Re: [R] Experiment Design
Thanks for the comments. I think i phrased my question incorrectly but the following gave me what I wanted. 18 treatments(plants) , 3 blocks, a total of 54 plots. I could not figure out exac library(agricolae) trt- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18) design.rcbd(trt, 3, serie = 2, seed = 0, kinds = Super-Duper, first=TRUE) $parameters $parameters$design [1] rcbd $parameters$trt [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 [18] 18 $parameters$r [1] 3 $parameters$serie [1] 2 $parameters$seed [1] 942260715 $parameters$kinds [1] Super-Duper $book plots block trt 1101 1 3 2102 1 17 3103 1 4 4104 1 18 5105 1 6 6106 1 5 7107 1 9 8108 1 2 9109 1 16 10 110 1 10 11 111 1 8 12 112 1 7 13 113 1 13 14 114 1 12 15 115 1 11 16 116 1 1 17 117 1 14 18 118 1 15 19 201 2 13 20 202 2 6 21 203 2 18 22 204 2 10 23 205 2 4 24 206 2 3 25 207 2 8 26 208 2 14 27 209 2 16 28 210 2 5 29 211 2 15 30 212 2 7 31 213 2 12 32 214 2 2 33 215 2 1 34 216 2 17 35 217 2 11 36 218 2 9 37 301 3 10 38 302 3 4 39 303 3 2 40 304 3 8 41 305 3 5 42 306 3 7 43 307 3 16 44 308 3 15 45 309 3 13 46 310 3 1 47 311 3 12 48 312 3 18 49 313 3 6 50 314 3 14 51 315 3 11 52 316 3 3 53 317 3 17 54 318 3 9 -- View this message in context: http://r.789695.n4.nabble.com/Experiment-Design-tp4692015p4692028.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] Experiment Design
I guessed you mean 2 treatment groups, N=18 and 3 blocks (so n=6 per block):
blocks - 3
ngroups - 2
groupsize - 3
group - NULL;
block - NULL;
set.seed(911)
for (i in 1:blocks){
foo - sample(rep(c('A', 'B'), groupsize))
group - c(group, foo)
block - c(block, rep(i, groupsize*ngroups))
}
cbind(group, block, rep(1:(groupsize*ngroups), blocks))
Denis
*_Denis *Haine
+1-514-572-7174
2014-06-12 13:59 GMT-04:00 Sarah Goslee sarah.gos...@gmail.com:
Sorry, try this again.
Does this help:
http://www2.warwick.ac.uk/fac/sci/moac/degrees/moac/ch923/lectures/moac_-_designing_experiments_in_r.pdf
If not, we need some more specifics about what you've tried and where
you are running into problems.
Sarah
On Thu, Jun 12, 2014 at 11:34 AM, Chatcher cnyi...@vols.utk.edu wrote:
Good Morning,
I would like help with code that creates a RBD experiment. That is I
have 18
treatments and I will need three blocks. I cannot seem to find an easy
way
to do this in R.
I think it is a relatively easy thing to do but i tried Agricolae and
seem
to be getting so lost I cannot even get started.
Thanks.
--
Sarah Goslee
http://www.sarahgoslee.com
http://www.functionaldiversity.org
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[R] Simulating spatio-temporal marked ETAS point process
Hello all I am searching for a package or code that is used for generating data for spatio-temporal marked ETAS processes. I saw that spatial ETAS function in PtProcess package doesn't work with simulate function... Thank you in advance, Ferra [[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] no x label using axis.Date
I have tried multiple different methods to figure out how to get a date axis of my preference (start date of each month). Any assistance would be appreciated. The below section is not producing a date axis: plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Emax),xaxt=n) *xlabels-(strptime(totaldays$totaldays,%Y-%m-%d,tz=)) xlabels-xlabels[xlabels$mday==1] axis.Date(1,at=xlabels,format=%b-%Y)* The above section is used in the following plot: par(mfrow=c(4,1)) par(mar=c(0.8,0,0,0)) par(oma=c(2,4.5,3,2)) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Dmax),xaxt=n) points(D$date,D$TSP,col=Dcol,type=p,pch=16,lwd=2,xaxt=n) text(mdate,Dmax-20,labels=DM1',cex=1) legend(topright,pch=16,cex=0.8,col=colLegend, legend=ALegend) abline(h=Target,col=red,lwd=2) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Bmax),xaxt=n) points(B$date,B$TSP,col=Bcol,type=p,pch=16,lwd=2,xaxt=n) text(mdate,Bmax-20,labels=DM2,cex=1) abline(h=Target,col=red,lwd=2) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Amax),xaxt=n) points(A$date,A$TSP,col=Acol,type=p,pch=16,lwd=2,xaxt=n) text(mdate,Amax-20,labels=DM3,cex=1) abline(h=Target,col=red,lwd=2) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Emax),xaxt=n) *xlabels-(strptime(totaldays$totaldays,%Y-%m-%d,tz=)) xlabels-xlabels[xlabels$mday==1] axis.Date(1,at=xlabels,format=%b-%Y)* points(E$date,E$TSP,col=Ecol,type=p,pch=16,lwd=2) text(mdate,Emax-20,labels=DM4',cex=1) abline(h=Target,col=red,lwd=2) title(main=Amain, ylab=Labely, outer=TRUE,cex.lab=1, cex.main=1.5) Below is an excerpt of the total days data: totaldays y 1 2013-11-01 1 2 2013-11-02 1 3 2013-11-03 1 4 2013-11-04 1 5 2013-11-05 1 6 2013-11-06 1 7 2013-11-07 1 8 2013-11-08 1 9 2013-11-09 1 10 2013-11-10 1 -- View this message in context: http://r.789695.n4.nabble.com/no-x-label-using-axis-Date-tp4692034.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] Extracting data from density plots
I have a dataframe âdf â with 3 columns. Details âof df are â as follows summary(df) DateTestVartype Min. :2002-05-10 00:00:00 Min. :-3.8531 Bottom: 313 1st Qu.:2005-05-09 12:00:00 1st Qu.:-0.7773 Other :2501 Median :2008-05-07 00:00:00 Median : 0.2482 Top : 313 Mean :2008-05-07 00:00:00 Mean : 0.1980 3rd Qu.:2011-05-05 12:00:00 3rd Qu.: 1.2250 Max. :2014-05-05 00:00:00 Max. : 3.6633 str(df) 'data.frame': 3127 obs. of 3 variables: $ Date : POSIXlt, format: 2002-05-10 2002-05-13 2002-05-14 ... $ TestVar: num 1.34 2.02 1.39 1.54 2.45 ... $ type : Factor w/ 3 levels Bottom,Other,..: 3 3 3 3 3 3 3 3 2 2 ... I have plotted the following density plot for column 2 i.e. TestVar and then color coded according to factor variable type http://r.789695.n4.nabble.com/file/n4691999/Rplot01.jpeg ggplot(df)+ geom_density(aes(x=TestVar, color=type)) I now want to filter out data samples such that the density âcurve of Top is higher than Bottom. Basically I need TestVar values for which blue line is higher than red line. How can this be achieved? Is there a way i can extract density values out of the plot? [[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] barp {plotrix} Start bars at 0 with a vector of positive values
Dear list, Please consider the following example: library(plotrix) barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) How to force the bars to start at 0? I could not find the way to do it. Regards, Pascal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Add points to subplots
Thanks David, I already did this, but in case the code gets updated I will have to re-add the annotation, which I do not think it is ideal. I was just wondering if there is an easy solution to this. Thanks a lot for the help, Cheers, Luca 2014-06-13 3:12 GMT+02:00 David Winsemius dwinsem...@comcast.net: On Jun 12, 2014, at 8:06 AM, Luca Cerone wrote: Dear all, I am running some analysis using the pamr package (available on CRAN). One of the plots I produce is made using the function pamr.plotcv. This displays two plots in the same figure (using par(mfrow=c(2,1)). When the figure is created, I would like to be able to add some points and lines, to the top plot. After producing the plot with pamr.cvplot, I have tried to add a line doing something like: par(mfg=c(1,1)) lines(c(3,3), c(0,1), col = blue, lty = 3) However this doesn't work and the line is shown in the bottom plot. How can I add points and lines to the top plot? I think you would be better off hacking the code and inserting your desired annotations before the focus is moved to the second region. There are split screen methods that allow changing focus for base graphics, but I'm not aware of ones that use the par() controls. -- David Winsemius Alameda, CA, USA -- Luca Cerone Tel: +34 692 06 71 28 Skype: luca.cerone __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] barp {plotrix} Start bars at 0 with a vector of positive values
On Fri, 13 Jun 2014 06:50:59 PM Pascal Oettli wrote: Dear list, Please consider the following example: library(plotrix) barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) How to force the bars to start at 0? I could not find the way to do it. Hi Pascal, You are right, I had not considered the idea of setting a negative y limit when there were no negative values. I think you can fix this by changing: negy-any(height0,na.rm=TRUE) to negy-any(height0,na.rm=TRUE) || ylim[1] 0 I'll check this later (I'm just going out) and if it doesn't work, I'll send another email with something that does. 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] barp {plotrix} Start bars at 0 with a vector of positive values
Hi Pascal, Perhaps I am missing something, but what about changing passing ylim = c(0, 10) to barp()? Best, Jorge.- On Fri, Jun 13, 2014 at 7:50 PM, Pascal Oettli kri...@ymail.com wrote: Dear list, Please consider the following example: library(plotrix) barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) How to force the bars to start at 0? I could not find the way to do it. Regards, Pascal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] barp {plotrix} Start bars at 0 with a vector of positive values
Hi Jim, Thank you for the suggestion. I will try it. Pascal On Fri, Jun 13, 2014 at 7:09 PM, Jim Lemon j...@bitwrit.com.au wrote: On Fri, 13 Jun 2014 06:50:59 PM Pascal Oettli wrote: Dear list, Please consider the following example: library(plotrix) barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) How to force the bars to start at 0? I could not find the way to do it. Hi Pascal, You are right, I had not considered the idea of setting a negative y limit when there were no negative values. I think you can fix this by changing: negy-any(height0,na.rm=TRUE) to negy-any(height0,na.rm=TRUE) || ylim[1] 0 I'll check this later (I'm just going out) and if it doesn't work, I'll send another email with something that does. Jim -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] barp {plotrix} Start bars at 0 with a vector of positive values
Hi Jorge, It is in order to have different barplots with the same range of limits, to keep them easily comparable. Some have negative values, some others no. Pascal On Fri, Jun 13, 2014 at 7:11 PM, Jorge I Velez jorgeivanve...@gmail.com wrote: Hi Pascal, Perhaps I am missing something, but what about changing passing ylim = c(0, 10) to barp()? Best, Jorge.- On Fri, Jun 13, 2014 at 7:50 PM, Pascal Oettli kri...@ymail.com wrote: Dear list, Please consider the following example: library(plotrix) barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) How to force the bars to start at 0? I could not find the way to do it. Regards, Pascal __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Defining default method for S3, S4 and R5 classes
Dear all, I am writing a script implementing a pipeline to analyze some of the data we receive. One of the steps in this pipeline involves clustering the data, and I am interested in studying the effects of different clustering algorithms on the final results. I am having issues making my code general enough because the clustering algorithms we are interested all return different types of objects (S3, S4 and R5 classes, as well as simple named lists). From the output of these algorithms I need to extract a list with as many elements as the number of clusters and such that each element contains the ids of the elements in each cluster. I have easily done this for each of the cluster algorithms, the problem is: how can I make so that rather than having to check for classes and types this is done automatically? For example, for the algorithms that return S3 classes I have defined a method get_cluster_list.default and then created the methods for the individual classes, which is used in the main body of the pipeline. I have no idea how I can do this for S4 and R5 classes and, more importantly, I would like an approach that works when using all S3, S4 and R5 classes. Do you know how I could do this? Thanks for the help, Luca __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Rscript fails where Rterm works
Hadley, You are a genius. Stephen B -Original Message- From: Hadley Wickham [mailto:h.wick...@gmail.com] Sent: Thursday, June 12, 2014 5:18 PM To: Bond, Stephen Cc: r-help@r-project.org Subject: Re: [R] Rscript fails where Rterm works Explicitly load the methods package: library(methods) Hadley On Thu, Jun 12, 2014 at 2:22 PM, Bond, Stephen stephen.b...@cibc.com wrote: I have a script which loads library(XLConnect) wb - loadWorkbook(wbname) the code works without errors when run from ESS which uses R version 3.0.1 (2013-05-16) -- Good Sport Copyright (C) 2013 The R Foundation for Statistical Computing Platform: i386-w64-mingw32/i386 (32-bit) But fails when run from Rscript which Rscript C:\Program Files\R\R-3.0.1\bin\i386\Rscript.EXE Rscript updatevols.R Warning message: package 'RODBC' was built under R version 3.0.2 XLConnect 0.2-7 by Mirai Solutions GmbH http://www.mirai-solutions.com , http://miraisolutions.wordpress.com Warning message: package 'XLConnect' was built under R version 3.0.3 Error in .jfield(x, Ljava/lang/Class;, TYPE) : could not find function getClass Calls: loadWorkbook ... ._java_class_list - lapply - FUN - .jfield - .Call Execution halted Warning message: closing unused RODBC handle 1 Any clues? Stephen B [[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. -- 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] Release of R 3.1.1 scheduled for July 10
Just a quick notice that we intend to have a patch release on July 10. Nickname is still undecided. Due to calendar conflicts, the binaries for some platforms may be delayed by a week or so. -pd -- 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-annou...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting data from density plots
I don't know if you can get the information from the plot, but you can certainly get it from the density() function directly. For example, # fake data TestVar - rnorm(60) type - rep(c(Bottom, Other, Top), 20) # density of TestVar for each type, estimated at 100 points along range of TestVar d - tapply(TestVar, type, density, from=min(TestVar), to=max(TestVar), n=100) results - data.frame(TestVar=d$Top$x, sapply(d, [[, 2)) # select only those records where the Top density is greater than the Bottom density results[results$Top results$Bottom, ] Jean On Thu, Jun 12, 2014 at 10:25 PM, Parimal Lokhande parimallokha...@gmail.com wrote: I have a dataframe âdf â with 3 columns. Details âof df are â as follows summary(df) DateTestVartype Min. :2002-05-10 00:00:00 Min. :-3.8531 Bottom: 313 1st Qu.:2005-05-09 12:00:00 1st Qu.:-0.7773 Other :2501 Median :2008-05-07 00:00:00 Median : 0.2482 Top : 313 Mean :2008-05-07 00:00:00 Mean : 0.1980 3rd Qu.:2011-05-05 12:00:00 3rd Qu.: 1.2250 Max. :2014-05-05 00:00:00 Max. : 3.6633 str(df) 'data.frame': 3127 obs. of 3 variables: $ Date : POSIXlt, format: 2002-05-10 2002-05-13 2002-05-14 ... $ TestVar: num 1.34 2.02 1.39 1.54 2.45 ... $ type : Factor w/ 3 levels Bottom,Other,..: 3 3 3 3 3 3 3 3 2 2 ... I have plotted the following density plot for column 2 i.e. TestVar and then color coded according to factor variable type http://r.789695.n4.nabble.com/file/n4691999/Rplot01.jpeg ggplot(df)+ geom_density(aes(x=TestVar, color=type)) I now want to filter out data samples such that the density âcurve of Top is higher than Bottom. Basically I need TestVar values for which blue line is higher than red line. How can this be achieved? Is there a way i can extract density values out of the plot? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] data format setting
Dear R family, I hope you all be doing great. I have a dataset of following format. The data file is of the following format. st year month day discharge 1 A 2004 1 1 6.752828 2 A 2004 1 2 7.602053 3 A 2004 1 3 5.583619 4 A 2004 1 4 5.019562 5 A 2004 1 5 4.804489 6 A 2004 1 6 4.363541 7 A 2004 1 7 3.801333 8 A 2004 1 8 3.455991 9 A 2004 1 9 3.402634 10 A 2004 1 10 3.250693 .. .. continue .. .. st year month day discharge 2AY 196710 3 0.56 20001AY 196710 4 0.56 20002AY 196710 5 0.48 20003AY 196710 6 0.56 20004AY 196710 7 0.48 20005AY 196710 8 0.40 20006AY 196710 9 0.40 20007AY 196710 10 0.56 20008AY 196710 11 0.56 20009AY 196710 12 0.65 20010AY 196710 13 0.85 you can see that there are five columns. The first column has the name of the station. I want to split the data w.r.t the names of the stations. Each station has data for certain years. for example A has data for years from 2004 to 2010 and for AY its from 1967 to 2000.similarly for other years there is data for different number of years. I want to make a list of matrices each containing the data for that station in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000 each column should have 365 to 366 values depending on whether there is a leap year or not. obviously for non-leap years 366th row should be an NA. kindly help me on it. Thankyou very much in advance. Eliza [[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] data format setting
?sort, ?unique, and subset come to mind. Clint BowmanINTERNET: cl...@ecy.wa.gov Air Quality Modeler INTERNET: cl...@math.utah.edu Department of Ecology VOICE: (360) 407-6815 PO Box 47600FAX:(360) 407-7534 Olympia, WA 98504-7600 USPS: PO Box 47600, Olympia, WA 98504-7600 Parcels:300 Desmond Drive, Lacey, WA 98503-1274 On Fri, 13 Jun 2014, eliza botto wrote: Dear R family, I hope you all be doing great. I have a dataset of following format. The data file is of the following format. st year month day discharge 1 A 2004 1 1 6.752828 2 A 2004 1 2 7.602053 3 A 2004 1 3 5.583619 4 A 2004 1 4 5.019562 5 A 2004 1 5 4.804489 6 A 2004 1 6 4.363541 7 A 2004 1 7 3.801333 8 A 2004 1 8 3.455991 9 A 2004 1 9 3.402634 10 A 2004 1 10 3.250693 .. .. continue .. .. st year month day discharge 2AY 196710 3 0.56 20001AY 196710 4 0.56 20002AY 196710 5 0.48 20003AY 196710 6 0.56 20004AY 196710 7 0.48 20005AY 196710 8 0.40 20006AY 196710 9 0.40 20007AY 196710 10 0.56 20008AY 196710 11 0.56 20009AY 196710 12 0.65 20010AY 196710 13 0.85 you can see that there are five columns. The first column has the name of the station. I want to split the data w.r.t the names of the stations. Each station has data for certain years. for example A has data for years from 2004 to 2010 and for AY its from 1967 to 2000.similarly for other years there is data for different number of years. I want to make a list of matrices each containing the data for that station in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000 each column should have 365 to 366 values depending on whether there is a leap year or not. obviously for non-leap years 366th row should be an NA. kindly help me on it. Thankyou very much in advance. Eliza [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] no x label using axis.Date
I'd suggest a different strategy for calculating your axis. You may also
be using the axis.Date() function incorrectly. See this example:
tmpd - data.frame(dt=seq.Date( as.Date('2014-1-7'),
as.Date('2014-08-13'), len=15),
y=runif(15)
)
xax - seq.Date(min(tmpd$dt), max(tmpd$dt), by='mon')
plot(tmpd$dt, tmpd$y, xaxt='n')axis(1, at=xax, lab=format(xax,'%b-%Y'))
By the way, most of the information in your post does not appear to be
relevant to your question (all the stuff starting with and following the
first par() statement), and it is distracting. It's always best to pare
things down to the minimum, as it makes it easier for potential helpers.
Also, it's hard to figure out why your attempt doesn't work, because you
haven't provided a reproducible example, which is easily done.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 6/12/14 7:47 PM, MKN matthewnewl...@gauge.com.au wrote:
I have tried multiple different methods to figure out how to get a date
axis
of my preference (start date of each month). Any assistance would be
appreciated.
The below section is not producing a date axis:
plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Emax),xaxt=n)
*xlabels-(strptime(totaldays$totaldays,%Y-%m-%d,tz=))
xlabels-xlabels[xlabels$mday==1]
axis.Date(1,at=xlabels,format=%b-%Y)*
The above section is used in the following plot:
par(mfrow=c(4,1))
par(mar=c(0.8,0,0,0))
par(oma=c(2,4.5,3,2))
plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Dmax),xaxt=n)
points(D$date,D$TSP,col=Dcol,type=p,pch=16,lwd=2,xaxt=n)
text(mdate,Dmax-20,labels=DM1',cex=1)
legend(topright,pch=16,cex=0.8,col=colLegend,
legend=ALegend)
abline(h=Target,col=red,lwd=2)
plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Bmax),xaxt=n)
points(B$date,B$TSP,col=Bcol,type=p,pch=16,lwd=2,xaxt=n)
text(mdate,Bmax-20,labels=DM2,cex=1)
abline(h=Target,col=red,lwd=2)
plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Amax),xaxt=n)
points(A$date,A$TSP,col=Acol,type=p,pch=16,lwd=2,xaxt=n)
text(mdate,Amax-20,labels=DM3,cex=1)
abline(h=Target,col=red,lwd=2)
plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Emax),xaxt=n)
*xlabels-(strptime(totaldays$totaldays,%Y-%m-%d,tz=))
xlabels-xlabels[xlabels$mday==1]
axis.Date(1,at=xlabels,format=%b-%Y)*
points(E$date,E$TSP,col=Ecol,type=p,pch=16,lwd=2)
text(mdate,Emax-20,labels=DM4',cex=1)
abline(h=Target,col=red,lwd=2)
title(main=Amain, ylab=Labely,
outer=TRUE,cex.lab=1, cex.main=1.5)
Below is an excerpt of the total days data:
totaldays y
1 2013-11-01 1
2 2013-11-02 1
3 2013-11-03 1
4 2013-11-04 1
5 2013-11-05 1
6 2013-11-06 1
7 2013-11-07 1
8 2013-11-08 1
9 2013-11-09 1
10 2013-11-10 1
--
View this message in context:
http://r.789695.n4.nabble.com/no-x-label-using-axis-Date-tp4692034.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] data format setting
Thanks dennis, It worked but I had to do some simple modifications to get to the ultimate format. Now I have a list in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000... some columns had 365 rows and some 366. those with 365 rows had their 366 row as NA. Now I want to apply approx. command to interpolate 366 values to 365, but when I apply approx command I gives out something which is with $x and $y, and frankly speaking it messed up everything. Is their a way that i do it neatly without getting the format deteriorated? In any way, thank-you very much indeed. Eliza Date: Fri, 13 Jun 2014 11:11:37 -0700 Subject: Re: [R] data format setting From: djmu...@gmail.com To: eliza_bo...@hotmail.com Hi: Maybe something like this: library(reshape2) L - split(DF, DF$year) L2 - llply(L, function(d) dcast(d, month + day ~ year, value.var = discharge)) Obviously untested, so caveat emptor. The idea is to use the dcast function to reshape the data from long to wide format within year. Dennis On Fri, Jun 13, 2014 at 8:55 AM, eliza botto eliza_bo...@hotmail.com wrote: Dear R family, I hope you all be doing great. I have a dataset of following format. The data file is of the following format. st year month day discharge 1 A 2004 1 1 6.752828 2 A 2004 1 2 7.602053 3 A 2004 1 3 5.583619 4 A 2004 1 4 5.019562 5 A 2004 1 5 4.804489 6 A 2004 1 6 4.363541 7 A 2004 1 7 3.801333 8 A 2004 1 8 3.455991 9 A 2004 1 9 3.402634 10 A 2004 1 10 3.250693 .. .. continue .. .. st year month day discharge 2AY 196710 3 0.56 20001AY 196710 4 0.56 20002AY 196710 5 0.48 20003AY 196710 6 0.56 20004AY 196710 7 0.48 20005AY 196710 8 0.40 20006AY 196710 9 0.40 20007AY 196710 10 0.56 20008AY 196710 11 0.56 20009AY 196710 12 0.65 20010AY 196710 13 0.85 you can see that there are five columns. The first column has the name of the station. I want to split the data w.r.t the names of the stations. Each station has data for certain years. for example A has data for years from 2004 to 2010 and for AY its from 1967 to 2000.similarly for other years there is data for different number of years. I want to make a list of matrices each containing the data for that station in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000 each column should have 365 to 366 values depending on whether there is a leap year or not. obviously for non-leap years 366th row should be an NA. kindly help me on it. Thankyou very much in advance. Eliza [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error message in varimp function: Party package
Dear R community, Iâm trying to use the âpartyâ package to view variables that are important in my data frame (FIM_RS4). My data has a combination of ordinal, nominal, categorical, and continuous data. Iâve written my code below, with my predictor variables being 2:36, and my response is 37 (Mussel). If anyone could point me in the direction of how to get rid of this error message, I would very much appreciate the guidance. FIM_RS4-read.csv(FIM_RS_4.csv, header=TRUE, stringsAsFactors=T) #all ordinal variables have been ordered in my complete code as well mycontrols - cforest_control(ntree=5000,mtry=4, minsplit=5) cforest.rf - cforest(as.factor(Mussel)~., data=FIM_RS4[,c(37,2:36)], controls=mycontrols) # indata=FIM_RS4[,2:36] # rf.model=cforest.rf temp - varimp.function(indata=FIM_RS4[,2:6], rf.model=cforest.rf,num.iterations=5, seedset = 2000,condition.varimp = T,write_csv=F) Ã Error in model@fit(data, ...) : error code 1 from Lapack routine 'dgesdd' #The reason Iâm assigning a varimp.function to temp is to run iterations in the next piece of code. #Same error occurs if I do not assign the iteration function varimp.function to temp, and simply do: varimp(cforest.rf, conditional = TRUE) Ã Error in model@fit(data, ...) : error code 1 from Lapack routine 'dgesdd' Many thanks in advance, Rox [[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] data format setting
As always, you are requested to post in plain text and to provide a reproducible example. Messed things up is quite vague. FWIW: In general, processing in sequence is best done BEFORE you cast your data to wide format. --- 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. On June 13, 2014 12:46:01 PM MDT, eliza botto eliza_bo...@hotmail.com wrote: Thanks dennis, It worked but I had to do some simple modifications to get to the ultimate format. Now I have a list in the following format $A 2004 200520062007200820092010 .. ... ... .. ... $AY 1967 19682000... some columns had 365 rows and some 366. those with 365 rows had their 366 row as NA. Now I want to apply approx. command to interpolate 366 values to 365, but when I apply approx command I gives out something which is with $x and $y, and frankly speaking it messed up everything. Is their a way that i do it neatly without getting the format deteriorated? In any way, thank-you very much indeed. Eliza Date: Fri, 13 Jun 2014 11:11:37 -0700 Subject: Re: [R] data format setting From: djmu...@gmail.com To: eliza_bo...@hotmail.com Hi: Maybe something like this: library(reshape2) L - split(DF, DF$year) L2 - llply(L, function(d) dcast(d, month + day ~ year, value.var = discharge)) Obviously untested, so caveat emptor. The idea is to use the dcast function to reshape the data from long to wide format within year. Dennis On Fri, Jun 13, 2014 at 8:55 AM, eliza botto eliza_bo...@hotmail.com wrote: Dear R family, I hope you all be doing great. I have a dataset of following format. The data file is of the following format. st year month day discharge 1 A 2004 1 1 6.752828 2 A 2004 1 2 7.602053 3 A 2004 1 3 5.583619 4 A 2004 1 4 5.019562 5 A 2004 1 5 4.804489 6 A 2004 1 6 4.363541 7 A 2004 1 7 3.801333 8 A 2004 1 8 3.455991 9 A 2004 1 9 3.402634 10 A 2004 1 10 3.250693 .. .. continue .. .. st year month day discharge 2AY 196710 3 0.56 20001AY 196710 4 0.56 20002AY 196710 5 0.48 20003AY 196710 6 0.56 20004AY 196710 7 0.48 20005AY 196710 8 0.40 20006AY 196710 9 0.40 20007AY 196710 10 0.56 20008AY 196710 11 0.56 20009AY 196710 12 0.65 20010AY 196710 13 0.85 you can see that there are five columns. The first column has the name of the station. I want to split the data w.r.t the names of the stations. Each station has data for certain years. for example A has data for years from 2004 to 2010 and for AY its from 1967 to 2000.similarly for other years there is data for different number of years. I want to make a list of matrices each containing the data for that station in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000 each column should have 365 to 366 values depending on whether there is a leap year or not. obviously for non-leap years 366th row should be an NA. kindly help me on it. Thankyou very much in advance. Eliza [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] no x label using axis.Date
Dates are a problem in any language because they are irregular As you have not provided a reproducible example here are 2 solutions totaldays - seq.Date(as.Date(2013-11-01), by = 7, length = 10) str(totaldays) Date[1:10], format: 2013-11-01 2013-11-08 2013-11-15 2013-11-22 2013-11-29 2013-12-06 2013-12-13 2013-12-20 2013-12-27 2014-01-03 plot(totaldays, rep(1,10), axes = F) axis(1, at = seq.Date(as.Date(2013-11-01), by = month, length = 3), labels = format(seq.Date(as.Date(2013-11-01), by = month, length = 3), %b %Y)) axis(2, at = seq(0,2,0.5), labels = paste(seq(0,2,0.5))) box() library(lattice) xyplot(rep(1,10)~ totaldays, scales = list(x= list(at = seq.Date(as.Date(2013-11-01), by = month, length = 3), labels = format(seq.Date(as.Date(2013-11-01), by = month, length = 3), %b %Y If you have more data you may get away without having to format the xaxis in lattice. xyplot(rep(1,10)~ totaldays, xlim =c(as.Date(2013-10-01), as.Date(2014-02-01)) ) HTH Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of MKN Sent: Friday, 13 June 2014 12:47 To: r-help@r-project.org Subject: [R] no x label using axis.Date I have tried multiple different methods to figure out how to get a date axis of my preference (start date of each month). Any assistance would be appreciated. The below section is not producing a date axis: plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Emax),xaxt=n) *xlabels-(strptime(totaldays$totaldays,%Y-%m-%d,tz=)) xlabels-xlabels[xlabels$mday==1] axis.Date(1,at=xlabels,format=%b-%Y)* The above section is used in the following plot: par(mfrow=c(4,1)) par(mar=c(0.8,0,0,0)) par(oma=c(2,4.5,3,2)) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Dmax),xaxt=n) points(D$date,D$TSP,col=Dcol,type=p,pch=16,lwd=2,xaxt=n) text(mdate,Dmax-20,labels=DM1',cex=1) legend(topright,pch=16,cex=0.8,col=colLegend, legend=ALegend) abline(h=Target,col=red,lwd=2) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Bmax),xaxt=n) points(B$date,B$TSP,col=Bcol,type=p,pch=16,lwd=2,xaxt=n) text(mdate,Bmax-20,labels=DM2,cex=1) abline(h=Target,col=red,lwd=2) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Amax),xaxt=n) points(A$date,A$TSP,col=Acol,type=p,pch=16,lwd=2,xaxt=n) text(mdate,Amax-20,labels=DM3,cex=1) abline(h=Target,col=red,lwd=2) plot(totaldays$totaldays,totaldays$y,type=n,ylim=c(0,Emax),xaxt=n) *xlabels-(strptime(totaldays$totaldays,%Y-%m-%d,tz=)) xlabels-xlabels[xlabels$mday==1] axis.Date(1,at=xlabels,format=%b-%Y)* points(E$date,E$TSP,col=Ecol,type=p,pch=16,lwd=2) text(mdate,Emax-20,labels=DM4',cex=1) abline(h=Target,col=red,lwd=2) title(main=Amain, ylab=Labely, outer=TRUE,cex.lab=1, cex.main=1.5) Below is an excerpt of the total days data: totaldays y 1 2013-11-01 1 2 2013-11-02 1 3 2013-11-03 1 4 2013-11-04 1 5 2013-11-05 1 6 2013-11-06 1 7 2013-11-07 1 8 2013-11-08 1 9 2013-11-09 1 10 2013-11-10 1 -- View this message in context: http://r.789695.n4.nabble.com/no-x-label-using-axis-Date-tp4692034.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] Add points to subplots
On Fri, 13 Jun 2014 12:02:34 PM Luca Cerone wrote: Thanks David, I already did this, but in case the code gets updated I will have to re-add the annotation, which I do not think it is ideal. I was just wondering if there is an easy solution to this. Thanks a lot for the help, Hi Luca, I decided to try using the split.screen function, which I had previously tried without getting what I wanted. This example: split.screen(c(2,1)) screen(1) screen() plot(1:7) screen(2) plot(1:6) screen(1,FALSE) screen() points(x=7:1,y=1:7,col=red) when run with no graphics devices currently open, seems to work and do what you want. When I first tried it on an open device (x11) it would not switch screens properly, remaining on the top (first) screen despite calls to screen() that reported that the screen had changed. I tried this after opening a png device and it also worked correctly, so I think that if you are careful to initialize the graphics device immediately before split.screen it should do it. 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] barp {plotrix} Start bars at 0 with a vector of positive values
Hi Jim, I tried your fix. This one works: barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) This one fails: barp(c(2,3,4,5,6,7,8)) Regards, Pascal On Fri, Jun 13, 2014 at 7:09 PM, Jim Lemon j...@bitwrit.com.au wrote: On Fri, 13 Jun 2014 06:50:59 PM Pascal Oettli wrote: Dear list, Please consider the following example: library(plotrix) barp(c(2,3,4,5,6,7,8), ylim=c(-10,10)) How to force the bars to start at 0? I could not find the way to do it. Hi Pascal, You are right, I had not considered the idea of setting a negative y limit when there were no negative values. I think you can fix this by changing: negy-any(height0,na.rm=TRUE) to negy-any(height0,na.rm=TRUE) || ylim[1] 0 I'll check this later (I'm just going out) and if it doesn't work, I'll send another email with something that does. Jim -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data format setting
Hi Eliza To me it seems like that you're not thinking before you messing about with the data before an analysis. The years with data for 366 days is leap years. It happens every fourth year and the extra day falls on the 29th of februar. I guess it is the results from the dcast function that screws things up to make you believe that it's day number 366. The best thing to do is to do your analysis on the complete data with some missing values for February 29th between leap years. Or you can discard the leap year day for leap years and do the analysis for all years of 365 days. What is the rationale by imputing missing data using the approx function? I mean the no leap year has only 365 days. If you for some unknown reasons you want to fill in value for the NAs one natural way is to substitute the NAs on February 29th by the mean of the values on February 28th and Marts 1st. I think there is a na.approx function for that in some package (perhaps zoo). Other metods are available in R: google for R + impute. Best Regards Frede Sendt fra Samsung mobil Oprindelig meddelelse Fra: eliza botto Dato:13/06/2014 20.48 (GMT+01:00) Til: r-help@r-project.org Emne: Re: [R] data format setting Thanks dennis, It worked but I had to do some simple modifications to get to the ultimate format. Now I have a list in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000... some columns had 365 rows and some 366. those with 365 rows had their 366 row as NA. Now I want to apply approx. command to interpolate 366 values to 365, but when I apply approx command I gives out something which is with $x and $y, and frankly speaking it messed up everything. Is their a way that i do it neatly without getting the format deteriorated? In any way, thank-you very much indeed. Eliza Date: Fri, 13 Jun 2014 11:11:37 -0700 Subject: Re: [R] data format setting From: djmu...@gmail.com To: eliza_bo...@hotmail.com Hi: Maybe something like this: library(reshape2) L - split(DF, DF$year) L2 - llply(L, function(d) dcast(d, month + day ~ year, value.var = discharge)) Obviously untested, so caveat emptor. The idea is to use the dcast function to reshape the data from long to wide format within year. Dennis On Fri, Jun 13, 2014 at 8:55 AM, eliza botto eliza_bo...@hotmail.com wrote: Dear R family, I hope you all be doing great. I have a dataset of following format. The data file is of the following format. st year month day discharge 1 A 2004 1 1 6.752828 2 A 2004 1 2 7.602053 3 A 2004 1 3 5.583619 4 A 2004 1 4 5.019562 5 A 2004 1 5 4.804489 6 A 2004 1 6 4.363541 7 A 2004 1 7 3.801333 8 A 2004 1 8 3.455991 9 A 2004 1 9 3.402634 10 A 2004 1 10 3.250693 .. .. continue .. .. st year month day discharge 2AY 196710 3 0.56 20001AY 196710 4 0.56 20002AY 196710 5 0.48 20003AY 196710 6 0.56 20004AY 196710 7 0.48 20005AY 196710 8 0.40 20006AY 196710 9 0.40 20007AY 196710 10 0.56 20008AY 196710 11 0.56 20009AY 196710 12 0.65 20010AY 196710 13 0.85 you can see that there are five columns. The first column has the name of the station. I want to split the data w.r.t the names of the stations. Each station has data for certain years. for example A has data for years from 2004 to 2010 and for AY its from 1967 to 2000.similarly for other years there is data for different number of years. I want to make a list of matrices each containing the data for that station in the following format $A 2004200520062007200820092010 .. ... ... .. ... $AY 196719682000 each column should have 365 to 366 values depending on whether there is a leap year or not. obviously for non-leap years 366th row should be an NA. kindly help me on it. Thankyou very much in advance. Eliza [[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. [[alternative HTML version deleted]]
Re: [R] barp {plotrix} Start bars at 0 with a vector of positive values
On Sat, 14 Jun 2014 12:57:12 PM you wrote:
Hi Jim,
I tried your fix.
This one works:
barp(c(2,3,4,5,6,7,8), ylim=c(-10,10))
This one fails:
barp(c(2,3,4,5,6,7,8))
Regards,
Pascal
Hi Pascal,
Right again. This seems to work for both and I think handles the problem
correctly:
if(is.null(ylim)) {
negy-any(height0,na.rm=TRUE)
if(negy) miny-min(height,na.rm=TRUE)*1.05
else miny-ifelse(ylog,min(height)/10,0)
ylim-c(miny,max(height,na.rm=TRUE)*1.05)
}
else {
miny-ylim[1]
negy-miny0
}
Remove the line:
negy-any(height0,na.rm=TRUE)
and modify the succeeding conditional clause as above. This will appear
in version 3.5-8.
Jim
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Re: [R] I have a question
On Sat, 14 Jun 2014 10:59:07 AM 오건희 wrote: Hi I got a wrong graph when I typed qplot(names(termFrequency),termFrequency, geom=bar, xlab=Terms,stat=bin)+coord_flip() The frequency weren't displayed on the graph. what's the problem? I attached the screen shot to this e-mail. Hi 오건희 , You seem to be supplying a character vector - names(termFrequency) - as the x argument to qplot. Making a wild guess that these names are actually numbers, perhaps: as.numeric(names(termFrequency)) will work. If the wild guess was wrong, perhaps you want to specify termFrequency as the first argument. When something like this happens, it is often helpful to explicitly name your arguments: qplot(x=names(termFrequency),y=termFrequency, geom=bar, xlab=Terms,stat=bin)+coord_flip() This will sometimes give a meaningful error message that helps to identify what has gone wrong. 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-es] p values con LMER
Para obtener el pvalue de un t-test, lo que puedes hacer es (1-pt(abs(tval), df=dftval))*2 donde tval es el valor de tu estadístico t (el que te sale en el summary) y dftval son los grados de libertad del test. La discusión sobre los p-valores en los modelos mixtos tiene que ver con cómo estimar adecuadamente los grados de libertad. Una vez decidas cuántos grados de libertad tienes, ya lo tienes solucionado ... ;-) Saludos, Marcelino El 13/06/2014 14:05, José Luis Cañadas escribió: Existe discusión sobre el uso de los p-valores en modelos mixtos. Como se ha dicho antes, para mi lo más adecuado es comparar modelos mediante la función anova. Por Internet se puede encontrar un buen libro de Douglas Bates y en español, busca modelos mixtos con R de Luis Cayuela, enfocado hacia ecologÃa, pero está muy bien El 13/06/2014 14:00, Jorge I Velez jorgeivanve...@gmail.com escribió: Hola Miguel, 2014-06-13 21:50 GMT+10:00 Miguel Lázaro lazar...@yahoo.es: Hola Manuel lo he tratado de hacer pero me sale Error: unexpected string constante in: Creo que te falto una anova(a,as,test=Chisq) ^ debe ser anova(a, as, test = Chisq) Saludos, Jorge.- no tengo ni idea de por qué... Me resulta alucinante no poder contar ya con pvals.fnc. ¿Será imposible hacerse con ello? Saludos, Miguel El vie, 13/6/14, Manuel Azcárate mazcarategar...@gmail.com escribió: Asunto: Re: [R-es] p values con LMER Para: Miguel Lázaro lazar...@yahoo.es CC: r-help-es@r-project.org Fecha: viernes, 13 de junio, 2014 13:21 Hola Miguel, Aunque algo más arduo que algún paquete que lo calcule directamente, yo lo que hago es crear un modelo reducido sin la variable de la que quiero saber el pvalor y compararlos mediante un test anova. El valor obtenido por esta comparación puede utilizarse con el pvalor de esa variable. Por ejemplo: Lm1=lmer(rt_ln ~ (fre_ln * Z_nsize * frebase_ln + (1|word), data = x) Lm2= lmer(rt_ln ~ (Z_nsize * frebase_ln + (1|word), data = x) anova(Lm1,Lm2, test=Chisq) #Obtiens el pvalor de la variable fre_ln Lm3=lmer(rt_ln ~ (fre_ln * frebase_ln + (1|word), data = x) anova(Lm1,Lm3, test=Chisq) #Obtiens el pvalor de la variable Z_nsize Espero que te sea de utilidad, Un saludo Manuel El 13 de junio de 2014, 12:25, Miguel Lázaro lazar...@yahoo.es escribió: Hola a todos, querÃa preguntaros un medio para obtener los valores p usando lmer. He tratado con pvals.fnc, que es lo que me habÃan recomendado, pero por algún motivo no está ya disponible etc. Ésta es la función que tengo, pero da las t, sin los valores p. Aunque Baayen indica que valores por encima de 2 son significativos necesito saber las p. resultado = lmer(rt_ln ~ (fre_ln * Z_nsize * frebase_ln + (1|word) data = x) (abreviado) = print (resultado, c=F) Fixed effects: Estimate Std. Error t value (Intercept)6.640496 0.034490 192.54 fre_ln-0.046880 0.008278 -5.66 Z_nsize0.005787 0.0088490.65 frebase_ln-0.009938 0.006670 -1.49 Z_wlength 14.239570 20.1025360.71 Z_slength 0.011011 0.0066921.65 Z_TF 0.009903 0.0088011.13 Z_prodctvsufij-0.005079 0.009052 -0.56 Z_rootlenght -17.961932 25.420022 -0.71 Zaverageres_1 0.018265 0.0091951.99 Zrootlengthres_1 10.954681 15.5111460.71 Saludos, Miguel ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] p values con LMER
Sí, en efecto, aunque es muy extraño porque lo ahbía mirado. Me faltaba unas comillas. Ahora lo veo ahí bien claro. Como decís hay una enorme discusión sobre esto de los valores y supongo que si no los dan directamente en lmer será porque decidieron no hacerlo. En todo caso el tema de pvals.fnc, para lo que usamos la estadística a nivel usuario, estaba fenomenal. Utilizo el procedimiento que me habéis mostrado y quedo a la expectativa de que vuelva a estar disponible, o algo similar, lo anterior. ¡muchas gracias! Miguel El vie, 13/6/14, José Luis Cañadas canadasre...@gmail.com escribió: Asunto: Re: [R-es] p values con LMER Para: Jorge I Velez jorgeivanve...@gmail.com CC: Miguel Lázaro lazar...@yahoo.es, r-help-es r-help-es@r-project.org Fecha: viernes, 13 de junio, 2014 14:05 Existe discusión sobre el uso de los p-valores en modelos mixtos. Como se ha dicho antes, para mi lo más adecuado es comparar modelos mediante la función anova. Por Internet se puede encontrar un buen libro de Douglas Bates y en español, busca modelos mixtos con R de Luis Cayuela, enfocado hacia ecología, pero está muy bien El 13/06/2014 14:00, Jorge I Velez jorgeivanve...@gmail.com escribió: Hola Miguel, 2014-06-13 21:50 GMT+10:00 Miguel Lázaro lazar...@yahoo.es: Hola Manuel lo he tratado de hacer pero me sale Error: unexpected string constante in: Creo que te falto una anova(a,as,test=Chisq) ^ debe ser anova(a, as, test = Chisq) Saludos, Jorge.- no tengo ni idea de por qué... Me resulta alucinante no poder contar ya con pvals.fnc. ¿Será imposible hacerse con ello? Saludos, Miguel El vie, 13/6/14, Manuel Azcárate mazcarategar...@gmail.com escribió: Asunto: Re: [R-es] p values con LMER Para: Miguel Lázaro lazar...@yahoo.es CC: r-help-es@r-project.org Fecha: viernes, 13 de junio, 2014 13:21 Hola Miguel, Aunque algo más arduo que algún paquete que lo calcule directamente, yo lo que hago es crear un modelo reducido sin la variable de la que quiero saber el pvalor y compararlos mediante un test anova. El valor obtenido por esta comparación puede utilizarse con el pvalor de esa variable. Por ejemplo: Lm1=lmer(rt_ln ~ (fre_ln * Z_nsize * frebase_ln + (1|word), data = x) Lm2= lmer(rt_ln ~ (Z_nsize * frebase_ln + (1|word), data = x) anova(Lm1,Lm2, test=Chisq) #Obtiens el pvalor de la variable fre_ln Lm3=lmer(rt_ln ~ (fre_ln * frebase_ln + (1|word), data = x) anova(Lm1,Lm3, test=Chisq) #Obtiens el pvalor de la variable Z_nsize Espero que te sea de utilidad, Un saludo Manuel El 13 de junio de 2014, 12:25, Miguel Lázaro lazar...@yahoo.es escribió: Hola a todos, quería preguntaros un medio para obtener los valores p usando lmer. He tratado con pvals.fnc, que es lo que me habían recomendado, pero por algún motivo no está ya disponible etc. Ésta es la función que tengo, pero da las t, sin los valores p. Aunque Baayen indica que valores por encima de 2 son significativos necesito saber las p. resultado = lmer(rt_ln ~ (fre_ln * Z_nsize * frebase_ln + (1|word) data = x) (abreviado) = print (resultado, c=F) Fixed effects: Estimate Std. Error t value (Intercept) 6.640496 0.034490 192.54 fre_ln -0.046880 0.008278 -5.66 Z_nsize 0.005787 0.008849 0.65 frebase_ln -0.009938 0.006670 -1.49 Z_wlength 14.239570 20.102536 0.71 Z_slength 0.011011 0.006692 1.65 Z_TF 0.009903 0.008801 1.13 Z_prodctvsufij -0.005079 0.009052 -0.56 Z_rootlenght -17.961932 25.420022 -0.71 Zaverageres_1 0.018265 0.009195 1.99 Zrootlengthres_1 10.954681 15.511146 0.71 Saludos, Miguel ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es ___ R-help-es mailing list R-help-es@r-project.org
[R-es] Crear matriz binaria
Hola, Estoy tratando de crear una matriz cuadrada, binaria y no tengo idea por dónde comenzar. Los datos que tengo son patentes e inventores por patente, las patentes las identifico por su código y a los inventores por nombre y también los tengo codificados. Deseo hacer una matriz de nxn en donde los renglones son inventores de la A a la Z y columnas son los mismos inventores de la A a la Z, las entradas de la matriz corresponderán a 1 si el inventor A tiene una patente con el inventor K…L…M, cualquier otra situación será cero. Evidentemente la matriz debe tener ceros en la diagonal pues el inventor B no puede patentar algo con el inventor B. Pues bien, intenté hacerlo en excel con tablas dinámicas pero no me permite poner la misma variable (inventores) en renglones y columnas. Pensé que se puede hacer en R, pero no tengo idea por dónde empezar, nisiquiera cuál sería la mejor forma de ordenar los datos. Si pudieran ayudarme sugiréndome algún manual o algún ejemplo para comenzar, se los agradezco de antemano. Saludos. Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] Crear matriz binaria
Hola, Esta es una tabla dinámica que cree, en donde los renglones son los inventores y las columnas son patentes, las entradas de la matriz tienen 1 en aquellos inventores que participan por patente. https://www.dropbox.com/s/b8gx0hei6w91ko0/tabla_din.xlsx Esta es otra tabla es un arreglo parecido al de arriba, sólo que los renglones son patentes y las columnas son número de inventores, tienen 1 en donde participan inventores en la patente. https://www.dropbox.com/s/xpi0hescm1xm58c/tabla_din2.xlsx Este es otro arreglo donde sólo hay tres columnas, patentes, número de inventor e inventor. https://www.dropbox.com/s/rmmgiswmbppt68w/datos_col.csv El 13/06/2014, a las 10:27, Jorge I Velez jorgeivanve...@gmail.com escribió: Hola Rolando, Podrias enviarnos una submatriz de esos datos? Quizas tusdatos[1:10, 1:10]? Creo que es mas facil trabajar sobre ello. Gracias, Jorge.- 2014-06-14 1:10 GMT+10:00 Rolando Valdez rvald...@gmail.com: Hola, no encuentro el momento en el que se relacionan los inventores, es decir, lo que relaciona a dos o más inventores es la patente. Primero, ¿cómo debo ordenar los datos? Tengo 1813 número de patentes, es decir, la matriz es de dimension 1813 x 1813. Tengo varios órdenes en los datos, por ejemplo, el primer arreglo es por patente, en las columnas tengo “no. de patente”, “inventor 1”, “inventor 2”, “inventor 3”…… “inventor 13”, y en las filas tengo números de patente. En las entradas tengo nombre del inventor. Otro arreglo que tengo es mediante una tabla dinámica que hice en donde tengo por columnas “nombre de inventor”, “inventor 1”, “inventor 2”……. “ inventor 13”. En los renglones tengo nombres de inventor y en las entradas tengo 1 y 0. Saludos. El 13/06/2014, a las 09:14, Jorge I Velez jorgeivanve...@gmail.com escribió: Hola Rolando, Podrias adaptar lo siguiente: set.seed(10) X - matrix(sample(0:1, 100, replace = TRUE), ncol = 10) colnames(X) - LETTERS[1:10] X Y - X = 1 Y - apply(Y, 2, as, numeric) #boolean matrix rownames(Y) - rownames(X) Z - t(Y) %*% Y #adjacency matrix Z[Z = 1] - 1 diag(Z) - 0 Z Saludos, Jorge.- 2014-06-14 0:08 GMT+10:00 Rolando Valdez rvald...@gmail.com: Hola, Estoy tratando de crear una matriz cuadrada, binaria y no tengo idea por dónde comenzar. Los datos que tengo son patentes e inventores por patente, las patentes las identifico por su código y a los inventores por nombre y también los tengo codificados. Deseo hacer una matriz de nxn en donde los renglones son inventores de la A a la Z y columnas son los mismos inventores de la A a la Z, las entradas de la matriz corresponderán a 1 si el inventor A tiene una patente con el inventor K…L…M, cualquier otra situación será cero. Evidentemente la matriz debe tener ceros en la diagonal pues el inventor B no puede patentar algo con el inventor B. Pues bien, intenté hacerlo en excel con tablas dinámicas pero no me permite poner la misma variable (inventores) en renglones y columnas. Pensé que se puede hacer en R, pero no tengo idea por dónde empezar, nisiquiera cuál sería la mejor forma de ordenar los datos. Si pudieran ayudarme sugiréndome algún manual o algún ejemplo para comenzar, se los agradezco de antemano. Saludos. Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
