[R] Michaelson-Morley Speed of Light Data
URL: http://finzi.psych.upenn.edu/R/library/datasets/html/morley.html The classical data of Michaelson and Morley on the speed of light Can you provide more information about the data? How were they obtained, etc.? I do not have the book Genstat Primer and the nearest location where it is available is University of York which is rather far from my location. Note that the data for the Michelson-Morley experiments [1] consist of 6 experiments of 17 runs each, not of 5 series of 20 runs each. Best regards, Christopher Yeleighton ___ [1] URL: http://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether __ 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] taylor.diagram from plotrix package
On 04/10/2012 11:50 PM, anaraster wrote: Is there a way to access the numeric results (standard deviation and correlation) obtained with the taylor.diagram ? Hi anaraster, That wouldn't be too difficult, just alter the code to return a list like this: return(list(oldpar,R,sd.r,sd.f)) instead of: return(oldpar) Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] clock24.plot
On 04/11/2012 12:59 AM, Nick Fankhauser wrote: I've got the strange problem with clock24.plot that only the first data point (phase = 23.38, size = 0.44) from the phases/sizes numeric vectors is plotted. Does anyone have an idea why this could be? library(plotrix) phases- c(23.38, 22.29, 22.71) sizes- c(0.44, 0.30, 0.30) clock24.plot(sizes,phases) Hi Nick, Try this: clock24.plot(sizes,phases,radial.lim=c(0,max(sizes)) By default, clock24.plot only plots the range of values in lengths. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is a better way to deal with lag/difference and loops in time series using R?
Hello, Thank you very much. How can I create an empty time series with the same dates as existing time series? If x is a ts object, then it would be easy: y-ts(NA, start=start(x), end=end(x),frequency=frequency(x)) What can I do if x is a zoo or xts object? I come up with a cumbersome way of doing this: xts-as.ts(x) y1-ts(NA, start=start(y), end=end(y),frequency=frequency(y)) y-as.xts(y1) Is there any easier way to do it? Thanks miao 2012/4/11 R. Michael Weylandt michael.weyla...@gmail.com Two ways around this: I = Easy) Just use zoo/xts objects. ts objects a real pain in the proverbial donkey because of things like this. Something like: library(xts) PI1.yq - as.xts(PI1) # Specialty class for quarterly data (or regular zoo works) lag(PI1.yq) II = Hard) lag on a ts actually changes the time indices while keeping all the data, [so the fourth data point is the same value -- just a different time point] what you may want to do is cbind() the objects to see how they line up now. cbind(PI1, lag(PI1,4)) Hope this helps, Michael On Tue, Apr 10, 2012 at 11:21 PM, jpm miao miao...@gmail.com wrote: Hello, I am writing codes for time series computation but encountering some problems Given the quarterly data from 1983Q1 to 1984Q2 PI1-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485, -1.190061246, -0.553031799, 0.686874720, 0.953911035), start=c(1983,1), frequency=4) PI1 Qtr1 Qtr2 Qtr3 Qtr4 1983 2.747365190 2.791594762 -0.009953715 -0.015059485 1984 -1.190061246 -0.553031799 0.686874720 0.953911035 If I would like to create a time series vector containing the data in 4 quarters ahead PI4-lag(PI1,4) PI4 Qtr1 Qtr2 Qtr3 Qtr4 1982 2.747365190 2.791594762 -0.009953715 -0.015059485 1983 -1.190061246 -0.553031799 0.686874720 0.953911035 Confusingly, PI1[1] and PI4[1] are exactly the same! I usually would like to calculate the difference between the vector of interest and the corresponding values 4 quarters ahead, but it is zero! PI1[1][1] 2.747365 PI4[1][1] 2.747365 One remedy that comes into my mind is to use window,but a warning message emerges PI4w-window(PI4, start=start(PI1), end=end(PI1))Warning message:In window.default(x, ...) : 'end' value not changed PI4w Qtr1 Qtr2 Qtr3 Qtr4 1983 -1.1900612 -0.5530318 0.6868747 0.9539110 Similar problems happen with the usage of the function diff, which calculate the difference. I wonder if it is better to work with the dates (1983Q1, 1983Q2,.) directly? If I want to write a loop, say, to conduct some computation from 1983Q1 to 2011Q4, the only way I know is to convert the dates to the ordinal indices, 1, 2, 3.. Can we work with the dates? Is there any built-in equality that provides the computation like 1983Q1 +1 equals 1983Q2? In EViews, it is easy to do that. We can let %s run from 1983Q1 to 2011Q4, and he knows that 1983Q1+1 is exactly 1983Q2. Thanks very much for your reply! miao [[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] how to save multiple work space
Hi Petr and Michael, Thank you very much for the response. It worked! Now I can use multiple R sessions simultaneously, I just need to save their workspace separately. And If I want those objects in the single session, I just load them all in this session, and save the workspace again together:) Thank you very much. This save me lots of time! Best regards, ya On 2012-4-10 15:14, Petr PIKAL wrote: Hi You'll need to save them manually to avoid name conflicts -- save.image() is the function to do so but you need to give a file name. Or it is necessary have separate folder for each R session. Regards Petr Michael On Apr 10, 2012, at 7:41 AM, yaxinxi...@163.com wrote: Hi guys, I have a question. I am running 3 R sessions simultaneously for different analysis. I found out that when R quit, only objects in one of these sessions was saved in the work space. How can I save objects of all 3 R sessions? Thank you very much. YA __ 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] how to save multiple work space
Hi Petr and Michael, Thank you very much for the response. It worked! Now I can use multiple R sessions simultaneously, I just need to save their workspace separately. And If I want those objects in the single session, I just load them all in this session, and save the workspace again together:) Thank you very much. This save me lots of time! Best regards, ya On 2012-4-10 15:14, Petr PIKAL wrote: Hi You'll need to save them manually to avoid name conflicts -- save.image() is the function to do so but you need to give a file name. Or it is necessary have separate folder for each R session. Regards Petr Michael On Apr 10, 2012, at 7:41 AM, yaxinxi...@163.com wrote: Hi guys, I have a question. I am running 3 R sessions simultaneously for different analysis. I found out that when R quit, only objects in one of these sessions was saved in the work space. How can I save objects of all 3 R sessions? Thank you very much. YA __ 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.
[R] Read .idat Illumina files in R
Dear Bioc and R List Users, I am having trouble analysing illumine data generated from BeadScan. I have .idat files and JPEG images. I realise that i need bead-level summary data to be able to begin quality control followed by normalization. Is there a way i can read .idat files for expression analysis or do i need to go back to BeadScan and generate .txt files/tiff files ? Appreciate any help here. Many Thanks, Ekta Jain The information contained in this electronic message and in any attachments to this message is confidential, legally privileged and intended only for use by the person or entity to which this electronic message is addressed. If you are not the intended recipient, and have received this message in error, please notify the sender and system manager by return email and delete the message and its attachments and also you are hereby notified that any distribution, copying, review, retransmission, dissemination or other use of this electronic transmission or the information contained in it is strictly prohibited. Please note that any views or opinions presented in this email are solely those of the author and may not represent those of the Company or bind the Company. Any commitments made over e-mail are not financially binding on the company unless accompanied or followed by a valid purchase order. This message has been scanned for viruses and dangerous content by Mail Scanner, a! nd is believed to be clean. The Company accepts no liability for any damage caused by any virus transmitted by this email. www.jubl.com [[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] Markov-switching VAR estimation and simulation
Dear R users, I need to fit and then simulate the fitted 2 state Markov-switching VAR model. My google search shows that there is a package called MSVAR for problems of this kind. But it seems the package is removed from the CRAN repository. May I get a help from someone here please? Thanks Mamush. [[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] Problem with effects package
sessionInfo() R version 2.15.0 (2012-03-30) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] grid datasets splines utils stats [6] graphics grDevices methods base other attached packages: [1] effects_2.1-0 colorspace_1.1-1 [3] nnet_7.3-1 nlme_3.1-103 [5] lattice_0.20-6 reshape_0.8.4 [7] plyr_1.7.1 catdata_1.0 [9] cacheSweave_0.6-1 stashR_0.3-5 [11] filehash_2.2-1 BiocInstaller_1.4.3 [13] ctv_0.7-4 sos_1.3-1 [15] brew_1.0-6 Hmisc_3.9-3 [17] survival_2.36-12MASS_7.3-17 loaded via a namespace (and not attached): [1] cluster_1.14.2 digest_0.5.2 tools_2.15.0 # copy and paste the following: library( catdata ) data( unemployment ) unempt - unemployment unempt$durbin - unempt$durbin - 1 library( reshape ) unempt - melt( table( unempt ) ) unempw - cast( unempt, age ~ durbin ) names( unempw ) - c( 'age', 'short', 'long' ) modt - glm( durbin ~ age, weights = value, family = binomial, data = unempt ) modw - glm( cbind( short, long ) ~ age, family = binomial, data = unempw ) library( effects ) modt.ef - effect( 'age', modt ) # works! modw.ef - effect( 'age', modw ) # doesn't work! # Error in eval(expr, envir, enclos) : object 'age' not found # end __ Professor Michael Kubovy University of Virginia Department of Psychology for mail add: for FedEx or UPS add: P.O.Box 400400 Gilmer Hall, Room 102 Charlottesville, VA 22904-4400 485 McCormick Road USA Charlottesville, VA 22903 roomphone Office:B011 +1-434-982-4729 Lab:B019+1-434-982-4751 WWW:http://www.people.virginia.edu/~mk9y/ [[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] What is a better way to deal with lag/difference and loops in time series using R?
On Tue, Apr 10, 2012 at 11:21 PM, jpm miao miao...@gmail.com wrote: Hello, I am writing codes for time series computation but encountering some problems Given the quarterly data from 1983Q1 to 1984Q2 PI1-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485, -1.190061246, -0.553031799, 0.686874720, 0.953911035), start=c(1983,1), frequency=4) If I would like to create a time series vector containing the data in 4 quarters ahead Similar problems happen with the usage of the function diff, which calculate the difference. I wonder if it is better to work with the dates (1983Q1, 1983Q2,.) directly? If I want to write a loop, say, to conduct some computation from 1983Q1 to 2011Q4, the only way I know is to convert the dates to the ordinal indices, 1, 2, 3.. Can we work with the dates? Is there any built-in equality that provides the computation like 1983Q1 +1 equals 1983Q2? In EViews, it is easy to do that. We can let %s run from 1983Q1 to 2011Q4, and he knows that 1983Q1+1 is exactly 1983Q2. This takes the difference between each point and the point 4 quarters back and then lags the result forward. lag(diff(PI1, 4), 4) In zoo you can do this: library(zoo) z - as.zoo(PI1) time(z) - as.yearqtr(time(z)) # quarter after '83 Q2 z[as.yearqtr(1983 Q2)+1/4] You might also want to look at rollapply in the zoo package. xts which is related to zoo has flexible subscripting. If your data is regular as in your example then you could also use the tis package which is related to the fame package but can be used without it. zoo has several vignettes which give many examples of processing time series. The timeSeries package is another alternative. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with effects package
Dear Michael, From ?effect (under Warnings and Limitations): Binomial generalized linear models cannot have a matrix of successes and failures on the left-hand side of the model formula; instead specify the proportion of successes (i.e., successes/(successes + failures)) as the response, and give the number of binomial trials (i.e., successes + failures) in the weights argument to glm. Best, John John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Wed, 11 Apr 2012 07:20:51 -0400 Michael Kubovy kub...@virginia.edu wrote: sessionInfo() R version 2.15.0 (2012-03-30) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] grid datasets splines utils stats [6] graphics grDevices methods base other attached packages: [1] effects_2.1-0 colorspace_1.1-1 [3] nnet_7.3-1 nlme_3.1-103 [5] lattice_0.20-6 reshape_0.8.4 [7] plyr_1.7.1 catdata_1.0 [9] cacheSweave_0.6-1 stashR_0.3-5 [11] filehash_2.2-1 BiocInstaller_1.4.3 [13] ctv_0.7-4 sos_1.3-1 [15] brew_1.0-6 Hmisc_3.9-3 [17] survival_2.36-12MASS_7.3-17 loaded via a namespace (and not attached): [1] cluster_1.14.2 digest_0.5.2 tools_2.15.0 # copy and paste the following: library( catdata ) data( unemployment ) unempt - unemployment unempt$durbin - unempt$durbin - 1 library( reshape ) unempt - melt( table( unempt ) ) unempw - cast( unempt, age ~ durbin ) names( unempw ) - c( 'age', 'short', 'long' ) modt - glm( durbin ~ age, weights = value, family = binomial, data = unempt ) modw - glm( cbind( short, long ) ~ age, family = binomial, data = unempw ) library( effects ) modt.ef - effect( 'age', modt ) # works! modw.ef - effect( 'age', modw ) # doesn't work! # Error in eval(expr, envir, enclos) : object 'age' not found # end __ Professor Michael Kubovy University of Virginia Department of Psychology for mail add: for FedEx or UPS add: P.O.Box 400400Gilmer Hall, Room 102 Charlottesville, VA 22904-4400485 McCormick Road USA Charlottesville, VA 22903 roomphone Office:B011 +1-434-982-4729 Lab:B019 +1-434-982-4751 WWW:http://www.people.virginia.edu/~mk9y/ __ 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] Michaelson-Morley Speed of Light Data
On 12-04-11 12:43 AM, Křištof Želechovski wrote: URL: http://finzi.psych.upenn.edu/R/library/datasets/html/morley.html The classical data of Michaelson and Morley on the speed of light Can you provide more information about the data? How were they obtained, etc.? I do not have the book Genstat Primer and the nearest location where it is available is University of York which is rather far from my location. If you can't find the cited reference, I'd try Google. For instance, it led me to this page http://en.wikipedia.org/wiki/File:Michelsonmorley-boxplot.svg which appears to show five series. Duncan Murdoch Note that the data for the Michelson-Morley experiments [1] consist of 6 experiments of 17 runs each, not of 5 series of 20 runs each. Best regards, Christopher Yeleighton ___ [1]URL: http://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether __ 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] Michaelson-Morley Speed of Light Data
On 11/04/2012 12:42, Duncan Murdoch wrote: On 12-04-11 12:43 AM, Křištof Želechovski wrote: URL: http://finzi.psych.upenn.edu/R/library/datasets/html/morley.html The classical data of Michaelson and Morley on the speed of light Can you provide more information about the data? How were they obtained, etc.? I do not have the book Genstat Primer and the nearest location where it is available is University of York which is rather far from my location. If you can't find the cited reference, I'd try Google. For instance, it led me to this page http://en.wikipedia.org/wiki/File:Michelsonmorley-boxplot.svg which appears to show five series. Yes, but that is derived from R. AFAIR the history, Bill Venables got this from Weekes (1986), a book I have only ever seen in Adelaide. A better reference is S. M. Stigler (1977) Do robust estimators work with real data? Annals of Statistics 5, 1055–1098. (See Table 6.) -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to increase the limit for max.print in R
Hi Pooja, You must use options command, something like this options(max.print=5.5E5) For more information type? ?options -- Bernardo Rangel Tura, M.D,MPH,Ph.D National Institute of Cardiology Brazil Thanks Bernardo; cut and paste above line done.. -- View this message in context: http://r.789695.n4.nabble.com/how-to-increase-the-limit-for-max-print-in-R-tp886412p4548611.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 compare C-index in Cox model
Dear List, If I calculate the C-index after 10-fold cross-validation based on Cox model, which statistic test should I used to compare two C-statistc values.I means I use cross-validation to obtain a predictive probability for each individual, then combine all the probabilities to calculate the C-index.If I get two C-index values(0.73 vs. 0.80) based on two different model, then how to test the difference is significant. For the C-index values got from a single dataset, we may use bootstrap to get the standard error, and use wald test to measure whether they are significant different,but if c-index got from cross- validation,whether I can still use this method to compare them. -- View this message in context: http://r.789695.n4.nabble.com/how-to-compare-C-index-in-Cox-model-tp4548251p4548251.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] 7 arguments passed to .Internal(identical) which requires 6
Hello, I've just installed R-2.15.0 (I've had R 2.13.2 before it and I've deleted everything before I started to install new version). When I've tried to run my script by command source() I received this message: Error in source(script.R) : 7 arguments passed to .Internal(identical) which requires 6 I know I should delete my $R_HOME but it contain only one string: [1] C:/PROGRA~1/R/R-215~1.0 I don't know what's happened. Thank you! -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4548460.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] xyplot lattice fine control of axes limits and thick marks (with log scale)
hi, I just realised I want to go a little further in the control of the chart
appearance and I would like to have the same number of ticks displayed in
both axes of all panels
given this code
xyplot(tv ~ ms | sub_family, data=tm,
#as.table=TRUE,
aspect=xy,
xlab = expression(paste('ms [ fg/', m^3, ' ]', sep = '')),
ylab = expression(paste('tv [ fg/', m^3, ' ]', sep = '')),
scales= list(x=list(relation=free, log=10, cex=0.8),
y=list(relation=free, log=10, cex=0.8)),
prepanel = function(x, y, subscripts) {
rr- range(cbind(x,y))
list(xlim = rr, ylim= rr)
},
panel = function(x, y ,subscripts,...) {
panel.xyplot(x, y, cex=0.8,...)
panel.abline(a = 0, b = 1, lty = 2, col =gray)
panel.text(x, y, labels=tm$name_short[subscripts], cex = 0.8, pos=3,
offset=0.5, srt=0, adj=c(1,1))
},
subscripts=TRUE,
xscale.components = xscale.components.logpower,
yscale.components = yscale.components.logpower
)
...I have been trying to insert in the 'prepanel' and also in the 'panel'
the statement 'tick.number=5' but this does not seem to have any effect
some useful hints for this?
thanks a lot
--
View this message in context:
http://r.789695.n4.nabble.com/xyplot-lattice-fine-control-of-axes-limits-and-thick-marks-with-log-scale-tp4511897p4548502.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] xyplot lattice fine control of axes limits and thick marks (with log scale)
On Apr 11, 2012, at 6:28 AM, maxbre wrote:
hi, I just realised I want to go a little further in the control of
the chart
appearance and I would like to have the same number of ticks
displayed in
both axes of all panels
given this code
xyplot(tv ~ ms | sub_family, data=tm,
#as.table=TRUE,
aspect=xy,
xlab = expression(paste('ms [ fg/', m^3, ' ]', sep = '')),
ylab = expression(paste('tv [ fg/', m^3, ' ]', sep = '')),
scales= list(x=list(relation=free, log=10, cex=0.8),
y=list(relation=free, log=10, cex=0.8)),
I'm wondering if you should be using relation=free when you have
already set a panel specific range for the x and y limits? I'm
thinking that the panel function may be reversing your earlier
prepanel efforts. (No data offered ... why don't you use one of the
many test datasets in the examples of the lattice package?)
prepanel = function(x, y, subscripts) {
rr- range(cbind(x,y))
list(xlim = rr, ylim= rr)
},
panel = function(x, y ,subscripts,...) {
panel.xyplot(x, y, cex=0.8,...)
panel.abline(a = 0, b = 1, lty = 2, col =gray)
panel.text(x, y, labels=tm$name_short[subscripts], cex = 0.8,
pos=3,
offset=0.5, srt=0, adj=c(1,1))
},
subscripts=TRUE,
xscale.components = xscale.components.logpower,
yscale.components = yscale.components.logpower
)
...I have been trying to insert in the 'prepanel' and also in the
'panel'
the statement 'tick.number=5' but this does not seem to have any
effect
some useful hints for this?
thanks a lot
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] What is a better way to deal with lag/difference and loops in time series using R?
Here's one: x - xts(1:5, Sys.Date() + 1:5) y - xts(rep(NA, length(x)), time(x)) or another, less direct but shorter: y - cbind(x, NA)[,2] Michael On Wed, Apr 11, 2012 at 4:16 AM, jpm miao miao...@gmail.com wrote: Hello, Thank you very much. How can I create an empty time series with the same dates as existing time series? If x is a ts object, then it would be easy: y-ts(NA, start=start(x), end=end(x),frequency=frequency(x)) What can I do if x is a zoo or xts object? I come up with a cumbersome way of doing this: xts-as.ts(x) y1-ts(NA, start=start(y), end=end(y),frequency=frequency(y)) y-as.xts(y1) Is there any easier way to do it? Thanks miao 2012/4/11 R. Michael Weylandt michael.weyla...@gmail.com Two ways around this: I = Easy) Just use zoo/xts objects. ts objects a real pain in the proverbial donkey because of things like this. Something like: library(xts) PI1.yq - as.xts(PI1) # Specialty class for quarterly data (or regular zoo works) lag(PI1.yq) II = Hard) lag on a ts actually changes the time indices while keeping all the data, [so the fourth data point is the same value -- just a different time point] what you may want to do is cbind() the objects to see how they line up now. cbind(PI1, lag(PI1,4)) Hope this helps, Michael On Tue, Apr 10, 2012 at 11:21 PM, jpm miao miao...@gmail.com wrote: Hello, I am writing codes for time series computation but encountering some problems Given the quarterly data from 1983Q1 to 1984Q2 PI1-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485, -1.190061246, -0.553031799, 0.686874720, 0.953911035), start=c(1983,1), frequency=4) PI1 Qtr1 Qtr2 Qtr3 Qtr4 1983 2.747365190 2.791594762 -0.009953715 -0.015059485 1984 -1.190061246 -0.553031799 0.686874720 0.953911035 If I would like to create a time series vector containing the data in 4 quarters ahead PI4-lag(PI1,4) PI4 Qtr1 Qtr2 Qtr3 Qtr4 1982 2.747365190 2.791594762 -0.009953715 -0.015059485 1983 -1.190061246 -0.553031799 0.686874720 0.953911035 Confusingly, PI1[1] and PI4[1] are exactly the same! I usually would like to calculate the difference between the vector of interest and the corresponding values 4 quarters ahead, but it is zero! PI1[1][1] 2.747365 PI4[1][1] 2.747365 One remedy that comes into my mind is to use window,but a warning message emerges PI4w-window(PI4, start=start(PI1), end=end(PI1))Warning message:In window.default(x, ...) : 'end' value not changed PI4w Qtr1 Qtr2 Qtr3 Qtr4 1983 -1.1900612 -0.5530318 0.6868747 0.9539110 Similar problems happen with the usage of the function diff, which calculate the difference. I wonder if it is better to work with the dates (1983Q1, 1983Q2,.) directly? If I want to write a loop, say, to conduct some computation from 1983Q1 to 2011Q4, the only way I know is to convert the dates to the ordinal indices, 1, 2, 3.. Can we work with the dates? Is there any built-in equality that provides the computation like 1983Q1 +1 equals 1983Q2? In EViews, it is easy to do that. We can let %s run from 1983Q1 to 2011Q4, and he knows that 1983Q1+1 is exactly 1983Q2. Thanks very much for your reply! miao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] convex nonnegative basis vectors in nullspace of matrix
Dear all, I want to explore the nullspace of a matrix S: I currently use the function Null from the MASS package to get a basis for the null space: S = matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1)); S MASS::Null(t(S)) My problem is that I actually need a nonnegative basis for the null space of S. There should be a unique set of convex basis vectors spanning a vector space in which each vector v satisfies sum (S %*% v) == 0 and min(v)=0. Is there maybe an R function that can calculate that for me? I would appreciate any help, Thanks in advance, Hannes -- View this message in context: http://r.789695.n4.nabble.com/convex-nonnegative-basis-vectors-in-nullspace-of-matrix-tp4548822p4548822.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] 7 arguments passed to .Internal(identical) which requires 6
Well, obviously the interface to .Internal(identical) changed between 2.13.x and 2.15 -- as, ?.Internal says: Only true R wizards should even consider using this function... Anyways, find where you use it in your script and then we can help -- there's not much we can do without seeing the relevant line of code. Michael On Wed, Apr 11, 2012 at 5:58 AM, krtek marshal...@mail.ru wrote: Hello, I've just installed R-2.15.0 (I've had R 2.13.2 before it and I've deleted everything before I started to install new version). When I've tried to run my script by command source() I received this message: Error in source(script.R) : 7 arguments passed to .Internal(identical) which requires 6 I know I should delete my $R_HOME but it contain only one string: [1] C:/PROGRA~1/R/R-215~1.0 I don't know what's happened. Thank you! -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4548460.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] Merging multiple .csv files
Simply pass all = FALSE to merge_all merge_all(list_of_files, by = Name, all = FALSE) Michael On Wed, Apr 11, 2012 at 1:09 AM, Chintanu chint...@gmail.com wrote: Thanks to David and Michael Michael: That works, however with a glitch. Each of my 24 files has got two columns: Name, and Rank/score. file_list - list.files() list_of_files - lapply(file_list, read.csv) # Read in each file # I can see the 2-columns at this stage. However, the following line: merge_all(list_of_files, by = Name) # produces some NAs for the 2nd column (except the beginning 1/3rd of the columns which have values). Not sure about the reason - the original files don't have any NAs. Further, I understand that it gives the union of (rows of) files based on Name. Is there a way to look for intersection, i.e., similar to using: merge ( ,by=Name, all=FALSE) ? David: It came up with an error : do.call(merge, list_of_files, by=Name) Error in do.call(merge, list_of_files, by = Name) : unused argument(s) (by = Name) Cheers, Chintanu __ 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] xyplot lattice fine control of axes limits and thick marks (with log scale)
On Apr 11, 2012, at 9:03 AM, David Winsemius wrote:
On Apr 11, 2012, at 6:28 AM, maxbre wrote:
hi, I just realised I want to go a little further in the control of
the chart
appearance and I would like to have the same number of ticks
displayed in
both axes of all panels
given this code
xyplot(tv ~ ms | sub_family, data=tm,
#as.table=TRUE,
aspect=xy,
xlab = expression(paste('ms [ fg/', m^3, ' ]', sep = '')),
ylab = expression(paste('tv [ fg/', m^3, ' ]', sep = '')),
scales= list(x=list(relation=free, log=10, cex=0.8),
y=list(relation=free, log=10, cex=0.8)),
I'm wondering if you should be using relation=free when you have
already set a panel specific range for the x and y limits? I'm
thinking that the panel function may be reversing your earlier
prepanel efforts. (No data offered ... why don't you use one of the
many test datasets in the examples of the lattice package?)
On further meandering up this thread I see that you omitted the
context of earlier data offerings, so not I in turn offer what I think
is a your request. Change relation from free to sliced
scales= list(x=list(relation=sliced, log=10, cex=0.8, tick.number=5),
y=list(relation=sliced, log=10, cex=0.8, tick.number=5))
--
David.
prepanel = function(x, y, subscripts) {
rr- range(cbind(x,y))
list(xlim = rr, ylim= rr)
},
panel = function(x, y ,subscripts,...) {
panel.xyplot(x, y, cex=0.8,...)
panel.abline(a = 0, b = 1, lty = 2, col =gray)
panel.text(x, y, labels=tm$name_short[subscripts], cex = 0.8,
pos=3,
offset=0.5, srt=0, adj=c(1,1))
},
subscripts=TRUE,
xscale.components = xscale.components.logpower,
yscale.components = yscale.components.logpower
)
...I have been trying to insert in the 'prepanel' and also in the
'panel'
the statement 'tick.number=5' but this does not seem to have any
effect
some useful hints for this?
thanks a lot
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Question on Counting Factors
Hi,
I hope this is not too trivial, but I've had this recurring problem
and I think there is super easy solution, just not sure what it is.
Please see short example below. I would like to get the frequency
(counts) of all the variables in a single column (that is easy), but I
would also like to return the value 0 for the absence of variables
defined in another column.
For example:
animals =
matrix(c('cat','tiger','cat','tiger','fish','fish','dog','dog'),ncol=2,
byrow=F)
animals = as.data.frame(animals)
table(animals$V1)
# Returns the count for the variables in the first column as
cat tiger
2 2
# But I would like to have table(animals$V1) return
cat tiger dog fish
2 2 0 0
Any help is much appreciated.
Cheers,
Dan
__
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] 7 arguments passed to .Internal(identical) which requires 6
On 11/04/2012 14:08, R. Michael Weylandt wrote: Well, obviously the interface to .Internal(identical) changed between 2.13.x and 2.15 -- as, ?.Internal says: Only true R wizards should even consider using this function... Anyways, find where you use it in your script and then we can help -- there's not much we can do without seeing the relevant line of code. It may not be in his code. What is odd is that R 2.15.0's identical() does require 6 arguments, whereas in R = R-patched it accepts 6 or 7. So it looks like something has been done under a later version of R, and the usual culprit here is an S4-using package which has captured the definition of identical() from a later version of R. Updating to R-patched may solve the problem. Michael On Wed, Apr 11, 2012 at 5:58 AM, krtekmarshal...@mail.ru wrote: Hello, I've just installed R-2.15.0 (I've had R 2.13.2 before it and I've deleted everything before I started to install new version). When I've tried to run my script by command source() I received this message: Error in source(script.R) : 7 arguments passed to .Internal(identical) which requires 6 I know I should delete my $R_HOME but it contain only one string: [1] C:/PROGRA~1/R/R-215~1.0 I don't know what's happened. Thank you! -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4548460.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on Counting Factors
On Apr 11, 2012, at 9:21 AM, Daniel Gabrieli wrote:
Hi,
I hope this is not too trivial, but I've had this recurring problem
and I think there is super easy solution, just not sure what it is.
Please see short example below. I would like to get the frequency
(counts) of all the variables in a single column (that is easy), but I
would also like to return the value 0 for the absence of variables
defined in another column.
For example:
animals =
matrix
(c('cat','tiger','cat','tiger','fish','fish','dog','dog'),ncol=2,
byrow=F)
animals = as.data.frame(animals)
table(animals$V1)
# Returns the count for the variables in the first column as
cat tiger
2 2
# But I would like to have table(animals$V1) return
cat tiger dog fish
2 20 0
You need to add the missing levels to that factor.
levels(animals$V1) -
unique( c('cat','tiger','cat','tiger','fish','fish','dog','dog'))
table(animals$V1)
cat tiger fish dog
2 2 0 0
Any help is much appreciated.
Cheers,
Dan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on Counting Factors
This would do it in your example: levels(animals$V1) - c(cat,tiger,dog,fish) table(animals) cattigerdog fish 2 200 HTH David cat tiger dog fish 2 2 0 0 On 11 April 2012 14:21, Daniel Gabrieli daniel.gabri...@gmail.com wrote: animals = as.data.frame(animals) table(animals$V1) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on Counting Factors
I would like to get the frequency
(counts) of all the variables in a single column (that is
easy), but I would also like to return the value 0 for the
absence of variables defined in another column.
If you use factor() on your columns and include all the animals in the factor
levels, you should get what you want.
For example
animal.names - sort(c(fish, dog, tiger, cat))
V1 - sample(c('cat', 'tiger'), 10, replace=TRUE)
V1 - factor(V1, levels=animal.names)
table(V1)
For your data frame, you can get animal.names from your existing data set
directly rather than specify in advance. If they are all already factors (as
they will be if you have used as.data.frame on a character matrix) you can get
all the levels using rapply. Re-using factor will again get you what you're
after:
animals -
matrix(c('cat','tiger','cat','tiger','fish','fish','dog','dog'),ncol=2, byrow=F)
animals - as.data.frame(animals)
animal.names - sort(rapply(animals, levels))
animals2 - as.data.frame( lapply(animals, factor, levels=animal.names))
table(animals2$V1)
For extra safety, you might want to wrap the second factor() round an
as.character:
animal.names - sort(rapply(animals, levels))
animals2 - as.data.frame( lapply(animals, function(x, l)
factor(as.character(x), levels=l), l=animal.names))
table(animals2$V1)
S Ellison
***
This email and any attachments are confidential. Any use...{{dropped:8}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on Counting Factors - PS
-Original Message-
This would do it in your example:
levels(animals$V1) - c(cat,tiger,dog,fish)
table(animals)
cattigerdog fish
2 200
But be very wary of
levels(animals$V2)- c(cat,tiger,dog,fish)
table(animals$V2)
cat tiger dog fish
2 2 0 0 # !!
levels overwrites factor levels... factor(x, levels = ...) doesn't.
S***
This email and any attachments are confidential. Any use...{{dropped:8}}
__
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] inference for customized regression in R?
Hi all, Are there functions in R that could help me do the following? We have a special type of regression which is called Geometric Mean Regression. We have done some search and found the following: https://stat.ethz.ch/pipermail/r-help/2011-July/285022.html The question is: how to do the statistical inference on GMR results? More specifically, we are looking for the prediction interval: Lets say we regress y1, y2, ..., yn onto x1, x2, ..., xn: we would like to know what's the prediction interval for a new data point: x_new=x1+x2+x3 (i.e. the new data point is the sum of the existing first three data points) In ordinary linear regression, we could derive prediction interval for an in-sample data point as well as a new data point... For our x_new=x1+x2+x3, we can derive formulas for the prediction interval. But for the above customized regression, how do we obtain the prediction intervals? -- Are there functions in R that can help us do this? We are thinking of using bootstrapping, etc. Are there functions in R help us on this? Thanks a lot! [[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] 7 arguments passed to .Internal(identical) which requires 6
On 11.04.2012 15:27, Prof Brian Ripley wrote: On 11/04/2012 14:08, R. Michael Weylandt wrote: Well, obviously the interface to .Internal(identical) changed between 2.13.x and 2.15 -- as, ?.Internal says: Only true R wizards should even consider using this function... Anyways, find where you use it in your script and then we can help -- there's not much we can do without seeing the relevant line of code. It may not be in his code. What is odd is that R 2.15.0's identical() does require 6 arguments, whereas in R = R-patched it accepts 6 or 7. I suspect he uses a mixture of R-patched and R-release and / or R-2.13.2 so that internal code and R code in the package does not match anymore, probably because some (outdated?) base package or so is also in an additional library on the search path. uwe So it looks like something has been done under a later version of R, and the usual culprit here is an S4-using package which has captured the definition of identical() from a later version of R. Updating to R-patched may solve the problem. Michael On Wed, Apr 11, 2012 at 5:58 AM, krtekmarshal...@mail.ru wrote: Hello, I've just installed R-2.15.0 (I've had R 2.13.2 before it and I've deleted everything before I started to install new version). When I've tried to run my script by command source() I received this message: Error in source(script.R) : 7 arguments passed to .Internal(identical) which requires 6 I know I should delete my $R_HOME but it contain only one string: [1] C:/PROGRA~1/R/R-215~1.0 I don't know what's happened. Thank you! -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4548460.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] X11 display problem
I use ssh -Y in this situation. You can test with other X client applications, such as xclock. Try running xclock after your login. If you get the same unable to open message then it's not an R problem. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 4/10/12 6:48 AM, carol white wht_...@yahoo.com wrote: Hi, I run R on a unix server and login from a Mac with ssh -X. When I want to run a graphics function like hist, I get the following x11 message: Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma, d$colortype, : unable to start device X11cairo In addition: Warning message: In function (display = , width, height, pointsize, gamma, bg, : unable to open connection to X11 display '' Do I need any library like cairo installed on my local Mac? I also set LC_CTYPE=en_US.UTF-8 in my profile Cheers, Carol [[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] inference for customized regression in R?
Given the caveats Ted describes here: http://tolstoy.newcastle.edu.au/R/help/05/06/5992.html it seems that bootstrapping might be the only way to get (somewhat) credible prediction intervals: the boot package on CRAN can help to facilitate getting these. Here's some documentation for CI: http://www.statmethods.net/advstats/bootstrapping.html but you'll need to adopt it for a prediction interval, which might entail hacking boot.ci(). You might also see if this question, by someone who most certainly isn't you because cross-posting is discouraged, gets some answers: http://stats.stackexchange.com/questions/26277/how-to-bootstrap-prediction-intervals-for-customized-regression-models-in-r Michael Weylandt On Wed, Apr 11, 2012 at 10:29 AM, Michael comtech@gmail.com wrote: Hi all, Are there functions in R that could help me do the following? We have a special type of regression which is called Geometric Mean Regression. We have done some search and found the following: https://stat.ethz.ch/pipermail/r-help/2011-July/285022.html The question is: how to do the statistical inference on GMR results? More specifically, we are looking for the prediction interval: Lets say we regress y1, y2, ..., yn onto x1, x2, ..., xn: we would like to know what's the prediction interval for a new data point: x_new=x1+x2+x3 (i.e. the new data point is the sum of the existing first three data points) In ordinary linear regression, we could derive prediction interval for an in-sample data point as well as a new data point... For our x_new=x1+x2+x3, we can derive formulas for the prediction interval. But for the above customized regression, how do we obtain the prediction intervals? -- Are there functions in R that can help us do this? We are thinking of using bootstrapping, etc. Are there functions in R help us on this? Thanks a lot! [[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] For loops
In addition to what Jeff and David have said...
If you really want to create a separate data frame for each subgroup then
you need use the assign function, and also data.split[[i]] instead of
data.split[1].
for (i in 1:3) {
assign( paste('sub',i,sep='') , data.split[[i]]
}
Jeff showed you how to skip the splitting, by subsetting within the loop.
'class' is also a R built-in function name, and thus not a good choice for
one's on use.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 4/9/12 2:33 PM, Christopher Desjardins cddesjard...@gmail.com wrote:
Hi,
I am having trouble with syntax for a for loop. Here is what I am trying
to
do.
class=c(rep(1,3),rep(2,3),rep(3,3))
out1=rnorm(length(class))
out2=rnorm(length(class))
out3=rnorm(length(class))
data=data.frame(class,out1,out2,out3)
dat.split=split(data,data$class)
for(i in 1:3){
sub[i]=dat.split[i]
}
However, the for loop doesn't work. I want to assign each split to a
different data object. Better yet, how I could assign each class to a
separate object and skip the splitting?
Thanks,
Chris
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Need help with scraping Google Insight
Hi all, I appreciate that if someone please help me with an R solution/code/library for scraping Google Insight data. It seems difficult to get hold of the URL corresponding to the resulting CSV file that is appropriate for *read.csv*or another similar function. Best, R [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] mosaic 0.4 on CRAN
One of the products of Project MOSAIC (funded by an NSF CCLI grant) has been the development of an R package with the goal of making it easier to use R, especially in teaching situations. We're not quite ready to declare that we've reached version 1.0, but version 0.4 does represent a fairly large step in that direction. You can find out more about the package on CRAN or by installing it, but here are some of the highlights (some example code appears at the end of this message): * extensions of syntax to promote consistency across functions and make wider use of the formula interface * simplified ways of creating and plotting functions, including extracting model fits as functions * a tally() function that combines features of table() and xtabs() and more in a common syntax * expanded syntax for summary functions like mean(), median(), max(), sd(), var(), etc. that accepts formulas and data frames * a do() function that simplifies resampling-based statistical analysis * numerical integration and differentiation to support using calculus techniques in R * first drafts of vignettes on teaching resampling and calculus in R * some functions that add extra features to familiar functions (e.g., xchisq.test(), xhistogram(), xpnorm(), ...) * some data sets If you are using mosaic and discover bugs, or have suggestions for future development, consider submitting an issue on our github development site: http://github.com/rpruim/mosaic/issues/ You can also look there to see what's already on our to-do list. ---rjp (on behalf of the development team that includes Danny Kaplan and Nick Horton) Randall Pruimphone: 616.526.7113 Dept. of Mathematics and Statistics email: rpr...@calvin.edu Calvin College office: NH 284 1740 Knollcrest Circle SE URL: http://www.calvin.edu/~rpruim/ Grand Rapids, MI 49546-4403FAX: 616.526.6501 - Here are the promised code examples to give you a feel for what mosaic makes possible: mean(age, data=HELPrct) [1] 35.65342 mean(~age, data=HELPrct) [1] 35.65342 mean(age ~ sex, data=HELPrct) female male 36.25234 35.46821 mean(age ~ sex treat, data=HELPrct) female.nomale.no female.yes male.yes 37.56364 35.90173 34.86538 35.03468 interval(binom.test( ~ eruptions 3, faithful)) probability of success lower upper 0.6433824 0.5832982 0.7003038 pval(binom.test( ~ eruptions 3, faithful)) p.value 2.608528e-06 xchisq.test(phs) # physicians health study example (data entry omitted) Pearson's Chi-squared test with Yates' continuity correction data: phs X-squared = 24.4291, df = 1, p-value = 7.71e-07 104.00 10933.00 ( 146.52) (10890.48) [12.34] [ 0.17] -3.51 0.41 189.00 10845.00 ( 146.48) (10887.52) [12.34] [ 0.17] 3.51 -0.41 key: observed (expected) [contribution to X-squared] residual model - lm(length ~ width + sex, KidsFeet) L - makeFun(model) L( 9.0, 'B') 1 24.80017 L( 9.0, 'B', interval='confidence') fit lwr upr 1 24.80017 24.30979 25.29055 xyplot( length ~ width, groups= sex, KidsFeet ) # scatter plot with different symbols for boys and girls plotFun(L(x,'B') ~ x, add=TRUE) # add model fit (for boys) to plot plotFun(L(x,'G') ~ x, add=TRUE, lty=2) # add model fit (for girls) to plot rflip(10) # flip a coin 10 times Flipping 10 coins [ Prob(Heads) = 0.5 ] ... T H T H H H T H H T Result: 6 heads. do(2) * rflip(10) # do that twice; notice that do() extracts interesting info n heads tails 1 10 4 6 2 10 6 4 ladyTastingTea - do(5000) * rflip(10) # simulate 5000 ladies tasting tea tally(~heads, ladyTastingTea) 0 1 2 3 4 5 6 7 8 910 Total 552 221 573 1032 1227 1027 606 19852 7 5000 tally(~heads, ladyTastingTea, format='proportion') 0 1 2 3 4 5 6 7 8 9 10 Total 0.0010 0.0104 0.0442 0.1146 0.2064 0.2454 0.2054 0.1212 0.0396 0.0104 0.0014 1. # do() extracts useful information from lm objects so that randomization tests are easy. do(2) * lm( length ~ width + shuffle(sex), data=KidsFeet ) Interceptwidth sexGsigma r-squared 1 9.646822 1.693137 -0.3057453 1.026824 0.4246224 2 11.416739 1.453416 0.4860068 1.013323 0.4396534 tally( ~ sex substance, HELPrct ) substance sex alcohol cocaine heroin Total female 36 41 30 107 male 141 111 94 346 Total 177 152124 453 tally( ~ sex | substance, HELPrct ) # auto switch to proportions for
Re: [R] inference for customized regression in R?
Hi Michael, Thank you for your help! I did some googling and researching... Reading the following article, http://www.ecd.bnl.gov/pubs/BNL-79819-2008-JA.pdf It seems that once we estimate the parameters of the bivariate normal distribution, then we can plug into the formula of conditional distribution of Y|X=x1+x2+x3 ? http://en.wikipedia.org/wiki/Multivariate_normal_distribution My question is: Is it a correct procedure to do the following: Step 1: estimate the parameters of the bivariate normal distribution; Step 2: plug the estimated parameters into the Y|X=x1+x2+x3 formula and get the 95% quantile of it? Do I need to repeat Step 2 many times following the bootstrapping procedure? Or one shot of Step 2 is enough? I got very much confused... Any thoughts? Thanks a lot! On Wed, Apr 11, 2012 at 10:12 AM, R. Michael Weylandt michael.weyla...@gmail.com wrote: Given the caveats Ted describes here: http://tolstoy.newcastle.edu.au/R/help/05/06/5992.html it seems that bootstrapping might be the only way to get (somewhat) credible prediction intervals: the boot package on CRAN can help to facilitate getting these. Here's some documentation for CI: http://www.statmethods.net/advstats/bootstrapping.html but you'll need to adopt it for a prediction interval, which might entail hacking boot.ci(). You might also see if this question, by someone who most certainly isn't you because cross-posting is discouraged, gets some answers: http://stats.stackexchange.com/questions/26277/how-to-bootstrap-prediction-intervals-for-customized-regression-models-in-r Michael Weylandt On Wed, Apr 11, 2012 at 10:29 AM, Michael comtech@gmail.com wrote: Hi all, Are there functions in R that could help me do the following? We have a special type of regression which is called Geometric Mean Regression. We have done some search and found the following: https://stat.ethz.ch/pipermail/r-help/2011-July/285022.html The question is: how to do the statistical inference on GMR results? More specifically, we are looking for the prediction interval: Lets say we regress y1, y2, ..., yn onto x1, x2, ..., xn: we would like to know what's the prediction interval for a new data point: x_new=x1+x2+x3 (i.e. the new data point is the sum of the existing first three data points) In ordinary linear regression, we could derive prediction interval for an in-sample data point as well as a new data point... For our x_new=x1+x2+x3, we can derive formulas for the prediction interval. But for the above customized regression, how do we obtain the prediction intervals? -- Are there functions in R that can help us do this? We are thinking of using bootstrapping, etc. Are there functions in R help us on this? Thanks a lot! [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multicore/mcparallel error
Michael,
Thanks for your help!
I may not be understanding you. I've upgraded to 2.15. From your text
below, I thought the parallel package was included with R 2.15, but I
cannot find an mclapply function there. One exists through the multicore
package, which I've now installed. I also went looking for a parallel
package, but was unable to find one.
Is there an mclapply included in 2.15? Is there a parallel package I'm
missing? Or am I completely misunderstanding your response?
Thanks!
Wyatt
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Tuesday, April 10, 2012 12:07 PM
To: Wyatt McMahon
Cc: R-help@r-project.org
Subject: Re: [R] multicore/mcparallel error
I don't know the multicore package, but if possible, it might be easier to
upgrade to 2.15 and use the new built-in parallel package that was
introduced in R 2.14.
Then your syntax would be something like
mclapply(files, illumqc)
Michael
On Tue, Apr 10, 2012 at 11:33 AM, Wyatt McMahon wmcma...@vbi.vt.edu
wrote:
Hello everyone,
I'm trying to parallelize an R script I have written. To do this, I
am first trying to use the multicore package, because I've had some
previous success with that.
The function I'm trying to parallelize is illumqc. I'd like to create
a separate process for each of 8 files, contained in the vector files.
Below is my code:
for(i in 1:length(files)){
mcparallel(illumqc(files[i]))}
I get the following error:
Error in sendMaster(serialize(try(eval(expr, env), silent = TRUE),
NULL,
:
ignoring SIGPIPE signal
Calls: mcparallel - sendMaster - .Call
In addition: Warning message:
In min(which(alf != 0)) : no non-missing arguments to min; returning
Inf
However, if I try and make a simpler example, everything works
correctly, so I'm not sure what's going wrong with this function. Do
I need to post the contents of the function as well? I'm hoping
someone can recognize this error and give me a clue as to what is
going wrong since the function is fairly long.
Thanks in advance,
Wyatt
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ShortRead_1.10.4 Rsamtools_1.4.3 lattice_0.19-30
[4] Biostrings_2.20.4 GenomicRanges_1.4.8 IRanges_1.10.6
[7] multicore_0.1-7
loaded via a namespace (and not attached):
[1] Biobase_2.12.2 grid_2.13.1 hwriter_1.3
[[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] For loops
This has nothing to do with your question, but instead of using class=c(rep(1,3),rep(2,3),rep(3,3)) It's probably easier to use class = rep(1:3, each =3) [[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] Survreg output - interpretation
Hello R users, I am analizing survival data (mostly uncensored) and want to extract the most out of it. Since I have more than one factor, I´ve read that the survival regression can help to test the interactions between factors, and then decide how to do the comparisons using the Log-rank test (survdiff). 1- if I chose the Weibull distribution, does the output inform the goodness of fit to it? perhaps in this part of the output... Weibull distribution Loglik(model)= -1302.8 Loglik(intercept only)= -1311 Chisq= 16.49 on 11 degrees of freedom, p= 0.12 Number of Newton-Raphson Iterations: 7 n= 873 2- one of my factors is gender (2 levels). With survreg, it appears as significant, but if I compare them with log-rank it turns not significant. Are they comparing different things? or is it a test power issue? thank you very much Eugenia Lic. M. E. Utgés INMeT Argentina -- View this message in context: http://r.789695.n4.nabble.com/Survreg-output-interpretation-tp4549368p4549368.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] nls function
nls() often gives this message, which is misleading in that it is the Jacobian that is not of full rank in the solution of J * delta ~ - residuals or in more conventional Gauss-Newton J' * J delta = -g = - J' * residuals. My view is that the gradient itself cannot be singular. It's just the slope of the sum of squares w.r.t. the parameters. I'm separately sending you an experimental code that does get an answer. It is in the package nlmrt on R-forge https://r-forge.r-project.org/R/?group_id=395 which is under development. Note that the I() function doesn't seem to be defined. I left it out to get an answer. John Nash On 04/11/2012 06:00 AM, r-help-requ...@r-project.org wrote: Message: 83 Date: Tue, 10 Apr 2012 13:03:58 -0700 (PDT) From: nerak13 karen.vandep...@gmail.com To: r-help@r-project.org Subject: [R] nls function Message-ID: 1334088238773-4546791.p...@n4.nabble.com Content-Type: text/plain; charset=us-ascii Hi, I've got the following data: x-c(1,3,5,7) y-c(37.98,11.68,3.65,3.93) penetrationks28-dataframe(x=x,y=y) now I need to fit a non linear function so I did: fit - nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start = list(a=0,b = 1,c=1), trace = T) The error message I get is: Error in nls(y ~ I(a + b * exp(1)^(-c * x)), data = penetrationks28, start = list(a = 0, : singular gradient I've tried to change the startervalues but it always gives the same error I've also tried the following adjustment hoping that the c value would be negative: fit - nls(y ~ I(a+b*exp(1)^(c * x)), data = penetrationks28, start = list(a = 1,b = 1,c=1), trace = T) but then the error message is: Error in nls(y ~ I(a + b * exp(1)^(c * x)), data = penetrationks28, start = list(a = 1, : number of iterations exceeded maximum of 50 What can I do ? Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/nls-function-tp4546791p4546791.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] strsplit help
Dear all, I want to use string split to parse column names, however, I am having some errors that I don't understand. I see a problem when I try to rbind the output from strsplit. please let me know if I'm missing something obvious, thanks, alison here are my commands: strsplit-strsplit(as.character(Rumino_Reps_agreeWalign$geneid),\\.) Rumino_Reps_agreeWalignTR-transform(Rumino_Reps_agreeWalign,taxid=do.call(rbind, strsplit)) Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) here is my data: head(Rumino_Reps_agreeWalign) geneid count_Conser count_NonCons count_ConsSubst 1 657313.locus_tag:RTO_089407 5 5 2 457412.2518480181 4 3 3 657314.locus_tag:CK5_206302 4 1 4 657323.locus_tag:CK1_330601 0 1 5 657313.locus_tag:RTO_096903 0 3 6 471875.1972971060 2 1 count_NCSubst 1 1 2 0 3 0 4 0 5 1 6 1 here are the results from strsplit: head(strsplit) [[1]] [1] 657313 locus_tag:RTO_08940 [[2]] [1] 457412251848018 [[3]] [1] 657314 locus_tag:CK5_20630 [[4]] [1] 657323 locus_tag:CK1_33060 [[5]] [1] 657313 locus_tag:RTO_09690 [[6]] [1] 471875197297106 __ 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] Date formate “ %y-%d-%m “ or “ %m-%d-%y “ ?
Hello, I would like to know for each variable what is date format? Is it “ %y-%d-%m “ or “ %m-%d-%y “ It's 4 digits, with uppercase 'Y' Try Data - data.frame(var1=c(12-15-2011, 12-15-2011, 12-17-2001), var2=c(2001-15-12, 2001-15-01, 2001-15-01), stringsAsFactors=FALSE ) str(Data) yr - sapply(sapply(Data, strsplit, -), function(x) which(nchar(x) == 4)) yr [1] 3 3 3 1 1 1 To have the result with the same dimensions as your original data.frame (but with a different class, I don't believe it matters) matrix(yr, ncol=ncol(Data)) Hope this helps, Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Date-formate-y-d-m-or-m-d-y-tp4548736p4549342.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] [R-SIG-Mac] plot 2 graphs on the same x-y plane
Hi, A line can be added to an existing plot using the abline function. For example, if a is the intercept and b is the slope, the command would be abline(a=a, b=b) To overlay a new plot on an existing one, use the command: par(new=TRUE). For example: plot(1:10, 1:10) par(new=TRUE) plot(log(1:10), 1:10) This approach often leads to issues with the scale of the axes, tick marks, and labels. Fortunately, R provides the flexibility to deal with all of them. I recommend you carefully read the help pages for plot, par, and axis, and have some fun playing around with all the options. --Susan On Apr 10, 2012, at 11:55 PM, Tawee Laoitichote wrote: Dear Michael (and Davis), Your answer is not what I want to know. My question is to find any command to plot the data I got from the field; such as a set of (x,y) data ( I actually have these data) and together withe the derived ones . I brought these data to plot on x-y plane, getting a graph showing some relation. Then, I wanted to find some linear relation, I would use least square method to solve having a simple function such as; y = ax + b, solving the a and b. So, I could plot a straight line using this function, or perhaps forecast some data of y which I know the value x. My problem is when I did a scatter plot by command plot(x,y), I got a graph. Whilst I plotted another graph using the above function the existing graph disappeared replaced be the latter function. Unfortunately, after searching a while to find the solution command, I can not find the command. I asked the question as to request some help not the example you shown. Any way, thanks for you effort. ! Ta! wee Mac OS10.7.3 From: michael.weyla...@gmail.com Date: Tue, 10 Apr 2012 23:31:08 -0400 Subject: Re: [R] plot 2 graphs on the same x-y plane To: ohowow2...@hotmail.com CC: r-help@r-project.org This is the same malformatted message you posted on R-SIG-Mac even after David specifically asked for clarification not to reward bad behavior, but perhaps this will enlighten: # Minimal reproducible data! x - runif(15, 0, 5) y - 3*x - 2 + runif(15) dat - data.frame(x = x, y = y) rm(list = c(x, y)) # Base graphics plot using the formula interface plot(y ~ x, data = dat) abline(lm(y~x, data = dat), col = red3, lwd = 2) Alternatively library(ggplot2) ggplot(dat, aes(x = x, y = y)) + geom_point() + stat_smooth(method = lm, color = I(red3)) which is perhaps overkill in this situation. Michael On Tue, Apr 10, 2012 at 11:03 PM, Tawee Laoitichote ohowow2...@hotmail.com wrote: hi, I'm doing some data on least square curve fitting. What I like to have is to compare the scatter plot whilst having the fitting curve on the same coordinates. Any suggestting command besides plot(x,y). TaweeMac OSX 10.7.3 [[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-SIG-Mac mailing list r-sig-...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-sig-mac __ 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-help; generating censored data
Hello, can i implement this as 10% censored data where t gives me failure and x censored. Thank you p=2;b=120 n=50 set.seed(132); r-sample(1:50,45) t-rweibull(r,shape=p,scale=b) t set.seed(123); cens - sample(1:50, 5) x-runif(cens,shape=p,scale=b) x Chris Guure Researcher, Institute for Mathematical Research UPM __ 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] [BioC] Read .idat Illumina files in R
Hi Ekta, You can use the bioconductor package methylumi to read in .idat files. The function methylumIDAT() will read in .idat files and provide you with beta values along with the M and U values as a MethyLumiSet object. You can then use the functions methylumi.bgcorr() and normalizeMethyLumiSet() to background correct and normalize your data. See the vignette for more details. Best, Moiz On Apr 11, 2012, at 1:46 AM, Ekta Jain wrote: Dear Bioc and R List Users, I am having trouble analysing illumine data generated from BeadScan. I have .idat files and JPEG images. I realise that i need bead-level summary data to be able to begin quality control followed by normalization. Is there a way i can read .idat files for expression analysis or do i need to go back to BeadScan and generate .txt files/tiff files ? Appreciate any help here. Many Thanks, Ekta Jain The information contained in this electronic message and in any attachments to this message is confidential, legally privileged and intended only for use by the person or entity to which this electronic message is addressed. If you are not the intended recipient, and have received this message in error, please notify the sender and system manager by return email and delete the message and its attachments and also you are hereby notified that any distribution, copying, review, retransmission, dissemination or other use of this electronic transmission or the information contained in it is strictly prohibited. Please note that any views or opinions presented in this email are solely those of the author and may not represent those of the Company or bind the Company. Any commitments made over e-mail are not financially binding on the company unless accompanied or followed by a valid purchase order. This message has been scanned for viruses and dangerous content by Mail Scanner,! a! nd is believed to be clean. The Company accepts no liability for any damage caused by any virus transmitted by this email. www.jubl.com [[alternative HTML version deleted]] ___ Bioconductor mailing list bioconduc...@r-project.org https://stat.ethz.ch/mailman/listinfo/bioconductor Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor __ 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] [BioC] Read .idat Illumina files in R
Unfortunately, this won't help for expression arrays. Last time I checked, those IDATs appeared to be encrypted. crlmm, methylumi, and minfi can all read IDAT files... *if* they are version 3 or later (e.g. genotyping, methylation, etc). Otherwise you are probably stuck with GenomeStudio if it is expression data you're dealing with. On Wed, Apr 11, 2012 at 7:14 AM, Moiz Bootwalla msbootwa...@gmail.comwrote: Hi Ekta, You can use the bioconductor package methylumi to read in .idat files. The function methylumIDAT() will read in .idat files and provide you with beta values along with the M and U values as a MethyLumiSet object. You can then use the functions methylumi.bgcorr() and normalizeMethyLumiSet() to background correct and normalize your data. See the vignette for more details. Best, Moiz On Apr 11, 2012, at 1:46 AM, Ekta Jain wrote: Dear Bioc and R List Users, I am having trouble analysing illumine data generated from BeadScan. I have .idat files and JPEG images. I realise that i need bead-level summary data to be able to begin quality control followed by normalization. Is there a way i can read .idat files for expression analysis or do i need to go back to BeadScan and generate .txt files/tiff files ? Appreciate any help here. Many Thanks, Ekta Jain The information contained in this electronic message and in any attachments to this message is confidential, legally privileged and intended only for use by the person or entity to which this electronic message is addressed. If you are not the intended recipient, and have received this message in error, please notify the sender and system manager by return email and delete the message and its attachments and also you are hereby notified that any distribution, copying, review, retransmission, dissemination or other use of this electronic transmission or the information contained in it is strictly prohibited. Please note that any views or opinions presented in this email are solely those of the author and may not represent those of the Company or bind the Company. Any commitments made over e-mail are not financially binding on the company unless accompanied or followed by a valid purchase order. This message has been scanned for viruses and dangerous content by Mail Scanner,! a! nd is believed to be clean. The Company accepts no liability for any damage caused by any virus transmitted by this email. www.jubl.com [[alternative HTML version deleted]] ___ Bioconductor mailing list bioconduc...@r-project.org https://stat.ethz.ch/mailman/listinfo/bioconductor Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor ___ Bioconductor mailing list bioconduc...@r-project.org https://stat.ethz.ch/mailman/listinfo/bioconductor Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor -- *A model is a lie that helps you see the truth.* * * Howard Skipperhttp://cancerres.aacrjournals.org/content/31/9/1173.full.pdf [[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] xyplot lattice fine control of axes limits and thick marks (with log scale)
Oh yes, I see now the problem... thank you max -- View this message in context: http://r.789695.n4.nabble.com/xyplot-lattice-fine-control-of-axes-limits-and-thick-marks-with-log-scale-tp4511897p4549180.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] 7 arguments passed to .Internal(identical) which requires 6
Thank you! Updating to R-patched really helped me. My problem were not been into my code. I've tried to run source() and the error had occurred again. I didn't use a mixture of different versions of R, version 2.13.2 has been deleted. -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4549183.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] bayesian gene network construction
Hello: I have looked at the bnlearn and deal packages for infering bayesian network. Can anyone suggest any other suitable package for constructing bayesian gene regulatory network using gene expression data? Thanks! John [[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] 7 arguments passed to .Internal(identical) which requires 6
Can you give the line of code that gives the error? That'd make it much easier to see what's there that should/shouldn't be. Michael On Wed, Apr 11, 2012 at 11:28 AM, krtek marshal...@mail.ru wrote: Thank you! Updating to R-patched really helped me. My problem were not been into my code. I've tried to run source() and the error had occurred again. I didn't use a mixture of different versions of R, version 2.13.2 has been deleted. -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4549183.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.
[R] row.names in dunes and dunes.env?
Hello, I've got a small dataset on box turtle shell measurements that I would like to perform a detrended correspondence analysis on. I thought that it would be interesting to examine the morphometrics for each species in the area of overlap and in areas where neither species occurs. I've taken a look at the dune and dune.env datasets in vegan. Using the str() command gives me str(dune) 'data.frame': 20 obs. of 30 variables: $ Belper: num 3 0 2 0 0 0 0 2 0 0 ... $ Empnig: num 0 0 0 0 0 0 0 0 0 0 ... $ Junbuf: num 0 3 0 0 0 0 0 0 0 0 ... $ Junart: num 0 0 0 3 0 0 4 0 0 3 ... ... However, when I try looking directly at the data frame using the edit command I see that there is a column called row.names to the left of Belper. Likewise, when I use the str() command on dune.env I get str(dune.env) 'data.frame': 20 obs. of 5 variables: $ A1 : num 3.5 6 4.2 5.7 4.3 2.8 4.2 6.3 4 11.5 ... $ Moisture : Ord.factor w/ 4 levels 1245: 1 4 2 4 1 1 4 1 2 4 ... $ Management: Factor w/ 4 levels BF,HF,NM,..: 1 4 4 4 2 4 2 2 3 3 ... $ Use : Ord.factor w/ 3 levels HayfieldHaypastu..: 2 2 2 3 2 2 3 1 1 2 ... $ Manure : Ord.factor w/ 5 levels 0123..: 3 4 5 4 3 5 4 3 1 1 ... but using the edit() command shows a column named row.names. I assume that the the row.names column is used to link the two files together. My turtle data is saved as a *.csv, and I've added a column called row.names, so that it looks like this row.names,CL,CCL,CW,CCW,CH,CCH 1,104.4,131.8,89.887,137.4,43.391,89.7 2,108.79,135.9,87.78,118.1,50.72,71.2 3,114.12,126.1,89.33,132.8,142.39,78.3 4,102.87,128.2,84.2,125,45.42,72.4 5,84.6,104.8,72.61,111.8,41.1,57.3 I've called this file turtles_dca.csv. I've also created a file called turtles_dca_env.csv that looks like this row.names,Species,Sex,Distribution,Concatenated,Species_overlap 1,Terrapene_ornata,Female,overlap,TO_F_Overlap,TO_Overlap 2,Terrapene_ornata,Female,overlap,TO_F_Overlap,TO_Overlap 3,Terrapene_ornata,Female,overlap,TO_F_Overlap,TO_Overlap 4,Terrapene_ornata,Female,overlap,TO_F_Overlap,TO_Overlap 5,Terrapene_ornata,Female,overlap,TO_F_Overlap,TO_Overlap However, when I read the data into R using this command turtles.env = read.csv(turtles_dca_env.csv, header = TRUE) and then using the str() command I get str(turtles) 'data.frame': 67 obs. of 7 variables: $ row.names: int 1 2 3 4 5 6 7 8 9 10 ... $ CL : num 104.4 108.8 114.1 102.9 84.6 ... $ CCL : num 132 136 126 128 105 ... $ CW : num 89.9 87.8 89.3 84.2 72.6 ... $ CCW : num 137 118 133 125 112 ... $ CH : num 43.4 50.7 142.4 45.4 41.1 ... $ CCH : num 89.7 71.2 78.3 72.4 57.3 73.4 67 57 68.8 68 ... When I run decorana() on this dataset, it appears that the column row.names is included in the analysis, which isn't what I'm looking for. If I go ahead and delete the column row.names from my data frames (i.e. removing it from turtles and turtles.env), I don't believe that the analysis is performed correctly. The two species differ significantly in most of their measurements, but the ordihull() and ordispider() commands show them overlapping almost completely. I think that I'm missing something pretty basic about inputting and formatting this data for this analysis. Can anyone offer a suggestion on where I'm going astray? I can send a copy of the data if anyone wants to look at it. Best wishes, Chris University of Central Oklahoma [[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] multicore/mcparallel error
Parallel is one of those packages (like tools or grid) that is not
loaded by default, but comes with the standard installation.
Running library(parallel) will make mclapply() available. Then (at
least for me) it's easy to parallelize some code:
system.time(lapply(1:5, function(x) max(rnorm(500 # ~ 8 seconds
system.time(mclapply(1:5, function(x) max(rnorm(500 # ~ 4.4
seconds [I have two processors]
Michael
On Wed, Apr 11, 2012 at 12:34 PM, Wyatt McMahon wmcma...@vbi.vt.edu wrote:
Michael,
Thanks for your help!
I may not be understanding you. I've upgraded to 2.15. From your text
below, I thought the parallel package was included with R 2.15, but I
cannot find an mclapply function there. One exists through the multicore
package, which I've now installed. I also went looking for a parallel
package, but was unable to find one.
Is there an mclapply included in 2.15? Is there a parallel package I'm
missing? Or am I completely misunderstanding your response?
Thanks!
Wyatt
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Tuesday, April 10, 2012 12:07 PM
To: Wyatt McMahon
Cc: R-help@r-project.org
Subject: Re: [R] multicore/mcparallel error
I don't know the multicore package, but if possible, it might be easier to
upgrade to 2.15 and use the new built-in parallel package that was
introduced in R 2.14.
Then your syntax would be something like
mclapply(files, illumqc)
Michael
On Tue, Apr 10, 2012 at 11:33 AM, Wyatt McMahon wmcma...@vbi.vt.edu
wrote:
Hello everyone,
I'm trying to parallelize an R script I have written. To do this, I
am first trying to use the multicore package, because I've had some
previous success with that.
The function I'm trying to parallelize is illumqc. I'd like to create
a separate process for each of 8 files, contained in the vector files.
Below is my code:
for(i in 1:length(files)){
mcparallel(illumqc(files[i]))}
I get the following error:
Error in sendMaster(serialize(try(eval(expr, env), silent = TRUE),
NULL,
:
ignoring SIGPIPE signal
Calls: mcparallel - sendMaster - .Call
In addition: Warning message:
In min(which(alf != 0)) : no non-missing arguments to min; returning
Inf
However, if I try and make a simpler example, everything works
correctly, so I'm not sure what's going wrong with this function. Do
I need to post the contents of the function as well? I'm hoping
someone can recognize this error and give me a clue as to what is
going wrong since the function is fairly long.
Thanks in advance,
Wyatt
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ShortRead_1.10.4 Rsamtools_1.4.3 lattice_0.19-30
[4] Biostrings_2.20.4 GenomicRanges_1.4.8 IRanges_1.10.6
[7] multicore_0.1-7
loaded via a namespace (and not attached):
[1] Biobase_2.12.2 grid_2.13.1 hwriter_1.3
[[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] multicore/mcparallel error
Thanks a ton, Michael! Everything is running much faster now!!
Wyatt
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Wednesday, April 11, 2012 1:41 PM
To: Wyatt McMahon
Cc: R-help@r-project.org
Subject: Re: [R] multicore/mcparallel error
Parallel is one of those packages (like tools or grid) that is not loaded
by default, but comes with the standard installation.
Running library(parallel) will make mclapply() available. Then (at least
for me) it's easy to parallelize some code:
system.time(lapply(1:5, function(x) max(rnorm(500 # ~ 8 seconds
system.time(mclapply(1:5, function(x) max(rnorm(500 # ~ 4.4
seconds [I have two processors]
Michael
On Wed, Apr 11, 2012 at 12:34 PM, Wyatt McMahon wmcma...@vbi.vt.edu
wrote:
Michael,
Thanks for your help!
I may not be understanding you. I've upgraded to 2.15. From your text
below, I thought the parallel package was included with R 2.15, but I
cannot find an mclapply function there. One exists through the
multicore package, which I've now installed. I also went looking for a
parallel
package, but was unable to find one.
Is there an mclapply included in 2.15? Is there a parallel package
I'm missing? Or am I completely misunderstanding your response?
Thanks!
Wyatt
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Tuesday, April 10, 2012 12:07 PM
To: Wyatt McMahon
Cc: R-help@r-project.org
Subject: Re: [R] multicore/mcparallel error
I don't know the multicore package, but if possible, it might be
easier to upgrade to 2.15 and use the new built-in parallel package
that was introduced in R 2.14.
Then your syntax would be something like
mclapply(files, illumqc)
Michael
On Tue, Apr 10, 2012 at 11:33 AM, Wyatt McMahon wmcma...@vbi.vt.edu
wrote:
Hello everyone,
I'm trying to parallelize an R script I have written. To do this, I
am first trying to use the multicore package, because I've had some
previous success with that.
The function I'm trying to parallelize is illumqc. I'd like to
create a separate process for each of 8 files, contained in the vector
files.
Below is my code:
for(i in 1:length(files)){
mcparallel(illumqc(files[i]))}
I get the following error:
Error in sendMaster(serialize(try(eval(expr, env), silent = TRUE),
NULL,
:
ignoring SIGPIPE signal
Calls: mcparallel - sendMaster - .Call
In addition: Warning message:
In min(which(alf != 0)) : no non-missing arguments to min; returning
Inf
However, if I try and make a simpler example, everything works
correctly, so I'm not sure what's going wrong with this function. Do
I need to post the contents of the function as well? I'm hoping
someone can recognize this error and give me a clue as to what is
going wrong since the function is fairly long.
Thanks in advance,
Wyatt
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ShortRead_1.10.4 Rsamtools_1.4.3 lattice_0.19-30
[4] Biostrings_2.20.4 GenomicRanges_1.4.8 IRanges_1.10.6
[7] multicore_0.1-7
loaded via a namespace (and not attached):
[1] Biobase_2.12.2 grid_2.13.1 hwriter_1.3
[[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] strsplit help
Alison, Your code works fine on the first six lines of the data that you provided. Rumino_Reps_agreeWalign - data.frame( geneid = c(657313.locus_tag:RTO_08940, 457412.251848018, 657314.locus_tag:CK5_20630, 657323.locus_tag:CK1_33060, 657313.locus_tag:RTO_09690, 471875.197297106), count_Conser = c(7, 1, 2, 1, 3, 0), count_NonCons = c(5, 4, 4, 0, 0, 2), count_ConsSubst = c(5, 3, 1, 1, 3, 1), count_NCSubst = c(1, 0, 0, 0, 1, 1)) gene.list - strsplit(as.character(Rumino_Reps_agreeWalign$geneid), \\.) Rumino_Reps_agreeWalignTR - transform(Rumino_Reps_agreeWalign, taxid=do.call(rbind, gene.list)) Perhaps in later rows of the data there are cases where there is no . in geneid? If not, can you provide a subset of your data that results in the warning? Use the dput() function. It's not a good idea to create an object named strsplit. That will only mask the function strsplit() in later runs. If time is an issue, a slightly faster way to do this, after the strsplit() function is: Rumino_Reps_agreeWalign$geneid.prefix - sapply(gene.list, [, 1) Rumino_Reps_agreeWalign$geneid.suffix - sapply(gene.list, [, 2) Jean alison waller wrote on 04/11/2012 08:23:29 AM: Dear all, I want to use string split to parse column names, however, I am having some errors that I don't understand. I see a problem when I try to rbind the output from strsplit. please let me know if I'm missing something obvious, thanks, alison here are my commands: strsplit-strsplit(as.character(Rumino_Reps_agreeWalign$geneid),\\.) Rumino_Reps_agreeWalignTR-transform (Rumino_Reps_agreeWalign,taxid=do.call(rbind, strsplit)) Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) here is my data: head(Rumino_Reps_agreeWalign) geneid count_Conser count_NonCons count_ConsSubst 1 657313.locus_tag:RTO_089407 5 5 2 457412.2518480181 4 3 3 657314.locus_tag:CK5_206302 4 1 4 657323.locus_tag:CK1_330601 0 1 5 657313.locus_tag:RTO_096903 0 3 6 471875.1972971060 2 1 count_NCSubst 1 1 2 0 3 0 4 0 5 1 6 1 here are the results from strsplit: head(strsplit) [[1]] [1] 657313 locus_tag:RTO_08940 [[2]] [1] 457412251848018 [[3]] [1] 657314 locus_tag:CK5_20630 [[4]] [1] 657323 locus_tag:CK1_33060 [[5]] [1] 657313 locus_tag:RTO_09690 [[6]] [1] 471875197297106 [[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] Lattice densityplot with semitransparent filled regions
Hello,
I'm doing some graphics for a paper and a need customize such with filled
region above the density curve. My attempts I get something very near what
I need, but I don't solve the problem of use semitransparent filled. Below
a minimal reproducible code. Someone has any idea?
require(lattice)
# toy data...
dt - expand.grid(A=1:2, B=1:3, y=1:50)
dt$y - rnorm(nrow(dt), dt$B, dt$A)
# regular plot...
densityplot(~y|B, groups=A, data=dt, plot.points=rug)
# the actual panel...
panel.densityplot
# so, I edit this...
my.panel.densityplot -
function (x, darg = list(n = 30), plot.points = jitter, ref = FALSE,
groups = NULL, weights = NULL, jitter.amount = 0.01 *
diff(current.panel.limits()$ylim),
type = p, ..., identifier = density)
{
if (ref) {
reference.line - trellis.par.get(reference.line)
panel.abline(h = 0, col = reference.line$col, lty =
reference.line$lty,
lwd = reference.line$lwd, identifier = paste(identifier,
abline))
}
if (!is.null(groups)) {
panel.superpose(x, darg = darg, plot.points = plot.points,
ref = FALSE, groups = groups, weights = weights,
panel.groups = panel.densityplot, jitter.amount =
jitter.amount, # alterei para my.panel
type = type, ...)
}
else {
switch(as.character(plot.points), `TRUE` = panel.xyplot(x = x,
y = rep(0, length(x)), type = type, ..., identifier =
identifier),
rug = panel.rug(x = x, start = 0, end = 0, x.units = c(npc,
native), type = type, ..., identifier = paste(identifier,
rug)), jitter = panel.xyplot(x = x, y = jitter(rep(0,
length(x)), amount = jitter.amount), type = type,
..., identifier = identifier))
density.fun - function(x, weights, subscripts = TRUE,
darg, ...) {
do.call(density, c(list(x = x, weights =
weights[subscripts]),
darg))
}
if (sum(!is.na(x)) 1) {
h - density.fun(x = x, weights = weights, ..., darg = darg)
lim - current.panel.limits()$xlim
id - h$x min(lim) h$x max(lim)
panel.lines(x = h$x[id], y = h$y[id], ..., identifier =
identifier)
## line above was added
panel.polygon(x=h$x[id], y = h$y[id], ..., identifier =
identifier, alpha=0.2)
}
}
}
# my customized plot, I want semitransparent colors
# and use the colors of trellis.par.set(superpose.polygon) to fill
densityplot(~y|B, groups=A, data=dt,
plot.points=rug, col=2:3,
panel=panel.superpose,
panel.groups=my.panel.densityplot)
Thanks!
Walmes.
==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes
linux user number: 531218
==
[[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] 7 arguments passed to .Internal(identical) which requires 6
FYI, whenever getting an error, run traceback() *immediately after* (i.e. before any other commands) and include that in your error report. Also include the output of sessionInfo(). It helps tremendously and spares lots of second guessing. /Henrik On Wed, Apr 11, 2012 at 8:28 AM, krtek marshal...@mail.ru wrote: Thank you! Updating to R-patched really helped me. My problem were not been into my code. I've tried to run source() and the error had occurred again. I didn't use a mixture of different versions of R, version 2.13.2 has been deleted. -- View this message in context: http://r.789695.n4.nabble.com/7-arguments-passed-to-Internal-identical-which-requires-6-tp4548460p4549183.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] strsplit help
On Apr 11, 2012, at 2:01 PM, Jean V Adams wrote: Alison, Your code works fine on the first six lines of the data that you provided. Rumino_Reps_agreeWalign - data.frame( geneid = c(657313.locus_tag:RTO_08940, 457412.251848018, 657314.locus_tag:CK5_20630, 657323.locus_tag:CK1_33060, 657313.locus_tag:RTO_09690, 471875.197297106), count_Conser = c(7, 1, 2, 1, 3, 0), count_NonCons = c(5, 4, 4, 0, 0, 2), count_ConsSubst = c(5, 3, 1, 1, 3, 1), count_NCSubst = c(1, 0, 0, 0, 1, 1)) gene.list - strsplit(as.character(Rumino_Reps_agreeWalign$geneid), \\.) Rumino_Reps_agreeWalignTR - transform(Rumino_Reps_agreeWalign, taxid=do.call(rbind, gene.list)) Perhaps in later rows of the data there are cases where there is no . in geneid? If not, can you provide a subset of your data that results in the warning? Use the dput() function. It's not a good idea to create an object named strsplit. That will only mask the function strsplit() in later runs. There is not a problem with masking the function unless the new name is replaced with a language object (which wasn't the case here). The potential confusion is in minds of users. Function names are stored separately from non-language object names so you can have a data object named 'strsplit' and it will not mask the function 'strsplit'. -- David. If time is an issue, a slightly faster way to do this, after the strsplit() function is: Rumino_Reps_agreeWalign$geneid.prefix - sapply(gene.list, [, 1) Rumino_Reps_agreeWalign$geneid.suffix - sapply(gene.list, [, 2) Jean alison waller wrote on 04/11/2012 08:23:29 AM: Dear all, I want to use string split to parse column names, however, I am having some errors that I don't understand. I see a problem when I try to rbind the output from strsplit. please let me know if I'm missing something obvious, thanks, alison here are my commands: strsplit-strsplit(as.character(Rumino_Reps_agreeWalign$geneid),\ \.) Rumino_Reps_agreeWalignTR-transform (Rumino_Reps_agreeWalign,taxid=do.call(rbind, strsplit)) Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) here is my data: head(Rumino_Reps_agreeWalign) geneid count_Conser count_NonCons count_ConsSubst 1 657313.locus_tag:RTO_089407 5 5 2 457412.2518480181 4 3 3 657314.locus_tag:CK5_206302 4 1 4 657323.locus_tag:CK1_330601 0 1 5 657313.locus_tag:RTO_096903 0 3 6 471875.1972971060 2 1 count_NCSubst 1 1 2 0 3 0 4 0 5 1 6 1 here are the results from strsplit: head(strsplit) [[1]] [1] 657313 locus_tag:RTO_08940 [[2]] [1] 457412251848018 [[3]] [1] 657314 locus_tag:CK5_20630 [[4]] [1] 657323 locus_tag:CK1_33060 [[5]] [1] 657313 locus_tag:RTO_09690 [[6]] [1] 471875197297106 [[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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strsplit help
David, Right you are! Thanks for pointing that out. strsplit - 1:10 strsplit(With spaces, NULL) strsplit Jean David Winsemius dwinsem...@comcast.net wrote on 04/11/2012 01:17:07 PM: [image removed] Re: [R] strsplit help David Winsemius to: Jean V Adams 04/11/2012 01:19 PM Cc: alison waller, r-help On Apr 11, 2012, at 2:01 PM, Jean V Adams wrote: Alison, Your code works fine on the first six lines of the data that you provided. Rumino_Reps_agreeWalign - data.frame( geneid = c(657313.locus_tag:RTO_08940, 457412.251848018, 657314.locus_tag:CK5_20630, 657323.locus_tag:CK1_33060, 657313.locus_tag:RTO_09690, 471875.197297106), count_Conser = c(7, 1, 2, 1, 3, 0), count_NonCons = c(5, 4, 4, 0, 0, 2), count_ConsSubst = c(5, 3, 1, 1, 3, 1), count_NCSubst = c(1, 0, 0, 0, 1, 1)) gene.list - strsplit(as.character(Rumino_Reps_agreeWalign$geneid), \\.) Rumino_Reps_agreeWalignTR - transform(Rumino_Reps_agreeWalign, taxid=do.call(rbind, gene.list)) Perhaps in later rows of the data there are cases where there is no . in geneid? If not, can you provide a subset of your data that results in the warning? Use the dput() function. It's not a good idea to create an object named strsplit. That will only mask the function strsplit() in later runs. There is not a problem with masking the function unless the new name is replaced with a language object (which wasn't the case here). The potential confusion is in minds of users. Function names are stored separately from non-language object names so you can have a data object named 'strsplit' and it will not mask the function 'strsplit'. -- David. If time is an issue, a slightly faster way to do this, after the strsplit() function is: Rumino_Reps_agreeWalign$geneid.prefix - sapply(gene.list, [, 1) Rumino_Reps_agreeWalign$geneid.suffix - sapply(gene.list, [, 2) Jean alison waller wrote on 04/11/2012 08:23:29 AM: Dear all, I want to use string split to parse column names, however, I am having some errors that I don't understand. I see a problem when I try to rbind the output from strsplit. please let me know if I'm missing something obvious, thanks, alison here are my commands: strsplit-strsplit(as.character(Rumino_Reps_agreeWalign$geneid),\ \.) Rumino_Reps_agreeWalignTR-transform (Rumino_Reps_agreeWalign,taxid=do.call(rbind, strsplit)) Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) here is my data: head(Rumino_Reps_agreeWalign) geneid count_Conser count_NonCons count_ConsSubst 1 657313.locus_tag:RTO_089407 5 5 2 457412.2518480181 4 3 3 657314.locus_tag:CK5_206302 4 1 4 657323.locus_tag:CK1_330601 0 1 5 657313.locus_tag:RTO_096903 0 3 6 471875.1972971060 2 1 count_NCSubst 1 1 2 0 3 0 4 0 5 1 6 1 here are the results from strsplit: head(strsplit) [[1]] [1] 657313 locus_tag:RTO_08940 [[2]] [1] 457412251848018 [[3]] [1] 657314 locus_tag:CK5_20630 [[4]] [1] 657323 locus_tag:CK1_33060 [[5]] [1] 657313 locus_tag:RTO_09690 [[6]] [1] 471875197297106 [[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. David Winsemius, MD West Hartford, CT [[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] nls function
On Apr 11, 2012, at 16:51 , John C Nash wrote: nls() often gives this message, which is misleading in that it is the Jacobian that is not of full rank in the solution of J * delta ~ - residuals or in more conventional Gauss-Newton J' * J delta = -g = - J' * residuals. My view is that the gradient itself cannot be singular. It's just the slope of the sum of squares w.r.t. the parameters. That's one view, John, but the more common one is that it is the gradient matrix of the fitted values with respect to the parameters that is singular. I.e., we are locally approximating the nonlinear relation y=f(theta0+delta) with the linear one y-f(theta0)= (Df)delta, in which Df can be singular in the usual matrix sense. That said, I completely agree that it is silly to let this condition be a show-stopper when it happens early in the iteration sequence. As long as you can move in _any_ direction and improve the sum of squares, why not go there and retry? I'm separately sending you an experimental code that does get an answer. It is in the package nlmrt on R-forge https://r-forge.r-project.org/R/?group_id=395 which is under development. Note that the I() function doesn't seem to be defined. I left it out to get an answer. John Nash On 04/11/2012 06:00 AM, r-help-requ...@r-project.org wrote: Message: 83 Date: Tue, 10 Apr 2012 13:03:58 -0700 (PDT) From: nerak13 karen.vandep...@gmail.com To: r-help@r-project.org Subject: [R] nls function Message-ID: 1334088238773-4546791.p...@n4.nabble.com Content-Type: text/plain; charset=us-ascii Hi, I've got the following data: x-c(1,3,5,7) y-c(37.98,11.68,3.65,3.93) penetrationks28-dataframe(x=x,y=y) now I need to fit a non linear function so I did: fit - nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start = list(a=0,b = 1,c=1), trace = T) The error message I get is: Error in nls(y ~ I(a + b * exp(1)^(-c * x)), data = penetrationks28, start = list(a = 0, : singular gradient I've tried to change the startervalues but it always gives the same error I've also tried the following adjustment hoping that the c value would be negative: fit - nls(y ~ I(a+b*exp(1)^(c * x)), data = penetrationks28, start = list(a = 1,b = 1,c=1), trace = T) but then the error message is: Error in nls(y ~ I(a + b * exp(1)^(c * x)), data = penetrationks28, start = list(a = 1, : number of iterations exceeded maximum of 50 What can I do ? Thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/nls-function-tp4546791p4546791.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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice densityplot with semitransparent filled regions
densityplot(~y|B, groups=A, data=dt,
plot.points=rug,
col=trellis.par.get(superpose.polygon)$col, alpha=.5,
panel=panel.superpose,
panel.groups=my.panel.densityplot)
Worked for me (i.e. semi-transparent superpose.polygon colors). Is
that not what you are seeing ?
On Wed, Apr 11, 2012 at 12:05 PM, Walmes Zeviani
walmeszevi...@gmail.com wrote:
Hello,
I'm doing some graphics for a paper and a need customize such with filled
region above the density curve. My attempts I get something very near what
I need, but I don't solve the problem of use semitransparent filled. Below
a minimal reproducible code. Someone has any idea?
require(lattice)
# toy data...
dt - expand.grid(A=1:2, B=1:3, y=1:50)
dt$y - rnorm(nrow(dt), dt$B, dt$A)
# regular plot...
densityplot(~y|B, groups=A, data=dt, plot.points=rug)
# the actual panel...
panel.densityplot
# so, I edit this...
my.panel.densityplot -
function (x, darg = list(n = 30), plot.points = jitter, ref = FALSE,
groups = NULL, weights = NULL, jitter.amount = 0.01 *
diff(current.panel.limits()$ylim),
type = p, ..., identifier = density)
{
if (ref) {
reference.line - trellis.par.get(reference.line)
panel.abline(h = 0, col = reference.line$col, lty =
reference.line$lty,
lwd = reference.line$lwd, identifier = paste(identifier,
abline))
}
if (!is.null(groups)) {
panel.superpose(x, darg = darg, plot.points = plot.points,
ref = FALSE, groups = groups, weights = weights,
panel.groups = panel.densityplot, jitter.amount =
jitter.amount, # alterei para my.panel
type = type, ...)
}
else {
switch(as.character(plot.points), `TRUE` = panel.xyplot(x = x,
y = rep(0, length(x)), type = type, ..., identifier =
identifier),
rug = panel.rug(x = x, start = 0, end = 0, x.units = c(npc,
native), type = type, ..., identifier = paste(identifier,
rug)), jitter = panel.xyplot(x = x, y = jitter(rep(0,
length(x)), amount = jitter.amount), type = type,
..., identifier = identifier))
density.fun - function(x, weights, subscripts = TRUE,
darg, ...) {
do.call(density, c(list(x = x, weights =
weights[subscripts]),
darg))
}
if (sum(!is.na(x)) 1) {
h - density.fun(x = x, weights = weights, ..., darg = darg)
lim - current.panel.limits()$xlim
id - h$x min(lim) h$x max(lim)
panel.lines(x = h$x[id], y = h$y[id], ..., identifier =
identifier)
## line above was added
panel.polygon(x=h$x[id], y = h$y[id], ..., identifier =
identifier, alpha=0.2)
}
}
}
# my customized plot, I want semitransparent colors
# and use the colors of trellis.par.set(superpose.polygon) to fill
densityplot(~y|B, groups=A, data=dt,
plot.points=rug, col=2:3,
panel=panel.superpose,
panel.groups=my.panel.densityplot)
Thanks!
Walmes.
==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes
linux user number: 531218
==
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Partial Dependence and RandomForest
Hello all~ I am interested in clarifying something more conceptual, so I won't be providing any data or code here. From what I understand, partial dependence plots can help you understand the relative dependence on a variable, and the subsequent values of that variable, after averaging out the effects of the other input variables. This is great, but what I am interested in knowing is how that relates to each predictor class, not just the overall prediction. Is it possible to plot partial dependence per class? Specifically, I'd like to know the important threshold values of my most important variables. Thank you for your time, -- View this message in context: http://r.789695.n4.nabble.com/Partial-Dependence-and-RandomForest-tp4549705p4549705.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] Michaelson-Morley Speed of Light Data
Dnia środa, 11 kwietnia 2012 13:26:43 Prof Brian Ripley pisze: On 11/04/2012 12:42, Duncan Murdoch wrote: On 12-04-11 12:43 AM, Křištof Želechovski wrote: URL: http://finzi.psych.upenn.edu/R/library/datasets/html/morley.html The classical data of Michaelson and Morley on the speed of light Can you provide more information about the data? How were they obtained, etc.? I do not have the book Genstat Primer and the nearest location where it is available is University of York which is rather far from my location. If you can't find the cited reference, I'd try Google. For instance, it led me to this page http://en.wikipedia.org/wiki/File:Michelsonmorley-boxplot.svg which appears to show five series. Yes, but that is derived from R. AFAIR the history, Bill Venables got this from Weekes (1986), a book I have only ever seen in Adelaide. A better reference is S. M. Stigler (1977) Do robust estimators work with real data? Annals of Statistics 5, 1055–1098. (See Table 6.) The data in R are identical with Table 6 but they were not obtained by Michaelson and Morley; they were obtained by Michelson. The description is wrong. It is also a problem in R itself, as evidenced by the following instruction: data(morley) Of course, you can name a data set whatever you like, even data(marilyn) would be all right, but may I suggest that this data set be renamed to michelson? Please fix, Chris __ 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] Lattice densityplot with semitransparent filled regions
Thank you Ilai. Problem solved. There is a small detail, alpha affects the rug and the curve line opacity too. Is possible to specify it just to polygon? Bests. == Walmes Marques Zeviani LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W) Departamento de Estatística - Universidade Federal do Paraná fone: (+55) 41 3361 3573 VoIP: (3361 3600) 1053 1173 e-mail: wal...@ufpr.br twitter: @walmeszeviani homepage: http://www.leg.ufpr.br/~walmes linux user number: 531218 == [[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] Faceted bar plot shows wrong counts (ggplot2)
And it's now fixed in the dev version. Hadley On Tue, Mar 13, 2012 at 11:37 AM, Helios de Rosario helios.derosa...@ibv.upv.es wrote: Michael, Thanks for the pointer to the discussion in the ggplot list. It seems that the reason of this behaviour of facet_grid() is already known and being discussed by the developers of ggplot2. facet_grid() reduces the original data frame with unique() before applying the stats. If the data frame has any other column that prevents duplicated rows, counts are correctly computed. E.g. diamonds25 - droplevels(diamonds[1:25,]) # Keep all columns # Everything else as before: base - ggplot(diamonds25, aes(fill = cut)) + geom_bar(position = dodge) + opts(legend.position = none) base + aes(x = cut) + facet_grid(. ~ color) Helios El día 12/03/2012 a las 20:59, R. Michael Weylandt michael.weyla...@gmail.com escribió: You get the good behavior with base + aes(x = cut) + facet_wrap(~ color, ncol = 5) so this seems buggy to me. If someone here doesn't step forward with more insight, I'd forward it to the ggplot list to see if one of the developers there can give an explanation or possibly make the official call that it's a bug. There was another report of a possible bug in facet_grid() today that could be related: https://groups.google.com/group/ggplot2/browse_thread/thread/5213ac35da6b36d 4 Michael On Mon, Mar 12, 2012 at 7:16 AM, Helios de Rosario helios.derosa...@ibv.upv.es wrote: I have encountered a problem with faceted bar plots. I have tried to create something like the example explained in the ggplot2 book (see pp. 126-128): library(ggplot2) mpg4 - subset(mpg, manufacturer %in% c(audi, volkswagen, jeep)) mpg4$manufacturer - as.character(mpg4$manufacturer) mpg4$model - as.character(mpg4$model) base - ggplot(mpg4, aes(fill = model)) + geom_bar(position = dodge) + opts(legend.position = none) base + aes(x = model) + facet_grid(. ~ manufacturer) That example works fine; the bar heights are just the same as the counts in the table: table(mpg4[,1:2]) model manufacturer a4 a4 quattro a6 quattro grand cherokee 4wd gti jetta new beetle audi 7 8 3 0 0 0 0 jeep 0 0 0 8 0 0 0 volkswagen 0 0 0 0 5 9 6 model manufacturer passat audi 0 jeep 0 But in other cases this does not occur. For instance, take a small subset of data(diamonds): diamonds25 - droplevels(diamonds[1:25,2:3]) table(diamonds25) color cut E F H I J Fair 1 0 0 0 0 Good 1 0 0 1 4 Very Good 1 0 3 1 4 Premium 3 1 0 1 0 Ideal 1 0 0 1 2 And change the variables mapped in the previous plot: base - ggplot(diamonds25, aes(fill = cut)) + geom_bar(position = dodge) + opts(legend.position = none) base + aes(x = cut) + facet_grid(. ~ color) I see all bars with height = 1. I have ovserved this problem (wrong bar heights, but not always = 1), in other cases when all counts are very small or zero. What's wrong here? Regards, Helios sessionInfo() R version 2.14.2 (2012-02-29) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Spanish_Spain.1252 LC_CTYPE=Spanish_Spain.1252 [3] LC_MONETARY=Spanish_Spain.1252 LC_NUMERIC=C [5] LC_TIME=Spanish_Spain.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] ggplot2_0.9.0 loaded via a namespace (and not attached): [1] colorspace_1.1-1 dichromat_1.2-4 digest_0.5.1 grid_2.14.2 [5] MASS_7.3-17 memoise_0.1 munsell_0.3 plyr_1.7.1 [9] proto_0.3-9.2 RColorBrewer_1.0-5 reshape2_1.2.1 scales_0.2.0 [13] stringr_0.6 INSTITUTO DE BIOMECÁNICA DE VALENCIA Universidad Politécnica de Valencia ● Edificio 9C Camino de Vera s/n ● 46022 VALENCIA (ESPAÑA) Tel. +34 96 387 91 60 ● Fax +34 96 387 91 69 www.ibv.org Antes de imprimir este e-mail piense bien si es necesario hacerlo. En cumplimiento de la Ley Orgánica 15/1999 reguladora de la Protección de Datos de Carácter Personal, le informamos de que el presente mensaje contiene información confidencial, siendo para uso exclusivo del destinatario arriba indicado. En caso de no ser usted el destinatario del mismo le informamos que su recepción no le autoriza a su divulgación o reproducción por cualquier medio, debiendo destruirlo de inmediato, rogándole lo notifique al remitente. __ 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. INSTITUTO DE BIOMECÁNICA DE VALENCIA Universidad Politécnica
Re: [R] xyplot lattice fine control of axes limits and thick marks (with log scale)
On Wed, Apr 11, 2012 at 7:16 AM, David Winsemius dwinsem...@comcast.net wrote:
On Apr 11, 2012, at 9:03 AM, David Winsemius wrote:
On Apr 11, 2012, at 6:28 AM, maxbre wrote:
hi, I just realised I want to go a little further in the control of the
chart
appearance and I would like to have the same number of ticks displayed in
both axes of all panels
I'm wondering if you should be using relation=free when you have already
set a panel specific range for the x and y limits? I'm thinking that the
panel function may be reversing your earlier prepanel efforts. (No data
offered ... why don't you use one of the many test datasets in the examples
of the lattice package?)
On further meandering up this thread I see that you omitted the context of
earlier data offerings, so not I in turn offer what I think is a your
request. Change relation from free to sliced
David, you make a good point. Seems OP's long and winding road {end
quote} is slowly circling back to the origin (see the first couple of
messages in thread).
slice is better than free, but isn't tick.number just a suggestion
? i.e. a better choice of n in ?pretty will override ? For example
this data (below), barely noticeable, but see panel(2,1) has 7 ticks
compare with 6 for the others.
Any one please correct me (as I find I mess with these myself too
often... :) but I think if OP wants to force equal ticks (and lose the
pretty axis) there is no avoiding changes to x and yscale.components
?
tm - structure(list(name_short = structure(1:29, .Label = c(D4,
D5, D6a, D6b, D6c, D7, D8, F4, F5a, F5b, F6a,
F6b, F6c, F6d, F7a, F7b, F8, P105, P114, P118,
P123, P126, P156, P157, P167, P169, P189, P77,
P81), class = factor), sub_family = structure(c(3L, 3L, 3L,
3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L,
1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 2L), .Label = c(pcb mono-ortho subs,
pcb non-ortho subs, pcdd, pcdf), class = factor), tv = c(1.069,
6.461, 5.461, 12.764, 10.86, 117.912, 256.875, 452.204, 124.02,
327.856, 88.469, 61.539, 17.794, 84.117, 121.668, 13.414, 68.409,
3023.333, 428, 19454.667, 151.333, 324, 11478.667, 1220.667,
5335.333, 124.667, 1542.667, 594.667, 193.333), ms = c(1.787,
4.831, 3.456, 14.105, 10.808, 116.02, 296.957, 30.533, 21.821,
32.969, 33.767, 29.799, 12.812, 49.637, 126.522, 17.522, 106.087,
1787.5, 130, 6751.5, 81, 23, 370, 33.5, 147.5, 5.406, 18.5, 415,
69.906)), .Names = c(name_short, sub_family, tv, ms), class =
data.frame, row.names = c(NA, -29L))
# changing to sliced
xyplot(tv ~ ms | sub_family, data=tm,
#as.table=TRUE,
aspect=xy,
xlab = expression(paste('ms [ fg/', m^3, ' ]', sep = '')),
ylab = expression(paste('tv [ fg/', m^3, ' ]', sep = '')),
scales= list(x=list(relation=sliced, log=10, cex=0.8),
y=list(relation=sliced, log=10, cex=0.8)),
prepanel = function(x, y, subscripts) {
rr- range(cbind(x,y))
list(xlim = rr, ylim= rr)
},
panel = function(x, y ,subscripts,...) {
panel.xyplot(x, y, cex=0.8,...)
panel.abline(a = 0, b = 1, lty = 2, col =gray)
panel.text(x, y, labels=tm$name_short[subscripts], cex = 0.8, pos=3,
offset=0.5, srt=0, adj=c(1,1))
},
subscripts=TRUE,
xscale.components = xscale.components.logpower,
yscale.components = yscale.components.logpower
)
# Compare with
xyplot(tv ~ ms | sub_family, data=tm,
#as.table=TRUE,
aspect=xy,
xlab = expression(paste('ms [ fg/', m^3, ' ]', sep = '')),
ylab = expression(paste('tv [ fg/', m^3, ' ]', sep = '')),
scales= list(relation=free, log=10, cex=0.8),
prepanel = function(x, y, subscripts) {
rr- range(cbind(x,y))
list(xlim = rr, ylim= rr)
},
panel = function(x, y ,subscripts,...) {
panel.xyplot(x, y, cex=0.8,...)
panel.abline(a = 0, b = 1, lty = 2, col =gray)
panel.text(x, y, labels=tm$name_short[subscripts], cex = 0.8, pos=3,
offset=0.5, srt=0, adj=c(1,1))
},
subscripts=TRUE,
xscale.components = function(...) {
ans - xscale.components.logpower(...)
range - ans$num.limit
newtck - round(seq(range[1],range[2],l=7),1)
ans$bottom$ticks$at - newtck
ans$bottom$labels$at - newtck
ans$bottom$labels$labels -
parse(text=paste('10^',newtck,sep=''))
ans
} ,
yscale.components = function(...) {
ans - yscale.components.logpower(...)
range - ans$num.limit
newtck - round(seq(range[1],range[2],l=7),1)
ans$left$ticks$at - newtck
ans$left$labels$at - newtck
ans$left$labels$labels -
parse(text=paste('10^',newtck,sep=''))
ans
}
)
Cheers
scales= list(x=list(relation=sliced, log=10, cex=0.8, tick.number=5),
Re: [R] geom_plot creates Area Instead Of Lines
What I would have liked is something like a cloud of lines, similar to what I get when I convert the data into a matrix (why do I not just use a matrix? I come from MATLAB and this seems natural, however, my data is large and a data frame seems to be an advantageous way to handle that). It's hard to know without a reproducible example (https://github.com/hadley/devtools/wiki/Reproducibility), but perhaps you need to set the group aesthetic? Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ 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] Michaelson-Morley Speed of Light Data
On 11/04/2012 19:33, Křištof Želechovski wrote: Dnia środa, 11 kwietnia 2012 13:26:43 Prof Brian Ripley pisze: On 11/04/2012 12:42, Duncan Murdoch wrote: On 12-04-11 12:43 AM, Křištof Želechovski wrote: URL: http://finzi.psych.upenn.edu/R/library/datasets/html/morley.html The classical data of Michaelson and Morley on the speed of light Can you provide more information about the data? How were they obtained, etc.? I do not have the book Genstat Primer and the nearest location where it is available is University of York which is rather far from my location. If you can't find the cited reference, I'd try Google. For instance, it led me to this page http://en.wikipedia.org/wiki/File:Michelsonmorley-boxplot.svg which appears to show five series. Yes, but that is derived from R. AFAIR the history, Bill Venables got this from Weekes (1986), a book I have only ever seen in Adelaide. A better reference is S. M. Stigler (1977) Do robust estimators work with real data? Annals of Statistics 5, 1055–1098. (See Table 6.) The data in R are identical with Table 6 but they were not obtained by Michaelson and Morley; they were obtained by Michelson. The description is wrong. It is also a problem in R itself, as evidenced by the following instruction: data(morley) Of course, you can name a data set whatever you like, even data(marilyn) would be all right, but may I suggest that this data set be renamed to michelson? That is what it is called in package MASS Please fix, Chris -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extend data frame for plotting heat map in ggplot2
On Sun, Apr 1, 2012 at 9:16 AM, Till Bayer till.ba...@kaust.edu.sa wrote: Hi all! I want to generate a heat map from an all-vs-all comparison. I have the data, already scaled to 0-1. However, I have the values only for the comparisons in one way, and not for the comparisons between the same group (which are always 1), i.e. I have half the matrix and am missing the other half and the diagonal. What is a good way to get it into a form that ggplot2 can use for the heat map? This is an example of the data I have: A B value T1 T2 0.347 T1 T3 0.669 T2 T3 0.214 I assume the following is what I need for ggplot (or maybe I don't, if ggplot can somehow generate it?): A B value T1 T2 0.347 T1 T3 0.669 T2 T3 0.214 T2 T1 0.347 T3 T1 0.669 T3 T2 0.214 T1 T1 1 T2 T2 1 T3 T3 1 You can usually do something like: df - data.frame(A = 1:2, B = 3:4, value = runif(2)) all - expand.grid(unique(df[c(A, B)])) merge(all, df, all = T) Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ 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 graphing
can anybody tell me how i can draw x- y- axis and draw x^3 graph using R graph?? i need nice coordinate system with legends and coordinate numberings.. and nice graph of x^3 on it.. it will be nice if you tell me how i can center the graph.. i want the origin (0,0) to be right in the middle of the graph. thank you so much. __ 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 graphing
The easiest way is to just use ?curve (type ?curve at the prompt to get documentation for curve): e.g., curve(x^3, from = -5, to = 5) You could also build the plot yourself like: x - seq(-5, 5, length.out = 200) y - x^3 plot(x,y) Michael On Wed, Apr 11, 2012 at 4:41 PM, John Kim ktown4...@gmail.com wrote: can anybody tell me how i can draw x- y- axis and draw x^3 graph using R graph?? i need nice coordinate system with legends and coordinate numberings.. and nice graph of x^3 on it.. it will be nice if you tell me how i can center the graph.. i want the origin (0,0) to be right in the middle of the graph. thank you so much. __ 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 graphing
On Apr 11, 2012, at 4:41 PM, John Kim wrote: can anybody tell me how i can draw x- y- axis and draw x^3 graph using R graph?? i need nice coordinate system with legends and coordinate numberings.. Look at: ?curve and nice graph of x^3 on it.. it will be nice if you tell me how i can center the graph.. i want the origin (0,0) to be right in the middle of the graph. thank you so much. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] convex nonnegative basis vectors in nullspace of matrix
On Wed, Apr 11, 2012 at 06:04:28AM -0700, capy_bara wrote:
Dear all,
I want to explore the nullspace of a matrix S: I currently use the function
Null from the MASS package to get a basis for the null space:
S = matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1)); S
MASS::Null(t(S))
My problem is that I actually need a nonnegative basis for the null space of
S.
There should be a unique set of convex basis vectors spanning a vector space
in which each vector v satisfies sum (S %*% v) == 0 and min(v)=0.
Hi.
The null space of the above matrix has dimension 2. Its intersection
with nonnegative vectors in R^5 is an infinite cone. In order to restrict
it to a finite set, we can consider its intersection with the set
of vectors with the sum of coordinates equal to 1. Then, the solution
is a finite convex polytop and we can search for its vertices. The
following code searches for vertices in random directions and finds two
vertices. In this simple case, the polytop is in fact a line segment,
so we get its endpoints. These endpoints form a linear basis of the
original null space consisting of nonnegative vectors.
library(lpSolve)
S - matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1))
a - MASS::Null(t(S))
n - nrow(a)
a1 - rbind(a, colSums(a))
b - rep(0, times=n+1)
b[n+1] - 1
dir - c(rep(=, times=n), ==)
sol - matrix(nrow=100, ncol=n)
for (i in seq.int(length=nrow(sol))) {
crit - rnorm(ncol(a))
out - lp(objective.in=crit, const.mat=a1, const.dir=dir, const.rhs=b)
sol[i, ] - a %*% out$solution
}
unique(round(sol, digits=10))
[,1] [,2] [,3] [,4] [,5]
[1,] 0.250 0.250 0.000 0.250 0.250
[2,] 0.1642631 0.3357369 0.1714738 0.1642631 0.1642631
Use this with care, since for more complex cases, this method does not
guarantee that all vertices are found. So, it is not guaranteed that
every nonnegative vector in the null space is a nonnegative combination
of the obtained vectors.
Hope this helps.
Petr Savicky.
__
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-help; generating censored data
On 11-Apr-2012 16:28:31 Christopher Kelvin wrote: Hello, can i implement this as 10% censored data where t gives me failure and x censored. Thank you p=2;b=120 n=50 set.seed(132); r-sample(1:50,45) t-rweibull(r,shape=p,scale=b) t set.seed(123);_ cens - sample(1:50, 5)_ x-runif(cens,shape=p,scale=b)_ x Chris Guure Researcher, Institute for Mathematical Research UPM This query is obscure! First, its approach does not seem to conform to the standard notion of censored data. This refers to a situation where, for each item observed, either (a) it value falls within a certain range (which may itself depend on the item), in which case its value is recorded as a value in the data; or (b) its value falls outside that range, in which case that fact is recorded but the value is not recorded (thus being censored). Eaxmple: Patients who have been admitted to hospital for a particular disease are subsequently monitored for a period of time (days/months/years) which may vary from patient to patient. The reason for the time limitation may be that the design of the investigation set a limit, or may be haphazard as a result of the patient moving away and no longer being accessible. The value recorded (if available) is the time from admission to death. If not available, then all that can be recorded is that the event occurred later than the upper time limit for thaqt patient. As far as I can see, no element of your code above corresponds to this notion of censored. Next, your r-sample(1:50,45) selects 45 different values from (1:50), and then your t-rweibull(r,shape=p,scale=b) generates 45 values sampled from the Weibull distribution, ** regardless of the 45 values from (1:50) in r ** -- See under '?rweibull' where it says: n: number of observations. If 'length(n) 1', the length is taken to be the number required. So it would seem that your r-sample(1:50,45) is superfluous, and you could simply have written t-rweibull(45,shape=p,scale=b). Similar comments apply to your cens - sample(1:50, 5) x-runif(cens,shape=p,scale=b) where you could have equivalently written x-runif(5,shape=p,scale=b). Also, the parameters shape and scale would not be recognised by runif(), whose parameters are as in runif(n, min=..., max=...). Maybe you meant to write x-rweibull(cens,shape=p,scale=b), but then you would simply be sampling a further 5 values from the same Weibull distribution, along with your original 45. So how does censoring come into this? If you would explain, in plain words, what you are seeking to do, it would help to remove this obscurity and confusion! Hoping this helps, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 11-Apr-2012 Time: 23:23:36 This message was sent by XFMail __ 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] convex nonnegative basis vectors in nullspace of matrix
On Wed, Apr 11, 2012 at 06:04:28AM -0700, capy_bara wrote:
Dear all,
I want to explore the nullspace of a matrix S: I currently use the function
Null from the MASS package to get a basis for the null space:
S = matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1)); S
MASS::Null(t(S))
My problem is that I actually need a nonnegative basis for the null space of
S.
There should be a unique set of convex basis vectors spanning a vector space
in which each vector v satisfies sum (S %*% v) == 0 and min(v)=0.
Hi.
In my previous solution, i forgot that lp() assumes all variables
nonnegative. So, the code was searching only a subset of the true
set of solutions. A better alternative is as follows.
library(lpSolve)
S - matrix(nrow=3, ncol=5, c(1,0,0,-1,1,1,1,-1,-1,0,-1,0,0,0,-1))
a - MASS::Null(t(S))
a1 - cbind(a, -a)
n - nrow(a1)
a2 - rbind(a1, colSums(a1))
b - rep(0, times=n+1)
b[n+1] - 1
dir - c(rep(=, times=n), ==)
sol - matrix(nrow=100, ncol=n)
for (i in seq.int(length=nrow(sol))) {
crit - rnorm(ncol(a))
crit - c(crit, -crit)
out - lp(objective.in=crit, const.mat=a2, const.dir=dir, const.rhs=b)
sol[i, ] - a1 %*% out$solution
}
unique(round(sol, digits=10))
[,1] [,2] [,3] [,4] [,5]
[1,] 0.00 0.50 0.5 0.00 0.00
[2,] 0.25 0.25 0.0 0.25 0.25
Petr Savicky.
__
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 graphing
Please reply all to the list and don't send HTML. curve(x^3, from = -5, to = 5); grid() If you only want dotted lines at x = 0 y = 0 use grid(2,2) instead. Michael On Wed, Apr 11, 2012 at 6:49 PM, John Kim provicon2...@yahoo.com wrote: thanks for the reply.. i tried the code below.. it shows the graph.. but how do i make x and y axis show up in the middle?? i am teaching secondary school math.. and i need to produce graphs for learning purpose.. i need nice x- y-axis.. in the middle.. with arrows at the end of axis.. can you tell me how i can realize that?? john From: R. Michael Weylandt michael.weyla...@gmail.com To: John Kim ktown4...@gmail.com Cc: r-help@r-project.org; provicon2...@yahoo.com Sent: Wednesday, April 11, 2012 2:37 PM Subject: Re: [R] r graphing The easiest way is to just use ?curve (type ?curve at the prompt to get documentation for curve): e.g., curve(x^3, from = -5, to = 5) You could also build the plot yourself like: x - seq(-5, 5, length.out = 200) y - x^3 plot(x,y) Michael On Wed, Apr 11, 2012 at 4:41 PM, John Kim ktown4...@gmail.com wrote: can anybody tell me how i can draw x- y- axis and draw x^3 graph using R graph?? i need nice coordinate system with legends and coordinate numberings.. and nice graph of x^3 on it.. it will be nice if you tell me how i can center the graph.. i want the origin (0,0) to be right in the middle of the graph. thank you so much. __ 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 graphing
Alternatively, you can use curve(x^3, from = -5, to = 5); abline(h = 0, v = 0, lty = 2) which will work even if the axes aren't in the middle of the image. Michael On Wed, Apr 11, 2012 at 6:54 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: Please reply all to the list and don't send HTML. curve(x^3, from = -5, to = 5); grid() If you only want dotted lines at x = 0 y = 0 use grid(2,2) instead. Michael On Wed, Apr 11, 2012 at 6:49 PM, John Kim provicon2...@yahoo.com wrote: thanks for the reply.. i tried the code below.. it shows the graph.. but how do i make x and y axis show up in the middle?? i am teaching secondary school math.. and i need to produce graphs for learning purpose.. i need nice x- y-axis.. in the middle.. with arrows at the end of axis.. can you tell me how i can realize that?? john From: R. Michael Weylandt michael.weyla...@gmail.com To: John Kim ktown4...@gmail.com Cc: r-help@r-project.org; provicon2...@yahoo.com Sent: Wednesday, April 11, 2012 2:37 PM Subject: Re: [R] r graphing The easiest way is to just use ?curve (type ?curve at the prompt to get documentation for curve): e.g., curve(x^3, from = -5, to = 5) You could also build the plot yourself like: x - seq(-5, 5, length.out = 200) y - x^3 plot(x,y) Michael On Wed, Apr 11, 2012 at 4:41 PM, John Kim ktown4...@gmail.com wrote: can anybody tell me how i can draw x- y- axis and draw x^3 graph using R graph?? i need nice coordinate system with legends and coordinate numberings.. and nice graph of x^3 on it.. it will be nice if you tell me how i can center the graph.. i want the origin (0,0) to be right in the middle of the graph. thank you so much. __ 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 graphing
I repeat myself: Please reply all to the list and don't send HTML. I'm not aware of facilities to do that automatically (though others might now them) -- you can draw arrows manually with the arrows() function. Michael On Wed, Apr 11, 2012 at 7:08 PM, John Kim provicon2...@yahoo.com wrote: thank you so much.. is it possible to make arrows show up at the end of axis.. and make numberings show up in the middle axis?? From: R. Michael Weylandt michael.weyla...@gmail.com To: John Kim provicon2...@yahoo.com; r-help r-help@r-project.org Sent: Wednesday, April 11, 2012 4:06 PM Subject: Re: [R] r graphing Alternatively, you can use curve(x^3, from = -5, to = 5); abline(h = 0, v = 0, lty = 2) which will work even if the axes aren't in the middle of the image. Michael On Wed, Apr 11, 2012 at 6:54 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: Please reply all to the list and don't send HTML. curve(x^3, from = -5, to = 5); grid() If you only want dotted lines at x = 0 y = 0 use grid(2,2) instead. Michael On Wed, Apr 11, 2012 at 6:49 PM, John Kim provicon2...@yahoo.com wrote: thanks for the reply.. i tried the code below.. it shows the graph.. but how do i make x and y axis show up in the middle?? i am teaching secondary school math.. and i need to produce graphs for learning purpose.. i need nice x- y-axis.. in the middle.. with arrows at the end of axis.. can you tell me how i can realize that?? john From: R. Michael Weylandt michael.weyla...@gmail.com To: John Kim ktown4...@gmail.com Cc: r-help@r-project.org; provicon2...@yahoo.com Sent: Wednesday, April 11, 2012 2:37 PM Subject: Re: [R] r graphing The easiest way is to just use ?curve (type ?curve at the prompt to get documentation for curve): e.g., curve(x^3, from = -5, to = 5) You could also build the plot yourself like: x - seq(-5, 5, length.out = 200) y - x^3 plot(x,y) Michael On Wed, Apr 11, 2012 at 4:41 PM, John Kim ktown4...@gmail.com wrote: can anybody tell me how i can draw x- y- axis and draw x^3 graph using R graph?? i need nice coordinate system with legends and coordinate numberings.. and nice graph of x^3 on it.. it will be nice if you tell me how i can center the graph.. i want the origin (0,0) to be right in the middle of the graph. thank you so much. __ 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] Merging multiple .csv files
Thanks again. By the way, any idea why I get this error: merged3 - merge_all(list_of_files , by = Name) Error in `[.data.frame`(df, , match(names(dfs[[1]]), names(df))) : undefined columns selected sessionInfo() R version 2.14.2 (2012-02-29) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=English_Australia.1252 LC_CTYPE=English_Australia.1252 LC_MONETARY=English_Australia.1252 LC_NUMERIC=C LC_TIME=English_Australia.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] reshape_0.8.4 plyr_1.7.1 loaded via a namespace (and not attached): [1] tools_2.14.2 Cheers, Chintanu = On Wed, Apr 11, 2012 at 11:10 PM, R. Michael Weylandt michael.weyla...@gmail.com wrote: Simply pass all = FALSE to merge_all merge_all(list_of_files, by = Name, all = FALSE) Michael On Wed, Apr 11, 2012 at 1:09 AM, Chintanu chint...@gmail.com wrote: Thanks to David and Michael Michael: That works, however with a glitch. Each of my 24 files has got two columns: Name, and Rank/score. file_list - list.files() list_of_files - lapply(file_list, read.csv) # Read in each file # I can see the 2-columns at this stage. However, the following line: merge_all(list_of_files, by = Name) # produces some NAs for the 2nd column (except the beginning 1/3rd of the columns which have values). Not sure about the reason - the original files don't have any NAs. Further, I understand that it gives the union of (rows of) files based on Name. Is there a way to look for intersection, i.e., similar to using: merge ( ,by=Name, all=FALSE) ? David: It came up with an error : do.call(merge, list_of_files, by=Name) Error in do.call(merge, list_of_files, by = Name) : unused argument(s) (by = Name) Cheers, Chintanu [[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] Encoding of Sweave file error message
Hi
I ran the following sweave file in R2.14.1 and upgraded to R2.15
yesterday with the same setup
I got the following error message when I rand the following Sweave file
setwd(D:/Cic/Sweave/Parasite/Comb/12)
Sweave(D:/Cic/Sweave/Parasite/Comb/12/ParasiteComb12.Rnw)
Error: c('ParasiteComb12DS.Rnw', 'ParasiteComb12.Rnw') is not
ASCII and does not declare an encoding
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 15.0
year 2012
month 03
day30
svn rev58871
language R
version.string R version 2.15.0 (2012-03-30)
nickname
I checked that the files were encoded as ANSI
The file ParasiteComb12.Rnw is a master file and sets things up and
calls a number of Sweave files.
ParasiteComb12DS.Rnw which loads packages processes the data is the
first file to be called
I can run ParasiteComb12.Rnw by itself without any problems but when
I add other files then the error message occurs.
There are no references just plain text (no extended character set)
plain latex and R coding
The first few lines for the Sweave preamble is:
\documentclass[10pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage{textcomp}
\usepackage[latin1]{inputenc}
\usepackage{times}
\usepackage{courier}
\usepackage[scaled=.92]{helvet}
I can send the files if needed as the resultant pdf that is produced
is 42 pages.
Any pointers to solve the problem will be greatly appreciated
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
__
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] Encoding of Sweave file error message
I had the same problem! So, as I'm a linux user, I prefer use linux terminal. On terminal I type this to compile R CMD Sweave --encoding=utf-8 myfile.Rnw and the compilation is successful. Try to set the encoding option in Sweave(). Bests. Walmes. == Walmes Marques Zeviani LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W) Departamento de Estatística - Universidade Federal do Paraná fone: (+55) 41 3361 3573 VoIP: (3361 3600) 1053 1173 e-mail: wal...@ufpr.br twitter: @walmeszeviani homepage: http://www.leg.ufpr.br/~walmes linux user number: 531218 == [[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] deferred call
I must admit I'm a little ashamed to have been using R for so long,
and still lack a sound understanding of deferred calls, eval, deparse,
substitute, and friends.
I'm attempting to make a deferred call to a function which has default
arguments in the following way:
call.foo - function(f) {
x - f()
}
x - 1:10
f - function(x=x) { x^2 }
call.foo(f)
However, I'm getting the error:
Error in x^2 : 'x' is missing
Is there a common R idiom for calling 'formals' on the function, and
then grabbing the named default arguments from the enclosing frame?
I naively thought that since function 'f' was defined w/ a default
argument of 'x' and x is defined in the same envir as the function
itself, that the call would succeed.
-Whit
__
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] Definition of lag is opposite in ts and xts objects!
Example: Will ts objects be obsolete or modified? a[,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035 lag(a,4)[,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2 2.791594762 1984 Q3 -0.009953715 1984 Q4 -0.015059485 lag(as.ts(a, start=c(1983,1)),4) Qtr1 Qtr2 Qtr3 Qtr4 1982 2.747365190 2.791594762 -0.009953715 -0.015059485 1983 -1.190061246 -0.553031799 0.686874720 0.953911035 [[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] deferred call
On Wed, Apr 11, 2012 at 10:17 PM, Whit Armstrong
armstrong.w...@gmail.com wrote:
I must admit I'm a little ashamed to have been using R for so long,
and still lack a sound understanding of deferred calls, eval, deparse,
substitute, and friends.
I'm attempting to make a deferred call to a function which has default
arguments in the following way:
call.foo - function(f) {
x - f()
}
x - 1:10
f - function(x=x) { x^2 }
call.foo(f)
However, I'm getting the error:
Error in x^2 : 'x' is missing
Is there a common R idiom for calling 'formals' on the function, and
then grabbing the named default arguments from the enclosing frame?
I naively thought that since function 'f' was defined w/ a default
argument of 'x' and x is defined in the same envir as the function
itself, that the call would succeed.
f - function(x=x) x^2 is an endless recursion. Try
f - function(x.=x) { x^2 }
(note the dot)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Definition of lag is opposite in ts and xts objects!
BTW, zoo is like ts in the application of lag. In other words, zoo and xts are opposite in this issue. 2012/4/12 jpm miao miao...@gmail.com Example: Will ts objects be obsolete or modified? a[,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035 lag(a,4)[,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2 2.791594762 1984 Q3 -0.009953715 1984 Q4 -0.015059485 lag(as.ts(a, start=c(1983,1)),4) Qtr1 Qtr2 Qtr3 Qtr4 1982 2.747365190 2.791594762 -0.009953715 -0.015059485 1983 -1.190061246 -0.553031799 0.686874720 0.953911035 [[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] Definition of lag is opposite in ts and xts objects!
Yes, this is as documented. See ? lag.xts under details for the justification and how to change the default if desired. Michael On Wed, Apr 11, 2012 at 11:13 PM, jpm miao miao...@gmail.com wrote: BTW, zoo is like ts in the application of lag. In other words, zoo and xts are opposite in this issue. 2012/4/12 jpm miao miao...@gmail.com Example: Will ts objects be obsolete or modified? a [,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035 lag(a,4) [,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2 2.791594762 1984 Q3 -0.009953715 1984 Q4 -0.015059485 lag(as.ts(a, start=c(1983,1)),4) Qtr1 Qtr2 Qtr3 Qtr4 1982 2.747365190 2.791594762 -0.009953715 -0.015059485 1983 -1.190061246 -0.553031799 0.686874720 0.953911035 [[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] Definition of lag is opposite in ts and xts objects!
What makes you think those are the only two options? I happen to prefer the xts convention, but there is a lot of code out there that successfully uses ts just as it is, and I can't see breaking all of that to meet an arbitrary preference of sign convention. (It isn't my decision anyway...) --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. jpm miao miao...@gmail.com wrote: Example: Will ts objects be obsolete or modified? a[,1] 1983 Q1 2.747365190 1983 Q2 2.791594762 1983 Q3 -0.009953715 1983 Q4 -0.015059485 1984 Q1 -1.190061246 1984 Q2 -0.553031799 1984 Q3 0.686874720 1984 Q4 0.953911035 lag(a,4)[,1] 1983 Q1 NA 1983 Q2 NA 1983 Q3 NA 1983 Q4 NA 1984 Q1 2.747365190 1984 Q2 2.791594762 1984 Q3 -0.009953715 1984 Q4 -0.015059485 lag(as.ts(a, start=c(1983,1)),4) Qtr1 Qtr2 Qtr3 Qtr4 1982 2.747365190 2.791594762 -0.009953715 -0.015059485 1983 -1.190061246 -0.553031799 0.686874720 0.953911035 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Schedule R script using cron
I am trying to schedule my R script using cron, but it is not working. It seems R can not find packages in cron. Anyone can help me? Thanks. The following is my bash script # source my porfile . /home/winie/.profile # script.R will load packages R CMD BATCH /home/script.R -- View this message in context: http://r.789695.n4.nabble.com/Schedule-R-script-using-cron-tp4550729p4550729.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] phangorn and calculation of a rate matrix
Hi, I'm trying to calculate a ratematrix for a RNA aligment (U instead of T) in order to use it as a ratematrix in Profidst (a phylogenetic program which takes into account both the primary sequence and the secondary structure of the RNA, in my case rRNA). The sequence-structure aligment has been made in 4SALE (a java app) and saved as one-letter encoded (using a 12 letters alphabet, a,c,d,e,g,h,i,k,l,n,q,r instead of the conventional nucleotide codes). My intention is to calculate the ratematrix (which is a 12x12 matrix) for this special aligment with ape and phangorn, however I've repeatdly fail to do it. The aligment is in a phyDat object containing 30 sequences and 12 states (by using user-defined character states consisting on the 12 letters indicated above). I follow the steps described in the phangorn-specials vignette but the ratematrix (under a GTR substitution model) is not calculated. May the problem be that phangorn only accepts a,g,c and t as valid states for calculating the matrix? And in case phangorn could calculate the matrix, how could I do it? I have a very basic knowledge of R so please I would greatly appreciate (if possible) a step-by-step explanation. Thanks very much for the help. -- View this message in context: http://r.789695.n4.nabble.com/phangorn-and-calculation-of-a-rate-matrix-tp4550495p4550495.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] Converting php arrays to data frame
Hi, I am working on an implementation of R within some web application (R is just a part of a larger project, so R has to be incorporated within an existing setup) Here is the scenario, PHP grabs a csv dataset from some server and turns it into a 2 dimensional array (with each row being an array, and there are N of them, N being the number of columns) Now I want the PHP script to trigger Rscript to process the data with the exec() command. The idea is that R can access the data without actually fetching the csv from the server (through RMySQL or something like that) The question is, how do I turn the PHP array into a data frame or some kind of object that R can process? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Converting-php-arrays-to-data-frame-tp4550656p4550656.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 compare cross-validated C-index in Cox
If I calculate the C-index after 10-fold cross-validation based on Cox model, which statistic test should I used to compare two C-statistc values.I means I use cross-validation to obtain a predictive probability for each individual, then combine all the probabilities to calculate the C-index.If I get two C-index values(0.73 vs. 0.80) based on two different model, then how to test the difference is significant. Thanks -- View this message in context: http://r.789695.n4.nabble.com/how-to-compare-cross-validated-C-index-in-Cox-tp4550707p4550707.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] Could dynlm function work for xts objects?
It seems to only works for zoo or ts objects? I tried to run it for xts objects, and error message occurs. Once I coerce it to be an zoo object (by as.zoo), it works. Error message: Error in model.frame.default(formula = dynformula(PIh - PI ~ L(X, 0:i) + : variable lengths differ (found for 'L(X, 0:i)') In addition: Warning messages: 1: In zoo(coredata(x), order.by = index(x), ...) : some methods for “zoo” objects do not work if the index entries in ‘order.by’ are not unique 2: In zoo(coredata(x), order.by = index(x), ...) : some methods for “zoo” objects do not work if the index entries in ‘order.by’ are not unique __ 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] Encoding of Sweave file error message
At 12:03 12/04/2012, you wrote:
I had the same problem! So, as I'm a linux user,
I prefer use linux terminal. On terminal I type this to compile
R CMD Sweave --encoding=utf-8 myfile.Rnw
and the compilation is successful. Try to set the encoding option in Sweave().
Bests.
Walmes.
==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: mailto:wal...@ufpr.brwal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/%7Ewalmeshttp://www.leg.ufpr.br/~walmes
linux user number: 531218
==
Hi Walmes
Thank you very much. That appears to be the problem.
When I typed from the DOS terminal
R CMD Sweave --encoding=utf-8 ParasiteComb12.Rnw
it compiled the tex file without any error messages.
I have not really got into font encoding and
reading the Sweave manual I thought that what I had done would be sufficient.
I found an old note which gave a reference to
http://tolstoy.newcastle.edu.au/R/e10/help/10/05/4725.html
http://tolstoy.newcastle.edu.au/R/e10/help/10/05/4889.html
but that appears to be specific.
The ?Sweave and the Sweave manual appear to be
more specific about the latex side.
After having a look at iconvlist() and bearing
in mind Duncan Murdoch's comments about windows I tried
Sweave(D:/Cic/Sweave/Parasite/Comb/12/ParasiteComb12.Rnw,
encoding = UTF-8)
from the Rgui command window and compiled without any problems
When I had a look at the tex file there were a
few DOS Alt-248 (degree symbol ) within latex
comments which were added last running R2.14 before updating
Removing them and re running without the encoding
argument brought things back to normal.
I tried as a test
\SweaveOpts{encoding=UTF8}
but that appears not to work
All I have to do now is to put the extra argument
into my text editors clip library for Sweave for
next time when I cannot solve things.
Its been a long week !
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
[[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] deferred call
On Wed, Apr 11, 2012 at 8:12 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Apr 11, 2012 at 10:17 PM, Whit Armstrong
armstrong.w...@gmail.com wrote:
I must admit I'm a little ashamed to have been using R for so long,
and still lack a sound understanding of deferred calls, eval, deparse,
substitute, and friends.
I'm attempting to make a deferred call to a function which has default
arguments in the following way:
call.foo - function(f) {
x - f()
}
x - 1:10
f - function(x=x) { x^2 }
call.foo(f)
However, I'm getting the error:
Error in x^2 : 'x' is missing
Is there a common R idiom for calling 'formals' on the function, and
then grabbing the named default arguments from the enclosing frame?
I naively thought that since function 'f' was defined w/ a default
argument of 'x' and x is defined in the same envir as the function
itself, that the call would succeed.
f - function(x=x) x^2 is an endless recursion. Try
To amplify just a bit on Gabor's comment, section 4.3.3 of the R
language definition explicitly states:
One of the most important things to know about the evaluation of
arguments to a function is
that supplied arguments and default arguments are treated differently.
The supplied arguments
to a function are evaluated in the evaluation frame of the calling
function. The default arguments
to a function are evaluated in the evaluation frame of the function.
So foo - function (x = x) {...}
tries to define foo with the default argument x, which is evaluated in
the ** environment of the function ** not in the caller's environment,
where it is 1:10. So if x ever needs to be evaluated within foo
(i.e., its promise is forced), then it looks for the value of the rhs
of the x=x assignment within foo, which is (the promise for) x,
again, within foo, which is x again,...
HTH
-- Bert
f - function(x.=x) { x^2 }
(note the dot)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at 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.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
__
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.
