Re: [R] function to reduce resolution of a vector or matrix?
On Jun 3, 2011, at 17:42 , Carl Witthoft wrote: Hi, I feel dumb even asking, but isn't there an R function somewhere that I can use to reduce the resolution of a vector (or matrix) by summing terms in uniform blocks? That is, for a vector X, reduce it to some X.short as X.short[1]- sum(X[1:10]); X.short[2] - sum(X[11:20]), and so on. I did the following: X.short-colSums(matrix(X,16,2048/16)) # X is of length 2048 but surely there's already a function somewhere that does this in a more general case? And, my approach will get a bit painful for reducing a matrix in both dimensions. aggregate.ts, but then you need to convert to ts and back. I'd likely go for tapply(X, (seq_along(X) - 1) %/% N, sum) x [1] 1 1 0 -1 1 -1 0 0 1 -2 1 0 0 -1 1 as.numeric(aggregate(ts(x, frequency=3), 1, sum)) [1] 2 -1 1 -1 0 tapply(x, (seq_along(x) - 1) %/% 3, sum) 0 1 2 3 4 2 -1 1 -1 0 -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Can we prepare a questionaire in R
Is there a way to prepare a questionnaire in R like html forms whose data can be directly populated into 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] using stimulate(model) for parametric bootstrapping in lmer repeatabilities
Hi all,
I am currently doing a consistency analysis using an lmer model and
trying to use parametric bootstrapping for the confidence intervals.
My model is like this:
model-lmer(y~A+B+(1|C/D)+(1|E),binomial)
where E is the individual level for consistency analysis, A-D are
other fixed and random effects that I have to control for.
Following Nakagawa and Scheilzeth I can work out the repeatability
estimate using the following (as it is a binomial and the residual
variance is fixed at 1).
attr(lme4::VarCorr(model)$E, stddev)^2 /
(1*(pi^2)/3 + attr(lme4::VarCorr(model)$E, stddev)^2 +
attr(lme4::VarCorr(model)$C, stddev)^2 +
attr(lme4::VarCorr(model)$D, stddev)^2 )
My question is can I use stimulate(model) to generate values that I
can then use to do parametric bootstrap analysis and generate the
confidence intervals?
Something like this:
n-length(A)
niter-1000
y-matrix(nrow=n,ncol=niter*2)
for (i in 1:niter) {
y[,I(i*2-1):I(i*2)]-simulate(model)[,1] }
rvalues-numeric()
for (i in 1:niter) {
yboot-cbind(y[,I(i*2-1)],y[,I(i*2)])
mboot-lmer(y~A+B+(1|C/D)+(1|E),binomial)
rvalues[i]- attr(lme4::VarCorr(mboot)$E, stddev)^2/(1*(pi^2)/3 +
attr(lme4::VarCorr(mboot)$E, stddev)^2 +
attr(lme4::VarCorr(mboot)$C, stddev)^2 +
attr(lme4::VarCorr(mboot)$D, stddev)^2 )}
confidence.intervals-quantile(rvalues,c(0.05,0.95))
In the guide to lme4 it says that stimulate() generate simulations
based on the estimated fitted models (conditional on the estimated
values of both the random and fixed effects), which sounds like
exactly what I would need to generate values for parametric
bootstrapping but I can't find any examples of where anyone has done
this.
Any advice would be very much appreciated! Thank you very much.
Jenni
Jenni Sanderson
PhD student - Conflict and Cooperation in Vertebrate Societies
University of Exeter
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] determine frequencies in a matrix by row
here's the code to generate the matrix set.seed(1) types = c(1,2,3) n=5 p=1 pop.props = c(0.6,0.2,0.2) x=matrix(pop.props,nrow=n,ncol=length(pop.props), byrow=T) b=10 habs = rMultinom((x),b) print(habs) -- View this message in context: http://r.789695.n4.nabble.com/determine-frequencies-in-a-matrix-by-row-tp3581733p3581752.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] determine frequencies in a matrix by row
Hi, I have a matrix [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]131131123 2 [2,]131211122 1 [3,]122321112 1 [4,]321113121 2 [5,]112211311 2 and want to determine how many times 1s, 2s, and 3s occur per each row. I tried using 'table', but end up with frequencies for the whole matrix. Thanks in advance for your help -- View this message in context: http://r.789695.n4.nabble.com/determine-frequencies-in-a-matrix-by-row-tp3581733p3581733.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] About DCC-garch model...
Thank you all guys!! I have some other questionsFor garch model, for example, we have 10 time periods, and the function use MLE to get the parameters based on these 10 time periods. Then, the function calculates covariance matrix at each time period based on the estimated parameters. Is this right?? Moreover, does anyone know that whether the correlation matrix calculated by GARCH model is semidefinite positive? Thanks a lot!!! -- View this message in context: http://r.789695.n4.nabble.com/About-DCC-garch-model-tp3579140p3581851.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] determine frequencies in a matrix by row
Hi, Try apply function: apply(matrix,1,table) Regards M Le 08/06/11 08:23, the_big_kowalski a écrit : Hi, I have a matrix [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]131131123 2 [2,]131211122 1 [3,]122321112 1 [4,]321113121 2 [5,]112211311 2 and want to determine how many times 1s, 2s, and 3s occur per each row. I tried using 'table', but end up with frequencies for the whole matrix. Thanks in advance for your help -- View this message in context: http://r.789695.n4.nabble.com/determine-frequencies-in-a-matrix-by-row-tp3581733p3581733.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. -- Mohamed Lajnef,IE INSERM U955 eq 15# Pôle de Psychiatrie# Hôpital CHENEVIER # 40, rue Mesly # 94010 CRETEIL Cedex FRANCE # mohamed.laj...@inserm.fr # tel : 01 49 81 32 79 # Sec : 01 49 81 32 90 # fax : 01 49 81 30 99 # [[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] using stimulate(model) for parametric bootstrapping in lmer repeatabilities
Dear Jenni,
In the newest version of the doBy package there is a function called PBrefdist
(and PBrefdist.mer) for calculating the reference distribution of the
likelihood ratio statistic for comparing nested models. Looking into this
function may help you. Perhaps the functions PBmodcomp and BCmodcomp (and their
.mer methods) for calculating p-values based on parametric bootstrap can also
be of inspiration.
Regards
Søren
Fra: r-help-boun...@r-project.org [r-help-boun...@r-project.org] P#229; vegne
af Jenni Sanderson [jennifer.louise.sander...@googlemail.com]
Sendt: 8. juni 2011 09:56
Til: r-help@r-project.org; r-sig-mixed-mod...@r-project.org
Emne: [R] using stimulate(model) for parametric bootstrapping in lmer
repeatabilities
Hi all,
I am currently doing a consistency analysis using an lmer model and
trying to use parametric bootstrapping for the confidence intervals.
My model is like this:
model-lmer(y~A+B+(1|C/D)+(1|E),binomial)
where E is the individual level for consistency analysis, A-D are
other fixed and random effects that I have to control for.
Following Nakagawa and Scheilzeth I can work out the repeatability
estimate using the following (as it is a binomial and the residual
variance is fixed at 1).
attr(lme4::VarCorr(model)$E, stddev)^2 /
(1*(pi^2)/3 + attr(lme4::VarCorr(model)$E, stddev)^2 +
attr(lme4::VarCorr(model)$C, stddev)^2 +
attr(lme4::VarCorr(model)$D, stddev)^2 )
My question is can I use stimulate(model) to generate values that I
can then use to do parametric bootstrap analysis and generate the
confidence intervals?
Something like this:
n-length(A)
niter-1000
y-matrix(nrow=n,ncol=niter*2)
for (i in 1:niter) {
y[,I(i*2-1):I(i*2)]-simulate(model)[,1] }
rvalues-numeric()
for (i in 1:niter) {
yboot-cbind(y[,I(i*2-1)],y[,I(i*2)])
mboot-lmer(y~A+B+(1|C/D)+(1|E),binomial)
rvalues[i]- attr(lme4::VarCorr(mboot)$E, stddev)^2/(1*(pi^2)/3 +
attr(lme4::VarCorr(mboot)$E, stddev)^2 +
attr(lme4::VarCorr(mboot)$C, stddev)^2 +
attr(lme4::VarCorr(mboot)$D, stddev)^2 )}
confidence.intervals-quantile(rvalues,c(0.05,0.95))
In the guide to lme4 it says that stimulate() generate simulations
based on the estimated fitted models (conditional on the estimated
values of both the random and fixed effects), which sounds like
exactly what I would need to generate values for parametric
bootstrapping but I can't find any examples of where anyone has done
this.
Any advice would be very much appreciated! Thank you very much.
Jenni
Jenni Sanderson
PhD student - Conflict and Cooperation in Vertebrate Societies
University of Exeter
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Classifying boolean values
Thanks Sarah for the response; with the command str(echoknn.train) the coloumn class is a logi value (i think without any immagination that is a logical value ). So, how can I handle this type of data? Thanks a lot. P.S. Yes, is a course assignment and i was hoping to solve this problem (that i consider just a beginner problem) without asking my teacher . -- View this message in context: http://r.789695.n4.nabble.com/Classifying-boolean-values-tp3579993p3581980.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] library(SenoMineR)- Triangle Test Query
Vijayan, I cannot find an error in your code, but I had a look at the code of triangle.test -- unless I'm missing something, it contains a bug. If you study the way in which the matrix pref is updated, you find that the vector preference is compared to 1, 2 and 3 instead of X, Y and Z as it should be. That way, some of the non-diagonal entries of pref will always be zero, irrespective of the data, which does not make sense. I think it should work if you modify the code accordingly. Alternatively, a quick patch (untested!) might be to code preferences as 1, 2 and 3 instead of the letters (but I'm not sure whether this has any other implications). I CC the author of the function and maintainer of the package; he should correct me if needed or could otherwise update the code for the next release (I worked on SensoMineR 1.11). Hope this helps, Michael -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Vijayan Padmanabhan Sent: Saturday, June 04, 2011 9:21 To: r-help@r-project.org Subject: [R] library(SenoMineR)- Triangle Test Query Dear R Group I was trying to use the triangle.test function in SensoMineR and strangely i encounter a error in the output of preference matrix from the analysis. To illustrate, pl see the following dataframe of a design with the response and preference collected as shown below: design-structure(list(`Product X` = c(3, 1, 4, 2, 4, 2, 1, 3, 4, 2, 4, 2, 1, 3, 4, 2, 4, 2, 3, 1), `Product Y` = c(1, 1, 4, 4, 4, 3, 1, 1, 4, 4, 4, 3, 1, 1, 4, 4, 4, 3, 1, 1), `Product Z` = c(3, 2, 1, 2, 3, 3, 2, 3, 1, 2, 3, 3, 2, 3, 1, 2, 3, 3, 3, 2), Response = structure(c(1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L), .Label = c(X, Z), class = factor), Preference = structure(c(1L, 3L, 1L, 1L, 1L, 2L, 3L, 1L, 1L, 1L, 1L, 2L, 3L, 1L, 1L, 1L, 1L, 2L, 1L, 2L), .Label = c(X, Y, Z), class = factor)), .Names = c(Product X, Product Y, Product Z, Response, Preference), class = data.frame, row.names = c(Panelist1.Test1, Panelist1.Test2, Panelist2.Test1, Panelist2.Test2, Panelist3.Test1, Panelist3.Test2, Panelist4.Test1, Panelist4.Test2, Panelist5.Test1, Panelist5.Test2, Panelist6.Test1, Panelist6.Test2, Panelist7.Test1, Panelist7.Test2, Panelist8.Test1, Panelist8.Test2, Panelist9.Test1, Panelist9.Test2, Panelist10.Test1, Panelist10.Test2)) If you were to investigate the above dataframe, you would find that for the comparision of Product 2 Vs Product 3, the preference indicates product 3 is preferred over product 2 all the time. ## Read output from the following script to see what i mean above: subset(design,`Product X`==2`Product Y`==3`Product Z`==3) ##Output of above would be: . Product X Product Y Product Z Response Preference Panelist3.Test2 2 3 3X Y Panelist6.Test2 2 3 3X Y Panelist9.Test2 2 3 3X Y However when I analyse the design with the answers and preferences using the following script, I get the $pref output which shows that product 2 is preferred over 3 all the time. Can somebody explain what is wrong in my script? answer-as.vector(design$Response) preference-as.vector(design$Preference) triangle.test (design[,1:3], answer,preference) ##$pref output from the triangle.test function shows as follows: $pref 1 2 3 4 1 0 0 0 0 2 4 0 3 0 3 0 0 0 0 4 0 0 0 0 Any help in helping me identify what is going wrong here would be highly appreciated. Regards Vijayan Padmanabhan [[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] MBA the h value?
Hello everybody, I am trying different ways of interpolating a surface. Now, I am trying to interpolate these values using the MBA package. In order to do it I am using the surface option of the package. mba.surf(xyz, 20, 20, n = 1, m = 1, h = 8, extend=FALSE, sp=FALSE, ...) The function is giving me some results but not reliable at all. Maybe it's due to the n,m and h values which are not corresponding to my data. I don't understand what should be the value of h. Even I read the article cited in the pdf. Someone could maybe help me.Also what should be the value of n and m. My data are pasted below corresponding to xyz. Thank you very much Regards Pierre 4.00 45 -83.94753324 4.00 50 -115.8679098 4.00 55 -134.787 4.00 60 -144.6370393 4.00 65 -148.7164562 4.00 70 -149.4758468 4.00 75 -148.641352 4.05 45 -104.1007536 4.05 50 -128.2620683 4.05 55 -141.8810486 4.05 60 -148.3151116 4.05 65 -150.2664062 4.05 70 -149.7559503 4.05 75 -148.1990977 4.10 45 -118.9805984 4.10 50 -136.721655 4.10 55 -146.0525858 4.10 60 -149.7327347 4.10 65 -149.9930389 4.10 70 -148.4927918 4.10 75 -146.3873601 4.15 45 -129.467342 4.15 50 -141.998465 4.15 55 -147.8625806 4.15 60 -149.343257 4.15 65 -148.2761684 4.15 70 -146.0179853 4.15 75 -143.5069876 4.20 45 -136.3899382 4.20 50 -144.6996352 4.20 55 -147.7834847 4.20 60 -147.526368 4.20 65 -145.433249 4.20 70 -142.6081123 4.20 75 -139.8085943 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Decision Trees /Decision Analysis with R?
Hello, this question is a bit out of the blue. I am a big R fan and user and in my new job I do some decision modeling (mostly health economics). For that decision trees are often used (I guess the most classic example is the investment decision A, B, and C with different probabilities, what is the expected payoff). We use a specialized software called TreeAge that some might know. The basic setup of such simulations is actually very simple and I guess useful in many fields. So I was wondering whether there is already a package out there in R that is doing such a thing? Thanks for any hints! Best, Stefan PS (By decision tree I don't mean cluster-like analysis of a data set splitting by identifying decision nods, but the other way around: I have decision nodes, what is my expected outcome.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 and DBSCAN
Dear Paco, I tried dbscan on my computer with method=hybrid and a 155000*3 data matrix and it works. Needs some time though. (You can track the progress using something like countmode=c(1:10,100,1000,1,10).) Note that for some reason I don't exactly understand, it takes *much* longer for 1-dimensional data (I need to look into this), so if you tried only 1-d data yet, it may be worth a try to do the whole thing with the full 3-d dataset. So I'm not sure what goes wrong on your side. Perhaps look at str(sst2) in order to make sure that it is what you think it is. I can't advise you on how precisely to take longitude and latitude into account because this depends on your application and would probably require professional statistical advisory that is much more than just R-help. Note however that dbscan treats all variables equally. Best wishes, Christian On Tue, 7 Jun 2011, Paco Pastor wrote: Hello Christian Thanks for answering. Yes, I have tried dbscan from fpc but I'm still stuck on the memory problem. Regarding your answer, I'm not sure which memory parameter should I look at. Following is the code I tried with dbscan parameters, maybe you can see if there is any mistake. sstdat=read.csv(sst.dat,sep=;,header=F,col.names=c(lon,lat,sst)) library(fpc) sst1=subset(sstdat, sst50) sst2=subset(sst1, lon-6) sst2=subset(sst2, lon40) sst2=subset(sst2, lat46) dbscan(sst2$sst, 0.1, MinPts = 5, scale = FALSE, method = c(hybrid), seeds = FALSE, showplot = FALSE, countmode = NULL) Error: no se puede ubicar un vector de tamaño 858.2 Mb head(sst2) lon lat sst 1257 35.18 24.98 26.78 1258 35.22 24.98 26.78 1259 35.27 24.98 26.78 1260 35.31 24.98 26.78 1261 35.35 24.98 26.78 1262 35.40 24.98 26.85 In this example I only apply dbscan to temperature values, not lon/lat, so eps parameter is 0.1. As it is a gridded data set any point is surrounded by eight data points, then I thought that at least 5 of the surrounding points should be within the reachability distance. But I'm not sure I'm getting the right approach by only considering temperature value, maybe then I'm missing spatial information. How should I deal with longitude and latitude data? dimensions of sst2 are: 152243 rows x 3 columns Thanks again El 03/06/2011 18:24, Christian Hennig escribió: Have you considered the dbscan function in library fpc, or was it another one? dbscan in fpc doesn't have a distance parameter but several options, one of which may resolve your memory problem (look up the documentation of the memory parameter). Using a distance matrix for hundreds of thousands of points is a recipe for disaster (memory-wise). I'm not sure whether the function that you used did that, but dbscan in fpc can avoid it. It is true that dbscan requires tuning constants that the user has to provide. There is unfortunately no general rule how to do this; it would be necessary to understand the method and the meaning of the constants, and how this translates into the requirements of your application. You may try several different choices and do some cluster validation to see what works, but I can't explain this in general terms easily via email. Hope this helps at least a bit. Best regards, Christian On Fri, 3 Jun 2011, Paco Pastor wrote: Hello everyone, When looking for information about clustering of spatial data in R I was directed towards DBSCAN. I've read some docs about it and theb new questions have arisen. DBSCAN requires some parameters, one of them is distance. As my data are three dimensional, longitude, latitude and temperature, which distance should I use? which dimension is related to that distance? I suposse it should be temperature. How do I find such minimum distance with R? Another parameter is the minimum number of points neded to form a cluster. Is there any method to find that number? Unfortunately I haven't found. Searching thorugh Google I could not find an R example for using dbscan in a dataset similar to mine, do you know any website with such kind of examples? So I can read and try to adapt to my case. The last question is that my first R attempt with DBSCAN (without a proper answer to the prior questions) resulted in a memory problem. R says it can not allocate vector. I start with a 4 km spaced grid with 779191 points that ends in approximately 30 rows x 3 columns (latitude, longitude and temperature) when removing not valid SST points. Any hint to address this memory problem. Does it depend on my computer or in DBSCAN itself? Thanks for the patience to read a long and probably boring message and for your help. -- --- Francisco Pastor Meteorology department, Instituto Universitario CEAM-UMH http://www.ceam.es --- mail: p...@ceam.es skype: paco.pastor.guzman Researcher ID: http://www.researcherid.com/rid/B-8331-2008 Cosis profile: http://www.cosis.net/profile/francisco.pastor --- Parque Tecnologico,
[R] plotting boxplot based on frequency/count data
Hi all, I have a huge dataset of values and I have precalculated outside of R the frequency/count distribution (as for example counts returned by hist function or the output of table function). For example, x-hist(c(1,2,2,2,1,3), breaks=0:3, plot=F) x $breaks [1] 0 1 2 3 $counts [1] 2 3 1 Is there any function to calculate boxplot based on the counts? I guess I could expand the count vector to a values set and run boxplot but the problem is that I have a lot of data (300,000^2). Another way could be to calculate median and quartiles myself and custom draw the boxplot but seems complicated. Do you happen to know some more automated way, maybe a function in R? Many thanks for the help. Best regards, Ioannis [[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] 2.13 version doesn't load packages
Hi, I just installed the 2.13 version for Mac (without uninstalling the previous versions). I transferred the folder library containing the packages I normally use to the 2.13 library folder. When I require any package this is what I get: require(ape) Loading required package: ape Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared object '/Library/Frameworks/R.framework/Versions/2.13/Resources/library/gee/libs/x86_64/gee.so': dlopen(/Library/Frameworks/R.framework/Versions/2.13/Resources/library/gee/libs/x86_64/gee.so, 6): Library not loaded: /Library/Frameworks/R.framework/Versions/2.12/Resources/lib/libRblas.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.13/Resources/library/gee/libs/x86_64/gee.so Reason: image not found However when type library the packages are there. Any hints? Many thanks Mariana -- Mariana Varela, DVM, PhD MRC-University of Glasgow Centre for Virus Research Institute of Infection, Immunity and Inflammation College of Medical, Veterinary and Life Sciences Garscube Estate, 464 Bearsden Road Henry Wellcome Building, room 436 Glasgow, G61 1QH Scotland (UK) Phone: +44 (0) 141 330 2196 [[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] Can we prepare a questionaire in R
Date: Wed, 8 Jun 2011 12:37:33 +0530 From: ammasamri...@gmail.com To: r-help@r-project.org Subject: [R] Can we prepare a questionaire in R Is there a way to prepare a questionnaire in R like html forms whose data can be directly populated into R? I've started to use Rapache although Rserve would also be an option. When installed on server, your can point your html form to an rhtml page and get the form variables in R just as with other web languages. For writing html output, I had been using R2HTML ( see code excerpt below from rhtml page). I had also found Cairo works ok if you don't need X-11 for anything. For your specific situation however, it may be easier to just use whatever you already have and just use R for the data analysis. When a request for a results report is made, send that to Rapache for example. I would mention that I have gone to running two versions of Apache, one with R and one with PHP, to allow for security and easier development( I can write the php to fail nicely if the R apache server is not up and no new security issues are exposed). library(Cairo) library(R2HTML) library(RColorBrewer) You can of course also generate html from normal R commands, or for that matter bash scripts etc. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] determine frequencies in a matrix by row
If the result is limited to just the counts of 1,2 and 3, you can do
the following:
habs - matrix(sample(1:3, 50, TRUE, prob=c(.6,.2,.2)),5,10)
habs
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]132113111 2
[2,]123131111 1
[3,]123231211 3
[4,]313122111 2
[5,]112211313 2
result - apply(habs, 1, function(x){
+ c('1' = sum(x == 1), '2' = sum(x==2), '3' = sum(x == 3))
+ })
cbind(habs, t(result))
1 2 3
[1,] 1 3 2 1 1 3 1 1 1 2 6 2 2
[2,] 1 2 3 1 3 1 1 1 1 1 7 1 2
[3,] 1 2 3 2 3 1 2 1 1 3 4 3 3
[4,] 3 1 3 1 2 2 1 1 1 2 5 3 2
[5,] 1 1 2 2 1 1 3 1 3 2 5 3 2
On Wed, Jun 8, 2011 at 2:23 AM, the_big_kowalski bkowal...@csumb.edu wrote:
Hi,
I have a matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 3 1 1 3 1 1 2 3 2
[2,] 1 3 1 2 1 1 1 2 2 1
[3,] 1 2 2 3 2 1 1 1 2 1
[4,] 3 2 1 1 1 3 1 2 1 2
[5,] 1 1 2 2 1 1 3 1 1 2
and want to determine how many times 1s, 2s, and 3s occur per each row.
I tried using 'table', but end up with frequencies for the whole matrix.
Thanks in advance for your help
--
View this message in context:
http://r.789695.n4.nabble.com/determine-frequencies-in-a-matrix-by-row-tp3581733p3581733.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] RgoogleMaps Axes
Date: Tue, 7 Jun 2011 09:50:10 -0700 From: egregory2...@yahoo.com To: r-help@r-project.org Subject: [R] RgoogleMaps Axes R Help, I posted a question on StackOverflow yesterday regarding an issue I've been having with the RgoogleMaps packages' displaying of axes. Here is the text of that submission: http://stackoverflow.com/questions/6258408/rgooglemaps-axes I can't find any documentation of the following problem I'm having with the axis labels in RGoogleMaps: library(RgoogleMaps) datas -structure(list(LAT =c(37.875,37.925,37.775,37.875,37.875), LON =c(-122.225,-122.225,-122.075,-122.075,-122.025)), .Names=c(LAT,LON),class=data.frame, row.names =c(1418L,1419L,1536L,1538L,1578L)) # Get bounding box. boxt -qbbox(lat =datas$LAT,lon =datas$LON) MyMap-GetMap.bbox(boxt$lonR,boxt$latR,destfile =Arvin12Map.png, maptype =mobile) PlotOnStaticMap(MyMap,lat =datas$LAT,lon =datas$LON, axes =TRUE,mar =rep(4,4)) I haven't gotten too far as I had to download and install proj4 and rgdal but I did get a transofrm related warning. Did you get warnings? MyMap-GetMap.bbox(boxt$lonR,boxt$latR,destfile =Arvin12Map.png,maptype =m obile) [1] http://maps.google.com/maps/api/staticmap?center=37.85,-122.125zoom=12siz e=640x640maptype=mobileformat=png32sensor=true Loading required package: rgdal Loading required package: sp Geospatial Data Abstraction Library extensions to R successfully loaded Loaded GDAL runtime: GDAL 1.8.0, released 2011/01/12 Path to GDAL shared files: /usr/local/share/gdal Loaded PROJ.4 runtime: Rel. 4.7.1, 23 September 2009 Path to PROJ.4 shared files: (autodetected) Warning message: In readGDAL(destfile, silent = TRUE) : GeoTransform values not available PlotOnStaticMap(MyMap,lat =datas$LAT,lon =datas$LON, + axes =TRUE,mar =rep(4,4)) List of 6 $ lat.center: num 37.8 $ lon.center: num -122 $ zoom : num 12 $ myTile: int [1:640, 1:640] 968 853 855 969 1033 888 855 884 888 995 ... ..- attr(*, COL)= chr [1:1132] #00 #020201 #020202 #030302 ... ..- attr(*, type)= chr rgb $ BBOX :List of 2 ..$ ll: num [1, 1:2] 37.8 -122.2 .. ..- attr(*, dimnames)=List of 2 .. .. ..$ : chr Y .. .. ..$ : chr [1:2] lat lon ..$ ur: num [1, 1:2] 37.9 -122 .. ..- attr(*, dimnames)=List of 2 .. .. ..$ : chr Y .. .. ..$ : chr [1:2] lat lon $ url : chr google NULL [1] -291.2711 291.2711 [1] -276.5158 276.7972 When I run this on my computer the horizontal axis ranges from 300W to 60E, but the ticks in between aren't linearly spaced (300W, 200W, 100W, 0, 100E, 160W, 60W). Also, the vertical axis moves linearly from 300S to 300N. It seems that no matter what data I supply for datas, the axes are always labeled this way. My question is: 1. Does this problem occur on other machines using this code? 2. Does anyone have an explanation for it? and 3. Can anybody suggest a way to get the correct axes labels (assuming these are incorrect, but maybe i'm somehow misinterpreting the plot!)? Thank you for your time. There has been no answer other than that I ought to contact the package maintainer. Since it would be nice to have a publicly displayed solution, I opted to post here first before doing so. Does anyone have any insight to share? Thank you very much for your time, -Erik Gregory CSU Sacramento, Mathematics Student Assistant, California Environmental Protection Agency __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Can we prepare a questionaire in R
How can we create HTML forms in 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.
Re: [R] Decision Trees /Decision Analysis with R?
See packages rpart, randomForest, party. Also, typing R Decision Trees produced good google results. http://www.google.com/search?aq=fsourceid=chromeie=UTF-8q=R+Decision+Trees On Wed, Jun 8, 2011 at 7:02 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Hello, this question is a bit out of the blue. I am a big R fan and user and in my new job I do some decision modeling (mostly health economics). For that decision trees are often used (I guess the most classic example is the investment decision A, B, and C with different probabilities, what is the expected payoff). We use a specialized software called TreeAge that some might know. The basic setup of such simulations is actually very simple and I guess useful in many fields. So I was wondering whether there is already a package out there in R that is doing such a thing? Thanks for any hints! Best, Stefan PS (By decision tree I don't mean cluster-like analysis of a data set splitting by identifying decision nods, but the other way around: I have decision nodes, what is my expected outcome.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 2.13 version doesn't load packages
On Wed, 8 Jun 2011, Mariana Varela wrote: Hi, I just installed the 2.13 version for Mac (without uninstalling As the posting guide says, there is no such version of R Also, I think you did uninstall a previous version, for it is missing and the default behaviour of the Apple installer is to delete previous versions. the previous versions). I transferred the folder library containing the packages I normally use to the 2.13 library folder. When I require any package this is what I get: require(ape) Loading required package: ape Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared object '/Library/Frameworks/R.framework/Versions/2.13/Resources/library/gee/libs/x86_64/gee.so': dlopen(/Library/Frameworks/R.framework/Versions/2.13/Resources/library/gee/libs/x86_64/gee.so, 6): Library not loaded: /Library/Frameworks/R.framework/Versions/2.12/Resources/lib/libRblas.dylib Referenced from: /Library/Frameworks/R.framework/Versions/2.13/Resources/library/gee/libs/x86_64/gee.so Reason: image not found However when type library the packages are there. Any hints? Read the documentation: that is not the way to update! But given that you have done so, run update.packages(checkBuilt = TRUE) at the R prompt to download and install updated versions of your packages. We do ask that Mac questions are asked on r-sig-mac, at least if this is for the CRAN distribution of R (there are others, and it is substantially different from building R from the sources for yourself). One of the many things that is non-standard about that distribution is that it hardcodes library paths in its DSOs, so any package with compiled code is tied to a small range (e.g. 2.12.[012]) of R versions. You will find a lot of discussion about updating in the recent r-sig-mac archives. See also the rw-FAQ at http://cran.r-project.org/bin/windows/base/rw-FAQ.html#What_0027s-the-best-way-to-upgrade_003f Many thanks Mariana -- Mariana Varela, DVM, PhD MRC-University of Glasgow Centre for Virus Research Institute of Infection, Immunity and Inflammation College of Medical, Veterinary and Life Sciences Garscube Estate, 464 Bearsden Road Henry Wellcome Building, room 436 Glasgow, G61 1QH Scotland (UK) Phone: +44 (0) 141 330 2196 [[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. -- 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] Decision Trees /Decision Analysis with R?
Thank you so much for reply. But I am looking for the exact opposite. I do not have a data set which I want to partition. But already a sequence/tree-like set of decision rules and with which I want to simulate what is my expected outcome/pay-off given a particular scenario. As far as I understand it, those packages could calculate the expected outcome AFTER having fit them to a particular data set and not construct a synthetic tree with exogenously defined decision nods/rules. Or am I wrong? Thanks and best, Stefan On Wed, Jun 8, 2011 at 2:03 PM, Jonathan Daily biomathjda...@gmail.com wrote: See packages rpart, randomForest, party. Also, typing R Decision Trees produced good google results. http://www.google.com/search?aq=fsourceid=chromeie=UTF-8q=R+Decision+Trees On Wed, Jun 8, 2011 at 7:02 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Hello, this question is a bit out of the blue. I am a big R fan and user and in my new job I do some decision modeling (mostly health economics). For that decision trees are often used (I guess the most classic example is the investment decision A, B, and C with different probabilities, what is the expected payoff). We use a specialized software called TreeAge that some might know. The basic setup of such simulations is actually very simple and I guess useful in many fields. So I was wondering whether there is already a package out there in R that is doing such a thing? Thanks for any hints! Best, Stefan PS (By decision tree I don't mean cluster-like analysis of a data set splitting by identifying decision nods, but the other way around: I have decision nodes, what is my expected outcome.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] library(SenoMineR)- Triangle Test Query
Unless I missed it, neither the OP nor the list was CC'd on this, so for anyone
interested, I forward this solution (untested from my side) from the package
maintainer. Not sure whether the file comes through, so I include the updated
code in the message's body below.
Cheers, Michael
** updated code for triangle.test
triangle.test - function (design,answer,preference=NULL){
answer = gsub((\\w), \\U\\1, as.character(answer), perl=TRUE)
labprod =
levels(as.factor(c(as.character(design[,1]),as.character(design[,2]),as.character(design[,3]
nbprod = length(labprod)
nb.answer = nb.good = pref = matrix(0,nbprod,nbprod)
for (i in 1:nrow(design)){
for (j in 1:nbprod){
if (labprod[j] == design[i,1]) i1 = j
if (labprod[j] == design[i,2]) i2 = j
if (labprod[j] == design[i,3]) i3 = j
}
if (i1!=i2) nb.answer [i1,i2] = nb.answer[i1,i2]+1
if (i1==i2) nb.answer [i1,i3] = nb.answer[i1,i3]+1
if ((i1==i2)(answer[i]==Z)){
nb.good[i1,i3]=nb.good[i1,i3]+1
if (length(preference)0){
if (preference[i]!=Z) pref[i3,i1] = pref[i3,i1] +1
if (preference[i]==Z) pref[i1,i3] = pref[i1,i3] +1
}
}
if ((i1==i3)(answer[i]==Y)){
nb.good[i1,i2]=nb.good[i1,i2]+1
if (length(preference)0){
if (preference[i]!=Y) pref[i2,i1] = pref[i2,i1] +1
if (preference[i]==Y) pref[i1,i2] = pref[i1,i2] +1
}
}
if ((i2==i3)(answer[i]==X)){
nb.good[i1,i2]=nb.good[i1,i2]+1
if (length(preference)0){
if (preference[i]!=X) pref[i1,i2] = pref[i1,i2] +1
if (preference[i]==X) pref[i2,i1] = pref[i2,i1] +1
}
}
}
nb.good = nb.good + t(nb.good)
nb.answer = nb.answer + t(nb.answer)
diag(nb.answer)=1
prob = pbinom(nb.good-1,nb.answer,1/3,lower.tail=FALSE)
maxML = recognize = minimum = matrix(NA,nbprod,nbprod)
for (i in 1: (nbprod-1)){
for (j in (i+1):nbprod){
aux = matrix(0,nb.good[i,j]+1,1)
for (k in 0:nb.good[i,j]) aux[k] =
dbinom(nb.good[i,j]-k,nb.answer[i,j]-k,1/3)
maxML[i,j] = maxML[j,i] = max(aux)
recognize[i,j] = recognize[j,i] = rev(order(aux))[1]-1
mini = 0
for (k in round(nb.answer[i,j]/3):nb.answer[i,j]) if
((mini==0)(dbinom(k,nb.answer[i,j],1/3)0.05)) mini=k
minimum[i,j]=minimum[j,i]=mini
}
}
confusion = (nb.answer-recognize) / nb.answer
diag(nb.answer)=diag(recognize)=0
diag(maxML)=diag(confusion)=1
rownames(nb.answer) = colnames(nb.answer) = rownames(nb.good) =
colnames(nb.good) = labprod
rownames(prob) = colnames(prob)= rownames(confusion) = colnames(confusion)=
labprod
rownames(maxML) = colnames(maxML) = rownames(minimum) = colnames(minimum) =
rownames(recognize) = colnames(recognize) = labprod
if (length(preference)0) rownames(pref) = colnames(pref) = labprod
res = list()
res$nb.comp = nb.answer
res$nb.ident = nb.good
res$p.value = prob
res$nb.recognition = recognize
res$maxML = maxML
res$confusion = confusion
res$minimum = minimum
if (length(preference)0) res$pref = pref
##res$complete = result
return(res)
}
** end updated code for triangle.test
-Original Message-
From: Francois Husson
Sent: Wednesday, June 08, 2011 14:43
To: Meyners, Michael
Subject: Re: [R] library(SenoMineR)- Triangle Test Query
Dear Vijayan, dear Michael,
Indeed there was an error in the function triangle.test. I attach the new
function.
Thanks Michael for your answer.
Best
Francois
Le 08/06/2011 12:32, Meyners, Michael a écrit :
Vijayan,
I cannot find an error in your code, but I had a look at the code of
triangle.test -- unless I'm missing something, it contains a bug. If you
study the way in which the matrix pref is updated, you find that the vector
preference is compared to 1, 2 and 3 instead of X, Y and Z as it should
be. That way, some of the non-diagonal entries of pref will always be zero,
irrespective of the data, which does not make sense. I think it should work
if you modify the code accordingly. Alternatively, a quick patch (untested!)
might be to code preferences as 1, 2 and 3 instead of the letters (but I'm
not sure whether this has any other implications).
I CC the author of the function and maintainer of the package; he should
correct me if needed or could otherwise update the code for the next release
(I worked on SensoMineR 1.11).
Hope this helps,
Michael
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Vijayan Padmanabhan
Sent: Saturday, June 04, 2011 9:21
To: r-help@r-project.org
Subject: [R] library(SenoMineR)- Triangle Test Query
Dear R Group
I was trying to use the triangle.test function in SensoMineR and
strangely i
encounter a error in the output of preference matrix from the analysis.
To illustrate, pl see the following dataframe of a design with the
response
and preference collected as shown below:
design-structure(list(`Product X` = c(3, 1, 4, 2, 4, 2, 1, 3, 4, 2,
4, 2, 1, 3, 4, 2, 4, 2, 3, 1), `Product Y` = c(1, 1, 4, 4, 4,
3, 1, 1,
Re: [R] Decision Trees /Decision Analysis with R?
So TreeAge fits models but won't predict from them? That seems like bizarre behavior. I suppose I would recommend, then, looking at the source code from the aforementioned packages for how they store their split data. It sounds like you would have to write code to hack TreeAge outputs into another packages' format (e.g. look at ?rpart.object). Sorry I couldn't help more, Jon On Wed, Jun 8, 2011 at 9:47 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Thank you so much for reply. But I am looking for the exact opposite. I do not have a data set which I want to partition. But already a sequence/tree-like set of decision rules and with which I want to simulate what is my expected outcome/pay-off given a particular scenario. As far as I understand it, those packages could calculate the expected outcome AFTER having fit them to a particular data set and not construct a synthetic tree with exogenously defined decision nods/rules. Or am I wrong? Thanks and best, Stefan On Wed, Jun 8, 2011 at 2:03 PM, Jonathan Daily biomathjda...@gmail.com wrote: See packages rpart, randomForest, party. Also, typing R Decision Trees produced good google results. http://www.google.com/search?aq=fsourceid=chromeie=UTF-8q=R+Decision+Trees On Wed, Jun 8, 2011 at 7:02 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Hello, this question is a bit out of the blue. I am a big R fan and user and in my new job I do some decision modeling (mostly health economics). For that decision trees are often used (I guess the most classic example is the investment decision A, B, and C with different probabilities, what is the expected payoff). We use a specialized software called TreeAge that some might know. The basic setup of such simulations is actually very simple and I guess useful in many fields. So I was wondering whether there is already a package out there in R that is doing such a thing? Thanks for any hints! Best, Stefan PS (By decision tree I don't mean cluster-like analysis of a data set splitting by identifying decision nods, but the other way around: I have decision nodes, what is my expected outcome.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Cleveland dot plots
Colin Wahl wrote: I would rather use cleveland dot plots than bar charts to display my study results. I have not been able to find (or figure out) an R package that is capable of producing the publication quality dot charts Im looking for. I have either not been able to get error bars (lattice), cannot order the data display properly (latticeExtra), or cannot make adjustments to axes. Does anyone have a quick suggestion for a package that can handle cleveland dot plots well? Most of the things you mention can be accomplished within the lattice package, using panel functions and the scales= argument. The reorder() function will probably accomplish what you're looking for when you say order the data display properly. A few years ago, I wrote a short paper on dotplots that you might find helpful: Jacoby, William G. 2006. The Dotplot: A Graphical Display for Labeled Quantitative Values. The Political Methodologist 14(1): 6-14. A longer version of the paper is available http://polisci.msu.edu/jacoby/research/dotplots/ here . Both of these papers provide examples of the things you mention (error bars, reordering data, etc.). Hope this helps! -- View this message in context: http://r.789695.n4.nabble.com/Cleveland-dot-plots-tp3581122p3582719.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] Can we prepare a questionaire in R
As Mike had written, there are frameworks for web-development with R. RApache http://www.rapache.net is one. Also, see the R package Rook: http://cran.r-project.org/web/packages/Rook/index.html . On Wed, 2011-06-08 at 17:26 +0530, amrita gs wrote: How can we create HTML forms in R Wouldn't you rather create HTML forms in HTML? See the links above to use R for server-side scripting, for example, to receive form data from a web browser. [[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] residual checking for GAM (mgcv)
Dear list, i'm checking the residuals plots of a gam model after a processus of model selection. I found the best model, all my terms are significant, the r-square and the deviance explained are good, but I have strange residuals plots: http://dl.dropbox.com/u/1169100/gam.check.png http://dl.dropbox.com/u/1169100/residuals_vs_fitted.png The curve is caused by the zeroes in my data. I've also plotted each explanatory variables included in the model against residuals and everything looks fine. Is this curve does not allow me to accept this model? Does the use of an other family (eg negbin) would be the solution for fixing this problem? Currently I use the poisson family (and quasipoisson). I have a lot of 0 in my response variable, almost 65 % Should I use a specific function that allows me to use zero-inflated data?? Kind regards, Sam [[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] Can we prepare a questionaire in R
On Wed, Jun 8, 2011 at 12:56 PM, amrita gs ammasamri...@gmail.com wrote:
How can we create HTML forms in R
HTML is Just Plain Text, so you can create them using R's text output
'cat' function. E.g.
cat('form First name: input type=text name=firstname /br
/Last name: input type=text name=lastname //form',
file=test.html)
and job done. Open test.html in your web browser and there it is.
Other packages may help you construct these things - search CRAN for
html, and also 'brewer' which is a handy package for making templates
which you can render into HTML.
But to be honest, your question is way too general as stated to get a
decent response here.
Barry
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[R] install the “impute” package in unix
Hi, I am trying to install the impute package in unix. but I get the following error message. I followed the following steps. Do you know what is causing this and how I can solve this problem? source(http://www.bioconductor.org/biocLite.R;) biocLite(impute) Using R version 2.11.1, biocinstall version 2.6.10. Installing Bioconductor version 2.6 packages: [1] impute Please wait... Warning in install.packages(pkgs = pkgs, repos = repos, ...) : argument 'lib' is missing: using '/nfs/users/nfs_s/st5/R-modules' trying URL 'http://cran.ma.imperial.ac.uk/src/contrib/impute_1.26.0.tar.gz' Content type 'application/x-gzip' length 1191531 bytes (1.1 Mb) opened URL == downloaded 1.1 Mb ERROR: failed to lock directory â/nfs/users/nfs_s/st5/R-modulesâ for modifying Try removing â/nfs/users/nfs_s/st5/R-modules/00LOCKâ The downloaded packages are in â/tmp/RtmpgJur79/downloaded_packagesâ Warning message: In install.packages(pkgs = pkgs, repos = repos, ...) : installation of package 'impute' had non-zero exit status best, salih [[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] 3D-plotting a 2D-matrix that contains z-values (3rd dimension)
Hello, say I have a 2D-matrix (indexed by x and y), which contains z values, which I want to plot over x-y. Either dotted, or if possible as a landscape. I tried around with persp and plot3d (from rgl) and persp3d (from rgl). I sometimes get something that looks good and a while later, when trying some new data I need to worry about that again. Is there something lika a convenience function that can be used to feed the data into persp, rgl::plot3d and rgl::persp3d? At least persp3d is picky about the order of the input data, and I somehow always start again. (plot3d seems to be most mathcing how I think). Isn't that a very common case, where my z_x_y = mydata[x,y] ? Maybe I just don't know the right function that helps me. Any idea about that? Ciao, Oliver __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] predict with model (rms package)
This is a consequence of predict.ols calling predictrms which relies on model.frame which re-issues the expression of x.knots. You would have the same problem if using update(fit object) in another session. For that reason you have to keep an external knots vector available in your environment. Frank Mark Seeto wrote: Dear R-help, In the rms package, I have fitted an ols model with a variable represented as a restricted cubic spline, with the knot locations specified as a previously defined vector. When I save the model object and open it in another workspace which does not contain the vector of knot locations, I get an error message if I try to predict with that model. This also happens if only one workspace is used, but the vector of knot locations is removed: library(rms) set.seed(1) x - rnorm(100) y - 1 + x + x^2 + rnorm(100) x.knots - quantile(x, c(0.2, 0.5, 0.8)) ols1 - ols(y ~ rcs(x, x.knots)) predict(ols1, data.frame(x = 0)) # This works rm(x.knots) predict(ols1, data.frame(x = 0)) # Gives error The first predict gives 1 0.8340293 while the second predict gives Error in rcs(x, x.knots) : object 'x.knots' not found The same error happens if x.knots is simply defined as a vector like c(-1, 0, 1) (i.e. not using quantile). Is this the intended behaviour? The requirement that x.knots be in the workspace seems strange, given that the knot locations are stored in ols1$Design$parms. Thanks for any help you can give. Mark Seeto National Acoustic Laboratories, Australia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/predict-with-model-rms-package-tp3581229p3582758.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] install the “impute” package in unix
I manage to install the package. But I cant load it now. It says: library(CGHcall) Loading required package: impute Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared library '/nfs/users/nfs_s/st5/R-modules/impute/libs/impute.so': libgfortran.so.1: cannot open shared object file: No such file or directory In addition: Warning messages: 1: package 'CGHcall' was built under R version 2.11.1 2: package 'impute' was built under R version 2.11.1 Error: package 'impute' could not be loaded What is going wrong? Best, salih On Wed, Jun 8, 2011 at 3:52 PM, Salih Tuna saliht...@gmail.com wrote: Hi, I am trying to install the impute package in unix. but I get the following error message. I followed the following steps. Do you know what is causing this and how I can solve this problem? source(http://www.bioconductor.org/biocLite.R;) biocLite(impute) Using R version 2.11.1, biocinstall version 2.6.10. Installing Bioconductor version 2.6 packages: [1] impute Please wait... Warning in install.packages(pkgs = pkgs, repos = repos, ...) : argument 'lib' is missing: using '/nfs/users/nfs_s/st5/R-modules' trying URL 'http://cran.ma.imperial.ac.uk/src/contrib/impute_1.26.0.tar.gz ' Content type 'application/x-gzip' length 1191531 bytes (1.1 Mb) opened URL == downloaded 1.1 Mb ERROR: failed to lock directory â/nfs/users/nfs_s/st5/R-modulesâ for modifying Try removing â/nfs/users/nfs_s/st5/R-modules/00LOCKâ The downloaded packages are in â/tmp/RtmpgJur79/downloaded_packagesâ Warning message: In install.packages(pkgs = pkgs, repos = repos, ...) : installation of package 'impute' had non-zero exit status best, salih [[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] The simplest bar graph with ggplot is difficult to realize
Dear all, What is the simplest way of producing a bar graph using ggplot but avoiding calling qplot? That is, given: d - data.frame(x=seq(1,5), y=seq(1,5)) Why does the following line return an error? ggplot(d, aes(x=x, y=y)) + stat_identity() + geom_bar(bindwidth=1) Thanks in advance, jcb! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Error Installing or Updating Packages (Maybe because of a proxy)
Dear all,
I receive the very same error message on a Debian computer (testing)
with R 2.13.0 also.
install.packages('emu')
Installing package(s) into
‘/home/olivier/R/i486-pc-linux-gnu-library/2.13’ (as ‘lib’ is
unspecified)
Error in ret[i, ] - c(pkgs[i], lib, desc) :
number of items to replace is not a multiple of replacement length
I have no proxy settings on this computer (neither
in .bashrc / .bash_profile, nor in my desktop environment, and I'm
doing it at home where I'm not using any proxy).
I can download a file on the web from within R (using both download.file
() or download.packages
('emu','/home/olivier/R/i486-pc-linux-gnu-library/2.13')... and finally
I can also use install.packages() on this downloaded file and the
install works flawlessly.
I can't find any old R base package in the various directories
indicated by .libPaths().
.libPaths()
[1] /home/olivier/R/i486-pc-linux-gnu-library/2.13
[2] /usr/local/lib/R/site-library
[3] /usr/lib/R/site-library
[4] /usr/lib/R/library
The only place where I can think there may be one are the local trees
([1] and [2]) as the 2 others are (should be) updated automatically when
updating R with the Debian pkg mngmt system and there's nothing inside
them. Only [4] contains a base/ subdirectory (a single one) but I
suppose this is the current one for R 2.13.0
I can install a package once it's been downloaded locally (through R CMD
INSTALL pkg) but can't succeed in installing the same package from the
CRAN mirrors using install.packages(). I experience the very same issue
with all related instructions (old.packages(), update.packages())
I also could do that several months ago on a different Debian computer
(but with an older R version than the current one).
Any hints (including what kind of information I should give to enhance
the description of this issue)?
Olivier.
On Wed, 20 Apr 2011 10:29:17 +0200 Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
If the internet connection from R works, can you please verify that
you do not have any R base package from an old R version in a current
R library that you may have in the .libPaths() already?
Uwe Ligges
On 20.04.2011 09:25, Majid Einian wrote:
Dear R Helpers,
(I am using Ubuntu lucid and R 2.13.0
When I try to update packages I get this error:
update.packages()
--- Please select a CRAN mirror for use in this session ---
Loading Tcl/Tk interface ... done
Error in ret[i, ]- c(pkgs[i], lib, desc) :
number of items to replace is not a multiple of replacement
length
I had no problem before (case 1) but now (case 2) I cannot get it
to work, googleing did not help:
case 1:
* connecting directly without any proxy setting (at my
university)
* using R 2.12.2
case 2:
* connecting through proxy setting (at my workplace)
* using R 2.13.0
I set the proxy in terminal too but it does not help (echo
$http_proxy gives me http://192.168.0.1:8080/)
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--
Olivier Crouzet, PhD
Laboratoire de Linguistique -- EA3827
Département de Sciences du Langage
UFR Lettres et Langages
Université de Nantes
Chemin de la Censive du Tertre - BP 81227
44312 Nantes cedex 3
France
phone:(+33) 02 40 14 14 05 (lab.)
(+33) 02 40 14 14 36 (office)
fax: (+33) 02 40 14 13 27
e-mail: olivier.crou...@univ-nantes.fr
http://www.lling.fr/
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Re: [R] The simplest bar graph with ggplot is difficult to realize
Hi Juan, Each geom can have it's own stat, so stat_identity() doesn't change the stat used by geom_bar(). You need ggplot(d, aes(x=x, y=y)) + geom_bar(stat=identity) Best, Ista 2011/6/8 Juan Carlos Borrás jcbor...@gmail.com: Dear all, What is the simplest way of producing a bar graph using ggplot but avoiding calling qplot? That is, given: d - data.frame(x=seq(1,5), y=seq(1,5)) Why does the following line return an error? ggplot(d, aes(x=x, y=y)) + stat_identity() + geom_bar(bindwidth=1) Thanks in advance, jcb! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] CGHcall and normalization
Hi, i am trying to use the normalization function from CGHcall library. The command i use is normal.fullData - normalize(fullData, method = median, cellularity = 1) and i get the following error message. How can i solve this issue. Error in function (classes, fdef, mtable) : unable to find an inherited method for function copynumber, for signature data.frame best, salih [[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] PS to Taking Integral and Optimization using Integrate() and Optim()
Hello again.
Thank you for the comments. I have written these codes.
iy=function(x)
{
res=NULL
ress=0
for (i in (1:2))
{
for (xx in x[i])
{
fy=function(y)
(exp(-exp(y+log(xx)))*(-exp(y+log(xx)))^2)/(1-exp(-exp(y+log(xx
res=c(res,integrate(fy,-6.907,-1.246)$value)
ress=ress+res
}
}
return(ress)
}
iy(c(1,1))
integrate(fy,-6.907,-1.246)$value
In 1D optimize() works perfectly on iy(). However the problem is in 2D and
more, optimize() does not work and I need to apply optim(). I could not apply
optim() on iy().
Beside, I tried to use Ryacas, I faced with this error:
[1] Starting Yacas!
Error in socketConnection(host = 127.0.0.1, port = 9734, server = FALSE, :
cannot open the connection
In addition: Warning message:
In socketConnection(host = 127.0.0.1, port = 9734, server = FALSE, :
127.0.0.1:9734 cannot be opened
Cheers,
Maryam
[[alternative HTML version deleted]]
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Re: [R] Decision Trees /Decision Analysis with R?
Jon, So TreeAge fits models but won't predict from them? That seems like bizarre behavior. Nothing bizarre about TreeAge, just a different tool in a different disicpline. http://en.wikipedia.org/wiki/Decision_tree Graham [[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] Decision Trees /Decision Analysis with R?
TreeAge works just fine. But its commercial, thats all... On Wed, Jun 8, 2011 at 6:17 PM, Graham Smith myotis...@gmail.com wrote: Jon, So TreeAge fits models but won't predict from them? That seems like bizarre behavior. Nothing bizarre about TreeAge, just a different tool in a different disicpline. http://en.wikipedia.org/wiki/Decision_tree Graham __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] XML segfault on some architectures
Dear Prof Ripley, Apologies - I've re-sent that to Duncan Temple Lang, along with your note about lib versions. Version info was included in my original post - I gave full sessionInfo(). It's XML_3.4-0. I only have a very sketchy understanding of libraries and systems administration, but it looks like our libxml2 is version 2.6.26. I'll ask my sysadmin people whether they can update that, and try again. Janet On Jun 7, 2011, at 10:54 PM, Prof Brian Ripley wrote: On Tue, 7 Jun 2011, Janet Young wrote: Hi, I found an architecture-specific segfault problem with the XML package. I originally found the problem using the parseKGML2Graph function in the Bioconductor KEGGgraph package, but as far as I can tell the underlying issue seems to be with the xmlTreeParse which is called by parseKGML2Graph. I'm trying this piece of code, from the xmlTreeParse help page: library(XML) fileName - system.file(exampleData, test.xml, package=XML) x - xmlTreeParse(fileName) On my Mac and on nodes of one of the linux clusters I have access to, this works fine. But on another of the linux clusters I use, I get a segfault every time, on both 32-bit and 64-bit nodes of the cluster. The unames for those nodes are here: Linux kong053 2.6.18-194.17.1.el5xen #1 SMP Wed Sep 29 13:30:21 EDT 2010 x86_64 x86_64 x86_64 GNU/Linux Linux king049 2.6.18-194.26.1.el5xen #1 SMP Tue Nov 9 14:13:46 EST 2010 i686 i686 i386 GNU/Linux I think I've included all the relevant info below, but please let me know if there's anything else you'd like to see. As the posting guide says, report problems in contributed packages first to the maintainer, giving the 'at a minimum' information required (which includes the package version number). But note that package XML relies on libxml2, and it is entirely possible the fault is in the latter. Your kernel looks like RHEL 5 (and is an old version): that is well known for having very old versions of system software. One known issue with libxml2 is a mismatch between it and zlib 1.2.[45] prior to libxml2 2.7.7 (2.7.8 is current): from experience, that causes segfaults in package XML's examples. thanks, Janet --- Dr. Janet Young Fred Hutchinson Cancer Research Center 1100 Fairview Avenue N., C3-168, P.O. Box 19024, Seattle, WA 98109-1024, USA. tel: (206) 667 1471 fax: (206) 667 6524 email: jayoung ...at... fhcrc.org --- on 64-bit node library(XML) fileName - system.file(exampleData, test.xml, package=XML) fileName [1] /home/btrask/traskdata/lib_linux_64/R/library/XML/exampleData/test.xml sessionInfo() R version 2.13.0 (2011-04-13) 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-8LC_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] XML_3.4-0 system(uname -a) Linux kong053 2.6.18-194.17.1.el5xen #1 SMP Wed Sep 29 13:30:21 EDT 2010 x86_64 x86_64 x86_64 GNU/Linux x - xmlTreeParse(fileName) *** caught segfault *** address 0x51c4f, cause 'memory not mapped' Traceback: 1: .Call(RS_XML_ParseTree, as.character(file), handlers, as.logical(ignoreBlanks), as.logical(replaceEntities), as.logical(asText), as.logical(trim), as.logical(validate), as.logical(getDTD), as.logical(isURL), as.logical(addAttributeNamespaces), as.logical(useInternalNodes), FALSE, as.logical(isSchema), as.logical(fullNamespaceInfo), as.character(encoding), as.logical(useDotNames), xinclude, error, addFinalizer, PACKAGE = XML) 2: xmlTreeParse(fileName) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
[R] Help with plotting plsr loadings
Hi I am attempting to do a loadings plot from a plsr object. I have managed to do this using the gasoline data that comes with the pls package. However when I conduct this on my dataset i get the following error message. plot(BHPLS1, loadings, comps = 1:2, legendpos = topleft, labels = numbers, xlab = nm) Error in loadingplot.default(x, ...) : Could not convert variable names to numbers. str(BHPLS1_Loadings) loadings [1:8892, 1:60] -0.00717 0.00414 0.02611 0.00468 -0.00676 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:8892] PCIList1 PCIList2 PCIList3 PCIList4 ... ..$ : chr [1:60] Comp 1 Comp 2 Comp 3 Comp 4 ... - attr(*, explvar)= Named num [1:60] 2.67 4.14 4.41 3.55 2.59 ... ..- attr(*, names)= chr [1:60] Comp 1 Comp 2 Comp 3 Comp 4 ... Can anyone see the problem?? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to suppress factor labels
I am using ggplot2 to make a boxplot that overlays a scatterplot: pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) The x variable is a factor for every minute from 0:60. My problem is that ggplot2 labels every factor value on the x axis, and they overlap. How can I make ggplot2 label only, say every 5th factor value on the x axis? Thanks, Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] svmpath_0.95 uploaded to CRAN
This new version includes a plot method for plotting a particular instance along the path. Trevor Hastie has...@stanford.edu Professor, Department of Statistics, Stanford University Phone: (650) 725-2231 Fax: (650) 725-8977 URL: http://www.stanford.edu/~hastie address: room 104, Department of Statistics, Sequoia Hall 390 Serra Mall, Stanford University, CA 94305-4065 ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to suppress factor labels
Hi Jim, See ?scale_x_discrete Best, Ista On Wed, Jun 8, 2011 at 3:26 PM, James Rome jamesr...@gmail.com wrote: I am using ggplot2 to make a boxplot that overlays a scatterplot: pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) The x variable is a factor for every minute from 0:60. My problem is that ggplot2 labels every factor value on the x axis, and they overlap. How can I make ggplot2 label only, say every 5th factor value on the x axis? Thanks, 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. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Variable in file name png
Hi,
I'm having trouble with getting the png function to properly produce multiple
graphs. RIght now I have:
for (z in data) {
png(file=z,bg=white)
thisdf-data[[z]]
plot(thisdf$rc,thisdf$psi)
dev.off()
}
Which should take the data object, a list of data sets and produce a graph of
each with respect to the two variables rc and psi.
I want the names to change for each graph, but am not sure how to do it, any
help would be apreciated.
Thanks,
-Acoutino
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[R] Histogram
Hello , I am trying to create a histogram in order to compare between two groups and would like it to be similar to the figure attached. How can I generate this using R ? Thank you, Nandini http://r.789695.n4.nabble.com/file/n3582448/5634-15977-1-PB.gif -- View this message in context: http://r.789695.n4.nabble.com/Histogram-tp3582448p3582448.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] histogram help
Hello , I am trying to create a histogram in order to compare between two groups and would like it to be similar to the figure attached. How can I generate this using R ? Thank you, Nandini __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 command window
Hello. I'm a visually impaired statistician, working at the National Institute of Public Health in Denmark. I would like to use R for some analysis and have succesfully installed version 2.13.0 on my Windows XP labtop. I then would like to run R interactively but unfortunately the textfont of the command line in the R window is very hard for me to read. I use a special program called Zoomtext that among many other functionalities enables me to invert colurs. It is possible to change the prompt (using options(...)) but I cannot see that it is possible to change the font - eg. and most importantly changing its color or size. I have tried to check the Windows FAQ but couldn't find anything. Is this possible? And (of course) - if yes, how? I guess the best solution for me is to run R non-interactively because I can use an editor for my programs - but it is cumbersome. Can I as a DOS command write something like R input-file output-file? I'm very new in R so it's exciting for me to see if I have any response. Best wishes Michael Michael Davidsen National Institute of Public Health University of Southern Denmark Øster Farimagsgade 5A, 2. DK-1353 Copenhagen K Denmark E-mail: m...@sdu.dkmailto:m...@sdu.dk Web: www.si-folkesundhed.dkhttp://www.si-folkesundhed.dk [[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] Adding values to the end of a data frame
I'm trying to develop this function so that I can efficiently generate
all possibile combinations of the strings.
So I have certain roots, prefixes, and sufixes. I also have different
combinations of the data, some with two strings (roots, prefix) and others
with three strings (roots, prefix, suffix)
roots - c(car insurance, auto insurance)
roots2 - c(insurance)
prefix - c(cheap, budget)
prefix2 - c(low cost)
suffix - c(quote, quotes)
suffix2 - c(rate, rates)
suffix3 - c(comparison)
Here are just a few of the combinations I'm trying to generate.
# prefix, roots, suffix
# prefix, roots, suffix2
# prefix, roots, suffix3
# prefix2, roots, suffix
# prefix2, roots, suffix2
# prefix2, roots, suffix3
# prefix, roots2, suffix
# prefix, roots2, suffix2
# prefix, roots2, suffix3
# prefix2, roots2, suffix
# prefix2, roots2, suffix2
# prefix2, roots2, suffix3
# prefix, roots
# prefix2, roots
# prefix, roots2
# prefix2, roots2
# roots, suffix
# roots, suffix2
# roots, suffix3
# roots2, suffix
# roots2, suffix2
# roots2, suffix3
# state, roots
# city, roots
# cityst, roots
So instead of two functions for ones with two vs three parameters, I'm
wondering if it's possible to just develop one function.
one - function(x, y) {
nu - do.call(paste, expand.grid(x, y))
mydf - data.frame(nu)
}
two - function(x, y, z){
mu - do.call(paste, expand.grid(x, y, z))
mydf2 - data.frame(mu)
}
On Tue, Jun 7, 2011 at 6:54 PM, Dennis Murphy djmu...@gmail.com wrote:
Alas, you don't have a suffix2 object defined, but try this:
d1 - one(prefix, roots)
d2 - one(roots, suffix)
rbind(d1, d2)
To see a potential flaw in your function (as least as far as console
output is concerned), try
rbind(d1, one(roots, suffix))
HTH,
Dennis
On Tue, Jun 7, 2011 at 3:30 PM, Abraham Mathew abra...@thisorthat.com
wrote:
Let's say that I'm trying to write a functions that will allow me to
automate a process
where I examine all possible combinations of various string groupings.
Each
time I run
the one function, I want to include the new values to the end of a data
frame. The data
frame will basically be one column with a lot of rows.
roots - c(car insurance, auto insurance)
prefix - c(cheap, budget)
suffix - c(rate, rates)
one - function(x, y, z=0) {
nu - do.call(paste, expand.grid(x, y, z))
mydf - data.frame(nu)
print(mydf)
}
one(roots, suffix2)
one(prefix, roots)
one(prefix, roots, suffix2)
The code above just replaces each value in the data frame each time I run
the one function.
How can I add the new values to the end of the data frame?
Help!
I'm running R 2.13 on Ubuntu 10.10
WebRep
Overall rating
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[R] how can I??
st1=c(5,3) st2=c(-1,2) st3=c(1,3) st4=c(2,6) st = rbind(st1,st2,st3,st4) st dx=round(dist(st), digits=2) dx #À¯Å¬¸®µå °Å¸®Çà·Ä D1=dist(st, method=quot;euclideanquot;) D1 # À¯Å¬¸®µå°Å¸® ±¸Çϱâ D2=dist(st, method=quot;manhattanquot;) D2 # ¸ÇÇÏź °Å¸®Çà·Ä and I want know How can i use statistical distance and Mahalanobis distance [[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] package.skeleton() does not create 'data' folder
Hi again, I was using package.skeleton() function to create the
skeleton of my package in windows. Here is my attempt:
rm(list = ls())
setwd(F:/R_PackageBuild)
package.skeleton(trial1, namespace = TRUE, code_files =
F:/R_PackageBuild/trial.r)
In the trial.r file, there are 2 objects, one is a function and
another is data. Here they are:
fn1 - Vectorize(function(x,y,z) {
return(x + y +z)
}, SIMPLIFY = TRUE)
Data - rnorm(20)
However my problem is that package.skeleton() does not create any data
folder in the skeleton tree. However in the man folder there are 3 Rd
files (as expected), naming:
Data, fn1, trial1-package
However on the contrary if my code is like below then,
package.skeleton() creates data folder.
fn1 - Vectorize(function(x,y,z) {
+ return(x + y +z)
+ }, SIMPLIFY = TRUE)
Data - rnorm(20)
setwd(F:/R_PackageBuild)
package.skeleton(trial2)
So is it that if I use 'code_files ' argument then, R would not create
data folder?
Can somebody help me what I am missing in this process? Till now, I
create manually the data folder and within that folder manually put a
RData file where only object is that 'Data'. However I believe there
must be more elegant way to doing that.
While searching over net to settle this issue, I found this thread
'http://r.789695.n4.nabble.com/How-to-create-rda-file-to-be-used-in-package-building-td828148.html'
however this is answering my question.
Thanks,
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Re: [R] predict with model (rms package)
Thanks for your reply, Frank. I've noticed that the x.knots object doesn't actually have to be the vector of knots. Just having x.knots - 0 or even x.knots - a will allow predict to work. Mark Seeto Frank Harrell wrote: This is a consequence of predict.ols calling predictrms which relies on model.frame which re-issues the expression of x.knots. You would have the same problem if using update(fit object) in another session. For that reason you have to keep an external knots vector available in your environment. Frank Mark Seeto wrote: Dear R-help, In the rms package, I have fitted an ols model with a variable represented as a restricted cubic spline, with the knot locations specified as a previously defined vector. When I save the model object and open it in another workspace which does not contain the vector of knot locations, I get an error message if I try to predict with that model. This also happens if only one workspace is used, but the vector of knot locations is removed: library(rms) set.seed(1) x - rnorm(100) y - 1 + x + x^2 + rnorm(100) x.knots - quantile(x, c(0.2, 0.5, 0.8)) ols1 - ols(y ~ rcs(x, x.knots)) predict(ols1, data.frame(x = 0)) # This works rm(x.knots) predict(ols1, data.frame(x = 0)) # Gives error The first predict gives 1 0.8340293 while the second predict gives Error in rcs(x, x.knots) : object 'x.knots' not found The same error happens if x.knots is simply defined as a vector like c(-1, 0, 1) (i.e. not using quantile). Is this the intended behaviour? The requirement that x.knots be in the workspace seems strange, given that the knot locations are stored in ols1$Design$parms. Thanks for any help you can give. Mark Seeto National Acoustic Laboratories, Australia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/predict-with-model-rms-package-tp3581229p3583344.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] Decision Trees /Decision Analysis with R?
It is difficult for someone from a statistical frame of mind to understand what this is about --- you need to think a bit differently. It is mostly a simulation and decision analysis, with some use of statistical functions to draw random samples to simulate the fact that outcome of interest can take any value from a known or unknown distribution. For example, you may be comparing two interventions and a do-nothing decision to improve some health outcome of interest. The decision maker is interested in *relative* effectiveness and costs of the interventions to improve the outcome of interest. You have results from published literature that you can use as inputs into a simulation exercise to compare relative costs and benefits/effectiveness of the three options. A small decision tree can be easily simulated in a spreadsheet; for long trees with many decision nodes it is useful to have a specialized software. There are some Excel plugins that are sold about $100. Others are more expensive. I think R is not well suited for this kind of work. A decision analysis package in R may require user to write code like the one used in LaTeX or related programs (Metapost) to draw graphs of trees (e.g. complicated organizational trees, or hierarchical trees). However, in such a package there can be useful outputs, measures and graphs generated by R using code that may already exist for other packages. Look up journal Medical Decision Making to know what is being discussed. This method is used extensively in medicine and public health to study decisions. It even uses MCMC, though with a different flavor --- it may even be a different kind of food. Anupam. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jonathan Daily Sent: Wednesday, June 08, 2011 7:47 PM To: stefan.d...@gmail.com Cc: r-help@r-project.org Subject: Re: [R] Decision Trees /Decision Analysis with R? So TreeAge fits models but won't predict from them? That seems like bizarre behavior. I suppose I would recommend, then, looking at the source code from the aforementioned packages for how they store their split data. It sounds like you would have to write code to hack TreeAge outputs into another packages' format (e.g. look at ?rpart.object). Sorry I couldn't help more, Jon On Wed, Jun 8, 2011 at 9:47 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Thank you so much for reply. But I am looking for the exact opposite. I do not have a data set which I want to partition. But already a sequence/tree-like set of decision rules and with which I want to simulate what is my expected outcome/pay-off given a particular scenario. As far as I understand it, those packages could calculate the expected outcome AFTER having fit them to a particular data set and not construct a synthetic tree with exogenously defined decision nods/rules. Or am I wrong? Thanks and best, Stefan On Wed, Jun 8, 2011 at 2:03 PM, Jonathan Daily biomathjda...@gmail.com wrote: See packages rpart, randomForest, party. Also, typing R Decision Trees produced good google results. http://www.google.com/search?aq=fsourceid=chromeie=UTF-8q=R+Decisi on+Trees On Wed, Jun 8, 2011 at 7:02 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Hello, this question is a bit out of the blue. I am a big R fan and user and in my new job I do some decision modeling (mostly health economics). For that decision trees are often used (I guess the most classic example is the investment decision A, B, and C with different probabilities, what is the expected payoff). We use a specialized software called TreeAge that some might know. The basic setup of such simulations is actually very simple and I guess useful in many fields. So I was wondering whether there is already a package out there in R that is doing such a thing? Thanks for any hints! Best, Stefan PS (By decision tree I don't mean cluster-like analysis of a data set splitting by identifying decision nods, but the other way around: I have decision nodes, what is my expected outcome.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal,
[R] Results of CFA with Lavaan
I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent estimate that is set to 1 with no SD for each factor? Is that normal? When I've managed to get sem::sem to fit a model this has not been the case. Thanks, Sam Stewart HS.model - ' visual =~ x1 + x2 + x3 textual =~ x4 + x5 + x6 speed =~ x7 + x8 + x9 ' fit - sem(HS.model, data=HolzingerSwineford1939) summary(fit, fit.measures=TRUE) Lavaan (0.4-8) converged normally after 35 iterations Number of observations 301 Estimator ML Minimum Function Chi-square 85.306 Degrees of freedom24 P-value0.000 Chi-square test baseline model: Minimum Function Chi-square 918.852 Degrees of freedom36 P-value0.000 Full model versus baseline model: Comparative Fit Index (CFI)0.931 Tucker-Lewis Index (TLI) 0.896 Loglikelihood and Information Criteria: Loglikelihood user model (H0) -3737.745 Loglikelihood unrestricted model (H1) -3695.092 Number of free parameters 21 Akaike (AIC)7517.490 Bayesian (BIC) 7595.339 Sample-size adjusted Bayesian (BIC) 7528.739 Root Mean Square Error of Approximation: RMSEA 0.092 90 Percent Confidence Interval 0.071 0.114 P-value RMSEA = 0.05 0.001 Standardized Root Mean Square Residual: SRMR 0.065 Parameter estimates: Information Expected Standard Errors Standard Estimate Std.err Z-value P(|z|) Latent variables: visual =~ x11.000 x20.5540.1005.5540.000 x30.7290.1096.6850.000 textual =~ x41.000 x51.1130.065 17.0140.000 x60.9260.055 16.7030.000 speed =~ x71.000 x81.1800.1657.1520.000 x91.0820.1517.1550.000 Covariances: visual ~~ textual 0.4080.0745.5520.000 speed 0.2620.0564.6600.000 textual ~~ speed 0.1730.0493.5180.000 Variances: x10.5490.1144.8330.000 x21.1340.102 11.1460.000 x30.8440.0919.3170.000 x40.3710.0487.7780.000 x50.4460.0587.6420.000 x60.3560.0438.2770.000 x70.7990.0819.8230.000 x80.4880.0746.5730.000 x90.5660.0718.0030.000 visual0.8090.1455.5640.000 textual 0.9790.1128.7370.000 speed 0.3840.0864.4510.000 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Results of CFA with Lavaan
What do you mean by latent estimate? The table of variances has variances for each factors. Is there something different in the sem output that you don't see here? Yes, this looks normal. Jeremy On 8 June 2011 13:14, R Help rhelp.st...@gmail.com wrote: I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent estimate that is set to 1 with no SD for each factor? Is that normal? When I've managed to get sem::sem to fit a model this has not been the case. Thanks, Sam Stewart HS.model - ' visual =~ x1 + x2 + x3 textual =~ x4 + x5 + x6 speed =~ x7 + x8 + x9 ' fit - sem(HS.model, data=HolzingerSwineford1939) summary(fit, fit.measures=TRUE) Lavaan (0.4-8) converged normally after 35 iterations Number of observations 301 Estimator ML Minimum Function Chi-square 85.306 Degrees of freedom 24 P-value 0.000 Chi-square test baseline model: Minimum Function Chi-square 918.852 Degrees of freedom 36 P-value 0.000 Full model versus baseline model: Comparative Fit Index (CFI) 0.931 Tucker-Lewis Index (TLI) 0.896 Loglikelihood and Information Criteria: Loglikelihood user model (H0) -3737.745 Loglikelihood unrestricted model (H1) -3695.092 Number of free parameters 21 Akaike (AIC) 7517.490 Bayesian (BIC) 7595.339 Sample-size adjusted Bayesian (BIC) 7528.739 Root Mean Square Error of Approximation: RMSEA 0.092 90 Percent Confidence Interval 0.071 0.114 P-value RMSEA = 0.05 0.001 Standardized Root Mean Square Residual: SRMR 0.065 Parameter estimates: Information Expected Standard Errors Standard Estimate Std.err Z-value P(|z|) Latent variables: visual =~ x1 1.000 x2 0.554 0.100 5.554 0.000 x3 0.729 0.109 6.685 0.000 textual =~ x4 1.000 x5 1.113 0.065 17.014 0.000 x6 0.926 0.055 16.703 0.000 speed =~ x7 1.000 x8 1.180 0.165 7.152 0.000 x9 1.082 0.151 7.155 0.000 Covariances: visual ~~ textual 0.408 0.074 5.552 0.000 speed 0.262 0.056 4.660 0.000 textual ~~ speed 0.173 0.049 3.518 0.000 Variances: x1 0.549 0.114 4.833 0.000 x2 1.134 0.102 11.146 0.000 x3 0.844 0.091 9.317 0.000 x4 0.371 0.048 7.778 0.000 x5 0.446 0.058 7.642 0.000 x6 0.356 0.043 8.277 0.000 x7 0.799 0.081 9.823 0.000 x8 0.488 0.074 6.573 0.000 x9 0.566 0.071 8.003 0.000 visual 0.809 0.145 5.564 0.000 textual 0.979 0.112 8.737 0.000 speed 0.384 0.086 4.451 0.000 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Questions about building R packages
Hello R users, I have difficulties when trying to make R packages. I tried to read many tutorials but still could not find out the right way. Could any one help me out please? (I'm using Windows xp.) After running package.skeleton() and edit those RD files, I don't know how to use Rtools (or CMD shell?) to build the zip file. I installed the Rtools from Murdoch's link, but it doesn't look like a software... Anyone could give me a tutorial with more details about Rtools (or CMD shell)? I've already have MikTex installed. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Questions-about-building-R-packages-tp3583510p3583510.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] Results of CFA with Lavaan
Dear Sam, In each case, the first observed variable is treated as a reference indicator with its coefficient fixed to 1 to establish the metric of the corresponding factor and therefore to identify the model. If you didn't do the same thing (or something equivalent, such as fixing the factor variances to 1) in specifying the model to sem::sem(), that might account for the problems you encountered. Best, John John Fox Senator William McMaster Professor of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R Help Sent: June-08-11 4:15 PM To: r-help Subject: [R] Results of CFA with Lavaan I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent estimate that is set to 1 with no SD for each factor? Is that normal? When I've managed to get sem::sem to fit a model this has not been the case. Thanks, Sam Stewart HS.model - ' visual =~ x1 + x2 + x3 textual =~ x4 + x5 + x6 speed =~ x7 + x8 + x9 ' fit - sem(HS.model, data=HolzingerSwineford1939) summary(fit, fit.measures=TRUE) Lavaan (0.4-8) converged normally after 35 iterations Number of observations 301 Estimator ML Minimum Function Chi-square 85.306 Degrees of freedom24 P-value0.000 Chi-square test baseline model: Minimum Function Chi-square 918.852 Degrees of freedom36 P-value0.000 Full model versus baseline model: Comparative Fit Index (CFI)0.931 Tucker-Lewis Index (TLI) 0.896 Loglikelihood and Information Criteria: Loglikelihood user model (H0) -3737.745 Loglikelihood unrestricted model (H1) -3695.092 Number of free parameters 21 Akaike (AIC)7517.490 Bayesian (BIC) 7595.339 Sample-size adjusted Bayesian (BIC) 7528.739 Root Mean Square Error of Approximation: RMSEA 0.092 90 Percent Confidence Interval 0.071 0.114 P-value RMSEA = 0.05 0.001 Standardized Root Mean Square Residual: SRMR 0.065 Parameter estimates: Information Expected Standard Errors Standard Estimate Std.err Z-value P(|z|) Latent variables: visual =~ x11.000 x20.5540.1005.5540.000 x30.7290.1096.6850.000 textual =~ x41.000 x51.1130.065 17.0140.000 x60.9260.055 16.7030.000 speed =~ x71.000 x81.1800.1657.1520.000 x91.0820.1517.1550.000 Covariances: visual ~~ textual 0.4080.0745.5520.000 speed 0.2620.0564.6600.000 textual ~~ speed 0.1730.0493.5180.000 Variances: x10.5490.1144.8330.000 x21.1340.102 11.1460.000 x30.8440.0919.3170.000 x40.3710.0487.7780.000 x50.4460.0587.6420.000 x60.3560.0438.2770.000 x70.7990.0819.8230.000 x80.4880.0746.5730.000 x90.5660.0718.0030.000 visual0.8090.1455.5640.000 textual 0.9790.1128.7370.000 speed 0.3840.0864.4510.000 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
[R] Autocorrelation in R
Hi, I am trying to learn time series, and I am attending a colleague's course on Econometrics. However, he uses e-views, and I use R. I am trying to reproduce his examples in R, but I am having problems specifying a AR(1) model. Would anyone help me with my code? Thanks in advance! Reproducible code follows: download.file(https://sites.google.com/a/proxima.adm.br/main/ex_32.csv --no-check-certificate, ex_32.csv, method=wget) ex32=read.csv(ex_32.csv) lm_ex32=lm(gc ~ yd, data=ex32) summary(lm_ex32) # Durbin-Watson (slide 26) library(lmtest) dwtest(gc ~ yd, data=ex32) # or dwtest(lm_ex32) # Breusch-Godfrey bgtest(lm_ex32, order=2) # AR(1) # In e-views, the specification was: # GC = YD AR(1) # and the output was: # Dependent Variable: GC # Method: Least Squares # Sample: 1970Q2 1995Q2 # Included observations: 101 # Convergence achieved after 6 interations # = # Variable Coefficient Std.Error t-Statistic Prob. # C -56.99706 19.84692 -2.871835 0.0050 # YD 0.937035 0.006520 143.7170 0. # AR(1) 0.752407 0.066565 11.30338 0. # = # R-squared 0.999691 Mean dependent var 2345.867 # Adjusted R-squared 0.999685 S.D. dependent var 1284.675 # S.E. of regression 22.81029 Akaike info criterion 9.121554 # Sum squared resid 50990.32 Schwarz criterion 9.199231 # Log likelihood -457.6385 F-statistic 158548.1 # Durbin-Watson stat 2.350440 Prob(F-statistic) 0.00 # following code based on http://www.stat.pitt.edu/stoffer/tsa2/R_time_series_quick_fix.htm # And now for some regression with autocorrelated errors. # I've tried to follow the example in Pinheiro Bates (2004), p. 239-244, with no success. gc_ts = ts(ex32[66:166,gc]) yd_ts = ts(ex32[66:166,yd]) library(nlme) trend = time(gc_ts) fit_lm = lm(gc_ts ~ trend + yd_ts) acf(resid(fit_lm)) pacf(resid(fit_lm)) gls_ex32_ar1 = gls(gc_ts ~ trend + yd_ts, correlation = corAR1(form= ~yd_ts),method=ML) summary(gls_ex32_ar1) _ Dr. Iuri Gavronski Assistant Professor Programa de Pós-Graduação em Administração Universidade do Vale do Rio dos Sinos – UNISINOS Av. Unisinos, 950 – São Leopoldo – RS – Brasil Sala (Room) 5A 406 D 93022-000 www.unisinos.br TEL +55-51-3591-1122 ext. 1589 FAX +55-51-3590-8447 Email: igavron...@unisinos.br CV Lattes: http://lattes.cnpq.br/8843390959025944 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Decision Trees /Decision Analysis with R?
see inline below. On Wed, Jun 8, 2011 at 12:37 PM, Anupam anupa...@gmail.com wrote: It is difficult for someone from a statistical frame of mind to understand what this is about --- you need to think a bit differently. It is mostly a simulation and decision analysis, with some use of statistical functions to draw random samples to simulate the fact that outcome of interest can take any value from a known or unknown distribution. For example, you may be comparing two interventions and a do-nothing decision to improve some health outcome of interest. The decision maker is interested in *relative* effectiveness and costs of the interventions to improve the outcome of interest. You have results from published literature that you can use as inputs into a simulation exercise to compare relative costs and benefits/effectiveness of the three options. A small decision tree can be easily simulated in a spreadsheet; for long trees with many decision nodes it is useful to have a specialized software. There are some Excel plugins that are sold about $100. Others are more expensive. I think R is not well suited for this kind of work. A decision analysis Not necessarily! A desicion tree model is a kind of graphical model. See the CRAN task view gR (graphical models in R) and maybe ask on the special interest mailing list R-sig-gR kjetil package in R may require user to write code like the one used in LaTeX or related programs (Metapost) to draw graphs of trees (e.g. complicated organizational trees, or hierarchical trees). However, in such a package there can be useful outputs, measures and graphs generated by R using code that may already exist for other packages. Look up journal Medical Decision Making to know what is being discussed. This method is used extensively in medicine and public health to study decisions. It even uses MCMC, though with a different flavor --- it may even be a different kind of food. Anupam. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jonathan Daily Sent: Wednesday, June 08, 2011 7:47 PM To: stefan.d...@gmail.com Cc: r-help@r-project.org Subject: Re: [R] Decision Trees /Decision Analysis with R? So TreeAge fits models but won't predict from them? That seems like bizarre behavior. I suppose I would recommend, then, looking at the source code from the aforementioned packages for how they store their split data. It sounds like you would have to write code to hack TreeAge outputs into another packages' format (e.g. look at ?rpart.object). Sorry I couldn't help more, Jon On Wed, Jun 8, 2011 at 9:47 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Thank you so much for reply. But I am looking for the exact opposite. I do not have a data set which I want to partition. But already a sequence/tree-like set of decision rules and with which I want to simulate what is my expected outcome/pay-off given a particular scenario. As far as I understand it, those packages could calculate the expected outcome AFTER having fit them to a particular data set and not construct a synthetic tree with exogenously defined decision nods/rules. Or am I wrong? Thanks and best, Stefan On Wed, Jun 8, 2011 at 2:03 PM, Jonathan Daily biomathjda...@gmail.com wrote: See packages rpart, randomForest, party. Also, typing R Decision Trees produced good google results. http://www.google.com/search?aq=fsourceid=chromeie=UTF-8q=R+Decisi on+Trees On Wed, Jun 8, 2011 at 7:02 AM, stefan.d...@gmail.com stefan.d...@gmail.com wrote: Hello, this question is a bit out of the blue. I am a big R fan and user and in my new job I do some decision modeling (mostly health economics). For that decision trees are often used (I guess the most classic example is the investment decision A, B, and C with different probabilities, what is the expected payoff). We use a specialized software called TreeAge that some might know. The basic setup of such simulations is actually very simple and I guess useful in many fields. So I was wondering whether there is already a package out there in R that is doing such a thing? Thanks for any hints! Best, Stefan PS (By decision tree I don't mean cluster-like analysis of a data set splitting by identifying decision nods, but the other way around: I have decision nodes, what is my expected outcome.) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- === Jon Daily Technician === #!/usr/bin/env outside # It's great, trust me. -- === Jon Daily Technician
Re: [R] XML segfault on some architectures
Hi, Our sysadmin updated libxml2 to version 2.7.8, and now xmlTreeParse works fine with no segfault. Thank you very much - that was very helpful, Janet On Jun 8, 2011, at 11:59 AM, Janet Young wrote: Dear Prof Ripley, Apologies - I've re-sent that to Duncan Temple Lang, along with your note about lib versions. Version info was included in my original post - I gave full sessionInfo(). It's XML_3.4-0. I only have a very sketchy understanding of libraries and systems administration, but it looks like our libxml2 is version 2.6.26. I'll ask my sysadmin people whether they can update that, and try again. Janet On Jun 7, 2011, at 10:54 PM, Prof Brian Ripley wrote: On Tue, 7 Jun 2011, Janet Young wrote: Hi, I found an architecture-specific segfault problem with the XML package. I originally found the problem using the parseKGML2Graph function in the Bioconductor KEGGgraph package, but as far as I can tell the underlying issue seems to be with the xmlTreeParse which is called by parseKGML2Graph. I'm trying this piece of code, from the xmlTreeParse help page: library(XML) fileName - system.file(exampleData, test.xml, package=XML) x - xmlTreeParse(fileName) On my Mac and on nodes of one of the linux clusters I have access to, this works fine. But on another of the linux clusters I use, I get a segfault every time, on both 32-bit and 64-bit nodes of the cluster. The unames for those nodes are here: Linux kong053 2.6.18-194.17.1.el5xen #1 SMP Wed Sep 29 13:30:21 EDT 2010 x86_64 x86_64 x86_64 GNU/Linux Linux king049 2.6.18-194.26.1.el5xen #1 SMP Tue Nov 9 14:13:46 EST 2010 i686 i686 i386 GNU/Linux I think I've included all the relevant info below, but please let me know if there's anything else you'd like to see. As the posting guide says, report problems in contributed packages first to the maintainer, giving the 'at a minimum' information required (which includes the package version number). But note that package XML relies on libxml2, and it is entirely possible the fault is in the latter. Your kernel looks like RHEL 5 (and is an old version): that is well known for having very old versions of system software. One known issue with libxml2 is a mismatch between it and zlib 1.2.[45] prior to libxml2 2.7.7 (2.7.8 is current): from experience, that causes segfaults in package XML's examples. thanks, Janet --- Dr. Janet Young Fred Hutchinson Cancer Research Center 1100 Fairview Avenue N., C3-168, P.O. Box 19024, Seattle, WA 98109-1024, USA. tel: (206) 667 1471 fax: (206) 667 6524 email: jayoung ...at... fhcrc.org --- on 64-bit node library(XML) fileName - system.file(exampleData, test.xml, package=XML) fileName [1] /home/btrask/traskdata/lib_linux_64/R/library/XML/exampleData/test.xml sessionInfo() R version 2.13.0 (2011-04-13) 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-8LC_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] XML_3.4-0 system(uname -a) Linux kong053 2.6.18-194.17.1.el5xen #1 SMP Wed Sep 29 13:30:21 EDT 2010 x86_64 x86_64 x86_64 GNU/Linux x - xmlTreeParse(fileName) *** caught segfault *** address 0x51c4f, cause 'memory not mapped' Traceback: 1: .Call(RS_XML_ParseTree, as.character(file), handlers, as.logical(ignoreBlanks), as.logical(replaceEntities), as.logical(asText), as.logical(trim), as.logical(validate), as.logical(getDTD), as.logical(isURL), as.logical(addAttributeNamespaces), as.logical(useInternalNodes), FALSE, as.logical(isSchema), as.logical(fullNamespaceInfo), as.character(encoding), as.logical(useDotNames), xinclude, error, addFinalizer, PACKAGE = XML) 2: xmlTreeParse(fileName) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace Selection: __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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)
[R] accessing files from subfolders
Hi,
There must be an easy way to do this, but I'm not finding it..
I'd just like to know the syntax to move up and down folder levels, without
necessarily entering a full file path. Also, how to construct file and folder
paths using variables.
For example 1, if I wanted to print to the screen the contents of a file called
myFile.txt using the bash shell, I'd use the following:
cat ../myFile.txt
Also, for example 2, if I want to cd into a folder that contains my files, from
within a loop, where the counter serves as one of the folders in the path.
In bash:
for i in {1..5} A B C # A, B, and C are also folder names
do
cd ~/${i} # move into (change working directory to?) my folder of interest,
which is 1,2,3,4,5,A,B,or C
cat myFile.txt # print corresponding file of interest to screen
cd - # move back to the previous folder
done
I wonder if there's an easy corresponding way to accomplish this in R.
Any ideas most welcome!
Jonathan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 suppress factor labels
I think the issue is that the x axis is a factor. How would ggplot2 know which ones to drop? So it labels them all. If I do the following, the labels get better: pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) + scale_x_discrete( breaks = c(0, 5, 10, 15,20,25,30,35,40,45,50,55,60), labels=c(0, 5, 10, 15,20,25,30,35,40,45,50,55,60)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) But this ruins the boxplot--I get one box instead of a box at every minute. On 6/8/2011 3:59 PM, Ista Zahn wrote: Hi James, It's hard for me to see where the problem might be. Please post the data using dput() or even better, make a simplified example that illustrates the problem without all the other stuff going on. Chances are that in the process of making a simplified example you will find the problem yourself. Best, Ista On Wed, Jun 8, 2011 at 3:41 PM, James Rome jamesr...@passur.com wrote: I actually tried that, and get the same plot if I am using it properly: title=paste(Fitted RETA predictions for , airport, the week of , date, sep=) pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) + scale_x_discrete(breaks = seq(from=0, to=60, by=5), labels=seq(from=0, to=60, by=5)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) The x-axis is unchanged. Thanks, Jim On 6/8/2011 3:31 PM, Ista Zahn wrote: Hi Jim, See ?scale_x_discrete Best, Ista On Wed, Jun 8, 2011 at 3:26 PM, James Rome jamesr...@gmail.com wrote: I am using ggplot2 to make a boxplot that overlays a scatterplot: pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) The x variable is a factor for every minute from 0:60. My problem is that ggplot2 labels every factor value on the x axis, and they overlap. How can I make ggplot2 label only, say every 5th factor value on the x axis? Thanks, Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 suppress factor labels
On Wed, Jun 8, 2011 at 5:02 PM, James Rome jamesr...@passur.com wrote: I think the issue is that the x axis is a factor. Rather the opposite I think. In the data you sent, time is numeric, not a factor. This works for me: qplot(factor(time), error, data=times, size=I(1), geom=boxplot) + facet_wrap(~ runway, ncol=2) + scale_x_discrete(breaks = seq(from=0, to=60, by=10)) Best, Ista How would ggplot2 know which ones to drop? So it labels them all. If I do the following, the labels get better: pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) + scale_x_discrete( breaks = c(0, 5, 10, 15,20,25,30,35,40,45,50,55,60), labels=c(0, 5, 10, 15,20,25,30,35,40,45,50,55,60)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) But this ruins the boxplot--I get one box instead of a box at every minute. On 6/8/2011 3:59 PM, Ista Zahn wrote: Hi James, It's hard for me to see where the problem might be. Please post the data using dput() or even better, make a simplified example that illustrates the problem without all the other stuff going on. Chances are that in the process of making a simplified example you will find the problem yourself. Best, Ista On Wed, Jun 8, 2011 at 3:41 PM, James Rome jamesr...@passur.com wrote: I actually tried that, and get the same plot if I am using it properly: title=paste(Fitted RETA predictions for , airport, the week of , date, sep=) pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) + scale_x_discrete(breaks = seq(from=0, to=60, by=5), labels=seq(from=0, to=60, by=5)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) The x-axis is unchanged. Thanks, Jim On 6/8/2011 3:31 PM, Ista Zahn wrote: Hi Jim, See ?scale_x_discrete Best, Ista On Wed, Jun 8, 2011 at 3:26 PM, James Rome jamesr...@gmail.com wrote: I am using ggplot2 to make a boxplot that overlays a scatterplot: pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title, ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10), color=times$runway, ylim=c(-30,40)) pp2 = pp + with(times, facet_wrap(~ runway, ncol=2)) print(pp2 + geom_boxplot(alpha=.5, color=blue, outlier.colour=green, outlier.size=1)) The x variable is a factor for every minute from 0:60. My problem is that ggplot2 labels every factor value on the x axis, and they overlap. How can I make ggplot2 label only, say every 5th factor value on the x axis? Thanks, 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. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] accessing files from subfolders
On 11-06-08 5:35 PM, J wrote:
Hi,
There must be an easy way to do this, but I'm not finding it..
I'd just like to know the syntax to move up and down folder levels, without
necessarily entering a full file path. Also, how to construct file and folder
paths using variables.
For example 1, if I wanted to print to the screen the contents of a file called
myFile.txt using the bash shell, I'd use the following:
cat ../myFile.txt
Also, for example 2, if I want to cd into a folder that contains my files, from
within a loop, where the counter serves as one of the folders in the path.
In bash:
for i in {1..5} A B C # A, B, and C are also folder names
do
cd ~/${i} # move into (change working directory to?) my folder of interest,
which is 1,2,3,4,5,A,B,or C
cat myFile.txt # print corresponding file of interest to screen
cd - # move back to the previous folder
done
I wonder if there's an easy corresponding way to accomplish this in R.
The .. notation is supported on all platforms, as far as I know. So to
go to the parent of the current working dir, just use
setwd(..)
The ~ notation is not supported in all R functions, but setwd() supports
it on all platforms, as far as I know.
Use file.path() to construct longer paths.
For your second example it would simply be setwd(file.path(~, i)).
Duncan Murdoch
Any ideas most welcome!
Jonathan
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Results of CFA with Lavaan
Yes, that is the difference. For the last SEM I built I fixed the factor variances to 1, and I think that's what I want to do for the CFA I'm doing now. Does that make sense for a CFA? I'll try figuring out how to do that with lavaan later, but my model takes so long to fit that I can't try it right now. Thanks, Sam On Wed, Jun 8, 2011 at 5:58 PM, John Fox j...@mcmaster.ca wrote: Dear Sam, In each case, the first observed variable is treated as a reference indicator with its coefficient fixed to 1 to establish the metric of the corresponding factor and therefore to identify the model. If you didn't do the same thing (or something equivalent, such as fixing the factor variances to 1) in specifying the model to sem::sem(), that might account for the problems you encountered. Best, John John Fox Senator William McMaster Professor of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R Help Sent: June-08-11 4:15 PM To: r-help Subject: [R] Results of CFA with Lavaan I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent estimate that is set to 1 with no SD for each factor? Is that normal? When I've managed to get sem::sem to fit a model this has not been the case. Thanks, Sam Stewart HS.model - ' visual =~ x1 + x2 + x3 textual =~ x4 + x5 + x6 speed =~ x7 + x8 + x9 ' fit - sem(HS.model, data=HolzingerSwineford1939) summary(fit, fit.measures=TRUE) Lavaan (0.4-8) converged normally after 35 iterations Number of observations 301 Estimator ML Minimum Function Chi-square 85.306 Degrees of freedom 24 P-value 0.000 Chi-square test baseline model: Minimum Function Chi-square 918.852 Degrees of freedom 36 P-value 0.000 Full model versus baseline model: Comparative Fit Index (CFI) 0.931 Tucker-Lewis Index (TLI) 0.896 Loglikelihood and Information Criteria: Loglikelihood user model (H0) -3737.745 Loglikelihood unrestricted model (H1) -3695.092 Number of free parameters 21 Akaike (AIC) 7517.490 Bayesian (BIC) 7595.339 Sample-size adjusted Bayesian (BIC) 7528.739 Root Mean Square Error of Approximation: RMSEA 0.092 90 Percent Confidence Interval 0.071 0.114 P-value RMSEA = 0.05 0.001 Standardized Root Mean Square Residual: SRMR 0.065 Parameter estimates: Information Expected Standard Errors Standard Estimate Std.err Z-value P(|z|) Latent variables: visual =~ x1 1.000 x2 0.554 0.100 5.554 0.000 x3 0.729 0.109 6.685 0.000 textual =~ x4 1.000 x5 1.113 0.065 17.014 0.000 x6 0.926 0.055 16.703 0.000 speed =~ x7 1.000 x8 1.180 0.165 7.152 0.000 x9 1.082 0.151 7.155 0.000 Covariances: visual ~~ textual 0.408 0.074 5.552 0.000 speed 0.262 0.056 4.660 0.000 textual ~~ speed 0.173 0.049 3.518 0.000 Variances: x1 0.549 0.114 4.833 0.000 x2 1.134 0.102 11.146 0.000 x3 0.844 0.091 9.317 0.000 x4 0.371 0.048 7.778 0.000 x5 0.446 0.058 7.642 0.000 x6 0.356 0.043 8.277 0.000 x7 0.799 0.081 9.823 0.000 x8 0.488 0.074 6.573 0.000 x9 0.566 0.071 8.003 0.000 visual 0.809 0.145 5.564 0.000 textual 0.979 0.112 8.737 0.000 speed 0.384 0.086 4.451 0.000 __
Re: [R] Questions about building R packages
On 11-06-08 4:17 PM, Xia.Li wrote: Hello R users, I have difficulties when trying to make R packages. I tried to read many tutorials but still could not find out the right way. Could any one help me out please? (I'm using Windows xp.) Read the Writing R Extensions manual, not the many misleading tutorials. (You can read my tutorial, but don't trust any of the rest of them :-). Duncan Murdoch After running package.skeleton() and edit those RD files, I don't know how to use Rtools (or CMD shell?) to build the zip file. I installed the Rtools from Murdoch's link, but it doesn't look like a software... Anyone could give me a tutorial with more details about Rtools (or CMD shell)? I've already have MikTex installed. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Questions-about-building-R-packages-tp3583510p3583510.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] accessing files from subfolders
Try this:
lapply(dir(myPathDir, recursive = TRUE, pattern = myFile.txt$,
full.names = TRUE), readLines, warn = FALSE)
On Wed, Jun 8, 2011 at 6:35 PM, J jonsle...@gmail.com wrote:
Hi,
There must be an easy way to do this, but I'm not finding it..
I'd just like to know the syntax to move up and down folder levels, without
necessarily entering a full file path. Also, how to construct file and
folder paths using variables.
For example 1, if I wanted to print to the screen the contents of a file
called myFile.txt using the bash shell, I'd use the following:
cat ../myFile.txt
Also, for example 2, if I want to cd into a folder that contains my files,
from within a loop, where the counter serves as one of the folders in the
path.
In bash:
for i in {1..5} A B C # A, B, and C are also folder names
do
cd ~/${i} # move into (change working directory to?) my folder of interest,
which is 1,2,3,4,5,A,B,or C
cat myFile.txt # print corresponding file of interest to screen
cd - # move back to the previous folder
done
I wonder if there's an easy corresponding way to accomplish this in R.
Any ideas most welcome!
Jonathan
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
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25° 25' 40 S 49° 16' 22 O
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[R] return counts of elements on a table column depending on elements on another column
Hi, I am given the following table: head(hsa_refseq) chr genome regionstart stop nu strand nu.1nu.2 gene_id 1 chr1 hg19_refGeneCDS 6742 6751 0 +0 gene_id NM_032291 2 chr1 hg19_refGene exon 66999825 6751 0 +. gene_id NM_032291 3 chr1 hg19_refGeneCDS 67091530 67091593 0 +2 gene_id NM_032291 4 chr1 hg19_refGene exon 67091530 67091593 0 +. gene_id NM_032291 5 chr1 hg19_refGeneCDS 67098753 67098777 0 +1 gene_id NM_032291 6 chr2 hg19_refGene exon 67098753 67098777 0 +. gene_id NM_032291 What I've done is to find out how many of the elements on 3rd column are CDS, exon. sum(hsa_refseq$region==CDS) sum(hsa_refseq$region==exon) But what I would like is to print for each chromosome how many are exons and how many CDS. For example chr1 has 5 CDS and 2 exons chr2 has 10 CDS and 3 exons... Can you tell what should I add? Or if I am doing this wrong, how should I do it? Thank you, Regards, Nanami [[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] R command window
On 11-06-08 9:28 AM, Michael Davidsen wrote: Hello. I'm a visually impaired statistician, working at the National Institute of Public Health in Denmark. I would like to use R for some analysis and have succesfully installed version 2.13.0 on my Windows XP labtop. I then would like to run R interactively but unfortunately the textfont of the command line in the R window is very hard for me to read. I use a special program called Zoomtext that among many other functionalities enables me to invert colurs. There are two ways to run R in Windows. One runs Rterm in a CMD window. You'll need to use Windows procedures to change the font and colours in that window. The other way (what is the default on the shortcut) is to run RGui. It has a dialog accessed from File | GUI Preferences that lets you choose fonts and colors. (The font used in the dialog may be hard to read; if so, choose Save. That will write a text file, default name Rconsole, in which you can edit all of the settings. If you save it into RHOME/etc/Rconsole, it will be loaded each time you start R.) Duncan Murdoch It is possible to change the prompt (using options(...)) but I cannot see that it is possible to change the font - eg. and most importantly changing its color or size. I have tried to check the Windows FAQ but couldn't find anything. Is this possible? And (of course) - if yes, how? I guess the best solution for me is to run R non-interactively because I can use an editor for my programs - but it is cumbersome. Can I as a DOS command write something like Rinput-file output-file? I'm very new in R so it's exciting for me to see if I have any response. Best wishes Michael Michael Davidsen National Institute of Public Health University of Southern Denmark Øster Farimagsgade 5A, 2. DK-1353 Copenhagen K Denmark E-mail: m...@sdu.dkmailto:m...@sdu.dk Web: www.si-folkesundhed.dkhttp://www.si-folkesundhed.dk [[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] Histogram
I think the command you want is barplot x = rbinom(10,15,0.65) y = rbinom(10,15,0.25) barplot(rbind(x,y),beside=TRUE) Sam On Wed, Jun 8, 2011 at 10:14 AM, nandini_bn nandini...@hotmail.com wrote: Hello , I am trying to create a histogram in order to compare between two groups and would like it to be similar to the figure attached. How can I generate this using R ? Thank you, Nandini http://r.789695.n4.nabble.com/file/n3582448/5634-15977-1-PB.gif -- View this message in context: http://r.789695.n4.nabble.com/Histogram-tp3582448p3582448.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] Histogram
Have a look at: http://addictedtor.free.fr/graphiques/thumbs.php One of the graph examples they have is exactly what you are after. On Wed, Jun 8, 2011 at 11:14 PM, nandini_bn nandini...@hotmail.com wrote: Hello , I am trying to create a histogram in order to compare between two groups and would like it to be similar to the figure attached. How can I generate this using R ? Thank you, Nandini http://r.789695.n4.nabble.com/file/n3582448/5634-15977-1-PB.gif -- View this message in context: http://r.789695.n4.nabble.com/Histogram-tp3582448p3582448.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] error with geomap in googleVis
SNV Krishna krishna at primps.com.sg writes: Hi All, I am unable to get the plot geomap in googleVis package. data is as follows head(index.ret) countryytd 1 Argentina -10.18 2 Australia -3.42 3 Austria -2.70 4 Belgium 1.94 5Brazil -7.16 6Canada 0.56 map1 = gvisGeoMap(index.ret,locationvar = 'country', numvar = 'ytd') plot(map1) But it just displays a blank page, showing an error symbol at the right bottom corner. I tried demo(googleVis), it also had a similar problem. The demo showed all other plots/maps except for those geomaps. Could any one please hint me what/where could be the problem? Many thanks for the idea and support. Regards, SNV Krishna [[alternative HTML version deleted]] Hi All, I have also encountered this problem. I have tested the problem in Windows XP with R 2.13.0, Windows 7 with R 2.13.0, Ubuntu 11.04 32bit with R 2.13.0. I have latest java and flash and I have tried both Firefox and IE (both latest versions under Windows 7). All other plots exceot geomap in googleVis work just fine. I too would like to know how to solve this problem. Kind regards, Michael Phipps __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] return counts of elements on a table column depending on elements on another column
On 06/08/2011 03:06 PM, ads pit wrote: Hi, I am given the following table: head(hsa_refseq) chr genome regionstart stop nu strand nu.1nu.2 gene_id 1 chr1 hg19_refGeneCDS 6742 6751 0 +0 gene_id NM_032291 2 chr1 hg19_refGene exon 66999825 6751 0 +. gene_id NM_032291 3 chr1 hg19_refGeneCDS 67091530 67091593 0 +2 gene_id NM_032291 4 chr1 hg19_refGene exon 67091530 67091593 0 +. gene_id NM_032291 5 chr1 hg19_refGeneCDS 67098753 67098777 0 +1 gene_id NM_032291 6 chr2 hg19_refGene exon 67098753 67098777 0 +. gene_id NM_032291 What I've done is to find out how many of the elements on 3rd column are CDS, exon. sum(hsa_refseq$region==CDS) sum(hsa_refseq$region==exon) But what I would like is to print for each chromosome how many are exons and how many CDS. For example chr1 has 5 CDS and 2 exons chr2 has 10 CDS and 3 exons... Can you tell what should I add? Or if I am doing this wrong, how should I do it? Hi Nanami -- xtabs(~chr + region, hsa_refseq) might do the ticket. Martin Thank you, Regards, Nanami [[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. -- Computational Biology Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: M1-B861 Telephone: 206 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Results of CFA with Lavaan
Dear Sam, -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R Help Sent: June-08-11 5:57 PM To: John Fox Cc: r-help Subject: Re: [R] Results of CFA with Lavaan Yes, that is the difference. For the last SEM I built I fixed the factor variances to 1, and I think that's what I want to do for the CFA I'm doing now. Does that make sense for a CFA? Sure -- then the factor covariances are correlations. The point is that you have to do something to fix the metrics of the factors and identify the model. I'll try figuring out how to do that with lavaan later, but my model takes so long to fit that I can't try it right now. Maybe that should tell you something about the conditioning of the problem. Best, John Thanks, Sam On Wed, Jun 8, 2011 at 5:58 PM, John Fox j...@mcmaster.ca wrote: Dear Sam, In each case, the first observed variable is treated as a reference indicator with its coefficient fixed to 1 to establish the metric of the corresponding factor and therefore to identify the model. If you didn't do the same thing (or something equivalent, such as fixing the factor variances to 1) in specifying the model to sem::sem(), that might account for the problems you encountered. Best, John John Fox Senator William McMaster Professor of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of R Help Sent: June-08-11 4:15 PM To: r-help Subject: [R] Results of CFA with Lavaan I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent estimate that is set to 1 with no SD for each factor? Is that normal? When I've managed to get sem::sem to fit a model this has not been the case. Thanks, Sam Stewart HS.model - ' visual =~ x1 + x2 + x3 textual =~ x4 + x5 + x6 speed =~ x7 + x8 + x9 ' fit - sem(HS.model, data=HolzingerSwineford1939) summary(fit, fit.measures=TRUE) Lavaan (0.4-8) converged normally after 35 iterations Number of observations 301 Estimator ML Minimum Function Chi-square 85.306 Degrees of freedom 24 P-value 0.000 Chi-square test baseline model: Minimum Function Chi-square 918.852 Degrees of freedom 36 P-value 0.000 Full model versus baseline model: Comparative Fit Index (CFI) 0.931 Tucker-Lewis Index (TLI) 0.896 Loglikelihood and Information Criteria: Loglikelihood user model (H0) -3737.745 Loglikelihood unrestricted model (H1) -3695.092 Number of free parameters 21 Akaike (AIC) 7517.490 Bayesian (BIC) 7595.339 Sample-size adjusted Bayesian (BIC) 7528.739 Root Mean Square Error of Approximation: RMSEA 0.092 90 Percent Confidence Interval 0.071 0.114 P-value RMSEA = 0.05 0.001 Standardized Root Mean Square Residual: SRMR 0.065 Parameter estimates: Information Expected Standard Errors Standard Estimate Std.err Z-value P(|z|) Latent variables: visual =~ x1 1.000 x2 0.554 0.100 5.554 0.000 x3 0.729 0.109 6.685 0.000 textual =~ x4 1.000 x5 1.113 0.065 17.014 0.000 x6 0.926 0.055 16.703 0.000 speed =~ x7 1.000 x8 1.180 0.165 7.152 0.000 x9 1.082 0.151 7.155 0.000 Covariances: visual ~~ textual 0.408 0.074 5.552 0.000 speed 0.262 0.056 4.660 0.000 textual ~~ speed 0.173 0.049 3.518 0.000 Variances: x1 0.549 0.114 4.833 0.000 x2 1.134 0.102 11.146
Re: [R] Autocorrelation in R
On Wed, 8 Jun 2011, Iuri Gavronski wrote: Hi, I am trying to learn time series, and I am attending a colleague's course on Econometrics. However, he uses e-views, and I use R. I am trying to reproduce his examples in R, but I am having problems specifying a AR(1) model. Would anyone help me with my code? Thanks in advance! Reproducible code follows: download.file(https://sites.google.com/a/proxima.adm.br/main/ex_32.csv --no-check-certificate, ex_32.csv, method=wget) ex32=read.csv(ex_32.csv) lm_ex32=lm(gc ~ yd, data=ex32) summary(lm_ex32) # Durbin-Watson (slide 26) library(lmtest) dwtest(gc ~ yd, data=ex32) # or dwtest(lm_ex32) # Breusch-Godfrey bgtest(lm_ex32, order=2) # AR(1) # In e-views, the specification was: # GC = YD AR(1) # and the output was: # Dependent Variable: GC # Method: Least Squares # Sample: 1970Q2 1995Q2 # Included observations: 101 # Convergence achieved after 6 interations # = # Variable Coefficient Std.Error t-Statistic Prob. # C -56.99706 19.84692 -2.871835 0.0050 # YD 0.937035 0.006520 143.7170 0. # AR(1) 0.752407 0.066565 11.30338 0. # = # R-squared 0.999691 Mean dependent var 2345.867 # Adjusted R-squared 0.999685 S.D. dependent var 1284.675 # S.E. of regression 22.81029 Akaike info criterion 9.121554 # Sum squared resid 50990.32 Schwarz criterion 9.199231 # Log likelihood -457.6385 F-statistic 158548.1 # Durbin-Watson stat 2.350440 Prob(F-statistic) 0.00 I'm not sure what exactly E-Views does here, but an ARIMAX(1,0,0) model estimated by least squares seems to come rather close. ## create a time series object of your data ex32ts - ts(ex32[,-1], start = c(1954, 1), freq = 4) ## select relevant subset ex32ts1 - window(ex32ts, start = c(1970, 2)) ## fit ARIMAX(1,0,0) model m - arima(ex32ts1[,gc], order = c(1, 0, 0), xreg = ex32ts1[,yd], method = CSS) ## print output, coefficient tests, etc. m coeftest(m) logLik(m) It seems to be slightly different but that can well be due to different fitting algorithms... hth, Z # following code based on http://www.stat.pitt.edu/stoffer/tsa2/R_time_series_quick_fix.htm # And now for some regression with autocorrelated errors. # I've tried to follow the example in Pinheiro Bates (2004), p. 239-244, with no success. gc_ts = ts(ex32[66:166,gc]) yd_ts = ts(ex32[66:166,yd]) library(nlme) trend = time(gc_ts) fit_lm = lm(gc_ts ~ trend + yd_ts) acf(resid(fit_lm)) pacf(resid(fit_lm)) gls_ex32_ar1 = gls(gc_ts ~ trend + yd_ts, correlation = corAR1(form= ~yd_ts),method=ML) summary(gls_ex32_ar1) _ Dr. Iuri Gavronski Assistant Professor Programa de Pós-Graduação em Administração Universidade do Vale do Rio dos Sinos ? UNISINOS Av. Unisinos, 950 ? São Leopoldo ? RS ? Brasil Sala (Room) 5A 406 D 93022-000 www.unisinos.br TEL +55-51-3591-1122 ext. 1589 FAX +55-51-3590-8447 Email: igavron...@unisinos.br CV Lattes: http://lattes.cnpq.br/8843390959025944 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Automating a process
I have a series of strings and I am trying to find all combinations and then assign 1 or 0 to them based on whether they contain the words car or budged. I want the data to look like: car budget cheap car insurance quote10 budget car insurance quote 11 cheap auto insurance quote 00 budget auto insurance quote 01 cheap car insurance quotes 10 budget car insurance quotes 10 cheap auto insurance quotes 00 budget auto insurance quotes 01 I've created all the possible values using the following commands, but I wanted to know if I could automate the process whereby it would create the string combinations and assign binary values appropriately. roots - c(car insurance, auto insurance) prefix - c(cheap, budget) suffix - c(quote, quotes) d - do.call(paste, expand.grid(prefix, roots, suffix)) df = data.frame(keyword=c(d)) df I realize that this process can be done in two separate stages with the stringr package, but I'm trying to just get the thing done in one. Do I need to combine both the functions from above and the stringr package inside a function? Help! I'm using R 2.13 on Ubuntu 10.10 Abraham WebRep Overall rating [[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: missing values where TRUE/FALSE needed
I'm writing a function and keep getting the following error message.
myfunc - function(lst) {
lst - list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low cost), suffix = c(quote, quotes),
suffix2 = c(rate, rates), suffix3 = c(comparison))
myone - function(x, y) {
nu - do.call(paste, expand.grid(lst$x, lst$y))
mydf - data.frame(keyword=c(nu))
}
mytwo - function(x, y, z){
mu - do.call(paste, expand.grid(lst$x, lst$y, lst$z))
mydf2 - data.frame(keyword=c(mu))
}
d1 = mytwo(lst$prefix, lst$roots, lst$suffix)
d2 = mytwo(lst$prefix, lst$roots, lst$suffix2)
d3 = mytwo(lst$prefix, lst$roots, lst$suffix3)
d4 = mytwo(lst$prefix2, lst$roots, lst$suffix)
d5 = mytwo(lst$prefix2, lst$roots, lst$suffix2)
d6 = mytwo(prefix2, roots, suffix3)
d7 = mytwo(prefix, roots2, suffix)
d8 = mytwo(prefix, roots2, suffix2)
d9 = mytwo(prefix, roots2, suffix3)
d10 = mytwo(prefix2, roots2, suffix)
d11 = mytwo(prefix2, roots2, suffix2)
d12 = mytwo(prefix2, roots2, suffix3)
d13 = myone(prefix, roots)
d14 = myone(prefix2, roots)
d15 = myone(prefix, roots2)
d16 = myone(prefix2, roots2)
d17 = myone(roots, suffix)
d18 = myone(roots, suffix2)
d19 = myone(roots, suffix3)
d20 = myone(roots2, suffix)
d21 = myone(roots2, suffix2)
d22 = myone(roots2, suffix3)
d23 = myone(state, roots)
d24 = myone(city, roots)
d25 = myone(cityst, roots)
d26 = myone(inscompany, roots)
d27 = myone(state, roots2)
d28 = myone(city, roots2)
d29 = myone(cityst, roots2)
d30 = myone(inscompany, roots2)
d31 = mytwo(state, roots, suffix)
d32 = mytwo(city, roots, suffix)
d33 = mytwo(cityst, roots, suffix)
d34 = mytwo(inscompany, roots, suffix)
d35 = mytwo(state, roots, suffix2)
d36 = mytwo(city, roots, suffix2)
d37 = mytwo(cityst, roots, suffix2)
d38 = mytwo(inscompany, roots, suffix2)
d39 = mytwo(state, roots, suffix3)
d40 = mytwo(city, roots, suffix3)
d41 = mytwo(cityst, roots, suffix3)
d42 = mytwo(inscompany, roots, suffix3)
d43 = mytwo(state, roots2, suffix)
d44 = mytwo(city, roots2, suffix)
d45 = mytwo(cityst, roots2, suffix)
d46 = mytwo(inscompany, roots2, suffix)
d47 = mytwo(state, roots2, suffix2)
d48 = mytwo(city, roots2, suffix2)
d49 = mytwo(cityst, roots2, suffix2)
d50 = mytwo(inscompany, roots2, suffix2)
d51 = mytwo(state, roots2, suffix3)
d52 = mytwo(city, roots2, suffix3)
d53 = mytwo(cityst, roots2, suffix3)
d54 = mytwo(inscompany, roots2, suffix3)
d55 = mytwo(prefix, state, roots)
d56 = mytwo(prefix, city, roots)
d57 = mytwo(prefix, cityst, roots)
d58 = mytwo(prefix, inscompany, roots)
d59 = mytwo(prefix2, state, roots)
d60 = mytwo(prefix2, city, roots)
d61 = mytwo(prefix2, cityst, roots)
d62 = mytwo(prefix2, inscompany, roots)
d63 = mytwo(prefix, state, roots2)
d64 = mytwo(prefix, city, roots2)
d65 = mytwo(prefix, cityst, roots2)
d66 = mytwo(prefix, inscompany, roots2)
d67 = mytwo(prefix2, state, roots2)
d68 = mytwo(prefix2, city, roots2)
d69 = mytwo(prefix2, cityst, roots2)
d70 = mytwo(prefix2, inscompany, roots2)
mydf - rbind(d1, d2, d3, d4, d5, d6, d7, d8, d9, d10, d11, d12, d13, d14,
d15,
d16, d17, d18, d19, d20, d21, d22, d23, d24, d25, d26, d27, d28, d29,
d30, d31, d32, d33, d34, d35, d36, d37, d38, d39, d40, d41, d42, d43,
d44, d45, d46, d47, d48, d49, d50, d51, d52, d53, d54, d55, d56, d57,
d58, d59, d60, d61, d62, d63, d64, d65, d66, d67, d68, d69, d70)
library(stringr)
inscompany_match - str_c(inscompany, collapse = |)
state_match - str_c(state, collapse = |)
city_match - str_c(city, collapse = |)
mydf$inscompany - as.numeric(str_detect(mydf$keyword, inscompany_match))
mydf$state - as.numeric(str_detect(mydf$keyword, state_match))
mydf$city - as.numeric(str_detect(mydf$keyword, city_match))
for (i in 1:nrow(mydf)) {
Words = strsplit(as.character(mydf[i, 'keyword']), )[[1]]
if(any(Words == 'Colorado')){
if(Words[which(Words == 'Colorado') + 1] == 'Springs') mydf[i, 'state'] - 0
}
if(any(Words == 'Virginia')){
if(Words[which(Words == 'Virginia') + 1] == 'Beach') mydf[i, 'state'] - 0
}
if(any(Words == 'Oklahoma')){
if(Words[which(Words == 'Oklahoma') + 1] == 'City') mydf[i, 'state'] - 0
}
if(any(Words == 'Kansas')){
if(Words[which(Words == 'Kansas') + 1] == 'City') mydf[i, 'state'] - 0
}
if(any(Words == 'Washington')){
if(Words[which(Words == 'Washington') + 1] == 'DC') mydf[i, 'state'] - 0
}
if(any(Words == 'York')){
if(Words[which(Words == 'York') + 1] == 'City') mydf[i, 'state'] - 0
}
if(any(Words == Indianapolis)){
mydf[i, 'state'] - 0
}
if(any(Words == AARP)){
mydf[i, 'state'] - 0
}
if(any(Words == ANPAC)){
mydf[i, 'state'] - 0
}
if(any(Words == AMICA)){
mydf[i, 'state'] - 0
}
if(any(Words == GMAC)){
mydf[i, 'state'] - 0
}
if(any(Words == USAA)){
mydf[i, 'state'] - 0
}
return(mydf)
}
}
newdf - myfunc(lst)
newdf - myfunc(lst)
Error in if (any(Words == Colorado)) { :
missing value where TRUE/FALSE needed
What's going on?
[[alternative HTML version deleted]]
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[R] ANOVA with many IV's
I'd like to conduct one-way ANOVA's on multiple IV's. Is there a function for aov (y~all of my IV's)? Thank you! -- View this message in context: http://r.789695.n4.nabble.com/ANOVA-with-many-IV-s-tp3583788p3583788.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] Using a function inside a function
I'm trying to run a function inside a function but get an error message.
lst - list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low cost), suffix = c(quote, quotes),
suffix2 = c(rate, rates), suffix3 = c(comparison))
myfunc - function(lst) {
myone - function(x, y) {
nu - do.call(paste, expand.grid(x, y))
mydf - data.frame(keyword=c(nu))
}
mytwo - function(x, y, z){
mu - do.call(paste, expand.grid(x, y, z))
mydf2 - data.frame(keyword=c(mu))
}
d1 = mytwo(lst$prefix, lst$roots, lst$suffix)
d2 = mytwo(lst$prefix, lst$roots, lst$suffix2)
d3 = mytwo(lst$prefix, lst$roots, lst$suffix3)
d4 = mytwo(lst$prefix2, lst$roots, lst$suffix)
d5 = mytwo(lst$prefix2, lst$roots, lst$suffix2)
df = rbind(d1, d2, d3, d4, d5)
}
I get the following error message:
newdf - myfunc(lst)
Error in expand.grid(x, y) : object 'x' not found
Can anyone help!
I'm running R 2.13 on Ubuntu 10.10
Abraham
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Resources for utilizing multiple processors
Hello, I know of some various methods out there to utilize multiple processors but am not sure what the best solution would be. First some things to note: I'm running dependent simulations, so direct parallel coding is out (multicore, doSnow, etc). I'm on Windows, and don't know C. I don't plan on learning C or any of the *nix languages. My main concern deals with Multiple analyses on large data sets. By large I mean that when I'm done running 2 simulations R is using ~3G of RAM, the remaining ~3G is chewed up when I try to create the Gelman-Rubin statistic to compare the two resulting samples, grinding the process to a halt. I'd like to have separate cores simultaneously run each analysis. That will save on time and I'll have to ponder the BGR calculation problem another way. Can R temporarily use HD space to write calculations to instead of RAM? The second concern boils down to whether or not there is a way to split up dependent simulations. For example at iteration (t) I feed a(t-2) into FUN1 to generate a(t), then feed a(t), b(t-1) and c(t-1) into FUN2 to simulate b(t) and c(t). I'd love to have one core run FUN1 and another run FUN2, and better yet, a third to run all the pre-and post- processing tidbits! So if anyone has any suggestions as to a direction I can look into, it would be appreciated. Robin Jeffries MS, DrPH Candidate Department of Biostatistics UCLA 530-633-STAT(7828) [[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] ANOVA with many IV's
Hi:
You can try something like this: assuming the factor variables of
interest and the response variable are in a data frame named df,
ivset - c(comma separated vector of factor names)
myaovs - lapply(ivset, function(x) {
form - as.formula(substitute(yvar ~ foo, list(foo = as.name(x
aov(form, data = df)
} )
This should generate a list of aov objects, one per factor in ivset.
From there, R has functions to extract pieces of output as needed.
Since no example data was presented, the above is untested, so caveat emptor.
HTH,
Dennis
On Wed, Jun 8, 2011 at 3:25 PM, mandakaye mandak...@gmail.com wrote:
I'd like to conduct one-way ANOVA's on multiple IV's. Is there a function for
aov (y~all of my IV's)? Thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/ANOVA-with-many-IV-s-tp3583788p3583788.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using a function inside a function
Hi Abraham,
Your example runs fine for me. I get this as the newdf object (you
should be able to copy and paste into your console):
newdf - structure(list(keyword = structure(c(7L, 3L, 5L, 1L, 8L, 4L,
6L, 2L, 15L, 11L, 13L, 9L, 16L, 12L, 14L, 10L, 20L, 18L, 19L,
17L, 23L, 21L, 24L, 22L, 27L, 25L, 28L, 26L), .Label = c(budget auto
insurance quote,
budget auto insurance quotes, budget car insurance quote,
budget car insurance quotes, cheap auto insurance quote,
cheap auto insurance quotes, cheap car insurance quote, cheap car
insurance quotes,
budget auto insurance rate, budget auto insurance rates,
budget car insurance rate, budget car insurance rates, cheap auto
insurance rate,
cheap auto insurance rates, cheap car insurance rate, cheap car
insurance rates,
budget auto insurance comparison, budget car insurance comparison,
cheap auto insurance comparison, cheap car insurance comparison,
low cost auto insurance quote, low cost auto insurance quotes,
low cost car insurance quote, low cost car insurance quotes,
low cost auto insurance rate, low cost auto insurance rates,
low cost car insurance rate, low cost car insurance rates
), class = factor)), .Names = keyword, row.names = c(NA,
28L), class = data.frame)
Try re-running the example you gave in your email in a clean R
session, perhaps. If that does not work, you will need to provide
more information.
Cheers,
Josh
On Wed, Jun 8, 2011 at 5:43 PM, Abraham Mathew abmathe...@gmail.com wrote:
I'm trying to run a function inside a function but get an error message.
lst - list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low cost), suffix = c(quote, quotes),
suffix2 = c(rate, rates), suffix3 = c(comparison))
myfunc - function(lst) {
myone - function(x, y) {
nu - do.call(paste, expand.grid(x, y))
mydf - data.frame(keyword=c(nu))
}
mytwo - function(x, y, z){
mu - do.call(paste, expand.grid(x, y, z))
mydf2 - data.frame(keyword=c(mu))
}
d1 = mytwo(lst$prefix, lst$roots, lst$suffix)
d2 = mytwo(lst$prefix, lst$roots, lst$suffix2)
d3 = mytwo(lst$prefix, lst$roots, lst$suffix3)
d4 = mytwo(lst$prefix2, lst$roots, lst$suffix)
d5 = mytwo(lst$prefix2, lst$roots, lst$suffix2)
df = rbind(d1, d2, d3, d4, d5)
}
I get the following error message:
newdf - myfunc(lst)
Error in expand.grid(x, y) : object 'x' not found
Can anyone help!
I'm running R 2.13 on Ubuntu 10.10
Abraham
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: missing values where TRUE/FALSE needed
Hi Abraham,
mylist - list(roots = car, prefix = cheap)
myfoo - function(x) {
print(mylist$x)
}
myfoo(roots) ## fails, but in a sneaky way
## you actually extract variable x from mylist
## but there is no variable x (it is just NULL)
## so while no error is thrown, you get nothing
myfoo - function(x) {
print(mylist[[x]])
}
myfoo(roots) ## still fails but...
myfoo(roots) ## works
So I would rewrite 'myone' and 'mytwo' to use: lst[[x]] and then put
the quotes around the names
mytwo(prefix, roots2, suffix) etc.
Also look at ?debug to see ways to debug your function so you can more
easily find where the problem starts rather than what the final death
blow was. The usage is simple, just:
debug(myfunc) ## now next time myfunc() is called, it will be debugged
myfunc(lst) ## and off you go
Hope this helps,
Josh
On Wed, Jun 8, 2011 at 8:45 PM, Abraham Mathew abmathe...@gmail.com wrote:
I'm writing a function and keep getting the following error message.
myfunc - function(lst) {
lst - list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low cost), suffix = c(quote, quotes),
suffix2 = c(rate, rates), suffix3 = c(comparison))
myone - function(x, y) {
nu - do.call(paste, expand.grid(lst$x, lst$y))
mydf - data.frame(keyword=c(nu))
}
mytwo - function(x, y, z){
mu - do.call(paste, expand.grid(lst$x, lst$y, lst$z))
mydf2 - data.frame(keyword=c(mu))
}
d1 = mytwo(lst$prefix, lst$roots, lst$suffix)
d2 = mytwo(lst$prefix, lst$roots, lst$suffix2)
d3 = mytwo(lst$prefix, lst$roots, lst$suffix3)
d4 = mytwo(lst$prefix2, lst$roots, lst$suffix)
d5 = mytwo(lst$prefix2, lst$roots, lst$suffix2)
d6 = mytwo(prefix2, roots, suffix3)
d7 = mytwo(prefix, roots2, suffix)
d8 = mytwo(prefix, roots2, suffix2)
d9 = mytwo(prefix, roots2, suffix3)
d10 = mytwo(prefix2, roots2, suffix)
d11 = mytwo(prefix2, roots2, suffix2)
d12 = mytwo(prefix2, roots2, suffix3)
d13 = myone(prefix, roots)
d14 = myone(prefix2, roots)
d15 = myone(prefix, roots2)
d16 = myone(prefix2, roots2)
d17 = myone(roots, suffix)
d18 = myone(roots, suffix2)
d19 = myone(roots, suffix3)
d20 = myone(roots2, suffix)
d21 = myone(roots2, suffix2)
d22 = myone(roots2, suffix3)
d23 = myone(state, roots)
d24 = myone(city, roots)
d25 = myone(cityst, roots)
d26 = myone(inscompany, roots)
d27 = myone(state, roots2)
d28 = myone(city, roots2)
d29 = myone(cityst, roots2)
d30 = myone(inscompany, roots2)
d31 = mytwo(state, roots, suffix)
d32 = mytwo(city, roots, suffix)
d33 = mytwo(cityst, roots, suffix)
d34 = mytwo(inscompany, roots, suffix)
d35 = mytwo(state, roots, suffix2)
d36 = mytwo(city, roots, suffix2)
d37 = mytwo(cityst, roots, suffix2)
d38 = mytwo(inscompany, roots, suffix2)
d39 = mytwo(state, roots, suffix3)
d40 = mytwo(city, roots, suffix3)
d41 = mytwo(cityst, roots, suffix3)
d42 = mytwo(inscompany, roots, suffix3)
d43 = mytwo(state, roots2, suffix)
d44 = mytwo(city, roots2, suffix)
d45 = mytwo(cityst, roots2, suffix)
d46 = mytwo(inscompany, roots2, suffix)
d47 = mytwo(state, roots2, suffix2)
d48 = mytwo(city, roots2, suffix2)
d49 = mytwo(cityst, roots2, suffix2)
d50 = mytwo(inscompany, roots2, suffix2)
d51 = mytwo(state, roots2, suffix3)
d52 = mytwo(city, roots2, suffix3)
d53 = mytwo(cityst, roots2, suffix3)
d54 = mytwo(inscompany, roots2, suffix3)
d55 = mytwo(prefix, state, roots)
d56 = mytwo(prefix, city, roots)
d57 = mytwo(prefix, cityst, roots)
d58 = mytwo(prefix, inscompany, roots)
d59 = mytwo(prefix2, state, roots)
d60 = mytwo(prefix2, city, roots)
d61 = mytwo(prefix2, cityst, roots)
d62 = mytwo(prefix2, inscompany, roots)
d63 = mytwo(prefix, state, roots2)
d64 = mytwo(prefix, city, roots2)
d65 = mytwo(prefix, cityst, roots2)
d66 = mytwo(prefix, inscompany, roots2)
d67 = mytwo(prefix2, state, roots2)
d68 = mytwo(prefix2, city, roots2)
d69 = mytwo(prefix2, cityst, roots2)
d70 = mytwo(prefix2, inscompany, roots2)
mydf - rbind(d1, d2, d3, d4, d5, d6, d7, d8, d9, d10, d11, d12, d13, d14,
d15,
d16, d17, d18, d19, d20, d21, d22, d23, d24, d25, d26, d27, d28, d29,
d30, d31, d32, d33, d34, d35, d36, d37, d38, d39, d40, d41, d42, d43,
d44, d45, d46, d47, d48, d49, d50, d51, d52, d53, d54, d55, d56, d57,
d58, d59, d60, d61, d62, d63, d64, d65, d66, d67, d68, d69, d70)
library(stringr)
inscompany_match - str_c(inscompany, collapse = |)
state_match - str_c(state, collapse = |)
city_match - str_c(city, collapse = |)
mydf$inscompany - as.numeric(str_detect(mydf$keyword, inscompany_match))
mydf$state - as.numeric(str_detect(mydf$keyword, state_match))
mydf$city - as.numeric(str_detect(mydf$keyword, city_match))
for (i in 1:nrow(mydf)) {
Words = strsplit(as.character(mydf[i, 'keyword']), )[[1]]
if(any(Words == 'Colorado')){
if(Words[which(Words == 'Colorado') + 1] == 'Springs') mydf[i, 'state'] - 0
}
if(any(Words == 'Virginia')){
if(Words[which(Words == 'Virginia') + 1] == 'Beach') mydf[i, 'state'] - 0
}
if(any(Words ==
[R] a bug in heatmap.plus?
Hi Allen and list, See the code below. I've tried it on R2.13 and R2.8.0 using either heatmap.plus 1.3 or the latest. All gave the same results. The problem is in the last line: when I tried to plot two different color bars, the one corresponding to cm.colors(10) is not correct (it starts with one black and one red. Not sure where they're from?) Any ideas? Thanks! ...Tao library(heatmap.plus) set.seed(1234) x - matrix(rnorm(400), ncol=10) heatmap(x, ColSideColors= cm.colors(10), Colv=NA) heatmap.plus(x, ColSideColors=cbind(cm.colors(10), cm.colors(10)), Colv=NA) heatmap.plus(x, ColSideColors=cbind(rep(1:2,each=5), cm.colors(10)), Colv=NA) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Permission denied in Windows 7
I'm using package.skeleton() windows 7, 64 bit. When I try to specify the code_files package_skeleton(code_files = some directory) I get a warning that that the connection cannot be opened and I get a Permissions denied error. I'm running R as admin and I've given everybody full permissions on the folder. What am I missing [[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] Questions about building R packages
here i wrote a step by step tutorial. http://stevemosher.wordpress.com/2011/06/09/making-simple-packages-in-r-on-windows/ On Wed, Jun 8, 2011 at 1:17 PM, Xia.Li oddity...@gmail.com wrote: Hello R users, I have difficulties when trying to make R packages. I tried to read many tutorials but still could not find out the right way. Could any one help me out please? (I'm using Windows xp.) After running package.skeleton() and edit those RD files, I don't know how to use Rtools (or CMD shell?) to build the zip file. I installed the Rtools from Murdoch's link, but it doesn't look like a software... Anyone could give me a tutorial with more details about Rtools (or CMD shell)? I've already have MikTex installed. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Questions-about-building-R-packages-tp3583510p3583510.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ANOVA with many IV's
It is not nearly as complicated as Dennis Murphy makes out.
Just do
aov(y ~ ., data=X)
where X is your data frame with one column (the response) name ``y''
and then any number of other columns which will then form the
predictors (which may be either numeric predictors or factors).
cheers,
Rolf Turner
On 09/06/11 16:14, Dennis Murphy wrote:
Hi:
You can try something like this: assuming the factor variables of
interest and the response variable are in a data frame named df,
ivset- c(comma separated vector of factor names)
myaovs- lapply(ivset, function(x) {
form- as.formula(substitute(yvar ~ foo, list(foo = as.name(x
aov(form, data = df)
} )
This should generate a list of aov objects, one per factor in ivset.
From there, R has functions to extract pieces of output as needed.
Since no example data was presented, the above is untested, so caveat emptor.
HTH,
Dennis
On Wed, Jun 8, 2011 at 3:25 PM, mandakayemandak...@gmail.com wrote:
I'd like to conduct one-way ANOVA's on multiple IV's. Is there a function for
aov (y~all of my IV's)? Thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/ANOVA-with-many-IV-s-tp3583788p3583788.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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Histogram
Hi Sam,This is exactly what I wanted. Could you please explain the code ? what does 15, 0.65 and 0.25 stand for ?Nandini Date: Wed, 8 Jun 2011 15:16:45 -0700 From: ml-node+3583766-897200094-233...@n4.nabble.com To: nandini...@hotmail.com Subject: Re: Histogram I think the command you want is barplot x = rbinom(10,15,0.65) y = rbinom(10,15,0.25) barplot(rbind(x,y),beside=TRUE) Sam On Wed, Jun 8, 2011 at 10:14 AM, nandini_bn [hidden email] wrote: Hello , I am trying to create a histogram in order to compare between two groups and would like it to be similar to the figure attached. How can I generate this using R ? Thank you, Nandini http://r.789695.n4.nabble.com/file/n3582448/5634-15977-1-PB.gif -- View this message in context: http://r.789695.n4.nabble.com/Histogram-tp3582448p3582448.html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Histogram-tp3582448p3583766.html To unsubscribe from Histogram, click here. -- View this message in context: http://r.789695.n4.nabble.com/Histogram-tp3582448p3584425.html Sent from the R help mailing list archive at Nabble.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.
