,design.1)
print(est)
Estimate Std. Error p value
(0 0 0 1 0 0 0 0 0 0 0 0) 0.001824573 0.2106973 0.974
est - estimable(lme1,design.1)
print(est)
Estimate Std. Error p value
(0 0 0 1 0 0 0 0 0 0 0 0) 0.01276941 0.2154196 0.938
My SessionInfo is as follows:
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=English_New Zealand.1252;LC_CTYPE=English_New
Zealand.1252;LC_MONETARY=English_New
Zealand.1252;LC_NUMERIC=C;LC_TIME=English_New Zealand.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets tcltk
methods base
other attached packages:
[1] lme4_0.999375-32 Matrix_0.999375-31 lattice_0.17-25multcomp_1.1-2
survival_2.35-7mvtnorm_0.9-7 gmodels_2.15.0 svIDE_0.9-5
loaded via a namespace (and not attached):
[1] gdata_2.4.2 grid_2.9.2 gtools_2.6.1 MASS_7.2-48 svMisc_0.9-5
Many thanks for any help or advice that anyone can provide :)
Cheers,
Zaneta
Zaneta Park
Biometrician
T +64 6 351 8008
E zaneta.park...@agresearch.co.nz
AgResearch Limited
Grasslands Research Centre
Tennent Drive, Private Bag 11008, Palmerston North, New Zealand
T +64 6 356 8019 F +64 6 351 8032 www.agresearch.co.nz
Farming Food and Health. First
Te Ahuwhenua Te Kai me te Whai Ora. Tuatahi
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Re: [R] How to exclude certain columns by column names?
Try this;
x[,setdiff(colnames(x), 'a')]
On Mon, Nov 2, 2009 at 11:31 PM, Peng Yu pengyu...@gmail.com wrote:
I can exclude columns by column number using '-'. But I wondering if
there is an easy way to exclude some columns by column names.
x=cbind(c(1,2),c(3,4))
x
[,1] [,2]
[1,] 1 3
[2,] 2 4
colnames(x)=c('a','b')
x
a b
[1,] 1 3
[2,] 2 4
x[,-'a']
Error in -a : invalid argument to unary operator
x[,-1]
[1] 3 4
__
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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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Run R code by Vim-R-plugin (How do I to run two or more programs at the same time)?
On Tue, Nov 03, 2009 at 04:59:30PM +0800, Pan, Jia-chiun/潘家群 wrote: I use R on Ubuntu 9.04 by Vim editor with vim-R-plugin. I want to use Vim to run two R programs A.R and B.R. At beginning, I sent A.R to R prompt by pressing \rf and F5 on my keyboard, and it works well. Meanwhile, I want to sent another file B.R to another R window. When I press \rf, the R window shows up and still runs A.R. How could I run B.R in another R window by using vim-R-plugin? You may put the following line in your vimrc: let g:vimrplugin_nosingler = 1 Then, each Vim buffer will start its own R process. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expression substitution ...
Sander Timmer wrote: It's a bit strange: sub(/pc, '\\%', x) [1] % sub(/pc, '%', x) [1] \\% Also with fixed I'm not able to get a single \ as return value. Well, you get the printed representation (i.e. the quoted backslash - quoted with anotern backslash). Use cat(sub(/pc, '%', x)) to see the raw characters. Uwe Ligges Sander On 2 nov 2009, at 12:36, Roberto Brunelli wrote: I would like to replace all occurences of /pc with \% using something like sub(/pc, \%, x) but I am unable to make the '\' pass through. Any hint? Thanks, Roberto __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R2WinBUGS and R-2.10.0: The school example not running
Hi, I got another report about and will investigate shortly. uwe Luwis Tapiwa Diya wrote: Dear R users I have been encountering an error (in the Winbugs window: Blackbox - index out of range . ). So to me it seems R2WinBUGS is not linking up well with the new R version since the same model runs well under r-2.9.2. So is my assumption correct or am I doing something wrong. The data I used and the code are in the help of R2WinBUGS (?bugs). Regards, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [R-SIG-Finance] Additive decomposition method
Hi, Thank you for the function. I was trying to use the decompose function, but the program cannot run as there were several missing values in my data frame. Could someone advice me the way to handle on this matter? Thanks F __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] re ading tokens
On Tue, Nov 3, 2009 at 3:02 AM, Dieter Menne dieter.me...@menne-biomed.de wrote: j daniel wrote: I am not familiar with processing text in R. Can someone tell me how to read each line of words as separate elements in a list? FE, I would like to turn: word1 word2 word3 word2 word4 into a list of length two with three character elements in the first list and two elements in the second. I know that this should be easy, but I am a little confused by the text functions. You could use scan. Have a look at package gsubfn, where there is a demo, that show additional features you are going to use library(gsubfn) demo(gsubfn-gries) The example code is a bit overnested, but to better understand what is going on, unwrap it: So tail(sort(table(unlist(strapply(Lines1, \\w+, perl = TRUE) is: x1 = strapply(Lines1, \\w+, perl = TRUE) x1 x2 = ulist(x2) x2 x3 = table(x2) x3 x4 = sort(x3) x4 tail(x4) Just one small optimization. You don't actually need perl = TRUE here. By default, strapply uses tcl and the regex engine in tcl which does support \w and which can process the input faster although with the input of the above size it won't be material. If you do specify perl = TRUE (or engine = R) then it will use R and R's perl engine. There is a link to the tcl regex page on the gsubfn home page in the links box: http://gsubfn.googlecode.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package(vars) and generalized impulse response functions
Hi, I was wondering whether the package(vars) can estimate generalized impulse response functions for VARs / VECMs. The general option on package(vars) available is:- suppose I want to give a shock to 'a' and observe the response of 'b', then irf(x, impulse=a, response=b, n.ahead=10, boot=TRUE,ci=0.95,runs=100,seed=NULL...) The vignettes and manuals says that the default is 'ortho=TRUE' (and the other option is cumulative=NULL). Or is there any other package one would suggest? Thanks, Regards, Vimal [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting an interaction with error bars
Hello.
I need to plot a two-way interaction (5 levels X 3) with error bars.
The x.factor will be the five-levels var and the trace.factor will be
the three level var.
I was able to find functions that draw error bars, but still couldn't
find a way to draw an interaction plot that looks like what is common
in psychological (and related) journals: the lines (three in my need)
are plotted in parallel and to avoid overlapping of the error bars a
small jitter is added.
I was able to make plotmeans() {from gplots} draw the interaction but
with the lines plotted side by side (not parallel), resulting in three
panes of one line each. When I tried manually super-imposing (looping
a draw of one line for each level in the trace.factor) I got the
overall look, but the error bars overlap and are hard to distinguish.
Another attempt I made was with ci.plot() {from pda}. Inputting an
aov() model resulted in a perfect plot of interaction but which is
very limited to handle - I was unable to add a legend, change symbols
etc.
Does anybody knows if there is a way to manipulate either of the above
functions more then what I was able to?
Is there another function to draw interaction plots with error bars?
Thanks for any help or comment.
dror
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Re: [R] OpenOffice Calc ODBC equivalent
On Nov 2, 2009, at 10:10 PM, Kenneth Roy Cabrera Torres wrote: Hi R users: I am using RODBC to create some new .xls files each with several sheets (about 100) with sqlSave() from a data.frame inside R without any problem, but on windows XP platform. I would like to know if it is posible to make a similar solution on linux with openoffice? RODBC work only (for the moment) with .xls on windows, via the ODBC driver. Is it posible to make the same thing on openoffice? I mean, an .odt file with several sheets each, generated from a data.frame in R? Thank you for your help. Kenneth See this post from late 2007: http://tolstoy.newcastle.edu.au/R/e3/help/07/12/6478.html As far as I know, nothing has changed vis-a-via ODBC connectivity TO OpenOffice files. You can use ODBC FROM OpenOffice to connect to external data sources. A search of CRAN does not reveal any packages/functions to would appear to provide a facility to directly write data frames to a Calc file. However, you can always write to a XLS file, open it in Calc and re-save it as a native Calc file. Of course, anyone with Calc can open an XLS file, so you are not losing much in terms of portability by saving to an XLS file and distributing that. You also make it easier for folks who may be using other 'office' applications that do not have import filters for OO.org file formats. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] tapply function
Hi, I tried to use tapply function to find the mean of the data in each group as the following command, but the result are NA, as there are several missing values in each group. tapply(data,group,mean) Could someone please advice me the way to ignore the missing data in order for the fucntion to run successfully? Thanks Fir __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] hierarchical clustering with Jaccard index
Try Jaccard. :-) karuna m wrote: hi, I want to do hierarchical clustering with Jaccord index. I tried to do with vegan package for finding index and hierarchical clustering with hclust function. While doing clustering it is showing an error message as invalid distance method. I would be grateful if anyone tells how to rectify the error. Thanks in advance, kind regards, Ms.Karunambigai M PhD Scholar Dept. of Biostatistics NIMHANS Bangalore India Keep up with people you care about with Yahoo! India Mail. Learn how. http://in.overview.mail.yahoo.com/connectmore [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Don McKenzie Research Ecologist Pacific Wildland Fire Sciences Lab US Forest Service Affiliate Professor College of Forest Resources and CSES Climate Impacts Group University of Washington phone: 206-732-7824 cell: 206-321-5966 d...@u.washington.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] tapply function
you must have missing values in data. Try tapply(data, group, mean, na.rm = TRUE) If that's not the case, read the bottom of this email about the posting guide. HTH, --sundar On Tue, Nov 3, 2009 at 5:28 AM, FMH kagba2...@yahoo.com wrote: Hi, I tried to use tapply function to find the mean of the data in each group as the following command, but the result are NA, as there are several missing values in each group. tapply(data,group,mean) Could someone please advice me the way to ignore the missing data in order for the fucntion to run successfully? Thanks Fir __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fitting a confined mixture model
Hello all, I would like to fit a mixture model whose components are normal distributions confined in a closed interval. Since there are already several packages for EM, I would like to extend one of these instead of writing a new script from scratch. What would be the best way to customize such an existing package for doing that? Which one is the best in terms of extensibility? (e.g mixtools, flexmix ..) -- Giovanni L. Ciampaglia PhD Student University of Lugano, MACS Lab __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Exact String Compare in R?
Dear R users:
Here's a barebones example of what I can't make work.
As you can see, regexpr() does not perform an exact string match, which only
occurs in row 1 of these data frames. Instead, as it's supposed to do, it
finds b in bb and c in cc. Does anybody know what function I can use
such that only the first rows would be matched (ie, exact string match?)
I've also tried simply using the == operator, in which case i get the
error:level sets of factors are different
Thank you in advance,
B
## two toy data frames, containing character arrays
D1=as.data.frame(c(a,b,c))
D2=as.data.frame(c(a,bb,cc))
## loop through each comparing the strings in each row
i=1 #counter
while (regexpr(D1[i,1], D2[i,1]) == TRUE) {
+cat(identical match on row #, i, \n)
+i=i+1
+ if (i3) break
+ }
identical match on row # 1
identical match on row # 2
identical match on row # 3
--
View this message in context:
http://old.nabble.com/Exact-String-Compare-in-R--tp26160122p26160122.html
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and provide commented, minimal, self-contained, reproducible code.
[R] fixed color scale for a series of contour plots
Hello, I am having trouble figuring out how to create a color scale that will be consistent across multiple contour plots. What I would like is to plot a series of contour or image plots for different subsets of a data set but make sure that the color key stays fixed so that the plots are easily comparable. That is I need a map from the range of values of the entire data set to the color scale and keep then use this color scale for all plots + be able to display the color bar as a legend next to the plot. This sounds trivial, but I can't find useful hints in the documentation of the color arguments of the different plots. Thanks so much for any help Best Marion __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R2WinBUGS not working?
This weekend I noticed that my R2WinBUGS connection was no longer working on my Windows computer at work AND my Ubuntu linux computer at home. As soon as WinBUGS opens, the message reads Index Out Of Range. I have un-installed and re-installed R on both computers and the problem remains the same. At first I thought it was the new version of R, but I was able to overcome the problem (on Windows only) with the use of program=openbugs in R2WinBUGS. But that doesn't work on Linux. PS The smallest WinBUGS program I have, essentially a t-test, still works. But anything larger than that gets the error. Help. Thanks. -- View this message in context: http://old.nabble.com/R2WinBUGS-not-working--tp26160111p26160111.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re trieve scores using a cov matrix in factanal
Hi My goal is to do a weighted (by population) factor analysis (of votes) using R. For that I prepared the weigthed cov matrix, which matrix I pass in the factanal function. I'm facing troubles in calculating the scores of a factor analysis using as entry a cov matrix as Factanal is not able to get the scores (normal in the case of absence of any x). I'm looking for a method to calculate the scores manually afterward. Does anybody has the way I can retrieve that? Abram MFSA.ch -- View this message in context: http://old.nabble.com/Retrieve-scores-using-a-cov-matrix-in-factanal-tp26160114p26160114.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] random text added to names (bug with 2.10.0?)
I'm using 2.10.0 on Linux (64 bit), and I just noticed that random
numbers are occasionally added to the text of names in vectors. It's
happened to me in two separate, long-running R sessions, but I can't
find a way to reproduce it in a smaller setting.
The code I'm using is
diag.gam.2 - mdl.run.diag(fit.gam.2, ds.valid)
diag.gam.2
rmse mdae.1413751 mnae cv r2
0.023171639 0.007206221 0.012374469 1.259498479 0.385234756
mdl.run.diag
function (fit, ds) {
run.diagnostics(mdl.run(fit, ds))
}
run.diagnostics
function (ds) {
c(rmse=root.mean.squared.error(ds),
mdae=median.absolute.error(ds),
mnae=mean.absolute.error(ds),
cv=coefficient.of.variation(ds),
r2=coefficient.of.determination(ds))
}
The problem is that extra .1413751 that's mysteriously being added
to the mdae name.
Am I doing something obviously wrong, or is this a bug? I still have
the two sessions running, if there are any diagnostics would help.
Cheers,
Johann
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Re: [R] Exact String Compare in R?
On Tue, Nov 3, 2009 at 2:41 PM, bamsel benam...@gmail.com wrote:
Dear R users:
Here's a barebones example of what I can't make work.
As you can see, regexpr() does not perform an exact string match, which only
occurs in row 1 of these data frames. Instead, as it's supposed to do, it
finds b in bb and c in cc. Does anybody know what function I can use
such that only the first rows would be matched (ie, exact string match?)
I've also tried simply using the == operator, in which case i get the
error:level sets of factors are different
Because they are not strings, but factors. Convert them to strings
with as.character and use ==.
Best,
Gabor
Thank you in advance,
B
## two toy data frames, containing character arrays
D1=as.data.frame(c(a,b,c))
D2=as.data.frame(c(a,bb,cc))
## loop through each comparing the strings in each row
i=1 #counter
while (regexpr(D1[i,1], D2[i,1]) == TRUE) {
+ cat(identical match on row #, i, \n)
+ i=i+1
+ if (i3) break
+ }
identical match on row # 1
identical match on row # 2
identical match on row # 3
--
View this message in context:
http://old.nabble.com/Exact-String-Compare-in-R--tp26160122p26160122.html
Sent from the R help mailing list archive at Nabble.com.
__
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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.
--
Gabor Csardi gabor.csa...@unil.ch UNIL DGM
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Re: [R] fixed color scale for a series of contour plots
You don't tell us which function you use, but fixing the zlim argument in image or related functions should do the trick. M. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Marion Dumas Sent: Dienstag, 3. November 2009 14:38 To: r-help@r-project.org Subject: [R] fixed color scale for a series of contour plots Hello, I am having trouble figuring out how to create a color scale that will be consistent across multiple contour plots. What I would like is to plot a series of contour or image plots for different subsets of a data set but make sure that the color key stays fixed so that the plots are easily comparable. That is I need a map from the range of values of the entire data set to the color scale and keep then use this color scale for all plots + be able to display the color bar as a legend next to the plot. This sounds trivial, but I can't find useful hints in the documentation of the color arguments of the different plots. Thanks so much for any help Best Marion __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Exact String Compare in R?
Hi
what about
D2[,1] %in% D1[,1]
Regards
Petr
r-help-boun...@r-project.org napsal dne 03.11.2009 14:41:02:
Dear R users:
Here's a barebones example of what I can't make work.
As you can see, regexpr() does not perform an exact string match, which
only
occurs in row 1 of these data frames. Instead, as it's supposed to do,
it
finds b in bb and c in cc. Does anybody know what function I can
use
such that only the first rows would be matched (ie, exact string match?)
I've also tried simply using the == operator, in which case i get the
error:level sets of factors are different
Thank you in advance,
B
## two toy data frames, containing character arrays
D1=as.data.frame(c(a,b,c))
D2=as.data.frame(c(a,bb,cc))
## loop through each comparing the strings in each row
i=1 #counter
while (regexpr(D1[i,1], D2[i,1]) == TRUE) {
+cat(identical match on row #, i, \n)
+i=i+1
+ if (i3) break
+ }
identical match on row # 1
identical match on row # 2
identical match on row # 3
--
View this message in context:
http://old.nabble.com/Exact-String-Compare-in-
R--tp26160122p26160122.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] bargraph.ci - CI and color question.
On Nov 3, 2009, at 4:51 AM, Michael Just wrote: Hello, When using bargraph.CI in package sciplot can the bars for each group be different colors? Yes. ?bargraph.CI How do I select the color for each group? Read the help page. Fifth argument down. When I use this instead of the default (SD vs SE): bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) Am I getting 95% CI bars? Something +/- one SD(something will almost never be a 95% CI. But this raises the question: CI bars for what parameter or statistic? (And raises the question: Have you read the help page entry regarding that parameter?) I do think it's good that you are considering plotting standard deviations rather than std errors of the mean, because the more common practice of plotting sem's obscures the degree of variability in the population. It's just that you appear to be aiming for what might be called a prediction interval and for it to be a 95% one, would need to be +/- 2 SD's. And if the numbers in each group are small, the 2 would get replaced with a t-statistic. If on the other hand you want to know what would be the right label for mean +/- sd, then given that +/- 1.96 SD would be a 95% prediction interval (for large samples) then you can see from 1-2*pnorm(-1.96) [1] 0.9500042 1-2*pnorm(-1) [1] 0.6826895 ... that +/- one SD would be a 68% prediction interval. Thank you kindly, Michael Just -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package lme4
Thanks,Douglas, It really helps me a lot, but is there any other way if I want to show whether a random effect is significant in text file, like P value or other index. Thanks very much again. Wenjun. 2009/11/2 Douglas Bates ba...@stat.wisc.edu On Sun, Nov 1, 2009 at 9:01 AM, wenjun zheng wjzhen...@gmail.com wrote: Hi R Users, When I use package lme4 for mixed model analysis, I can't distinguish the significant and insignificant variables from all random independent variables. Here is my data and result: Data: Rice-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9), Variety=rep(rep(c(A1,A2,A3),each=3),3), Stand=rep(c(B1,B2,B3),9), Block=rep(1:3,each=9)) Rice.lmer-lmer(Yield ~ (1|Variety) + (1|Stand) + (1|Block) + (1|Variety:Stand), data = Rice) Result: Linear mixed model fit by REML Formula: Yield ~ (1 | Variety) + (1 | Stand) + (1 | Block) + (1 | Variety:Stand) Data: Rice AIC BIC logLik deviance REMLdev 96.25 104.0 -42.1285.33 84.25 Random effects: GroupsNameVariance Std.Dev. Variety:Stand (Intercept) 1.345679 1.16003 Block (Intercept) 0.00 0.0 Stand (Intercept) 0.89 0.94281 Variety (Intercept) 0.024691 0.15714 Residual 0.67 0.81650 Number of obs: 27, groups: Variety:Stand, 9; Block, 3; Stand, 3; Variety, 3 Fixed effects: Estimate Std. Error t value (Intercept) 7.1852 0.6919 10.38 Can you give me some advice for recognizing the significant variables among random effects above without other calculating. Well, since the estimate of the variance due to Block is zero, that's probably not one of the significant random effects. Why do you want to do this without other calculations? In olden days when each model fit involved substantial calculations by hand one did try to avoid fitting multiple models but now that is not a problem. You can get a hint of which random effects will be significant by looking at their precision in a caterpillar plot and then fit the reduced model and use anova to compare models. See the enclosed Any suggestions will be appreciated. Wenjun [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Scanning grep through huge files
Hi, I'm dealing which huge files I would like to index. On a linux system grep -buo PATTERN FILENAME hands me the byte offsets for PATTERN very quickly and I am looking to emulate that speed and ease with native R tools - for portability and elegance. gregexpr should be able to do that but I fail to combine it with scan or an equivalent to parse the whole file without having to read it all into memory. I'd be grateful for any hints on how to do this without a pipe(grep -buo PATTERN FILENAME). Thanks, Joh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 1 dimensional optimization with local minima
Hi,
One approach is to use the `multiStart' function in my BB package. You have
to provide a matrix of starting values for this.
Here is a simple, one-dimensional optimization problem with multiple peaks and
valleys. I show how to find all the peaks in the bounded interval [0, 1].
myfn - function(x) {
exp(-k1 * x) * sin(2*pi*k2*x)
}
k1 - 0.5
k2 - 5
x - seq(0, 1, length=1000)
plot(x, fn(x), type=l)
require(BB)
p0mat - matrix(runif(100), 100, 1) # random starting values
ans - multiStart(par=p0mat, fn=myfn, lower=0, upper=1, action=optimize,
control=list(maximize=TRUE))
pconv - ans$par[ans$conv] # converged solutions
rp - !duplicated(round(pconv, 3)) # unique, converged solutions
sort(pconv[rp]) # lists all the local maxima
Hope this helps,
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
- Original Message -
From: Jeroen Ooms jeroen.o...@stat.ucla.edu
Date: Tuesday, November 3, 2009 2:27 am
Subject: [R] 1 dimensional optimization with local minima
To: r-help@r-project.org
I am using numerical optimization to fit a 1 parameter model, in which
the
input parameter is bounded. I am currently using optimize(), however,
the
problem turns out to have local minima, and optimize does not always
seem to
find the global minimum. I could to write a wrapping function that tries
multiple intervals or starting values, but I would prefer a package
that has
built-in methods to make it more robust against local minima.
I checked the CRAN Task View for Optimization, however there seem to
be a
lot of alternatives, and not being an optimization expert, I could use
some
advice. What could be an appropriate package or function for one-dimensional
bounded optimization, that includes some protection against local
minima?
-
Jeroen Ooms * Dept. of Methodology and Statistics * Utrecht University
Visit www.jeroenooms.com to explore some of my
current projects.
--
View this message in context:
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] 1 dimensional optimization with local minima
Most methods for optimization in R seek local minima, though there are some -- mainly for 1 dimensions, that attempt to find the global minimum stochastically (optim method SANN, DEoptim). SANN does, I believe, handle 1 dimension, but does not have a convergence test, but runs a fixed number of function evaluations (I got bitten by this unusual behaviour, as it returns a converged flag after the specified number of evaluations). While there are some techniques that can generate global minima given conditions on the function, I would anticipate you will do better graphing your function(s) to learn if they have just a few (++) or very many (-- = very bad for you) optima. You should apply bounds to your domain -- a lot of computational time is wasted when routines take a wild excursion away from a region of interest. That may be enough. Or you may need to segment the domain to isolate the minima then select the best. That is, find a local minimum, choose a new interval that does not include this and apply optimize() again. Messy, but unless you have lots of local minima, should work OK. If you do have lots, it is likely you need to re-pose your problem. JN Message: 114 Date: Mon, 2 Nov 2009 23:00:40 -0800 (PST) From: Jeroen Ooms jeroen.o...@stat.ucla.edu Subject: [R] 1 dimensional optimization with local minima To: r-help@r-project.org Message-ID: 26160001.p...@talk.nabble.com Content-Type: text/plain; charset=us-ascii I am using numerical optimization to fit a 1 parameter model, in which the input parameter is bounded. I am currently using optimize(), however, the problem turns out to have local minima, and optimize does not always seem to find the global minimum. I could to write a wrapping function that tries multiple intervals or starting values, but I would prefer a package that has built-in methods to make it more robust against local minima. I checked the CRAN Task View for Optimization, however there seem to be a lot of alternatives, and not being an optimization expert, I could use some advice. What could be an appropriate package or function for one-dimensional bounded optimization, that includes some protection against local minima? - Jeroen Ooms * Dept. of Methodology and Statistics * Utrecht University Visit http://www.jeroenooms.com www.jeroenooms.com to explore some of my current projects. -- View this message in context: http://old.nabble.com/1-dimensional-optimization-with-local-minima-tp26160001p26160001.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to display a string containing greek chrs and variables
I'm trying something that I thought would be pretty simple, but it's proving quite frustrating... I want to display, for instance, the correlation coefficient rho in a graph. I can do something like: text(x, y, paste(rho =, cor)) where cor would be my previously calculated correlation coefficient, and x and y the coordinates. Obviously that displays rho, not the greek letter rho. I can do also this: text(x, y, paste(\\*r =,cor), vfont = c(sans serif,bold)) that works, but that doesn't look very pretty. What looks good is doing something like: text(x, y, expression(rho)) But I haven't managed to make it do what I need. I can get this: text(x, y, expression(rho == 0.8)) that looks good, but how can I substitute the 0.8 for teh name of a numeric variable whose value is 0.8? I'm going slightly mad... it can't be that hard. Thanks for any help! Jose -- Dr. Jose I. de las Heras Email: j.delashe...@ed.ac.uk The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131 6513374 Institute for Cell Molecular BiologyFax: +44 (0)131 6507360 Swann Building, Mayfield Road University of Edinburgh Edinburgh EH9 3JR UK * NEW EMAIL from July'09: nach.mcn...@gmail.com * -- The University of Edinburgh is a charitable body, registered in Scotland, with registration number SC005336. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Scanning grep through huge files
On 11/3/2009 9:29 AM, Johannes Graumann wrote: Hi, I'm dealing which huge files I would like to index. On a linux system grep -buo PATTERN FILENAME hands me the byte offsets for PATTERN very quickly and I am looking to emulate that speed and ease with native R tools - for portability and elegance. gregexpr should be able to do that but I fail to combine it with scan or an equivalent to parse the whole file without having to read it all into memory. I think you are going to have to write this yourself. R doesn't have very many stream oriented functions: almost everything is aimed at having the whole thing in memory. You will also have trouble with the byte offsets. The semantics of the -u option to grep are quite strange (at least according to the man page on Cygwin). What I'd do given your problem is use readLines to read the file, then post-process the result of gregexpr to give line and byte offset pairs for each match; those are more useful in R than the rather bizarre byte offsets that grep -buo will give. But for a huge file you'll probably have to do this in blocks, as the whole file may be too big. Duncan Murdoch I'd be grateful for any hints on how to do this without a pipe(grep -buo PATTERN FILENAME). Thanks, Joh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bargraph.ci - CI and color question.
On Tue, 2009-11-03 at 03:51 -0600, Michael Just wrote: Hello, When using bargraph.CI in package sciplot can the bars for each group be different colors? How do I select the color for each group? With the option err.col bargraph.CI(dose, len, group = supp, data =ToothGrowth, err.col=c(gray,black)) When I use this instead of the default (SD vs SE): bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) Am I getting 95% CI bars? No. You're plotting +/- 1 SD. Thank you kindly, Michael Just __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://mutualism.williams.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to display a string containing greek chrs and variables
Hi, try this, plot.new() x=0.8 text(0.5, 0.5, bquote(rho == .(x))) HTH, baptiste 2009/11/3 j.delashe...@ed.ac.uk: I'm trying something that I thought would be pretty simple, but it's proving quite frustrating... I want to display, for instance, the correlation coefficient rho in a graph. I can do something like: text(x, y, paste(rho =, cor)) where cor would be my previously calculated correlation coefficient, and x and y the coordinates. Obviously that displays rho, not the greek letter rho. I can do also this: text(x, y, paste(\\*r =,cor), vfont = c(sans serif,bold)) that works, but that doesn't look very pretty. What looks good is doing something like: text(x, y, expression(rho)) But I haven't managed to make it do what I need. I can get this: text(x, y, expression(rho == 0.8)) that looks good, but how can I substitute the 0.8 for teh name of a numeric variable whose value is 0.8? I'm going slightly mad... it can't be that hard. Thanks for any help! Jose -- Dr. Jose I. de las Heras Email: j.delashe...@ed.ac.uk The Wellcome Trust Centre for Cell Biology Phone: +44 (0)131 6513374 Institute for Cell Molecular Biology Fax: +44 (0)131 6507360 Swann Building, Mayfield Road University of Edinburgh Edinburgh EH9 3JR UK * NEW EMAIL from July'09: nach.mcn...@gmail.com * -- The University of Edinburgh is a charitable body, registered in Scotland, with registration number SC005336. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Change negative values in column
Hi all,
I'm trying to write a script that changes all negative values in a data frame
column to a small positive value, based on the the minimum value of the column.
However, I get the following error:
Error in if (x[i] 0) { : argument is of length zero
As well, I would minimum to be the smallest of the non-negative values...
Aa_non_neg=(fieldTrial0809$Aa) # Copy column from data frame to manipulate
nonNegative = function(x)
{
minimum=min(x) # Should only use positive minimum!
for (i in x)
{
if(x[i]0) # Found a negative value
{
x[i]=minimum/10 # Change to a new non-negative value
}
}
}
nonNegative(Aa_non_neg) # Apply function on column
_
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] package lme4
May be I can calculate p value by t testing approximately: 1-qnorm(Variance/Std.Dev.) But which function can help me to extract Variance and Std.Dev values from the results below: print(fm2 - lmer(Yield ~ 1 + (1|Stand) + (1|Variety) + (1|Variety:Stand),Rice)) Linear mixed model fit by REML Formula: Yield ~ 1 + (1 | Stand) + (1 | Variety) + (1 | Variety:Stand) Data: Rice AIC BIC logLik deviance REMLdev 94.25 100.7 -42.1285.33 84.25 Random effects: GroupsNameVariance Std.Dev. Variety:Stand (Intercept) 1.345679 1.16003 Variety (Intercept) 0.024692 0.15714 Stand (Intercept) 0.88 0.94281 Residual 0.67 0.81650 Number of obs: 27, groups: Variety:Stand, 9; Variety, 3; Stand, 3 Fixed effects: Estimate Std. Error t value (Intercept) 7.1852 0.6919 10.38 2009/11/2 Douglas Bates ba...@stat.wisc.edu On Sun, Nov 1, 2009 at 9:01 AM, wenjun zheng wjzhen...@gmail.com wrote: Hi R Users, When I use package lme4 for mixed model analysis, I can't distinguish the significant and insignificant variables from all random independent variables. Here is my data and result: Data: Rice-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9), Variety=rep(rep(c(A1,A2,A3),each=3),3), Stand=rep(c(B1,B2,B3),9), Block=rep(1:3,each=9)) Rice.lmer-lmer(Yield ~ (1|Variety) + (1|Stand) + (1|Block) + (1|Variety:Stand), data = Rice) Result: Linear mixed model fit by REML Formula: Yield ~ (1 | Variety) + (1 | Stand) + (1 | Block) + (1 | Variety:Stand) Data: Rice AIC BIC logLik deviance REMLdev 96.25 104.0 -42.1285.33 84.25 Random effects: GroupsNameVariance Std.Dev. Variety:Stand (Intercept) 1.345679 1.16003 Block (Intercept) 0.00 0.0 Stand (Intercept) 0.89 0.94281 Variety (Intercept) 0.024691 0.15714 Residual 0.67 0.81650 Number of obs: 27, groups: Variety:Stand, 9; Block, 3; Stand, 3; Variety, 3 Fixed effects: Estimate Std. Error t value (Intercept) 7.1852 0.6919 10.38 Can you give me some advice for recognizing the significant variables among random effects above without other calculating. Well, since the estimate of the variance due to Block is zero, that's probably not one of the significant random effects. Why do you want to do this without other calculations? In olden days when each model fit involved substantial calculations by hand one did try to avoid fitting multiple models but now that is not a problem. You can get a hint of which random effects will be significant by looking at their precision in a caterpillar plot and then fit the reduced model and use anova to compare models. See the enclosed Any suggestions will be appreciated. Wenjun [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mac question
Hi,
On Nov 3, 2009, at 2:41 AM, Antje wrote:
Hi there,
currently, I've updated R on my Mac (OS X) to version 2.10. I was
wondering if I have to install all additional packages again???
In Windows, I just needed to copy the library folder of the old
installation but how does it work with Mac?
This isn't really a mac or windows thing, actually. Although this
method may have worked for you in the past, it's generally not
recommended to use libraries compiled from one R version in another.
Using the same libraries between 2.9.X to 2.9.X+n is fine, but doing
the same when going from 2.9 to 2.10 is not a good idea.
So, in short, you should updated/reinstall your packages now that
you've updated R.
An easy way to do that would have been to save a file of all packages
installed from 2.9, then use that file to reinstall those packages
from within 2.10. Here's a recipe to do that now (assuming R 2.9 is
still installed on your system).
You can use RSwitch to switch which version of R is active at a
time, available here:
http://r.research.att.com/RSwitch-1.1.dmg
1. Fire up RSwitch and pick R 2.9
1.5. I have these two functions I use when I want to be lazy:
installedPackages - function(to=~/R.packages.dump.txt) {
writeLines(installed.packages()[,1], con=to)
}
reloadPackages - function(from=~/R.packages.dump.txt) {
source(http://bioconductor.org/biocLite.R;)
packages - readLines(from)
biocLite(packages)
}
2. Call installedPackages() from within R 2.9 and quit
3. Use rswitch to make R 2.10 active
4. Open R 2.10 and run reloadPackages()
biocLite is a function to install bioconductor packages, but it also
installs normal cran packages. You can likely replace the call to
biocLite with a call to install.packages(packages) if you don't have
any bioc packages to install, if you like, although it shouldn't be
necessary.
It might take a while to d/l all your packages again, but it should
just work
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] RandomForest PMML
I cannot seem to write a randomforest model in PMML - either through calling PMML(model) or through Rattle. It appears that it is not yet supported. Randomsurvivalforest is, but not randomforest. Any ideas on possible workarounds for this? Thanks ncs __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] about the cox result
coef(summary(fit_cox)) should give you what you wanted. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: 515-294-6609 Web: http://yihui.name Department of Statistics, Iowa State University 3211 Snedecor Hall, Ames, IA On Tue, Nov 3, 2009 at 12:31 AM, 孟欣 lm_meng...@163.com wrote: Hi all: I finished cox analysis like this: fit_cox-coxph(Surv(dat$Time, dat$death) ~ dat$CD4 + strata(dat$gender),data=dat); fit_cox Call: coxph(formula = Surv(data_ori$Time, data_ori$death) ~ data_ori$drug + strata(data_ori$gender), data = data_ori) coef exp(coef) se(coef) z p data_ori$drugddI 0.216 1.24 0.146 1.47 0.14 Likelihood ratio test=2.17 on 1 df, p=0.140 n= 467 I wanna extract the result: 0.216 1.24 0.146 1.47 0.14 and the corresponding name coef exp(coef) se(coef) z p from fit_cox. I use the command: str(fit_cox),but I can only find 0.216 of the result,but the other four results( 1.24 0.146 1.47 0.14) can't be found. Anyone can help me? Thanks a lot! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Calculate Averages for time data
Hello I have time data which is in hh:mm:ss format and I need to calculate averages for this data. I tried converting the data as integer so that I could calculate the average, but it doesn't serve the purpose. I tried using Chron package also. But I am stuck up as to how to deal with the time data. Please guide me as to how to deal with Time data in R. Thank you in advance Regards Sunita -- View this message in context: http://old.nabble.com/Calculate-Averages-for-time-data-tp26160202p26160202.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculate Averages for time data
Next time please provide sample input. library(chron) # input ch - c(02:24:00, 04:48:00, 07:12:00, 09:36:00, 12:00:00, 14:24:00, 16:48:00, 19:12:00, 21:36:00) # convert to times tt - times(ch) # calculate mean mean(tt) On Tue, Nov 3, 2009 at 10:35 AM, Sunita22 sunita...@gmail.com wrote: Hello I have time data which is in hh:mm:ss format and I need to calculate averages for this data. I tried converting the data as integer so that I could calculate the average, but it doesn't serve the purpose. I tried using Chron package also. But I am stuck up as to how to deal with the time data. Please guide me as to how to deal with Time data in R. Thank you in advance Regards Sunita -- View this message in context: http://old.nabble.com/Calculate-Averages-for-time-data-tp26160202p26160202.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change negative values in column
Hi,
On Nov 3, 2009, at 9:56 AM, Joel Fürstenberg-Hägg wrote:
Hi all,
I'm trying to write a script that changes all negative values in a
data frame column to a small positive value, based on the the
minimum value of the column.
However, I get the following error:
Error in if (x[i] 0) { : argument is of length zero
This is telling you that x[i] is a zero length object, so you're
indexing is wrong
As well, I would minimum to be the smallest of the non-negative
values...
Aa_non_neg=(fieldTrial0809$Aa) # Copy column from data frame to
manipulate
nonNegative = function(x)
{
minimum=min(x) # Should only use positive minimum!
You have to tell R to only use positive numbers, min(x[x 0]) would
have worked.
I'm not really sure what you're looping over below.
for (i in x)
{
if(x[i]0) # Found a negative value
{
x[i]=minimum/10 # Change to a new non-negative value
}
}
}
This will work over each column of an orig.df data.frame (assumes
all cols are numeric) and generate the new data.frame you're looking
for:
new.df - as.data.frame(apply(orig.df, 2, function(col) {
min.val - min(col[col 0])
col[col 0] - min.val
col
}))
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R2WinBUGS not working?
Paul Heinrich Dietrich wrote: This weekend I noticed that my R2WinBUGS connection was no longer working on my Windows computer at work AND my Ubuntu linux computer at home. As soon as WinBUGS opens, the message reads Index Out Of Range. I have un-installed Yes, thanks, problem known for R-2.10.0 and under investigation. and re-installed R on both computers and the problem remains the same. At first I thought it was the new version of R, but I was able to overcome the problem (on Windows only) with the use of program=openbugs in R2WinBUGS. But that doesn't work on Linux. That works because it just uses a wrapper for BRugs and the real is done be BRugs. Best, Uwe Ligges PS The smallest WinBUGS program I have, essentially a t-test, still works. But anything larger than that gets the error. Help. Thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculate Averages for time data
Thank you so much it did solve my purpose. Regards Our Thoughts have the Power to Change our Destiny. Sunita On Tue, Nov 3, 2009 at 9:15 PM, Gabor Grothendieck ggrothendi...@gmail.comwrote: Next time please provide sample input. library(chron) # input ch - c(02:24:00, 04:48:00, 07:12:00, 09:36:00, 12:00:00, 14:24:00, 16:48:00, 19:12:00, 21:36:00) # convert to times tt - times(ch) # calculate mean mean(tt) On Tue, Nov 3, 2009 at 10:35 AM, Sunita22 sunita...@gmail.com wrote: Hello I have time data which is in hh:mm:ss format and I need to calculate averages for this data. I tried converting the data as integer so that I could calculate the average, but it doesn't serve the purpose. I tried using Chron package also. But I am stuck up as to how to deal with the time data. Please guide me as to how to deal with Time data in R. Thank you in advance Regards Sunita -- View this message in context: http://old.nabble.com/Calculate-Averages-for-time-data-tp26160202p26160202.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to display full name for the coefficients/factors in summary()?
Hi, I am wondering if there is a way to display the full anme of the regression coeffients/factors in the summary? Suppose I have a bogus data set using weekday as factor which has 7 levels such as: mydata - sample(364) wk - rep(1:7, 52) weekday - factor(wk,1:7,c(Mon,Tue,Wed,Thu,Fri,Sat,Sun),ordered=T) test - data.frame(mydata,weekday) lm.test - lm(mydata ~ weekday, test) summary(lm.test) Call: lm(formula = mydata ~ weekday) Residuals: Min 1Q Median 3Q Max -205.615 -93.486 -2.212 88.851 199.423 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 182.5000 5.5137 33.100 2e-16 *** weekday.L -27.042614.5878 -1.854 0.0646 . weekday.Q-0.921114.5878 -0.063 0.9497 weekday.C-1.656514.5878 -0.114 0.9097 weekday^4 -16.644914.5878 -1.141 0.2546 weekday^5 9.943614.5878 0.682 0.4959 weekday^6 -14.397114.5878 -0.987 0.3243 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 105.2 on 357 degrees of freedom Multiple R-squared: 0.01705,Adjusted R-squared: 0.0005341 F-statistic: 1.032 on 6 and 357 DF, p-value: 0.4039 --- Why is the name for the regression coeffients/factors display as weekday.L, weekday.Q, weekday^6 instead of Tue, Wed, and such? Is there a simple way of changing options to fix it? Please advise and many thanks for your help! Jack Chang __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] bargraph.ci - CI and color question.
Hello, Thanks for the responses. Yes, I did try to use ?bargraph.CI for the colors. When I said bars, I meant the main bars on the graph not the error bars. However, this col=c(color, color) is what I was needed and while it didn't find in the ?bargraph.ci help, I suspect its a more fundamental R thing that I was unaware of. Regardless, thank you for the example. bargraph.CI(dose, len, group = supp, data =ToothGrowth, + col=c(gray,black)) Regarding the confidence intervals. Yes, I am looking for the 95% confidence interval of the mean. Using the ci.fun can I get +/- 1.96 SD? bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) I tried variations on this like, ci.fun= function(x) c(mean(x)-(1.96*sd(x))), mean(x) + (1.96*sd(x))) ) but this didn't work. Maybe this approach above seems silly, but I hope I am exhibiting that I do indeed what the 95% CI for the mean, but that I am unsure how to have the error bars reflect this using the ci.fun argument. Can I use this argument or is some pre-processing necessary? If this information is located somewhere besides the ?bargraph.ci help I would greatly appreciate an indication of where it is. Thank you kindly, Michael Just On Tue, Nov 3, 2009 at 8:53 AM, Manuel Morales manuel.a.mora...@williams.edu wrote: On Tue, 2009-11-03 at 03:51 -0600, Michael Just wrote: Hello, When using bargraph.CI in package sciplot can the bars for each group be different colors? How do I select the color for each group? With the option err.col bargraph.CI(dose, len, group = supp, data =ToothGrowth, err.col=c(gray,black)) When I use this instead of the default (SD vs SE): bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) Am I getting 95% CI bars? No. You're plotting +/- 1 SD. Thank you kindly, Michael Just __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://mutualism.williams.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to display a string containing greek chrs and variables
Quoting baptiste auguie baptiste.aug...@googlemail.com: Hi, try this, plot.new() x=0.8 text(0.5, 0.5, bquote(rho == .(x))) HTH, baptiste Aha! That does exactly what i wanted! Thanks! Jose -- The University of Edinburgh is a charitable body, registered in Scotland, with registration number SC005336. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change negative values in column
Hi Joel,
On Nov 3, 2009, at 11:30 AM, Joel Fürstenberg-Hägg wrote:
However, I get the following error:
Error in if (x[i] 0) { : argument is of length zero
This is telling you that x[i] is a zero length object, so you're
indexing is wrong
Doesn't x[i] means index i in vector x?
Yes.
Or could it be that there's a missing value at that index?
Don't think so: NA 0 returns NA, not an error regarding the
argument's length in your `if` clause.
How do I deal with them in that case? I need to keep the length of
the vector so I cannot remove NAs if that affects the length...
I'm not sure how to answer this. Just try to learn a bit more about
how to index using integers, vectors of integers, and vectors of
logical/boolean values.
This will work over each column of an orig.df data.frame (assumes
all cols are numeric) and generate the new data.frame you're looking
for:
new.df - as.data.frame(apply(orig.df, 2, function(col) {
min.val - min(col[col 0])
col[col 0] - min.val
col
}))
Ok, that's exactly what I was aming for! Thanks a lot!!
Great. Don't just use that code though, understand what it's doing ..
it'll help you to slice and dice your data in the future.
If you have any questions regarding the details of that code snippet
that you can't figure out after careful inspection, feel free to ask.
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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[R] Invitation to connect on LinkedIn
LinkedIn Konrad Banachewicz requested to add you as a connection on LinkedIn: -- Arnaud, I'd like to add you to my professional network on LinkedIn. - Konrad Accept invitation from Konrad Banachewicz http://www.linkedin.com/e/qlt6CtWzi7sEoE_As_C0_wMfEZD2c-gGLFL6LO/blk/I180262_2/pmpxnSRJrSdvj4R5fnhv9ClRsDgZp6lQs6lzoQ5AomZIpn8_cBYOdz8Me3kRdjkNiiZvliRQjQRTdOYOe3oQd3ATcPwLrCBxbOYWrSlI/EML_comm_afe/ View invitation from Konrad Banachewicz http://www.linkedin.com/e/qlt6CtWzi7sEoE_As_C0_wMfEZD2c-gGLFL6LO/blk/I180262_2/39vczoOc3wRdjkRckALqnpPbOYWrSlI/svi/ -- Why might connecting with Konrad Banachewicz be a good idea? Have a question? Konrad Banachewicz's network will probably have an answer: You can use LinkedIn Answers to distribute your professional questions to Konrad Banachewicz and your extended network. You can get high-quality answers from experienced professionals. http://www.linkedin.com/e/ash/inv19_ayn/ -- (c) 2009, LinkedIn Corporation [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to display full name for the coefficients/factors in summary()?
By default orthogonal polynomial contrasts are used for ordered factors. Drop the 'ordered = TRUE' and you will get treatment contrasts. hth, Kingsford Jones On Tue, Nov 3, 2009 at 8:52 AM, Jen-Chien Chang jcch...@andrew.cmu.edu wrote: Hi, I am wondering if there is a way to display the full anme of the regression coeffients/factors in the summary? Suppose I have a bogus data set using weekday as factor which has 7 levels such as: mydata - sample(364) wk - rep(1:7, 52) weekday - factor(wk,1:7,c(Mon,Tue,Wed,Thu,Fri,Sat,Sun),ordered=T) test - data.frame(mydata,weekday) lm.test - lm(mydata ~ weekday, test) summary(lm.test) Call: lm(formula = mydata ~ weekday) Residuals: Min 1Q Median 3Q Max -205.615 -93.486 -2.212 88.851 199.423 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 182.5000 5.5137 33.100 2e-16 *** weekday.L -27.0426 14.5878 -1.854 0.0646 . weekday.Q -0.9211 14.5878 -0.063 0.9497 weekday.C -1.6565 14.5878 -0.114 0.9097 weekday^4 -16.6449 14.5878 -1.141 0.2546 weekday^5 9.9436 14.5878 0.682 0.4959 weekday^6 -14.3971 14.5878 -0.987 0.3243 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 105.2 on 357 degrees of freedom Multiple R-squared: 0.01705, Adjusted R-squared: 0.0005341 F-statistic: 1.032 on 6 and 357 DF, p-value: 0.4039 --- Why is the name for the regression coeffients/factors display as weekday.L, weekday.Q, weekday^6 instead of Tue, Wed, and such? Is there a simple way of changing options to fix it? Please advise and many thanks for your help! Jack Chang __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] about the cox result
To extract various portions of the coxph standard printout, look at
summary.coxph
help('summary.coxph')
fit - coxph(...
sfit - summary(fit)
Terry Therneau
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Re: [R] how to display a string containing greek chrs and variables
j.delashe...@ed.ac.uk wrote: Quoting baptiste auguie baptiste.aug...@googlemail.com: Hi, try this, plot.new() x=0.8 text(0.5, 0.5, bquote(rho == .(x))) HTH, baptiste Aha! That does exactly what i wanted! Thanks! Jose But does it do what it should? It's customary to use rho for a _population_ correlation coefficient. If your (x,y) values represent a _sample_, then it would probably be more appropriate to use the more mundane r or perhaps rho-hat. -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] creating mulptiple new variables from one data.frame according to columns and rows in that frame
Dear R-helpers, I have a data.frame (bcpe.lat.m) containing 13 countries, ages 0-50yrs per month, and the corresponding musigma (see below). *I would like to limit the age range to include all 12 months for the 1st 5 years and only whole years for all ages thereafter for each of the countries present in the data frame. *I would like to create separate data.frames according to the country the data is from (Bolivia.bcpe.lat.m, brazil.bcpe.lat.m, etc) I have tried using: c(seq(0,5,1/12),seq(5,50,1) ) to select the desired ages but am unsure how to repeat that sequence for consecutive countries. I have tried using: split(bcpe.lat.m, bcpe.lat.m$country) But end up with a string which I am no longer to select the specific ages I want and all the data still remains in one variable Have also looked a 'by', 'apply' and things like 'for (i in 1:13)' Help with either or both steps would be greatly appreciated. Greetings from Formentera, Daniel Age(yrs) country mu sigma 1 0. Bolivia 11.42168 0.1014872 2 0.0833 Bolivia 11.33625 0.1053837 3 0.1667 Bolivia 11.28417 0.1070594 4 0.2500 Bolivia 11.21125 0.1083872 5 0. Bolivia 11.11637 0.1095305 ... 602 0. Brazil 11.54888 0.10839417 603 0.0833 Brazil 11.46345 0.11255592 604 0.1667 Brazil 11.41137 0.11434565 605 0.2500 Brazil 11.33844 0.11576378 606 0. Brazil 11.24357 0.11698489 ... [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Issue with %in% - not matching identical rows in data frames
Hi folks I have two data frames. I know that the nth (let's say the 7th) row in the first data frame (sequence) is there in the second (today.sequence). When I try to check that by doing 'sequence[7,] %in% today.sequence', I get all FALSE when it should be all TRUE. I'm certain I'm making some trivial mistake. Any solutions? The code to recreate the data frames and see for yourself is: sequence - structure(list(DATE = structure(c(14549, 14549, 14553, 14550, 14557, 14550, 14551, 14550), class = Date), DATASET = c(1L, 2L, 1L, 2L, 2L, 3L, 3L, 4L), REP = c(1L, 0L, 2L, 2L, 3L, 0L, 1L, 0L), WRONGS_ABS = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), WRONGS_RATIO = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), DONE = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), class = data.frame, row.names = c(NA, -8L)) today.sequence - structure(list(DATE = structure(c(14551, 14550), class = Date), DATASET = 3:4, REP = c(1L, 0L), WRONGS_ABS = c(0L, 0L), WRONGS_RATIO = c(0L, 0L), DONE = c(0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), row.names = 7:8, class = data.frame) sequence[7,] #You should see '2009-11-03 3 1 0 00' today.sequence #You can clearly see that sequence [7,] is the first row in today.sequence sequence[7,] %in% today.sequence #This should show 'TRUE TRUE TRUE TRUE TRUE TRUE'. Instead # it shows 'FALSE FALSE FALSE FALSE FALSE FALSE' Thanks -- Kaushik Krishnan (kaushik.s.krish...@gmail.com) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Stacked barplot: specifying individual bar hue/luminance
Hi all, I'm trying to generate barplots from simple but long (~10-row) data files, in which each bar will comprise two stacked 'sub-bars'. All the upper sub-bars will have the same hue, and all the lower bars will, likewise, have another uniform hue. However, I wish to specify the luminance (aka brightness) of each bar (i.e., each whole bar comprising two sub-bars) separately, and I wish the plot to render without spaces between successive bars. I've been trying to get this going in ggplot2. Two problems arise: first, I'm unable to eliminate the spaces between successive bars. And second (and more importantly), while I can vary transparency (i.e., alpha) of bars singly, I'm not sophisticated enough with the grammar to know how to define a color palette that lets me vary the luminance singly. Can anyone suggest how to do this, either in ggplot or in basic R (or some other package)? Many thanks, Nathan Pearson nathaniel.pear...@gmail.com -- View this message in context: http://old.nabble.com/Stacked-barplot%3A-specifying-individual-bar-hue-luminance-tp26160543p26160543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R 2.10.0: Error in gsub/calloc
I'm running R 2.10.0 under Mac OS X 10.5.8; however, I don't think this is a Mac-specific problem. I have a very large (158,908 possible sentences, ca. 58 MB) plain text document d which I am trying to tokenize: t - strapply(d, \\w+, perl = T). I am encountering the following error: Error in base::gsub(pattern, rs, x, ...) : Calloc could not allocate (-1398215180 of 1) memory This happens regardless of whether I run in 32- or 64-bit mode. The machine has 8 GB of RAM, so I can hardly believe that RAM is a problem. Thanks, Richard __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Stacked barplot: specifying individual bar hue/luminance
Hi all, I'm trying to generate barplots from simple but long (~10-row) data files, in which each bar will comprise two stacked 'sub-bars'. All the upper sub-bars will have the same hue, and all the lower bars will, likewise, have another uniform hue. However, I wish to specify the luminance (aka brightness) of each bar (i.e., each whole bar comprising two sub-bars) separately, and I wish the plot to render without spaces between successive bars. I've been trying to get this going in ggplot2. Two problems arise: first, I'm unable to eliminate the spaces between successive bars. And second (and more importantly), while I can vary transparency (i.e., alpha) of bars singly, I'm not sophisticated enough with the grammar to know how to define a color palette that lets me vary the luminance singly. Can anyone suggest how to do this, either in ggplot or in basic R (or some other package)? Many thanks, Nathan Pearson nathaniel.pear...@gmail.com -- View this message in context: http://old.nabble.com/Stacked-barplot%3A-specifying-individual-bar-hue-luminance-tp26160523p26160523.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Issue with %in% - not matching identical rows in data frames
?%in% says x and table must be vectors. You supplied data.frames. So %in% is coercing your today.sequence to a vector using as.character(today.sequence) Perhaps you should paste the columns together first: x - do.call(paste, c(sequence, sep = ::)) table - do.call(paste, c(today.sequence, sep = ::)) x[7] %in% table I'm not sure if this is what you want/need, but it does match your example. HTH, --sundar On Tue, Nov 3, 2009 at 7:53 AM, Kaushik Krishnan kaushik.s.krish...@gmail.com wrote: Hi folks I have two data frames. I know that the nth (let's say the 7th) row in the first data frame (sequence) is there in the second (today.sequence). When I try to check that by doing 'sequence[7,] %in% today.sequence', I get all FALSE when it should be all TRUE. I'm certain I'm making some trivial mistake. Any solutions? The code to recreate the data frames and see for yourself is: sequence - structure(list(DATE = structure(c(14549, 14549, 14553, 14550, 14557, 14550, 14551, 14550), class = Date), DATASET = c(1L, 2L, 1L, 2L, 2L, 3L, 3L, 4L), REP = c(1L, 0L, 2L, 2L, 3L, 0L, 1L, 0L), WRONGS_ABS = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), WRONGS_RATIO = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), DONE = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), class = data.frame, row.names = c(NA, -8L)) today.sequence - structure(list(DATE = structure(c(14551, 14550), class = Date), DATASET = 3:4, REP = c(1L, 0L), WRONGS_ABS = c(0L, 0L), WRONGS_RATIO = c(0L, 0L), DONE = c(0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), row.names = 7:8, class = data.frame) sequence[7,] #You should see '2009-11-03 3 1 0 0 0' today.sequence #You can clearly see that sequence [7,] is the first row in today.sequence sequence[7,] %in% today.sequence #This should show 'TRUE TRUE TRUE TRUE TRUE TRUE'. Instead # it shows 'FALSE FALSE FALSE FALSE FALSE FALSE' Thanks -- Kaushik Krishnan (kaushik.s.krish...@gmail.com) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] AR Simulation with non-normal innovations - Correct
Thanks Andreas.
This is just the start point I was needing.
Best,
Rick
From: Andreas Hary
Sent: Tuesday, November 03, 2009 7:19 AM
To: Ricardo Gonçalves Silva
Subject: Re: [R] AR Simulation with non-normal innovations - Correct
Have a look at function arima.sim. It allows you to specify a random number
generator, so you could try something like the following:
arModel - list(ar = 0, ma = 0, order = c(0, 1, 0))
tGarchGen - function(a, b, c) {
# your stuff here, must return a vector of random deviates
}
arima.sim(arModel, n = 100, rand.gen = tGarchGen)
If you would like to generate a bunch of series, say 200, all at once try
mySeries - replicate(200, arima.sim(arModel, n = 100, rand.gen = tGarchGen))
HTH,
Andreas
2009/11/2 Ricardo Gonçalves Silva ricard...@terra.com.br
Dear Users,
I would like to simulate an AR(1) (y_t=ct1+y_t-1+e_t) model in R where the
innovations are supposed to follow a t-GARCH(1,1) proccess.
By t-GARCH I want to mean that:
e_t=n_t*sqrt(h_t) and
h_t=ct2+a*(e_t)^2+b*h_t-1.
where n_t is a random variable with t-Student distribution.
If someone could give some guidelines, I can going developing the model.
I did it in matlab, but the loops are very slowly, so I would like to try R.
Thanks in advance,
Rick
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Re: [R] reference on p.adjust()
Peng Yu wrote: I'm wondering if there is a textbook that summarize the methods on adjusting p-values for multiple comparisons. Most of the references on p.adjust() are over 10 years old. Being 10 years old does not mean the calculus behind those methods has changed. I feel it would be better if somebody could recommend me a textbook on multiple hypothesis correction, so that I can quickly get a general idea of different methods. There are some around just search for your keywords. Best wishes, Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reference on p.adjust()
2009/11/3 Uwe Ligges lig...@statistik.tu-dortmund.de: Peng Yu wrote: I'm wondering if there is a textbook that summarize the methods on adjusting p-values for multiple comparisons. Most of the references on p.adjust() are over 10 years old. Being 10 years old does not mean the calculus behind those methods has changed. I never said that the calculus has been changed. I feel it would be better if somebody could recommend me a textbook on multiple hypothesis correction, so that I can quickly get a general idea of different methods. There are some around just search for your keywords. I prefer descriptions from books. Most online description are not very satisfactory, because they are not peer reviewed. I looked through Testing Statistical Hypotheses by Lehmann. This book only gives references to the recent works. If you happen to know a review article or a textbook that summarize all the methods in multiple hypothesis correction. Please let me know. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R 2.10.0: Error in gsub/calloc
richard@pueo-owl.ch wrote: I'm running R 2.10.0 under Mac OS X 10.5.8; however, I don't think this is a Mac-specific problem. I have a very large (158,908 possible sentences, ca. 58 MB) plain text document d which I am trying to tokenize: t - strapply(d, \\w+, perl = T). I am encountering the following error: What is strapply() and what is d? Uwe Ligges Error in base::gsub(pattern, rs, x, ...) : Calloc could not allocate (-1398215180 of 1) memory This happens regardless of whether I run in 32- or 64-bit mode. The machine has 8 GB of RAM, so I can hardly believe that RAM is a problem. Thanks, Richard __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with R and Webserver (SOAP)
Hello all... We are trying to use the analytical power of R in a Web based environment, and are seeking the optimal ways of accomplishing the same. We need R to accept requests from clients, perform analysis on the data as required by the users and return the results ( formatted text, graphs, etc) to the web browser. We have a Windows server with an IIS, Cold fusion app server that we want to talk to R. Can someone suggest an optimal way of accomplishing this? (I am a novice at this, so I hope you ask me for specific details that are necessary) Some of the things that we had unsuccessfully tried included playing around with: 1) CGIwithR package: This package appeared to be geared for Unix / Apache systems, so we couldn't get too far in the Windows Environment. 2) Biocep.net : We tried to use the tools from biocep.net to generate SOAP service definitions for R functions. We followed the instructions on biocep.net but we couldn't run the java program to access the R-soap from java. It would be great if someone in the group has had success in implementing R-Soap through biocep packages, that can guide us. Even otherwise, what is the best way to access R from the web? Will running R under Apache make things easier? Any help will be greatly appreciated. Thanks, Ravi -- The destiny of the river is not its shores but the ocean __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create Artificial Binary Matrix based on probability
Try matrix(rbinom(100, 1, prob = 0.048), nrow = 10) -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of bikemike42 Sent: Tuesday, November 03, 2009 11:49 AM To: r-help@r-project.org Subject: [R] Create Artificial Binary Matrix based on probability Dear All, I am trying to create an artificial binary matrix such that each cell has a probability of 0.048 of having a 1. So far the closest I've come is us by using a random poisson distribution with a mean of 0.048, but I can't figure out how to limit the max value to 1. Otherwise that would work fine it seems. Any suggestions? The main code I've got to create said matrix so far is: a-replicate(26,rpois(57,0.048)) Thanks in Advance, Mike -- View this message in context: http://old.nabble.com/Create-Artificial- Binary-Matrix-based-on-probability-tp26160622p26160622.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re place only first NA in column
Mike,
Is this what you're trying to do, it avoids the awkward use of a while loop,
taking advantage of the ability to vectorize R functions.
fun - function() {
z - matrix(rbinom(832, 1, prob = 0.048), nrow = 32)
sum(rowSums(z) == 0)
}
a - as.matrix(replicate(1000, fun()))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of bikemike42
Sent: Tuesday, November 03, 2009 2:58 PM
To: r-help@r-project.org
Subject: [R] Re place only first NA in column
Dear All,
I am trying to fill in a blank vector (a) with one value at a time, with
the value of the number of rows in a randomized dataset with rowSums=0.
Below is the code I've got so far, but what I want to be as the last line
is
a[1st NA,]=nz
such that this will run until all of my NAs are replaced with an integer
value based on randomizations.
a-matrix(nrow=1000,ncol=1)
while(sum(is.na(a)0))
{
z-matrix(rbinom(832, 1, prob = 0.048), nrow = 32)
rs-rowSums(z)
nz-length(rs[rs==0])
a[,]=nz
}
Thank you for your help!
Mike
--
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in-column-tp26163589p26163589.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] reference on p.adjust()
Peng Yu wrote: 2009/11/3 Uwe Ligges lig...@statistik.tu-dortmund.de: Peng Yu wrote: I'm wondering if there is a textbook that summarize the methods on adjusting p-values for multiple comparisons. Most of the references on p.adjust() are over 10 years old. Being 10 years old does not mean the calculus behind those methods has changed. I never said that the calculus has been changed. I feel it would be better if somebody could recommend me a textbook on multiple hypothesis correction, so that I can quickly get a general idea of different methods. There are some around just search for your keywords. I prefer descriptions from books. Most online description are not very satisfactory, because they are not peer reviewed. Sure, I meant to find the books. There is for example focussed on Bioinformatics applications: Dudoit, Sandrine and van der Laan, Mark J (2008): Multiple Testing Procedures with Applications to Genomics, Springer, New York. But that is just one example - and I tend not to mention books in answers to such requests since books are a matter of taste and may depend on the audience, community in which to use the method, and the application. I looked through Testing Statistical Hypotheses by Lehmann. This book only gives references to the recent works. If you happen to know a review article or a textbook that summarize all the methods in multiple hypothesis correction. Please let me know. For the good old methods, I only happen to know German textbooks. Should be summarized in any good overview textbook on testing, though. Best, Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R 2.10.0: Error in gsub/calloc
Try the patch version... Maybe is the same problem I had with large database when using gsub() HTH El mar, 03-11-2009 a las 20:31 +0100, Richard R. Liu escribió: I apologize for not being clear. d is a character vector of length 158908. Each element in the vector has been designated by sentDetect (package: openNLP) as a sentence. Some of these are really sentences. Others are merely groups of meaningless characters separated by white space. strapply is a function in the package gosubfn. It applies to each element of the first argument the regular expression (second argument). Every match is then sent to the designated function (third argument, in my case missing, hence the identity function). Thus, with strapply I am simply performing a white-space tokenization of each sentence. I am doing this in the hope of being able to distinguish true sentences from false ones on the basis of mean length of token, maximum length of token, or similar. Richard R. Liu Dittingerstr. 33 CH-4053 Basel Switzerland Tel.: +41 61 331 10 47 Email: richard@pueo-owl.ch On Nov 3, 2009, at 18:30 , Uwe Ligges wrote: richard@pueo-owl.ch wrote: I'm running R 2.10.0 under Mac OS X 10.5.8; however, I don't think this is a Mac-specific problem. I have a very large (158,908 possible sentences, ca. 58 MB) plain text document d which I am trying to tokenize: t - strapply(d, \\w+, perl = T). I am encountering the following error: What is strapply() and what is d? Uwe Ligges Error in base::gsub(pattern, rs, x, ...) : Calloc could not allocate (-1398215180 of 1) memory This happens regardless of whether I run in 32- or 64-bit mode. The machine has 8 GB of RAM, so I can hardly believe that RAM is a problem. Thanks, Richard __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. --Apple-Mail-8--203371287-- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Weird operator behaviour
Argument recycling is coming into play here with !=. I am guessing you want ! fish$Species %in% c(CRA, PHC) ? Best Regards, Erik Iverson -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of jimdare Sent: Tuesday, November 03, 2009 4:02 PM To: r-help@r-project.org Subject: [R] Weird operator behaviour Hi, I have a dataset called 'fish'. fish$Species returns extract 1. When I use fish$Species != c(CRA,PHC), i.e. I want all species except CRA and PHC, I get extract 2 which is blatantly wrong. Can anyone see what I'm doing wrong? Regards, James EXTRACT 1 fish$Species [1] ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ATO ATO [19] ATO ATO ATO ATO ATO ATO BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG [37] BIG BIG BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS CRA [55] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [73] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [91] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [109] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [127] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [145] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [163] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [181] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [199] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [217] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [235] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [253] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [271] CRA CRA CRA CRA CRA CRA CRA CRA MAK MAK MAK MAK MAK MAK MAK MAK MAK MAK [289] MAK MAK MAK MAK MAK MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO [307] MOO MOO PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC [325] PHC PHC PHC POS POS POS POS POS POS POS POS POS POS POS PTO PTO PTO PTO [343] PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO RBM RBM RBM RBM RBM RBM RBM [361] RBM RBM RBM RBM RBM RBM RBM RBM SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ [379] SKJ SKJ SKJ SKJ SKJ STM STM STM STM STM STM STM STM STN STN STN STN STN [397] STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN [415] SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO TOR TOR [433] TOR TOR TOR TOR TOR TOR YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN [451] YFN YFN YFN YFN 17 Levels: ALB ATO BIG BWS CRA MAK MOO PHC POS PTO RBM SKJ STM STN SWO ... YFN EXTRACT 2 fish$Species != c(CRA,PHC) [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [25] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [37] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [49] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [61] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [73] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [85] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [97] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [109] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [121] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [133] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [145] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [157] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [169] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [181] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [193] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [205] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [217] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [229] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [241] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [253] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [265] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [277] FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [289] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [301] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE [313] TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE [325] TRUE FALSE TRUE TRUE TRUE TRUE
Re: [R] Weird operator behaviour
Tena koe James
Have you tried something like !fish$Species%in%c('CRA','PHC')?
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of jimdare
Sent: Wednesday, 4 November 2009 11:02 a.m.
To: r-help@r-project.org
Subject: [R] Weird operator behaviour
Hi,
I have a dataset called 'fish'. fish$Species returns extract
1. When I use fish$Species != c(CRA,PHC), i.e. I want
all species except CRA and PHC, I get extract 2 which is
blatantly wrong. Can anyone see what I'm doing wrong?
Regards,
James
EXTRACT 1
fish$Species
[1] ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB
ALB ALB ATO ATO [19] ATO ATO ATO ATO ATO ATO BIG BIG BIG BIG
BIG BIG BIG BIG BIG BIG BIG BIG [37] BIG BIG BWS BWS BWS BWS
BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS CRA [55] CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA [73] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA [91] CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA [109] CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [127] CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA [145] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA [163] CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [181] CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [199]
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA [217] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA [235] CRA CRA CRA CRA CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [253] CRA CRA CRA
CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA
[271] CRA CRA CRA CRA CRA CRA CRA CRA MAK MAK MAK MAK MAK MAK
MAK MAK MAK MAK [289] MAK MAK MAK MAK MAK MOO MOO MOO MOO MOO
MOO MOO MOO MOO MOO MOO MOO MOO [307] MOO MOO PHC PHC PHC PHC
PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC [325] PHC PHC
PHC POS POS POS POS POS POS POS POS POS POS POS PTO PTO PTO
PTO [343] PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO RBM RBM
RBM RBM RBM RBM RBM [361] RBM RBM RBM RBM RBM RBM RBM RBM SKJ
SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ [379] SKJ SKJ SKJ SKJ SKJ
STM STM STM STM STM STM STM STM STN STN STN STN STN [397] STN
STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN
STN STN [415] SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO
SWO SWO SWO SWO TOR TOR [433] TOR TOR TOR TOR TOR TOR YFN YFN
YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN [451] YFN YFN YFN YFN
17 Levels: ALB ATO BIG BWS CRA MAK MOO PHC POS PTO RBM SKJ
STM STN SWO ...
YFN
EXTRACT 2
fish$Species != c(CRA,PHC)
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
TRUE TRUE TRUE [13] TRUE TRUE TRUE TRUE TRUE TRUE
TRUE TRUE TRUE TRUE TRUE TRUE [25] TRUE TRUE TRUE
TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [37]
TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
TRUE TRUE [49] TRUE TRUE TRUE TRUE TRUE TRUE FALSE
TRUE FALSE TRUE FALSE TRUE [61] FALSE TRUE FALSE TRUE
FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [73] FALSE
TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE
TRUE [85] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
FALSE TRUE FALSE TRUE [97] FALSE TRUE FALSE TRUE FALSE
TRUE FALSE TRUE FALSE TRUE FALSE TRUE [109] FALSE TRUE
FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
[121] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE
TRUE FALSE TRUE [133] FALSE TRUE FALSE TRUE FALSE TRUE
FALSE TRUE FALSE TRUE FALSE TRUE [145] FALSE TRUE FALSE
TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [157]
FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
FALSE TRUE [169] FALSE TRUE FALSE TRUE FALSE TRUE FALSE
TRUE FALSE TRUE FALSE TRUE [181] FALSE TRUE FALSE TRUE
FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [193] FALSE
TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE
TRUE [205] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
FALSE TRUE FALSE TRUE [217] FALSE TRUE FALSE TRUE FALSE
TRUE FALSE TRUE FALSE TRUE FALSE TRUE [229] FALSE TRUE
FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
[241] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE
TRUE FALSE TRUE [253] FALSE TRUE FALSE TRUE FALSE TRUE
FALSE TRUE FALSE TRUE FALSE TRUE [265] FALSE TRUE FALSE
TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [277]
FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
TRUE TRUE [289] TRUE TRUE TRUE TRUE TRUE TRUE TRUE
TRUE TRUE TRUE TRUE TRUE [301] TRUE TRUE TRUE TRUE
TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE [313] TRUE
FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
FALSE [325] TRUE FALSE TRUE TRUE TRUE TRUE
Re: [R] Re place only first NA in column
Hi,
On Nov 3, 2009, at 3:58 PM, bikemike42 wrote:
Dear All,
I am trying to fill in a blank vector (a) with one value at a
time, with
the value of the number of rows in a randomized dataset with
rowSums=0.
Below is the code I've got so far, but what I want to be as the last
line is
a[1st NA,]=nz
such that this will run until all of my NAs are replaced with an
integer
value based on randomizations.
a-matrix(nrow=1000,ncol=1)
while(sum(is.na(a)0))
{
z-matrix(rbinom(832, 1, prob = 0.048), nrow = 32)
rs-rowSums(z)
nz-length(rs[rs==0])
a[,]=nz
}
Thank you for your help!
It's not really clear to me what you want to actually replace each NA
value with. But it sounds like your 'general' question is how to find
the index of the first value of a vector that is NA.
Let's say your `a` vector has NA values you want to replace, one by one.
any.na - any(is.na(a))
while (any.na) {
first.na - which(is.na(a))[1]
a[first.na] - your.NA.replacing.function()
any.na - any(is.na(a))
}
HTH,
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
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Re: [R] help with SSOAP (can't find working examples)
Hi,
I can confirm this, just today
I tried to write a web service client.
Affected are both SSOAP-0.5-4 and SSOAP_0.4-6.
I can't access anonymous CVS atm. to check for recent fixes.
I am unable to map the error message to any of the items in
http://www.omegahat.org/SSOAP/Todo.html , is this already known ?
Yours,
Steffen
library(SSOAP)
Loading required package: XML
Loading required package: RCurl
Loading required package: bitops
w = processWSDL(http://www.massbank.jp/api/services/MassBankAPI?wsdl;)
Error: Cannot resolve ns:searchPeakDiff in SchemaCollection
In addition: Warning messages:
1: In function (node) :
skipping import node with no schemaLocation attribute
2: In processWSDL(http://www.massbank.jp/api/services/MassBankAPI?wsdl;) :
Ignoring additional serviceport ... elements
sessionInfo()
R version 2.8.1 (2008-12-22)
x86_64-pc-linux-gnu
locale:
LC_CTYPE=en_US.UTF-8;LC_NUMERIC=C;LC_TIME=en_US.UTF-8;LC_COLLATE=en_US.UTF-8;LC_MONETARY=C;LC_MESSAGES=en_US.UTF-8;LC_PAPER=en_US.UTF-8;LC_NAME=C;LC_ADDRESS=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] SSOAP_0.4-6RCurl_1.3-0bitops_1.0-4.1 XML_2.5-3
w = processWSDL(http://www.massbank.jp/api/services/MassBankAPI?wsdl;)
Error: Cannot resolve ns:searchPeakDiff in SchemaCollection
In addition: Warning messages:
1: In function (node) :
skipping import node with no schemaLocation attribute
2: In processWSDL(http://www.massbank.jp/api/services/MassBankAPI?wsdl;) :
Ignoring additional serviceport ... elements
Enter a frame number, or 0 to exit
1: processWSDL(http://www.massbank.jp/api/services/MassBankAPI?wsdl;)
2: lapply(tmp, processWSDLBindings, doc, types)
3: FUN(X[[1]], ...)
4: lapply(els, processWSDLOperation, types, doc, namespaceDefinitions, typeDef
5: FUN(X[[1]], ...)
6: xmlSApply(msg, function(x) {
7: xmlSApply.XMLNode(msg, function(x) {
8: sapply(xmlChildren(X), FUN, ...)
9: lapply(X, FUN, ...)
10: FUN(X[[1]], ...)
11: resolve(el, typeDefinitions)
12: resolve(el, typeDefinitions)
13: resolveError(Cannot resolve , obj, in , class(context))
--
IPB HalleAG Massenspektrometrie Bioinformatik
Dr. Steffen Neumann http://www.IPB-Halle.DE
Weinberg 3 http://msbi.bic-gh.de
06120 Halle Tel. +49 (0) 345 5582 - 1470
+49 (0) 345 5582 - 0
sneumann(at)IPB-Halle.DE Fax. +49 (0) 345 5582 - 1409
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Create Artificial Binary Matrix based on probability
Thanks a lot Erik and Tony! Both of your suggestions accomplish what I need. Cheers, Mike Erik Iverson wrote: Try matrix(rbinom(100, 1, prob = 0.048), nrow = 10) -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of bikemike42 Sent: Tuesday, November 03, 2009 11:49 AM To: r-help@r-project.org Subject: [R] Create Artificial Binary Matrix based on probability Dear All, I am trying to create an artificial binary matrix such that each cell has a probability of 0.048 of having a 1. So far the closest I've come is us by using a random poisson distribution with a mean of 0.048, but I can't figure out how to limit the max value to 1. Otherwise that would work fine it seems. Any suggestions? The main code I've got to create said matrix so far is: a-replicate(26,rpois(57,0.048)) Thanks in Advance, Mike -- View this message in context: http://old.nabble.com/Create-Artificial- Binary-Matrix-based-on-probability-tp26160622p26160622.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Michael L. Treglia Graduate Student Wildlife and Fisheries Sciences Texas AM University Lab: (979)862-7245 Cell: (917)841-5603 ml...@tamu.edu http://people.tamu.edu/~mlt35 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Issue with %in% - not matching identical rows in data frames
Kaushik, The documentation doesn't quite tell (me, anyway) how the function behaves when 'target' is a list (or data.frame). You'll need to dig into match.c or experiment with match() or %in% to see what it is actually doing. But it looks like it is matching whole columns of the data.frame rather than elements within each column : sequence %in% sequence [1] TRUE TRUE TRUE TRUE TRUE TRUE sequence %in% rev(sequence) [1] TRUE TRUE TRUE TRUE TRUE TRUE sequence[1,] %in% sequence [1] FALSE FALSE FALSE FALSE FALSE FALSE sequence[1,] %in% sequence[1,] [1] TRUE TRUE TRUE TRUE TRUE TRUE Maybe you wanted something like mapply( function(x,y) x%in%y , sequence[7, ], today.sequence ) ?? HTH, Chuck On Tue, 3 Nov 2009, Kaushik Krishnan wrote: Hi folks I have two data frames. I know that the nth (let's say the 7th) row in the first data frame (sequence) is there in the second (today.sequence). When I try to check that by doing 'sequence[7,] %in% today.sequence', I get all FALSE when it should be all TRUE. I'm certain I'm making some trivial mistake. Any solutions? The code to recreate the data frames and see for yourself is: sequence - structure(list(DATE = structure(c(14549, 14549, 14553, 14550, 14557, 14550, 14551, 14550), class = Date), DATASET = c(1L, 2L, 1L, 2L, 2L, 3L, 3L, 4L), REP = c(1L, 0L, 2L, 2L, 3L, 0L, 1L, 0L), WRONGS_ABS = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), WRONGS_RATIO = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), DONE = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), class = data.frame, row.names = c(NA, -8L)) today.sequence - structure(list(DATE = structure(c(14551, 14550), class = Date), DATASET = 3:4, REP = c(1L, 0L), WRONGS_ABS = c(0L, 0L), WRONGS_RATIO = c(0L, 0L), DONE = c(0L, 0L)), .Names = c(DATE, DATASET, REP, WRONGS_ABS, WRONGS_RATIO, DONE), row.names = 7:8, class = data.frame) sequence[7,] #You should see '2009-11-03 3 1 0 00' today.sequence #You can clearly see that sequence [7,] is the first row in today.sequence sequence[7,] %in% today.sequence #This should show 'TRUE TRUE TRUE TRUE TRUE TRUE'. Instead # it shows 'FALSE FALSE FALSE FALSE FALSE FALSE' Thanks -- Kaushik Krishnan (kaushik.s.krish...@gmail.com) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Create Artificial Binary Matrix based on probability
Dear All, I am trying to create an artificial binary matrix such that each cell has a probability of 0.048 of having a 1. So far the closest I've come is us by using a random poisson distribution with a mean of 0.048, but I can't figure out how to limit the max value to 1. Otherwise that would work fine it seems. Any suggestions? The main code I've got to create said matrix so far is: a-replicate(26,rpois(57,0.048)) Thanks in Advance, Mike -- View this message in context: http://old.nabble.com/Create-Artificial-Binary-Matrix-based-on-probability-tp26160620p26160620.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Create Artificial Binary Matrix based on probability
Dear All, I am trying to create an artificial binary matrix such that each cell has a probability of 0.048 of having a 1. So far the closest I've come is us by using a random poisson distribution with a mean of 0.048, but I can't figure out how to limit the max value to 1. Otherwise that would work fine it seems. Any suggestions? The main code I've got to create said matrix so far is: a-replicate(26,rpois(57,0.048)) Thanks in Advance, Mike -- View this message in context: http://old.nabble.com/Create-Artificial-Binary-Matrix-based-on-probability-tp26160622p26160622.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Passing Command to Optim in factanal
Hi, I am currently trying to execute the following command: f-factanal(factors=k$Components$nparallel,covmat=m,n.obs=2287,rotation=varimax,control=list(opt=list(method=c(BFGS but keep getting the error: L-BFGS-B needs finite values of 'fn' I can't figure out what I am doing wrong here, why isn't optim being told to use BFGS instead of L-BFGS-B... Any help would be very muc appreciated. Thanks, Ali __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re place only first NA in column
Dear All,
I am trying to fill in a blank vector (a) with one value at a time, with
the value of the number of rows in a randomized dataset with rowSums=0.
Below is the code I've got so far, but what I want to be as the last line is
a[1st NA,]=nz
such that this will run until all of my NAs are replaced with an integer
value based on randomizations.
a-matrix(nrow=1000,ncol=1)
while(sum(is.na(a)0))
{
z-matrix(rbinom(832, 1, prob = 0.048), nrow = 32)
rs-rowSums(z)
nz-length(rs[rs==0])
a[,]=nz
}
Thank you for your help!
Mike
--
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[R] ROC
My question would be related more to time-dependent ROC (survivalROC package): I would like to know how true positives (TP) and false positives (FP) are related to the order of survival times and diagnostic markers. That is, having the a set of TP and FP, each of them is calculated from which diagnostic marker and survival times. Let's take mayo data set. How could I match the TP and FP generated from survivalROC function to cut.values (output) and Stimes (input)? Best, Carol __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] OpenOffice Calc ODBC equivalent (SOLVED)
Thank you Dr. Schwartz: I did not read this post before, See this post from late 2007: http://tolstoy.newcastle.edu.au/R/e3/help/07/12/6478.html As far as I know, nothing has changed vis-a-via ODBC connectivity TO OpenOffice files. You can use ODBC FROM OpenOffice to connect to external data sources. Yes, I found that too. A search of CRAN does not reveal any packages/functions to would appear to provide a facility to directly write data frames to a Calc file. However, you can always write to a XLS file, Yes, I am going to do that. open it in Calc and re-save it as a native Calc file. Of course, anyone with Calc can open an XLS file, so you are not losing much in terms of portability by saving to an XLS file and distributing that. You also make it easier for folks who may be using other 'office' applications that do not have import filters for OO.org file formats. I was looking for xlsReadWrite package, but it seems that it is removed form the CRAN repository, then I found your WriteXLS package, which I am going to use. I see that dataframes2xls is other package that can helps me too. Thank you for your help. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re place only first NA in column
Erik and Steve,
Thanks again for your help, both solutions do exactly what I need.
Cheers,
Mike
Steve Lianoglou wrote:
Hi,
On Nov 3, 2009, at 3:58 PM, bikemike42 wrote:
Dear All,
I am trying to fill in a blank vector (a) with one value at a time,
with
the value of the number of rows in a randomized dataset with rowSums=0.
Below is the code I've got so far, but what I want to be as the last
line is
a[1st NA,]=nz
such that this will run until all of my NAs are replaced with an integer
value based on randomizations.
a-matrix(nrow=1000,ncol=1)
while(sum(is.na(a)0))
{
z-matrix(rbinom(832, 1, prob = 0.048), nrow = 32)
rs-rowSums(z)
nz-length(rs[rs==0])
a[,]=nz
}
Thank you for your help!
It's not really clear to me what you want to actually replace each NA
value with. But it sounds like your 'general' question is how to find
the index of the first value of a vector that is NA.
Let's say your `a` vector has NA values you want to replace, one by one.
any.na - any(is.na(a))
while (any.na) {
first.na - which(is.na(a))[1]
a[first.na] - your.NA.replacing.function()
any.na - any(is.na(a))
}
HTH,
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
--
Michael L. Treglia
Graduate Student
Wildlife and Fisheries Sciences
Texas AM University
Lab: (979)862-7245
Cell: (917)841-5603
ml...@tamu.edu
http://people.tamu.edu/~mlt35
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Re: [R] package lme4
On Tue, Nov 3, 2009 at 9:11 AM, wenjun zheng wjzhen...@gmail.com wrote: May be I can calculate p value by t testing approximately: 1-qnorm(Variance/Std.Dev.) That would be a z test, not a t test, wouldn't it? And it would only be meaningful if the distribution of the estimator is approximately normal, which we know it definitely is not. Estimates of variances have distributions like a chi-square, not like a normal. In particular, the estimators are not symmetrically distributed about the parameter value. But which function can help me to extract Variance and Std.Dev values from the results below: The VarCorr extractor produces the estimates of the variance. You would have to write your own functions to determine an approximate standard error of those estimates and I would not advise doing so. Firstly, the code would be rather complicated and secondly the answer would be of very limited usefulness, for the reasons discussed above. print(fm2 - lmer(Yield ~ 1 + (1|Stand) + (1|Variety) + (1|Variety:Stand),Rice)) Linear mixed model fit by REML Formula: Yield ~ 1 + (1 | Stand) + (1 | Variety) + (1 | Variety:Stand) Data: Rice AIC BIC logLik deviance REMLdev 94.25 100.7 -42.12 85.33 84.25 Random effects: Groups Name Variance Std.Dev. Variety:Stand (Intercept) 1.345679 1.16003 Variety (Intercept) 0.024692 0.15714 Stand (Intercept) 0.88 0.94281 Residual 0.67 0.81650 Number of obs: 27, groups: Variety:Stand, 9; Variety, 3; Stand, 3 Fixed effects: Estimate Std. Error t value (Intercept) 7.1852 0.6919 10.38 2009/11/2 Douglas Bates ba...@stat.wisc.edu On Sun, Nov 1, 2009 at 9:01 AM, wenjun zheng wjzhen...@gmail.com wrote: Hi R Users, When I use package lme4 for mixed model analysis, I can't distinguish the significant and insignificant variables from all random independent variables. Here is my data and result: Data: Rice-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9), Variety=rep(rep(c(A1,A2,A3),each=3),3), Stand=rep(c(B1,B2,B3),9), Block=rep(1:3,each=9)) Rice.lmer-lmer(Yield ~ (1|Variety) + (1|Stand) + (1|Block) + (1|Variety:Stand), data = Rice) Result: Linear mixed model fit by REML Formula: Yield ~ (1 | Variety) + (1 | Stand) + (1 | Block) + (1 | Variety:Stand) Data: Rice AIC BIC logLik deviance REMLdev 96.25 104.0 -42.12 85.33 84.25 Random effects: Groups Name Variance Std.Dev. Variety:Stand (Intercept) 1.345679 1.16003 Block (Intercept) 0.00 0.0 Stand (Intercept) 0.89 0.94281 Variety (Intercept) 0.024691 0.15714 Residual 0.67 0.81650 Number of obs: 27, groups: Variety:Stand, 9; Block, 3; Stand, 3; Variety, 3 Fixed effects: Estimate Std. Error t value (Intercept) 7.1852 0.6919 10.38 Can you give me some advice for recognizing the significant variables among random effects above without other calculating. Well, since the estimate of the variance due to Block is zero, that's probably not one of the significant random effects. Why do you want to do this without other calculations? In olden days when each model fit involved substantial calculations by hand one did try to avoid fitting multiple models but now that is not a problem. You can get a hint of which random effects will be significant by looking at their precision in a caterpillar plot and then fit the reduced model and use anova to compare models. See the enclosed Any suggestions will be appreciated. Wenjun [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to display a string containing greek chrs and variables
Quoting Peter Ehlers ehl...@ucalgary.ca: j.delashe...@ed.ac.uk wrote: Quoting baptiste auguie baptiste.aug...@googlemail.com: Hi, try this, plot.new() x=0.8 text(0.5, 0.5, bquote(rho == .(x))) HTH, baptiste Aha! That does exactly what i wanted! Thanks! Jose But does it do what it should? It's customary to use rho for a _population_ correlation coefficient. If your (x,y) values represent a _sample_, then it would probably be more appropriate to use the more mundane r or perhaps rho-hat. I'm talking rank correlation, Spearman's rho. Jose -- Dr. Jose I. de las Heras Email: j.delashe...@ed.ac.uk The Wellcome Trust Centre for Cell BiologyPhone: +44 (0)131 6513374 Institute for Cell Molecular BiologyFax: +44 (0)131 6507360 Swann Building, Mayfield Road University of Edinburgh Edinburgh EH9 3JR UK * NEW EMAIL from July'09: nach.mcn...@gmail.com * -- The University of Edinburgh is a charitable body, registered in Scotland, with registration number SC005336. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] one long column of data - three small columns
Tena koe ?matrix The exact syntax will depend on the class of your 'column of data'. If it is a dataframe, for example, then try matrix(yourData[,1], 3, 3) HTH ... Peter Alspach -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of frenchcr Sent: Wednesday, 4 November 2009 8:06 a.m. To: r-help@r-project.org Subject: [R] one long column of data - three small columns say i have a column of data like this... 2 3 4 2 1 6 6 4 7 and i want it in three columns like this 226 314 467 ...so i can make a contour plot. How do i do this? -- View this message in context: http://old.nabble.com/one-long-column-of-data--%3E-three-small -columns-tp26163165p26163165.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to display a string containing greek chrs and variables
j.delashe...@ed.ac.uk wrote: Quoting Peter Ehlers ehl...@ucalgary.ca: j.delashe...@ed.ac.uk wrote: Quoting baptiste auguie baptiste.aug...@googlemail.com: Hi, try this, plot.new() x=0.8 text(0.5, 0.5, bquote(rho == .(x))) HTH, baptiste Aha! That does exactly what i wanted! Thanks! Jose But does it do what it should? It's customary to use rho for a _population_ correlation coefficient. If your (x,y) values represent a _sample_, then it would probably be more appropriate to use the more mundane r or perhaps rho-hat. I'm talking rank correlation, Spearman's rho. Jose Rank correlation or not, it's still customary to use r or r[s] for samples and rho or rho[s] for populations. But it's up to you, of course. -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] one long column of data - three small columns
Hi french, Here is a suggestion: x - c(2, 3, 4, 2, 1, 6, 6, 4, 7) matrix(x, ncol = 3) # [,1] [,2] [,3] # [1,]226 # [2,]314 # [3,]467 with x the column of data you have. HTH, Jorge On Tue, Nov 3, 2009 at 2:05 PM, frenchcr wrote: say i have a column of data like this... 2 3 4 2 1 6 6 4 7 and i want it in three columns like this 226 314 467 ...so i can make a contour plot. How do i do this? -- View this message in context: http://old.nabble.com/one-long-column-of-data--%3E-three-small-columns-tp26163165p26163165.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Weird operator behaviour
Hi, I have a dataset called 'fish'. fish$Species returns extract 1. When I use fish$Species != c(CRA,PHC), i.e. I want all species except CRA and PHC, I get extract 2 which is blatantly wrong. Can anyone see what I'm doing wrong? Regards, James EXTRACT 1 fish$Species [1] ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ATO ATO [19] ATO ATO ATO ATO ATO ATO BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG [37] BIG BIG BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS CRA [55] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [73] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [91] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [109] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [127] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [145] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [163] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [181] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [199] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [217] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [235] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [253] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [271] CRA CRA CRA CRA CRA CRA CRA CRA MAK MAK MAK MAK MAK MAK MAK MAK MAK MAK [289] MAK MAK MAK MAK MAK MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO [307] MOO MOO PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC [325] PHC PHC PHC POS POS POS POS POS POS POS POS POS POS POS PTO PTO PTO PTO [343] PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO RBM RBM RBM RBM RBM RBM RBM [361] RBM RBM RBM RBM RBM RBM RBM RBM SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ [379] SKJ SKJ SKJ SKJ SKJ STM STM STM STM STM STM STM STM STN STN STN STN STN [397] STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN [415] SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO TOR TOR [433] TOR TOR TOR TOR TOR TOR YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN [451] YFN YFN YFN YFN 17 Levels: ALB ATO BIG BWS CRA MAK MOO PHC POS PTO RBM SKJ STM STN SWO ... YFN EXTRACT 2 fish$Species != c(CRA,PHC) [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [25] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [37] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [49] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [61] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [73] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [85] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [97] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [109] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [121] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [133] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [145] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [157] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [169] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [181] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [193] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [205] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [217] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [229] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [241] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [253] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [265] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [277] FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [289] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [301] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE [313] TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE [325] TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [337] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [349] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [361] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [373] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [385] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [397] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[R] Setting column widths in heatmap
Hello, I cannot figure out how to set the column widths (either relative or absolute) for the heatmap function - the full manual hints that layout and lcm will control this, but I haven't had any luck. This is in R 2.9.2 compiled for either FC10 or SUSE11.1 (x86_64)... and am sending the output of heatmap to the pdf driver. Thanks in advance! -kt _ Hotmail: Trusted email with Microsoft's powerful SPAM protection. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reshaping pairs of columns
Try this: require(reshape) # melt the data x - melt(maps, id='point') # split out the labels labels - do.call(rbind, strsplit(sub((.*)(.), \\1 \\2, x$variable), ' ')) x$source - labels[, 2] x$variable - labels[, 1] cast(x, point + source ~ variable) point sourcelat long 1 1 A 41.68 82.85 2 1 B 41.62 87.67 3 1 C -41.12 -83.03 4 1 D -41.03 -85.10 5 1 E 41.28 82.42 6 1 F 43.13 82.25 7 1 G 42.05 82.88 8 1 H 42.15 81.63 9 1 I 42.03 83.53 On Tue, Nov 3, 2009 at 5:26 PM, Michael Friendly frien...@yorku.ca wrote: Given a data frame consisting of a pointID and 12 pairs of (lat, long) variables, with names latA, longA, latB, longB, ... latL, longL, I want to reshape it into a data frame with the structure point source lat long 1 A ... ... 1 B ... ... I've looked at reshape and plyr, but can't figure out how to do this. Details of my data frame are below dput(head(maps)) structure(list(point = 1:6, latA = c(41.68, 41.9, 42.25, 42.55, 44.05, 43.65), longA = c(82.85, 82.52, 81.92, 80.03, 83.03, 83.95 ), latB = c(41.62, 42.12, 42.07, 42.47, 43.42, 43.12), longB = c(87.67, 87, 85.97, 84.17, 87.25, 87.92), latC = c(-41.12, 41.88, 42.13, 42.13, 44.23, 43.25), longC = c(-83.03, 82, 80.78, 79.55, 83.6, 84.48), latD = c(-41.03, 41.87, 41.98, 42, 44.13, 43.8), longD = c(-85.1, 85.37, 83.78, 83.18, 86.85, 87.1), latE = c(41.28, 42.08, 42.22, 42.07, 44.6, 43.72), longE = c(82.42, 81.17, 80.2, 80.27, 83.5, 83.23), latF = c(43.13, 42.35, 42.62, 42.83, 44.62, 43.62), longF = c(82.25, 81.77, 81.37, 79.92, 85.5, 85.25), latG = c(42.05, 42.23, 42.37, 42.37, 44.75, 44.03), longG = c(82.88, 82.35, 81.3, 80.03, 83.57, 83.47), latH = c(42.15, 42.38, 42.53, 42.87, 44.47, 44.03), longH = c(81.63, 80.78, 80.35, 78.73, 81.75, 81.72), latI = c(42.03, 42.25, 42.33, 42.33, 44.75, 44.03), longI = c(83.53, 82.45, 81.98, 80.65, 84.33, 84.22), latJ = c(42.02, 42.12, 42.5, 42.68, 44.35, 44.07), longJ = c(82.9, 82.42, 81.87, 80.15, 82.9, 83.6), latK = c(42, 42.08, 42.48, 42.67, 44.32, 44.03), longK = c(82.9, 82.37, 81.87, 80.18, 82.9, 83.57), latL = c(-41.12, 42, 42.32, 42.55, 44.27, -44.07), longL = c(-82.12, 82.08, 81.75, 80.08, 83, -83.12)), .Names = c(point, latA, longA, latB, longB, latC, longC, latD, longD, latE, longE, latF, longF, latG, longG, latH, longH, latI, longI, latJ, longJ, latK, longK, latL, longL), row.names = c(NA, 6L), class = data.frame) -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street http://www.math.yorku.ca/SCS/friendly.html Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] one long column of data - three small columns
frenchcr wrote: say i have a column of data like this... 2 3 4 2 1 6 6 4 7 and i want it in three columns like this 226 314 467 ...so i can make a contour plot. How do i do this? ok, so matrix(x, 3, 3) works. what if i have a b c a c a c and want a b c a c a c ?? -- View this message in context: http://old.nabble.com/one-long-column-of-data--%3E-three-small-columns-tp26163165p26163570.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Two questions about cloud().
(1) Is there a (simple) way of getting cloud() to do *both*
type=p and type=h? I.e. of getting it to plot the points
as points *and* drop a perpendicular line to the underlying plane?
(2) Is there a way of telling cloud() to drop its lines to the
floor of the bounding box, rather than to the plane z=0?
I thought that the zero.scaled argument (for panel.3dscatter())
might be the ticket here, but when I pass this argument to cloud()
I get an error ``Error using packet 1 formal argument zero.scaled
matched by multiple actual arguments.
Ta.
cheers,
Rolf Turner
##
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Re: [R] Weird operator behaviour
Hi Jim, Take a look at ?%in%, especially its first example. HTH, Jorge On Tue, Nov 3, 2009 at 5:02 PM, jimdare wrote: Hi, I have a dataset called 'fish'. fish$Species returns extract 1. When I use fish$Species != c(CRA,PHC), i.e. I want all species except CRA and PHC, I get extract 2 which is blatantly wrong. Can anyone see what I'm doing wrong? Regards, James EXTRACT 1 fish$Species [1] ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ALB ATO ATO [19] ATO ATO ATO ATO ATO ATO BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG BIG [37] BIG BIG BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS BWS CRA [55] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [73] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [91] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [109] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [127] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [145] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [163] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [181] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [199] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [217] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [235] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [253] CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA CRA [271] CRA CRA CRA CRA CRA CRA CRA CRA MAK MAK MAK MAK MAK MAK MAK MAK MAK MAK [289] MAK MAK MAK MAK MAK MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO MOO [307] MOO MOO PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC PHC [325] PHC PHC PHC POS POS POS POS POS POS POS POS POS POS POS PTO PTO PTO PTO [343] PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO PTO RBM RBM RBM RBM RBM RBM RBM [361] RBM RBM RBM RBM RBM RBM RBM RBM SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ SKJ [379] SKJ SKJ SKJ SKJ SKJ STM STM STM STM STM STM STM STM STN STN STN STN STN [397] STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN STN [415] SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO SWO TOR TOR [433] TOR TOR TOR TOR TOR TOR YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN YFN [451] YFN YFN YFN YFN 17 Levels: ALB ATO BIG BWS CRA MAK MOO PHC POS PTO RBM SKJ STM STN SWO ... YFN EXTRACT 2 fish$Species != c(CRA,PHC) [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [25] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [37] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [49] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [61] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [73] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [85] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [97] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [109] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [121] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [133] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [145] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [157] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [169] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [181] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [193] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [205] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [217] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [229] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [241] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [253] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [265] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE [277] FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [289] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [301] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE FALSE [313] TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE [325] TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [337] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [349] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [361] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
Re: [R] Create Artificial Binary Matrix based on probability
x - matrix(sample(0:1, 1200, replace=T, prob=c(0.952, 0.048)), ncol=30) table(x) x 01 1131 69 x - matrix(sample(0:1, 1200, replace=T, prob=c(0.952, 0.048)), ncol=30) table(x) x 01 1151 49 bikemike42 wrote: Dear All, I am trying to create an artificial binary matrix such that each cell has a probability of 0.048 of having a 1. So far the closest I've come is us by using a random poisson distribution with a mean of 0.048, but I can't figure out how to limit the max value to 1. Otherwise that would work fine it seems. Any suggestions? The main code I've got to create said matrix so far is: a-replicate(26,rpois(57,0.048)) Thanks in Advance, Mike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] avoiding loop
Thanks for your help.
Date: Mon, 2 Nov 2009 18:50:42 -0500
Subject: Re: [R] avoiding loop
From: jholt...@gmail.com
To: bbom...@hotmail.com
CC: mtmor...@fhcrc.org; r-help@r-project.org
The first thing I would suggest is convert your dataframes to matrices
so that you are not having to continually convert them in the calls to
the functions. Also I am not sure what the code:
realized_prob = with(DF, {
ind - (CHOSEN == 1)
n - tapply(theta_multiple[ind],
CS[ind], sum)
d - tapply(theta_multiple, CS, sum)
n / d
})
is doing. It looks like 'n' and 'd' might have different lengths
since they are being created by two different (CS CS[ind])
sequences. I have no idea why you are converting to the DF
dataframe. THere is no need for that. You could just leave the
vectors (e.g., theta_multiple, CS and ind) as they are and work with
them. This is probably where most of your time is being spent. So if
you start with matrices and leave the dataframes out of the main loop
you will probably see an increase in performance.
2009/11/2 parkbomee bbom...@hotmail.com:
This is the Rprof() report by self time.
Is it also possible that these routines, which take long self.time, are
causing the optim() to be slow?
$by.self
self.time self.pct total.time total.pct
FUN 94.16 16.5 94.16 16.5
unlist80.46 14.1 120.54 21.1
lapply76.94 13.5 255.48 44.7
match 60.76 10.6 60.88 10.7
as.matrix.data.frame 31.00 5.4 51.12 8.9
as.character 29.28 5.1 29.28 5.1
unique.default24.36 4.3 24.40 4.3
data.frame21.06 3.7 55.78 9.8
split.default 20.42 3.6 84.38 14.8
tapply13.84 2.4 414.28 72.5
structure 11.32 2.0 22.36 3.9
factor11.08 1.9 127.68 22.3
attributes- 11.00 1.9 11.00 1.9
==10.56 1.8 10.56 1.8
%*% 10.30 1.8 10.30 1.8
as.vector 10.22 1.8 10.22 1.8
as.integer 9.86 1.7 9.86 1.7
list 9.64 1.7 9.64 1.7
exp7.12 1.2 7.12 1.2
as.data.frame.integer 5.98 1.0 8.10 1.4
To: bbom...@hotmail.com
CC: jholt...@gmail.com; r-help@r-project.org
Subject: Re: [R] avoiding loop
From: mtmor...@fhcrc.org
Date: Sun, 1 Nov 2009 22:14:09 -0800
parkbomee bbom...@hotmail.com writes:
Thank you all.
What Chuck has suggested might not be applicable since the number of
different times is around 40,000.
The object of optimization in my function is the varying value,
which is basically data * parameter, of which parameter is the
object of optimization..
And from the r profiling with a subset of data,
I got this report..any idea what Anonymous is?
$by.total
total.time total.pct self.time self.pct
Anonymous 571.56 100.0 0.02 0.0
optim 571.56 100.0 0.00 0.0
fn 571.54 100.0 0.98 0.2
You're giving us 'by.total', so these are saying that all the time was
spent in these functions or the functions they called. Probably all
are in 'optim' and its arguments; since little self.time is spent
here, there isn't much to work with
eval 423.74 74.1 0.00 0.0
with.default 423.74 74.1 0.00 0.0
with 423.74 74.1 0.00 0.0
These are probably in the internals of optim, where the function
you're trying to optimize is being set up for evaluation. Again
there's little self.time, and all these say is that a big piece of the
time is being spent in code called by this code.
tapply 414.28 72.5 13.84 2.4
lapply 255.48 44.7 76.94 13.5
factor 127.68 22.3 11.08 1.9
unlist 120.54 21.1 80.46 14.1
FUN 94.16 16.5 94.16 16.5
these look like they are tapply-related calls (looking at the code for
tapply, it calls lapply, factor, and unlist, and FUN is the function
argument to tapply), perhaps from the function you're optimizing (did
you implement this as suggested below? it would really help to have a
possibly simplified version of the code you're calling).
There is material to work with here, as apparently a fairly large
amount of self.time is being spent in each of these functions. So
here's a sample data set
n - 10
set.seed(123)
df - data.frame(time=sort(as.integer(ceiling(runif(n)*n/5))),
value=ceiling(runif(n)*5))
