Re: [R] detect filetype (as in unix 'file')
On Sun, 3 Apr 2011, Jeroen Ooms wrote: Is there a way in R (in Linux) to detect the type of a file without invoking a shell? E.g to do this: system(file density.plot) density.plot: PDF document, version 1.4 but without using system()? I tried file() and file.info(), but both do display the information I am looking for. No, but what is wrong with using system()? 'file' is large and complex because it tries to be comprehensive (but it still does not know about some common systems, e.g. 64-bit Windows binaries). There simply is no point in replicating that in R: which is why we chose rather to port 'file' to Windows and provide in in Rools. -- 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.
[R] RmetricsTools.R
Hello I read on the Rmetrics webpage that all the development packages could be intalled using the following command source(RmetricsTools.R) install.RmetricsDev([Rmetrics package to be installed] I would like to know where I could get this RmetricsTools.R . I suppose it might be somewhere on R-Forge but I haven't been able to find it. Thank you Felipe Parra [[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] loading R object files on an RWeb server
jose romero-3 wrote: For one thing, i am using google docs to host the R object file and google docs has secure https URL's, which apparently cannot be handled by R's url(). So my questions are these: Try ?getURL in the RCurl package Dieter -- View this message in context: http://r.789695.n4.nabble.com/loading-R-object-files-on-an-RWeb-server-tp3424613p3424761.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] Support Counting
Hi,
I'm new to R and trying to some simple analysis. I have a data set with
about 88000 transactions and i want to perform a simple support count
analysis of an itemset which is say not a complete transaction but a subset
of a transaction.
say
{A,B,D} is a transaction and i want to find support of {A,B} even though it
never occurs as only A,B in the entire set
To this i needed to create a new itemsets class and then use the support
function but somehow the answers never seem to tally.
Thanks in advance
Srinivas
--
View this message in context:
http://r.789695.n4.nabble.com/Support-Counting-tp3424730p3424730.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] Structural equation modeling in R(lavaan,sem)
On 04/03/2011 09:38 PM, jouba wrote: Daer all, I have a question concerning longitudinal data: When we have a longitudinal data and we have to do sem analysis there is in the package lavaan some functions,options in this package that help to do this or we can treat these data like non longitudinal data The function 'growth' (in the lavaan package) can be used for (standard) growth modeling. Good material about growth modeling (using Mplus) can be found here: http://statistics.ats.ucla.edu/stat/mplus/seminars/gm/default.htm Next, you can read how to do growth modeling with lavaan by reading section 7 in the lavaan intro, which you can download from the documentation section on the lavaan website (http://lavaan.org). Yves. -- Yves Rosseel -- http://www.da.ugent.be Department of Data Analysis, Ghent University Henri Dunantlaan 1, B-9000 Gent, Belgium __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] multiple variables Y and X
Hello I have a model with several hundred Y variables, and also several 1000 X variables. The model is linear lm(Y ~ X). My questions are: 1.- how to avoid writing all Xs variables? is list() the right function? 2.- about the multiple Ys with dependence among some of them, how to incorporate that information in the linear model? Thank you Rosario __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] MySql Versus R
Dear All, Thank you to all of you for your fast reply. I will run the test and subscribe to the R-sig-db list. Cheers, Henri __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] RmetricsTools.R
LFP == Luis Felipe Parra felipe.pa...@quantil.com.co on Mon, 4 Apr 2011 14:32:32 +0800 writes: LFP Hello I read on the Rmetrics webpage that all the LFP development packages could be intalled using the LFP following command LFP source(RmetricsTools.R) LFP install.RmetricsDev([Rmetrics package to be installed] LFP I would like to know where I could get this LFP RmetricsTools.R . I suppose it might be somewhere on LFP R-Forge but I haven't been able to find it. Thank you It's top-level. With the standard subversion client, you can get it, e.g., by svn export svn://svn.r-forge.r-project.org/svnroot/rmetrics/pkg/RmetricsTools.R -- Martin Maechler, ETH Zurich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 in color2D.matplot : Error in plot.new() : figure margins too large
On 04/03/2011 10:02 PM, shahab wrote: Hi, I am using color2D.matplot (...) function of plotrix package. I used a matrix of size around 20*20 However, apparently it failed to visualize the matrix and gave the following exception, which I don't have any idea about possible source of this error. Error in plot.new() : figure margins too large It would be appreciated if someone points me to the right origin of this error. Hi Shahab, If you could send me the data you used (real or fake, as long as it produces the error), I'll try to work out what has happened. 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] svd
Dear list, I searched the libraries but could not find means to compute the svd of a coupled field. Is it possible in R Thanks nuncio -- Nuncio.M Research Scientist National Center for Antarctic and Ocean research Head land Sada Vasco da Gamma Goa-403804 [[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] power of 2 way ANOVA with interaction
On Apr 4, 2011, at 01:10 , Timothy Spier wrote: I've been searching for an answer to this for a while but no joy. I have a simple 2-way ANOVA with an interaction. I'd like to determine the power of this test for each factor (factor A, factor B, and the A*B interaction). How can I do this in R? I used to do this with proc Glmpower in SAS, but I can find no analogue in R. They're not massively hard to do by hand, if you know what you're doing (which, admittedly is a bit hard to be sure of in this case). The basic structure can be lifted from power.anova.test and the name of the game is to work out the noncentrality parameter of the relevant F tests. E.g., lifting an example from the SAS manual: twoway - cbind(expand.grid(ex=factor(1:2),var=factor(1:3)),x=c(14,10,16,15,21,16)) with(twoway,tapply(x,list(ex,var),mean)) 1 2 3 1 14 16 21 2 10 15 16 Now, you have 10 replicates of this with a specified SD of 5. If we do a skeleton analysis of the above table, we get anova(lm(x~ex*var,twoway)) Analysis of Variance Table Response: x Df Sum Sq Mean Sq F value Pr(F) ex 1 16.667 16.6667 var2 42.333 21.1667 ex:var 2 4.333 2.1667 Residuals 0 0.000 Warning message: In anova.lm(lm(x ~ ex * var, twoway)) : ANOVA F-tests on an essentially perfect fit are unreliable In a 10-fold replication, the SS would be 10 times bigger, and the residual Df would be 54; also, we need to take the error variance of 5^2 = 25 into account. The noncentrality for the interaction term is thus 43.333/25 and you can work out the power as pf(qf(.95,2,54),2,54,ncp=43.333/25,lower=F) [1] 0.1914457 Similarly, the main effect powers are pf(qf(.95,2,54),2,54,423.333/25,lower=F) [1] 0.956741 pf(qf(.95,1,54),1,54,166.7/25,lower=F) [1] 0.7176535 (whatever that means in the presence of interaction, but that is a different discussion) -- 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.
Re: [R] svd
Thanks juan, I got that, but what I have two matrices A and B, How can an svd be performed on the two together. Is it correct to get the covariance matrix and then perform the svd on the covariance matrix. If that is the case I have another doubt. I understand the covariance of A and B is t(A)%*%B. but this differs significantly from cov(A,B). Thanks nuncio 2011/4/4 Juan Carlos Borrás jcbor...@gmail.com m - matrix(c(1:12), nrow=3, ncol=4) svd(m) On Mon, Apr 4, 2011 at 11:51 AM, nuncio m nunci...@gmail.com wrote: Dear list, I searched the libraries but could not find means to compute the svd of a coupled field. Is it possible in R Thanks nuncio -- Nuncio.M Research Scientist National Center for Antarctic and Ocean research Head land Sada Vasco da Gamma Goa-403804 [[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. -- Cheers, jcb! ___ http://twitter.com/jcborras -- Nuncio.M Research Scientist National Center for Antarctic and Ocean research Head land Sada Vasco da Gamma Goa-403804 [[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] multiple variables Y and X
Hello, You can do that like in the example included in the function step() lm1 - lm(Fertility ~ ., data = swiss) But my advise is that prior to doing that, you should check this old thread in this list: http://tolstoy.newcastle.edu.au/R/e4/help/08/02/2842.html Regards, Carlos Ortega www.qualityexcellence.es On Mon, Apr 4, 2011 at 8:58 AM, Rosario Garcia Gil m.rosario.gar...@slu.sewrote: Hello I have a model with several hundred Y variables, and also several 1000 X variables. The model is linear lm(Y ~ X). My questions are: 1.- how to avoid writing all Xs variables? is list() the right function? 2.- about the multiple Ys with dependence among some of them, how to incorporate that information in the linear model? Thank you Rosario __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Support Counting
On Mon, Apr 04, 2011 at 01:11:37AM -0500, psombe wrote:
Hi,
I'm new to R and trying to some simple analysis. I have a data set with
about 88000 transactions and i want to perform a simple support count
analysis of an itemset which is say not a complete transaction but a subset
of a transaction.
say
{A,B,D} is a transaction and i want to find support of {A,B} even though it
never occurs as only A,B in the entire set
To this i needed to create a new itemsets class and then use the support
function but somehow the answers never seem to tally.
Hi.
The answer depends on the representation of the data set. Can you
describe the representation?
A possible representation of a data set for itemsets counting is a matrix
of 0/1. Using this representation, computing the support may be done
as follows.
db - matrix(0, nrow=5, ncol=5, dimnames=list(NULL, LETTERS[1:5]))
db[1, c(A, B, D)] - 1
db[2, c(A, B)] - 1
db[3, c(A, D, E)] - 1
db[4, c(B, C, D)] - 1
db[5, c(A, B, C)] - 1
db
A B C D E
[1,] 1 1 0 1 0
[2,] 1 1 0 0 0
[3,] 1 0 0 1 1
[4,] 0 1 1 1 0
[5,] 1 1 1 0 0
itemset - c(A, B)
# for each transaction, whether it contains c(A, B)
rowSums(db[, itemset]) == length(itemset)
[1] TRUE TRUE FALSE FALSE TRUE
# the number of transactions containing c(A, B)
sum(rowSums(db[, itemset]) == length(itemset))
[1] 3
Hope this helps.
Petr Savicky.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] another question on shapefiles and geom_point in ggplot2
Hi all Just as Pierre pointed out, i used rgeos instead of gpclib (the licence is less restrictive even though i'm still students). A since the last release the rgeos package seems to work well. I'm really disapointed by the fact that some packages (e.g adehabitatMA which is great thank's M. Calenge) still use gpclib and don't offer the possibility to switch to rgeos, and is really messy to have two package to do the same task and with the same function name in the path (eg when i use adehabitatMA and maptools). Manuel: May be you should use the *colour* aesthetic instead of* *the* fill *one* *in the call of *geom_path*. Something like that : p = ggplot(geo, aes(x, y)) p + geom_point(aes(size = ACE, colour = ACE)) + theme_bw() + scale_size(name = Número de especies, breaks = c(2, 4, 6, 8, 10, 12, 14, 16, 18, 20)) + scale_colour_gradientn(name = 'Número de especies', colours = heat.colors(10), breaks = c(2, 4, 6, 8, 10, 12, 14, 16, 18, 20))+ xlab(Longitud) + ylab(Latitud) + opts(axis.text.x = theme_text(size = 8, vjust = 1)) + opts(axis.text.y = theme_text(size = 8, hjust = 1)) + geom_path(aes(x=long,y=lat, group=group, *colour=id* ),data=fortify.ai_biotica) not tested, so i'm not really sure about this one 2011/4/4 Pierre Roudier pierre.roud...@gmail.com Hi all, 2011/4/4 Felipe Carrillo mazatlanmex...@yahoo.com: Manuel: As far as I know one needs gpclibPermit() in order to fortify see this: Note: polygon geometry computations in maptools depend on the package gpclib, which has a restricted licence. It is disabled by default; to enable gpclib, type gpclibPermit() I am going to guess that ahmadou dicko doesn't show gpclibPermit() on his code because he loaded it with Rprofile or some other way. I tried to run his code without gpclibPermit() and it wouldn't let me fortify, so not sure how he did it. On that specific point, Colin Arundel and Roger Bivant released the rgeos package on CRAN a few days [1]. This is a great achievement as it brings bindings to the GEOS C++ lib [2] - long story short, it makes the job the non-free [3] gpclib used to do. In its later release, maptools has an option to check if rgeos if present - if it is the case it is used instead of gpclib: library(maptools) Loading required package: foreign Loading required package: sp Loading required package: lattice Note: polygon geometry computations in maptools depend on the package gpclib, which has a restricted licence. It is disabled by default; to enable gpclib, type gpclibPermit() Checking rgeos availability as gpclib substitute: TRUE ?gpclibPermit Pierre [1] http://cran.r-project.org/web/packages/rgeos/ [2] http://trac.osgeo.org/geos/ [3] https://stat.ethz.ch/pipermail/r-sig-geo/2010-January/007400.html -- Scientist Landcare Research, New Zealand -- You received this message because you are subscribed to the ggplot2 mailing list. Please provide a reproducible example: http://gist.github.com/270442 To post: email ggpl...@googlegroups.com To unsubscribe: email ggplot2+unsubscr...@googlegroups.com More options: http://groups.google.com/group/ggplot2 [[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] Deriving formula with deriv
Dear list,
Hi,
I am trying to get the second derivative of a logistic formula, in R summary
the model is given as :
###
$nls
Nonlinear regression model
model: data ~ logistic(time, A, mu, lambda, addpar)
data: parent.frame()
A mu lambda
0.53243 0.03741 6.94296
###
but I know the formula used is
# y~'A'/(1+exp((4*'mu'/'A')*('lambda'-'time'))+2))# from the grofit (
package I am using to fit the model) documentation.
I have attempted to use the R function 'deriv' to get the
first derivative from which I can then reuse the deriv function to get the
second derivative unfortunately this does not seem to work
###
express-expression(y~'A'/(1+exp((4*'mu'/'A')*('lambda'-'time'))+2))
express
expression(y ~ A/(1 + exp((4 * mu/A) * (lambda - time)) +
2))
d1-deriv(express)
Error in deriv.default(express) : element 2 is empty;
the part of the args list of '.Internal' being evaluated was:
(expr, namevec, function.arg, tag, hessian)
Why is this not working and how do I get the second derivative?
Thank you for reading my post, all help is appreciated,
Kitty
[[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] add zero in front of numbers
Dear R users, I need to add 0 in front of a series of numbers, e.g. 1-001, 19-019, Is there a fast way of doing that? Many thanks yan [[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] add zero in front of numbers
Have a look at ?formatC On Mon, Apr 4, 2011 at 11:35 AM, Yan Jiao y.j...@ucl.ac.uk wrote: Dear R users, I need to add 0 in front of a series of numbers, e.g. 1-001, 19-019, Is there a fast way of doing that? Many thanks yan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] principal components
On Fri, 2011-04-01 at 11:52 +0530, nuncio m wrote: HI all, I am trying to compute the EOF of a matrix using prcomp but unable to get the expansion co-efficients. is it possible using prcomp or are there any other methods thanks nuncio *sigh* RSiteSearch(EOF) It is at times like this that an R equivalent of: http://lmgtfy.com/ would be handy ;-) Or it slightly cruder cousin... G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Clarks 2Dt function in R
Dear Ben,
you answerd to Nancy Shackelford about Clarks 2Dt function.
Since the thread ended just after your reply,
I would like to ask, if you have an idea how to use this function in R
I defined it the following way:
function(x , p, u) {
(p/(pi*u))*(1+(x^2/u))^(p+1)
}
and would like to fit this one to my obeservational data (count)
[,1] [,2]
[1,] 15 12
[2,] 45 13
[3,] 75 10
[4,] 1058
[5,] 135 16
[6,] 1655
[7,] 195 15
[8,] 2258
[9,] 2559
[10,] 285 12
[11,] 3155
[12,] 3454
[13,] 3751
[14,] 4051
[15,] 4351
[16,] 4650
[17,] 4951
[18,] 5252
[19,] 5550
[20,] 5850
[21,] 6150
[22,] 6450
[23,] 6750
but I am not able to fit anything.
Do you have an idea?
I guess there is something wrong in my formula for Clarks 2Dt
Thank you for reading
Ciao
Ronald Bialozyt
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Please help
Dear Sir/Madam, I am stuck with a nagging problem in using R for SVM regression. My data has 5 dimensions and 400 observations. The independent variables are : Peb, Ksub, Sub, and Xtt. The dependent variable is: Rexp. I tried using the svm.tune function to tune the hyper parameters: gamma, epsilon and C. I am getting the following error message: Error in predict.svm(ret, xhold, decision.values+TRUE): Model is empty! May you please help me! SADAF ZAIDI Associate Professor Department of Aligarh Muslim University Aligarh 202002. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem using svm.tune
Dear Sir, I am stuck with a nagging problem in using R for SVM regression. My data has 5 dimensions and 400 observations. The independent variables are : Peb, Ksub, Sub, and Xtt. The dependent variable is: Rexp. I tried using the svm.tune function as well as _tune(svm.), to tune the hyper parameters: gamma, epsilon and C. Since I am new to R, I am probably not using the svm.tune function properly. I am getting the following error message: Error in predict.svm(ret, xhold, decision.values=TRUE): Model is empty! May you please help me!SADAF ZAIDI [[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] ticklabs in scatterplot3d
Hello, I am having a problem to control ticklabs in scatterplot3d. When I am using it for small data then it is working fine. But with my big data all the labels are misplaced (one upon another). For example see the code below (I have just modified the scatterplot3d example to show my problem ). == my.mat - matrix(runif(625), nrow=25) dimnames(my.mat) - list(LETTERS[1:25], letters[1:25]) my.mat # this matrix I want to plot with the all 25 ticklabs scatterplot3d(s3d.dat, type=h, lwd=1, pch= ,x.ticklabs=colnames(my.mat), y.ticklabs=rownames(my.mat)) == I want all 25 labels to be written in y.ticklabs Can anyone please help me? Thank you very much, Suparna. [[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] add zero in front of numbers
Dear Yan, apart from formatC, you can also use sprintf, which works almost exactly like the C sprintf function. To convert an integer x to a string with 5 leading 0s, you do: sprintf( %05d, x ) Best regards, j. On Mon, Apr 4, 2011 at 12:35 PM, Yan Jiao y.j...@ucl.ac.uk wrote: Dear R users, I need to add 0 in front of a series of numbers, e.g. 1-001, 19-019, Is there a fast way of doing that? Many thanks yan [[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. -- Dr. January Weiner 3 -- Max Planck Institute for Infection Biology Charitéplatz 1 D-10117 Berlin, Germany Web : www.mpiib-berlin.mpg.de Tel : +49-30-28460514 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lattice: how to center a subtitle?
Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html). A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. Cheers, Marius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] system() command in R
Hi all, I have a local server insalled on my system and have to start that from within my R function. here is how I start it: cmd-sh start-server.sh system(cmd, wait=FALSE) My function has to start the server and proceed with further steps. The server starts but the further steps of the program are not executed.The cursor keeps waiting after the server is started. i tried removing the wait=FALSE, but it still keeps waiting. I also tried putting the start-server in a separate function and my further script in a separate function and then run them together, but it still waits. The transition from the start of server to next step is not happening. Please help. I have been stuck on this for quite some time now. -- Rasanpreet Kaur [[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] another question on shapefiles and geom_point in ggplot2
Thank you Ahmadou, I got an error when I change fill to colour: p + geom_point(aes(size = ACE, colour = ACE)) + theme_bw() + + scale_size(to = c(3, 15), name = Número de especies, breaks = c(10, 40, 70, 100, 130, 160, 190, 220, 250)) + + scale_colour_gradientn(name = 'Número de especies', colours = heat.colors(9), breaks = c(10, 40, 70, 100, 130, 160, 190, 220, 250)) + xlab(Longitud) + ylab(Latitud) + + opts(axis.text.x = theme_text(size = 8, vjust = 1)) + opts(axis.text.y = theme_text(size = 8, hjust = 1)) + + geom_path(aes(x=long,y=lat,group=group, colour=id),data=fortify.ai_biotica) Error: Non-continuous variable supplied to scale_colour_gradientn. Best, Manuel On 04/04/2011 03:37 a.m., ahmadou dicko wrote: Hi all Just as Pierre pointed out, i used rgeos instead of gpclib (the licence is less restrictive even though i'm still students). A since the last release the rgeos package seems to work well. I'm really disapointed by the fact that some packages (e.g adehabitatMA which is great thank's M. Calenge) still use gpclib and don't offer the possibility to switch to rgeos, and is really messy to have two package to do the same task and with the same function name in the path (eg when i use adehabitatMA and maptools). Manuel: May be you should use the *colour* aesthetic instead of**the*fill *one**in the call of *geom_path*. Something like that : p = ggplot(geo, aes(x, y)) p + geom_point(aes(size = ACE, colour = ACE)) + theme_bw() + scale_size(name = Número de especies, breaks = c(2, 4, 6, 8, 10, 12, 14, 16, 18, 20)) + scale_colour_gradientn(name = 'Número de especies', colours = heat.colors(10), breaks = c(2, 4, 6, 8, 10, 12, 14, 16, 18, 20))+ xlab(Longitud) + ylab(Latitud) + opts(axis.text.x = theme_text(size = 8, vjust = 1)) + opts(axis.text.y = theme_text(size = 8, hjust = 1)) + geom_path(aes(x=long,y=lat, group=group, *colour=id*),data=fortify.ai_biotica) not tested, so i'm not really sure about this one 2011/4/4 Pierre Roudier pierre.roud...@gmail.com mailto:pierre.roud...@gmail.com Hi all, 2011/4/4 Felipe Carrillo mazatlanmex...@yahoo.com mailto:mazatlanmex...@yahoo.com: Manuel: As far as I know one needs gpclibPermit() in order to fortify see this: Note: polygon geometry computations in maptools depend on the package gpclib, which has a restricted licence. It is disabled by default; to enable gpclib, type gpclibPermit() I am going to guess that ahmadou dicko doesn't show gpclibPermit() on his code because he loaded it with Rprofile or some other way. I tried to run his code without gpclibPermit() and it wouldn't let me fortify, so not sure how he did it. On that specific point, Colin Arundel and Roger Bivant released the rgeos package on CRAN a few days [1]. This is a great achievement as it brings bindings to the GEOS C++ lib [2] - long story short, it makes the job the non-free [3] gpclib used to do. In its later release, maptools has an option to check if rgeos if present - if it is the case it is used instead of gpclib: library(maptools) Loading required package: foreign Loading required package: sp Loading required package: lattice Note: polygon geometry computations in maptools depend on the package gpclib, which has a restricted licence. It is disabled by default; to enable gpclib, type gpclibPermit() Checking rgeos availability as gpclib substitute: TRUE ?gpclibPermit Pierre [1] http://cran.r-project.org/web/packages/rgeos/ [2] http://trac.osgeo.org/geos/ [3] https://stat.ethz.ch/pipermail/r-sig-geo/2010-January/007400.html -- Scientist Landcare Research, New Zealand -- You received this message because you are subscribed to the ggplot2 mailing list. Please provide a reproducible example: http://gist.github.com/270442 To post: email ggpl...@googlegroups.com mailto:ggpl...@googlegroups.com To unsubscribe: email ggplot2+unsubscr...@googlegroups.com mailto:ggplot2%2bunsubscr...@googlegroups.com More options: http://groups.google.com/group/ggplot2 -- *Manuel Spínola, Ph.D.* Instituto Internacional en Conservación y Manejo de Vida Silvestre Universidad Nacional Apartado 1350-3000 Heredia COSTA RICA mspin...@una.ac.cr mspinol...@gmail.com Teléfono: (506) 2277-3598 Fax: (506) 2237-7036 Personal website: Lobito de río https://sites.google.com/site/lobitoderio/ Institutional website: ICOMVIS http://www.icomvis.una.ac.cr/ [[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,
[R] moving mean and moving variance functions
Hello
Lets say as an example I have a dataframe with the following attributes:
rownum(1:405), colnum(1:287), year(2000:2009), daily(rownum x colnum x year)
and foragePotential (0:1, by 0.01). The data is actually stored in a netcdf
file and I'm trying to provide a conceptual version of the data.
Ok. I need to calculate a moving mean and a moving variance for each cell on
the following temporal
windows - 7 day, 14 day, and 28 day. So far I have code for the moving
average.
ma - function(x , n) {
filter(x, rep(1/n, n), sides = 1)
} # note that when the function is used, n is defined for the
temporal period (7, 14, and 28), and x is the input variable.
ma7 - ma(dat, 7) # where dat is accessing the foraging potential of the
birds.
ma14 - ma(dat, 14)
ma28 - ma(dat, 28)
This works fine. What I don't have is the code for a moving variance.
filter in the function above is included in the stats package and conducts a
linear filtering on a Time Series.
Is there comparable code some place in R for a moving variance?
Thanks in advance.
Thanks,
Steve
[[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] Examples of web-based Sweave use?
I appreciate that this is OT, but I'd be grateful for pointers to examples of where Sweave has been used for web-based applications. In particular, examples of where reports/analyses are produced automatically through submission of data to a web-sever. I am mostly interested in situations where pdf reports have been produced rather than, say, a plot/table etc shown on a web page. I've had limited success finding examples on this. Many thanks. David Carslaw Environmental Research Group MRC-HPA Centre for Environment and Health King's College London Franklin Wilkins Building Stamford Street London SE1 9NH david.cars...@kcl.ac.uk -- View this message in context: http://r.789695.n4.nabble.com/Examples-of-web-based-Sweave-use-tp3425324p3425324.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] hc2Newick is different than th hclust dendrogram
Hi R helpers... I am having troubles because of the discrepancy between the dendrogram plotted from hclust and what is wrote in the hc2Newick file. I've got a matrix C: hc - hclust(dist(C)) plot(hc) with the: write(hc2Newick(hc),file='test.newick') both things draw completely different trees... I have also tried with the raw distance matrix D and also the agnes function, but the same happens. The hclus and agnes dendrogram is logical, whereas the newick tree is not. Thanks for any help! PS attached C matrix that is a 112x112 matrix -- Jose Sergio Hleap Lozano, M. Sc. Ph. D. Student, Dalhousie University Researcher, SQUALUS Foundation __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] I think I just broke R
and time to upgrade R I'm still fighting to find out how to upgrade stuff on Ubuntu. After a repository update the newest available version was still 2.10.1. I'll figure it out, sooner or later :) That's simple. Just add $deb http://my.favorite.cran.mirror/bin/linux/ubuntu maverick/ to /etc/apt/sources.list (or whatever Ubuntu version you use) and type $sudo aptitude update $sudo aptitude safe-upgrade See http://cran.r-project.org/bin/linux/ubuntu/ Regards __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] moving mean and moving variance functions
On Mon, Apr 4, 2011 at 8:30 AM, Steve Friedman skfgla...@gmail.com wrote:
Hello
Lets say as an example I have a dataframe with the following attributes:
rownum(1:405), colnum(1:287), year(2000:2009), daily(rownum x colnum x year)
and foragePotential (0:1, by 0.01). The data is actually stored in a netcdf
file and I'm trying to provide a conceptual version of the data.
Ok. I need to calculate a moving mean and a moving variance for each cell on
the following temporal
windows - 7 day, 14 day, and 28 day. So far I have code for the moving
average.
ma - function(x , n) {
filter(x, rep(1/n, n), sides = 1)
} # note that when the function is used, n is defined for the
temporal period (7, 14, and 28), and x is the input variable.
ma7 - ma(dat, 7) # where dat is accessing the foraging potential of the
birds.
ma14 - ma(dat, 14)
ma28 - ma(dat, 28)
This works fine. What I don't have is the code for a moving variance.
filter in the function above is included in the stats package and conducts a
linear filtering on a Time Series.
Is there comparable code some place in R for a moving variance?
See rollmean and rollapply in the zoo package and runmean and runsd in
the caTools package.
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] RGtk2: How to populate an GtkListStore data model?
On Sun, Apr 3, 2011 at 8:29 PM, Cleber N. Borges kle...@yahoo.com.brwrote:
hello all
I am trying to learn how to use the RGtk2 package...
so, my first problem is: I don't get the right way for populate my
gtkListStore object!
any help is welcome... because I am trying several day to mount the code...
Thanks in advanced
Cleber N. Borges
---
# my testing code
library(RGtk2)
win - gtkWindowNew()
datamodel - gtkListStoreNew('gchararray')
treeview - gtkTreeViewNew()
renderer - gtkCellRendererText()
col_0 - gtkTreeViewColumnNewWithAttributes(title=TitleXXX,
cell=renderer, text=Bar)
nc_next - gtkTreeViewInsertColumn(object=treeview, column=col_0,
position=0)
gtkTreeViewSetModel( treeview, datamodel )
win$add( treeview ) # is there an alternative function for this?
# iter - gtkTreeModelGetIterFirst( datamodel )[[2]]
# this function don't give VALID iter
# gtkListStoreIterIsValid( datamodel, iter ) result in FALSE
iter - gtkListStoreInsert( datamodel, position=0 )[[2]]
gtkListStoreIterIsValid( datamodel, iter )
# the help of this function say to terminated in -1 value
# but -1 crash the R-pckage (or the gtk)...
Sorry about this. The help is autogenerated from the GTK+ C documentation.
gtkListStoreSet(object=datamodel, iter=iter, 0, textFoo)
# don't make any difference in the window... :-(
It would be much easier to use rGtkDataFrame for populating this list. See
the help for that function.
R version 2.13.0 alpha (2011-03-27 r55091)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=Portuguese_Brazil.1252
[2] LC_CTYPE=Portuguese_Brazil.1252
[3] LC_MONETARY=Portuguese_Brazil.1252
[4] LC_NUMERIC=C
[5] LC_TIME=Portuguese_Brazil.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods
[7] base
other attached packages:
[1] RGtk2_2.20.8
loaded via a namespace (and not attached):
[1] tools_2.13.0
my gtk version == 2.16.2
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Examples of web-based Sweave use?
I've written about a bunch of Web R interfaces here: * http://www.r-statistics.com/2010/04/jeroen-oomss-ggplot2-web-interface-a-new-version-released-v0-2/ * http://www.r-statistics.com/2010/04/r-node-a-web-front-end-to-r-with-protovis/ (And some other posts here: http://www.r-statistics.com/category/r-and-the-web/ ) http://www.r-statistics.com/category/r-and-the-web/I'm not sure which of them use Sweave behind them, but you could look around and check. Hope that helps, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Mon, Apr 4, 2011 at 2:31 PM, carslaw david.cars...@kcl.ac.uk wrote: I appreciate that this is OT, but I'd be grateful for pointers to examples of where Sweave has been used for web-based applications. In particular, examples of where reports/analyses are produced automatically through submission of data to a web-sever. I am mostly interested in situations where pdf reports have been produced rather than, say, a plot/table etc shown on a web page. I've had limited success finding examples on this. Many thanks. David Carslaw Environmental Research Group MRC-HPA Centre for Environment and Health King's College London Franklin Wilkins Building Stamford Street London SE1 9NH david.cars...@kcl.ac.uk -- View this message in context: http://r.789695.n4.nabble.com/Examples-of-web-based-Sweave-use-tp3425324p3425324.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to speed up grouping time series, help please
I retrieve for a few hundred times a group of time series (10-15 ts
with 1 values each), on every group I do some calculation, graphs
etc. I wonder if there is a faster method than what presented below to
get an appropriate timeseries object.
Making a query with RODBC for every group I get a data frame like this:
X
IDDATE VALUE
14 3 2000-01-01 00:00:03 0.5726334
4 1 2000-01-01 00:00:03 0.8830174
1 1 2000-01-01 00:00:00 0.2875775
15 3 2000-01-01 00:00:04 0.1029247
11 3 2000-01-01 00:00:00 0.9568333
9 2 2000-01-01 00:00:03 0.5514350
7 2 2000-01-01 00:00:01 0.5281055
6 2 2000-01-01 00:00:00 0.0455565
12 3 2000-01-01 00:00:01 0.4533342
8 2 2000-01-01 00:00:02 0.8924190
3 1 2000-01-01 00:00:02 0.4089769
13 3 2000-01-01 00:00:02 0.6775706
And I want to get a timeSeries object or xts object like this:
1 2 3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04NANA 0.1029247
Input data can be sorted or unsorted (the most complicated case is in
the example, unsorted and missing data) in the sense that I can sort
in query if I can take an advantage from this.
Some considerations:
- Xts is generally faster than timeSeries
- both accept a matrix so if I can create a matrix like the one
represented above and an array of characters representing dates faster
than what possible with xts:::cbind, for examole,I will have a faster
implementation (package data.table ?).
- create timeseries objects in multithread and then merge (package plyr ?)
- faster merge algorithms?
Below some code to generate the test case above:
set.seed(123)
N - 5 # number of observations
K - 3 # number of timeseries ID
X - data.frame(
ID = rep(1:K, each = N),
DATE = as.character(rep(as.POSIXct(2000-01-01, tz = GMT)+ 0:(N-1), K)),
VALUE = runif(N*K), stringsAsFactors = FALSE)
X - X[sample(1:(N*K), N*K),] # sample observations to get random
order (optional)
X - X[-(sample(1:nrow(X), floor(nrow(X)*0.2))),] # 20% missing
head(X, 15)
# use explicitly environments to avoid '-'
buildTimeSeriesFromDataFrame - function(x, env)
{
{
if(exists(xx, envir = env)) # if exist variable xx in env cbind
assign(xx,
cbind(get(xx, env), timeSeries(x$VALUE, x$DATE,
format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1]))),
envir = env)
else # create xx in env
assign(xx,
timeSeries(x$VALUE, x$DATE, format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1])),
envir = env)
return(TRUE)
}
}
# use package plyr, faster than 'by' function
tsDaply - function(...)
{
library(plyr)
e1 - new.env(parent = baseenv()) #create a new env
res - daply(X, ID, buildTimeSeriesFromDataFrame,
env = e1)
return(get(xx, e1)) # return xx from env
}
##replicate 100 times
#Time03 - replicate(100,
# system.time(tsDaply(X, X$ID))[[1]])
#median(Time03)
# result
tsDaply(X, X$ID)
Thanks in advance for any input, best regards,
Den
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] reading from text file that have different rowlength and create a data frame
Hi R-experts I have many text files to read and combined them into one into R that are output from other programs. My textfile have unbalanced number of rows for example: ;this is example ; r help Var1 Var2 Var3 Var4 Var5 0 0.05 0.0112 1 0.04 0.0618A 2 0.05 0.0814 3 0.01 0.0615 B 4 0.05 0.0714 C and so on Inames-as.data.frame(read.table(CLG1mpd.asc,header=T,comment=;)) Inames-as.matrix(read.table(example.txt,header=T,comment=;)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements In bestcase scenerio, I want to fill the blank space with NA's, with matrix or dataframe Var1 Var2 Var3 Var4 Var5 0 0.05 0.0112NA 1 0.04 0.0618A 2 0.05 0.0814 NA 3 0.01 0.0615 B 4 0.05 0.0714 C The minimum would be to remove the column Var5, so that my data.frame would look like the follows: Var1 Var2 Var3 Var4 0 0.05 0.0112 1 0.04 0.0618 2 0.05 0.0814 3 0.01 0.0615 4 0.05 0.0714 -- Thank you in advance for the help. Ram H [[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] gap.barplot doesn't support data arrays?
I am trying to make a barplot with a broken axis using gap.barplot (in the indispensable plotrix package). This works well when the data is a vector: twogrp-c(rnorm(10)+4,rnorm(10)+20) gap.barplot(twogrp,gap=c(8,16),xlab=Index,ytics=c(3,6,17,20),ylab=Group values,main=Barplot with gap) But when the data is an array (for a bar plot with multiple series) I get an error and a strange plot with no y-tics and bars stretching downwards, as if all the values were negative: twogrp2-array(twogrp, dim=c(2,5)) gap.barplot(twogrp2,gap=c(8,16),xlab=Index,ytics=c(3,6,17,20),ylab=Group values,main=Barplot with gap) Error in rect(xtics[bigones] - halfwidth, botgap, xtics[bigones] + halfwidth, : cannot mix zero-length and non-zero-length coordinates However, the main title and axis labels do appear correctly. Are data arrays unsupported for gap.barplot, or am I missing something? Thanks, Drew Steen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] reading from text file that have different rowlength and create a data frame
?read.table Then look at the 'fill' 'flush' parameters; this may do the trick On Mon, Apr 4, 2011 at 9:32 AM, Ram H. Sharma sharma.ra...@gmail.com wrote: Hi R-experts I have many text files to read and combined them into one into R that are output from other programs. My textfile have unbalanced number of rows for example: ;this is example ; r help Var1 Var2 Var3 Var4 Var5 0 0.05 0.01 12 1 0.04 0.06 18 A 2 0.05 0.08 14 3 0.01 0.06 15 B 4 0.05 0.07 14 C and so on Inames-as.data.frame(read.table(CLG1mpd.asc,header=T,comment=;)) Inames-as.matrix(read.table(example.txt,header=T,comment=;)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements In bestcase scenerio, I want to fill the blank space with NA's, with matrix or dataframe Var1 Var2 Var3 Var4 Var5 0 0.05 0.01 12 NA 1 0.04 0.06 18 A 2 0.05 0.08 14 NA 3 0.01 0.06 15 B 4 0.05 0.07 14 C The minimum would be to remove the column Var5, so that my data.frame would look like the follows: Var1 Var2 Var3 Var4 0 0.05 0.01 12 1 0.04 0.06 18 2 0.05 0.08 14 3 0.01 0.06 15 4 0.05 0.07 14 -- Thank you in advance for the help. Ram H [[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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Difference in mixture normals and one density
Hello, I am trying to find out if R can do the following: I have a mixture of normals say f = 0.2*Normal(2, 5) + 0.8*Normal(3,2) How do I find the difference in the densities at any particular point of f and at Normal(2,5)? -- Thanks, Jim. [[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] reading from text file that have different rowlength and create a data frame
try this: x - read.table(textConnection(;this is example + ; r help + Var1 Var2 Var3 Var4 Var5 + 0 0.05 0.0112 + 1 0.04 0.0618A + 2 0.05 0.0814 + 3 0.01 0.0615 B + 4 0.05 0.0714 C) + , comment = ';' + , fill = TRUE + , header = TRUE + , na.strings = '' + ) closeAllConnections() x Var1 Var2 Var3 Var4 Var5 10 0.05 0.01 12 NA 21 0.04 0.06 18A 32 0.05 0.08 14 NA 43 0.01 0.06 15B 54 0.05 0.07 14C On Mon, Apr 4, 2011 at 9:32 AM, Ram H. Sharma sharma.ra...@gmail.com wrote: Hi R-experts I have many text files to read and combined them into one into R that are output from other programs. My textfile have unbalanced number of rows for example: ;this is example ; r help Var1 Var2 Var3 Var4 Var5 0 0.05 0.01 12 1 0.04 0.06 18 A 2 0.05 0.08 14 3 0.01 0.06 15 B 4 0.05 0.07 14 C and so on Inames-as.data.frame(read.table(CLG1mpd.asc,header=T,comment=;)) Inames-as.matrix(read.table(example.txt,header=T,comment=;)) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 5 elements In bestcase scenerio, I want to fill the blank space with NA's, with matrix or dataframe Var1 Var2 Var3 Var4 Var5 0 0.05 0.01 12 NA 1 0.04 0.06 18 A 2 0.05 0.08 14 NA 3 0.01 0.06 15 B 4 0.05 0.07 14 C The minimum would be to remove the column Var5, so that my data.frame would look like the follows: Var1 Var2 Var3 Var4 0 0.05 0.01 12 1 0.04 0.06 18 2 0.05 0.08 14 3 0.01 0.06 15 4 0.05 0.07 14 -- Thank you in advance for the help. Ram H [[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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to speed up grouping time series, help please
Hi Dan,
On Mon, Apr 4, 2011 at 7:49 AM, Den Alpin den.al...@gmail.com wrote:
I retrieve for a few hundred times a group of time series (10-15 ts
with 1 values each), on every group I do some calculation, graphs
etc. I wonder if there is a faster method than what presented below to
get an appropriate timeseries object.
Making a query with RODBC for every group I get a data frame like this:
X
ID DATE VALUE
14 3 2000-01-01 00:00:03 0.5726334
4 1 2000-01-01 00:00:03 0.8830174
1 1 2000-01-01 00:00:00 0.2875775
15 3 2000-01-01 00:00:04 0.1029247
11 3 2000-01-01 00:00:00 0.9568333
9 2 2000-01-01 00:00:03 0.5514350
7 2 2000-01-01 00:00:01 0.5281055
6 2 2000-01-01 00:00:00 0.0455565
12 3 2000-01-01 00:00:01 0.4533342
8 2 2000-01-01 00:00:02 0.8924190
3 1 2000-01-01 00:00:02 0.4089769
13 3 2000-01-01 00:00:02 0.6775706
And I want to get a timeSeries object or xts object like this:
1 2 3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01 NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04 NA NA 0.1029247
Input data can be sorted or unsorted (the most complicated case is in
the example, unsorted and missing data) in the sense that I can sort
in query if I can take an advantage from this.
Some considerations:
- Xts is generally faster than timeSeries
- both accept a matrix so if I can create a matrix like the one
represented above and an array of characters representing dates faster
than what possible with xts:::cbind, for examole,I will have a faster
implementation (package data.table ?).
- create timeseries objects in multithread and then merge (package plyr ?)
- faster merge algorithms?
Below some code to generate the test case above:
set.seed(123)
N - 5 # number of observations
K - 3 # number of timeseries ID
X - data.frame(
ID = rep(1:K, each = N),
DATE = as.character(rep(as.POSIXct(2000-01-01, tz = GMT)+ 0:(N-1), K)),
VALUE = runif(N*K), stringsAsFactors = FALSE)
X - X[sample(1:(N*K), N*K),] # sample observations to get random
order (optional)
X - X[-(sample(1:nrow(X), floor(nrow(X)*0.2))),] # 20% missing
head(X, 15)
# use explicitly environments to avoid '-'
buildTimeSeriesFromDataFrame - function(x, env)
{
{
if(exists(xx, envir = env)) # if exist variable xx in env cbind
assign(xx,
cbind(get(xx, env), timeSeries(x$VALUE, x$DATE,
format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1]))),
envir = env)
else # create xx in env
assign(xx,
timeSeries(x$VALUE, x$DATE, format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1])),
envir = env)
return(TRUE)
}
}
# use package plyr, faster than 'by' function
tsDaply - function(...)
{
library(plyr)
e1 - new.env(parent = baseenv()) #create a new env
res - daply(X, ID, buildTimeSeriesFromDataFrame,
env = e1)
return(get(xx, e1)) # return xx from env
}
##replicate 100 times
#Time03 - replicate(100,
# system.time(tsDaply(X, X$ID))[[1]])
#median(Time03)
# result
tsDaply(X, X$ID)
Thanks in advance for any input, best regards,
Den
Here's how I would do it with xts:
x - xts(X[,c(ID,VALUE)], as.POSIXct(X[,DATE]))
do.call(merge, split(x$VALUE,x$ID))
My xts solution compares favorably to your solution:
Time03 - replicate(100,
+ system.time(tsDaply(X, X$ID))[[1]])
median(Time03)
[1] 0.02
xtsTime - replicate(100,
+ system.time(do.call(merge, split(x$VALUE,x$ID)))[[1]])
median(xtsTime)
[1] 0
Best,
--
Joshua Ulrich | FOSS Trading: www.fosstrading.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] How to speed up grouping time series, help please
On Mon, Apr 4, 2011 at 8:49 AM, Den Alpin den.al...@gmail.com wrote:
I retrieve for a few hundred times a group of time series (10-15 ts
with 1 values each), on every group I do some calculation, graphs
etc. I wonder if there is a faster method than what presented below to
get an appropriate timeseries object.
Making a query with RODBC for every group I get a data frame like this:
X
ID DATE VALUE
14 3 2000-01-01 00:00:03 0.5726334
4 1 2000-01-01 00:00:03 0.8830174
1 1 2000-01-01 00:00:00 0.2875775
15 3 2000-01-01 00:00:04 0.1029247
11 3 2000-01-01 00:00:00 0.9568333
9 2 2000-01-01 00:00:03 0.5514350
7 2 2000-01-01 00:00:01 0.5281055
6 2 2000-01-01 00:00:00 0.0455565
12 3 2000-01-01 00:00:01 0.4533342
8 2 2000-01-01 00:00:02 0.8924190
3 1 2000-01-01 00:00:02 0.4089769
13 3 2000-01-01 00:00:02 0.6775706
And I want to get a timeSeries object or xts object like this:
1 2 3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01 NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04 NA NA 0.1029247
Input data can be sorted or unsorted (the most complicated case is in
the example, unsorted and missing data) in the sense that I can sort
in query if I can take an advantage from this.
Some considerations:
- Xts is generally faster than timeSeries
- both accept a matrix so if I can create a matrix like the one
represented above and an array of characters representing dates faster
than what possible with xts:::cbind, for examole,I will have a faster
implementation (package data.table ?).
- create timeseries objects in multithread and then merge (package plyr ?)
- faster merge algorithms?
Below some code to generate the test case above:
set.seed(123)
N - 5 # number of observations
K - 3 # number of timeseries ID
X - data.frame(
ID = rep(1:K, each = N),
DATE = as.character(rep(as.POSIXct(2000-01-01, tz = GMT)+ 0:(N-1), K)),
VALUE = runif(N*K), stringsAsFactors = FALSE)
X - X[sample(1:(N*K), N*K),] # sample observations to get random
order (optional)
X - X[-(sample(1:nrow(X), floor(nrow(X)*0.2))),] # 20% missing
head(X, 15)
# use explicitly environments to avoid '-'
buildTimeSeriesFromDataFrame - function(x, env)
{
{
if(exists(xx, envir = env)) # if exist variable xx in env cbind
assign(xx,
cbind(get(xx, env), timeSeries(x$VALUE, x$DATE,
format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1]))),
envir = env)
else # create xx in env
assign(xx,
timeSeries(x$VALUE, x$DATE, format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1])),
envir = env)
return(TRUE)
}
}
# use package plyr, faster than 'by' function
tsDaply - function(...)
{
library(plyr)
e1 - new.env(parent = baseenv()) #create a new env
res - daply(X, ID, buildTimeSeriesFromDataFrame,
env = e1)
return(get(xx, e1)) # return xx from env
}
##replicate 100 times
#Time03 - replicate(100,
# system.time(tsDaply(X, X$ID))[[1]])
#median(Time03)
# result
tsDaply(X, X$ID)
Haven't checked how fast it is but using read.zoo its just one line of
code to produce the required matrix:
# set up input data frame, DF
Lines - ID,DATE,VALUE
3,2000-01-01 00:00:03,0.5726334
1,2000-01-01 00:00:03,0.8830174
1,2000-01-01 00:00:00,0.2875775
3,2000-01-01 00:00:04,0.1029247
3,2000-01-01 00:00:00,0.9568333
2,2000-01-01 00:00:03,0.5514350
2,2000-01-01 00:00:01,0.5281055
2,2000-01-01 00:00:00,0.0455565
3,2000-01-01 00:00:01,0.4533342
2,2000-01-01 00:00:02,0.8924190
1,2000-01-01 00:00:02,0.4089769
3,2000-01-01 00:00:02,0.6775706
DF - read.table(textConnection(Lines), header = TRUE, sep = ,)
# create zoo matrix
library(zoo)
z - read.zoo(DF, split = 1, index = 2, tz = )
The last line gives:
z
1 2 3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04NANA 0.1029247
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Difference in mixture normals and one density
Is something like this what you're looking for?
R library(nor1mix)
R nmix2 - norMix(c(2, 3), sig2=c(25, 4), w=c(.2, .8))
R dnorMix(1, nmix2) - dnorm(1, 2, 5)
[1] 0.03422146
Andy
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jim Silverton
Sent: Monday, April 04, 2011 10:01 AM
To: r-help@r-project.org
Subject: Re: [R] Difference in mixture normals and one density
Hello,
I am trying to find out if R can do the following:
I have a mixture of normals say f = 0.2*Normal(2, 5) + 0.8*Normal(3,2)
How do I find the difference in the densities at any
particular point of f
and at Normal(2,5)?
--
Thanks,
Jim.
[[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.
Notice: This e-mail message, together with any attachme...{{dropped:11}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Please help
Hi, On Mon, Apr 4, 2011 at 5:15 AM, Sadaf Zaidi s.zaidi...@amu.ac.in wrote: Dear Sir/Madam, I am stuck with a nagging problem in using R for SVM regression. My data has 5 dimensions and 400 observations. The independent variables are : Peb, Ksub, Sub, and Xtt. The dependent variable is: Rexp. I tried using the svm.tune function to tune the hyper parameters: gamma, epsilon and C. I am getting the following error message: Error in predict.svm(ret, xhold, decision.values+TRUE): Model is empty! May you please help me! You're not giving us much to go on, can you show us the code that you are using that gets you into this problem? (i) For example -- by svm.tune, do you mean the tune function from the e1071 package? (ii) What is the exact function call you are using that gives you this error. (iii) Can you build a normal svm model without tuning it. For instance, does svm(x,y,..) work with your data? (iv) Are you sure that the values you are inputting to the svm (and/or tune function) are of the correct type? With your follow up email that provides the code you tried and answers to some of the Q's above. Also provide a small bit of your data that we can use to help you debug. You can easily do so by using the `dput` function. Say your data (predictors and label) are in a variable `x`, paste the output of the following command in your follow up email: R dput(x[sample(nrow(x), 10),]) This will give us 10 random rows from your data that people trying to help you can use. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] gap.barplot doesn't support data arrays?
On 2011-04-04 06:39, Andrew D. Steen wrote: I am trying to make a barplot with a broken axis using gap.barplot (in the indispensable plotrix package). This works well when the data is a vector: twogrp-c(rnorm(10)+4,rnorm(10)+20) gap.barplot(twogrp,gap=c(8,16),xlab=Index,ytics=c(3,6,17,20),ylab=Group values,main=Barplot with gap) But when the data is an array (for a bar plot with multiple series) I get an error and a strange plot with no y-tics and bars stretching downwards, as if all the values were negative: twogrp2-array(twogrp, dim=c(2,5)) gap.barplot(twogrp2,gap=c(8,16),xlab=Index,ytics=c(3,6,17,20),ylab=Group values,main=Barplot with gap) Error in rect(xtics[bigones] - halfwidth, botgap, xtics[bigones] + halfwidth, : cannot mix zero-length and non-zero-length coordinates However, the main title and axis labels do appear correctly. Are data arrays unsupported for gap.barplot, or am I missing something? Looks like they're not supported, as you could easily see from the code. But do gapped stacked barplots even make sense? Not to me. Still, I think that the plotrix documentation is somewhat spotty. The help page for gap.barplot indicates that the input 'y' should be 'data values'; not overly informative. Peter Ehlers Thanks, Drew Steen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Creating multiple vector/list names-novice
Hi I have very simple issue as I am still new to the group of R I have basically vector of names for which i want to create mutliple combinations and then place them in different vectors. In some other language I can just place a third dimension to separate list (or matrix) but i do not know how to do it in R. My issue is simple I use cc-combn(colnames(DD),2) I would need to have this as vector1 or like vector[,,1] : cc-combn(colnames(DD),2) vector2or like vector[,,2] cc-combn(colnames(DD),3) etc..for up to k combinations something so then I can use for loop to go through the al of these combinations example: string-a, b , c ,d vector/list(1) ab ac ad bc bd be cd ce de vector/list (2) abc abd abe bcd bce bde cde Can you help me with this.. I know that it is a simple question for this thread thank you.. Michal -- View this message in context: http://r.789695.n4.nabble.com/Creating-multiple-vector-list-names-novice-tp3425616p3425616.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] Clarks 2Dt function in R
bialozyt at biologie.uni-marburg.de writes:
Dear Ben,
you answerd to Nancy Shackelford about Clarks 2Dt function.
Since the thread ended just after your reply,
I would like to ask, if you have an idea how to use this function in R
Dear Ronald,
I got started on your problem, but I didn't finish it.
I got a plausible answer to start with, but when checking
the answer I ran into some trouble. Unfortunately, fitting
these functions is a bit harder than one might expect ...
it takes quite a bit of fussing to get a good, reliable answer.
My partly-worked solution is below.
I defined it the following way:
You were multiplying instead of dividing by the second
term (I changed it by raising the term to a negative
power instead.
The lesson here: *always* do some sanity checks (graphical or otherwise)
of your functions. I actually did the whole fit before I tried to plot
the curves and found that they were increasing rather than decreasing ...
## fixed
clark2Dt - function(x , p, u=1) {
(p/(pi*u))/(1+(x^2/u))^(p+1)
}
It might be preferable to define this in terms of s=sqrt(u)
instead (then s would be a scale parameter with the same units
as x, more easily interpretable ...
Sanity checks:
par(las=1,bty=l) ## personal preferences
curve(clark2Dt(x,p=6),from=0,to=5)
curve(clark2Dt(x,p=4),col=2,add=TRUE)
curve(clark2Dt(x,p=2),col=4,add=TRUE)
legend(topright,paste(p,c(6,4,2),sep==),col=c(1,2,4),lty=1)
Grab data (in the future, if possible, please use dput(), which puts
your data in the most convenient form, or write out a statement like
this to define your data ...)
X - as.data.frame(matrix(
c(15,12,
45,13,
75,10,
105,8,
135,16,
165,5,
195,15,
225,8,
255,9,
285,12,
315,5,
345,4,
375,1,
405,1,
435,1,
465,0,
495,1,
525,2,
555,0,
585,0,
615,0,
645,0,
675,0),
ncol=2,byrow=TRUE,
dimnames=list(NULL,c(dist,count
## assume these are traps/samples with unit size
## (if not, it will get absorbed into the fecundity constant
library(bbmle)
m1 - mle2(count~dnbinom(mu=f*clark2Dt(dist,p,u),size=k),
data=X,start=list(f=20,u=10,p=5,k=2),
lower=rep(0.002,4),method=L-BFGS-B)
## we get a plausible-looking fit ...
with(X,plot(count~dist,pch=16,las=1,bty=l))
newdat - data.frame(dist=1:700) ## overkill but harmless
lines(newdat$dist,predict(m1,newdata=newdat))
## but the coefficients look funny, especially f
coef(m1)
## tried resetting parscale but it's bogus (gets stuck at a worse likelihood)
m2 - mle2(count~dnbinom(mu=f*clark2Dt(dist,p,u),size=k),
data=X,start=list(f=20,u=10,p=5,k=2),
control=list(parscale=abs(coef(m1))),
lower=rep(0.002,4),method=L-BFGS-B)
m3 - mle2(count~dnbinom(mu=exp(logf)*clark2Dt(dist,exp(logp),exp(logu)),
size=exp(logk)),
data=X,start=list(logf=log(20),logu=log(10),logp=log(5),
logk=log(2)),
method=Nelder-Mead)
exp(coef(m3))
coef(m1)
summary(m1)
## hmm. Redefine in terms of s instead of u and (more importantly)
## with f = seed density at r=0 rather the
cov2cor(vcov(m1)) ## shows that f and u are horribly correlated
newclark2Dt - function(x , p, s=1, eps=1e-70) {
d - (1+(x/s)^2)
r - 1/d^(p+1)
if (any(!is.finite(r))) browser()
r
}
dnbinom_pen - function(x,mu,size,pen=1000,log=TRUE) {
mu - rep(mu,length.out=length(x))
logval - ifelse(mu==0 x==0,pen*x^2,dnbinom(x,mu=mu,size=size,log=TRUE))
if (log) logval else exp(logval)
}
## needed for predict()
snbinom_pen - snbinom
m4 - mle2(count~dnbinom(mu=f*newclark2Dt(dist,p,s),size=k),
data=X,start=list(f=20,s=10,p=5,k=2),
lower=rep(0.002,4),method=L-BFGS-B)
m5 - mle2(count~dnbinom_pen(mu=f*newclark2Dt(dist,1/(pinv),s),size=exp(logk)),
data=X,start=list(f=15,s=10,pinv=100,logk=1),trace=TRUE,
## control=list(parscale=c(200,0.002,1.66,3600)),
lower=rep(0.002,4),method=L-BFGS-B)
with(X,plot(count~dist,pch=16,las=1,bty=l))
newdat - data.frame(dist=1:700) ## overkill but harmless
lines(newdat$dist,predict(m1,newdata=newdat))
lines(newdat$dist,predict(m5,newdata=newdat),col=2)
but I am not able to fit anything.
Do you have an idea?
I guess there is something wrong in my formula for Clarks 2Dt
Thank you for reading
Ciao
Ronald Bialozyt
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Creating multiple vector/list names-novice
Try this: lapply(2:3, FUN = combn, x = string, paste, collapse = '') On Mon, Apr 4, 2011 at 11:24 AM, michalseneca michalsen...@gmail.com wrote: Hi I have very simple issue as I am still new to the group of R I have basically vector of names for which i want to create mutliple combinations and then place them in different vectors. In some other language I can just place a third dimension to separate list (or matrix) but i do not know how to do it in R. My issue is simple I use cc-combn(colnames(DD),2) I would need to have this as vector1 or like vector[,,1] : cc-combn(colnames(DD),2) vector2or like vector[,,2] cc-combn(colnames(DD),3) etc..for up to k combinations something so then I can use for loop to go through the al of these combinations example: string-a, b , c ,d vector/list(1) ab ac ad bc bd be cd ce de vector/list (2) abc abd abe bcd bce bde cde Can you help me with this.. I know that it is a simple question for this thread thank you.. Michal -- View this message in context: http://r.789695.n4.nabble.com/Creating-multiple-vector-list-names-novice-tp3425616p3425616.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html) . A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] simulating a VARXls model using dse
Hello, Using the dse package I have estimated a VAR model using estVARXls(). I can perform forecasts using forecast() with no problems, but when I try to use simulate() with the same model, I get the following error: Error in diag(Cov, p) : 'nrow' or 'ncol' cannot be specified when 'x' is a matrix Can anyone shed some light on the meaning of this error? How can I simulate output using a model created with estVARXls()? Thanks, Alison __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] simulating a VARXls model using dse
Hi Paul, I am using R v. 2.12.2, and the dse package with build 2.12.2. I have attached some sample data to this email, and the R code I use to create the model and then forecast with it. Thanks, Alison On Mon, Apr 4, 2011 at 11:02 AM, Paul Gilbert pgilb...@bank-banque-canada.ca wrote: Could you please send me a reproducible example, and R and dse version info. Thanks, Paul -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Alison Callahan Sent: April 4, 2011 10:56 AM To: r-help@r-project.org Subject: [R] simulating a VARXls model using dse Hello, Using the dse package I have estimated a VAR model using estVARXls(). I can perform forecasts using forecast() with no problems, but when I try to use simulate() with the same model, I get the following error: Error in diag(Cov, p) : 'nrow' or 'ncol' cannot be specified when 'x' is a matrix Can anyone shed some light on the meaning of this error? How can I simulate output using a model created with estVARXls()? Thanks, Alison __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. La version française suit le texte anglais. This email may contain privileged and/or confidential information, and the Bank of Canada does not waive any related rights. Any distribution, use, or copying of this email or the information it contains by other than the intended recipient is unauthorized. If you received this email in error please delete it immediately from your system and notify the sender promptly by email that you have done so. Le présent courriel peut contenir de l'information privilégiée ou confidentielle. La Banque du Canada ne renonce pas aux droits qui s'y rapportent. Toute diffusion, utilisation ou copie de ce courriel ou des renseignements qu'il contient par une personne autre que le ou les destinataires désignés est interdite. Si vous recevez ce courriel par erreur, veuillez le supprimer immédiatement et envoyer sans délai à l'expéditeur un message électronique pour l'aviser que vous avez éliminé de votre ordinateur toute copie du courriel reçu. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 speed up grouping time series, help please
I did some tests on Your and Gabor solutions, below my findings:
- Your solution is fast as my solution in xts (below) but MUCH MORE
READABLE, in particular I think your test should take into account xts
creation from the data.frame (see below);
- Gabor's solution with read.zoo is fast as xts but give an xts object
that has some problems with time zones.
Any better idea to speed up grouping time series?
Thanks!
Below few line of codes to test (I suggest to grow X size to get
better comparison results):
xtsSplit - function(x)
{
x - xts(x[,c(ID,VALUE)], as.POSIXct(x[,DATE]))
x - do.call(merge, split(x$VALUE,x$ID))
return(x)
}
xtsSplitTime - replicate(100,
system.time(xtsSplit(X))[[1]])
median(xtsTime)
zooReadTime - replicate(100,
system.time(z - read.zoo(X, split = 1, index = 2, tz = ))[[1]])
median(zooReadTime)
And my (unreadable) solution:
library(xts)
buildXtsFromDataFrame - function(x, env)
{
{
if(exists(xx, envir = env))
{
VALUE - as.matrix(x$VALUE)
colnames(VALUE) - as.character(x$ID[1])
assign(xx,
cbind(get(xx, env), xts(VALUE,
as.POSIXct(x$DATE, tz = GMT,
format = '%Y-%m-%d %H:%M:%S'),
tzone = 'GMT')),
envir = env)
} else
{
VALUE - as.matrix(x$VALUE)
colnames(VALUE) - as.character(x$ID[1])
assign(xx,
xts(VALUE, as.POSIXct(x$DATE, tz = GMT,
format = '%Y-%m-%d %H:%M:%S'),
tzone = 'GMT'),
envir = env)
}
return(TRUE)
}
}
xtsDaply - function(...)
{
e1 - new.env(parent = baseenv())
res - daply(X, ID, buildXtsFromDataFrame,
env = e1)
return(get(xx, e1))
}
Time04 - replicate(100,
system.time(xtsDaply(X, X$ID))[[1]])
2011/4/4 Joshua Ulrich josh.m.ulr...@gmail.com:
Hi Dan,
On Mon, Apr 4, 2011 at 7:49 AM, Den Alpin den.al...@gmail.com wrote:
I retrieve for a few hundred times a group of time series (10-15 ts
with 1 values each), on every group I do some calculation, graphs
etc. I wonder if there is a faster method than what presented below to
get an appropriate timeseries object.
Making a query with RODBC for every group I get a data frame like this:
X
ID DATE VALUE
14 3 2000-01-01 00:00:03 0.5726334
4 1 2000-01-01 00:00:03 0.8830174
1 1 2000-01-01 00:00:00 0.2875775
15 3 2000-01-01 00:00:04 0.1029247
11 3 2000-01-01 00:00:00 0.9568333
9 2 2000-01-01 00:00:03 0.5514350
7 2 2000-01-01 00:00:01 0.5281055
6 2 2000-01-01 00:00:00 0.0455565
12 3 2000-01-01 00:00:01 0.4533342
8 2 2000-01-01 00:00:02 0.8924190
3 1 2000-01-01 00:00:02 0.4089769
13 3 2000-01-01 00:00:02 0.6775706
And I want to get a timeSeries object or xts object like this:
1 2 3
2000-01-01 00:00:00 0.2875775 0.0455565 0.9568333
2000-01-01 00:00:01 NA 0.5281055 0.4533342
2000-01-01 00:00:02 0.4089769 0.8924190 0.6775706
2000-01-01 00:00:03 0.8830174 0.5514350 0.5726334
2000-01-01 00:00:04 NA NA 0.1029247
Input data can be sorted or unsorted (the most complicated case is in
the example, unsorted and missing data) in the sense that I can sort
in query if I can take an advantage from this.
Some considerations:
- Xts is generally faster than timeSeries
- both accept a matrix so if I can create a matrix like the one
represented above and an array of characters representing dates faster
than what possible with xts:::cbind, for examole,I will have a faster
implementation (package data.table ?).
- create timeseries objects in multithread and then merge (package plyr ?)
- faster merge algorithms?
Below some code to generate the test case above:
set.seed(123)
N - 5 # number of observations
K - 3 # number of timeseries ID
X - data.frame(
ID = rep(1:K, each = N),
DATE = as.character(rep(as.POSIXct(2000-01-01, tz = GMT)+ 0:(N-1), K)),
VALUE = runif(N*K), stringsAsFactors = FALSE)
X - X[sample(1:(N*K), N*K),] # sample observations to get random
order (optional)
X - X[-(sample(1:nrow(X), floor(nrow(X)*0.2))),] # 20% missing
head(X, 15)
# use explicitly environments to avoid '-'
buildTimeSeriesFromDataFrame - function(x, env)
{
{
if(exists(xx, envir = env)) # if exist variable xx in env cbind
assign(xx,
cbind(get(xx, env), timeSeries(x$VALUE, x$DATE,
format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1]))),
envir = env)
else # create xx in env
assign(xx,
timeSeries(x$VALUE, x$DATE, format = '%Y-%m-%d %H:%M:%S',
zone = 'GMT', units = as.character(x$ID[1])),
envir = env)
return(TRUE)
}
}
# use package plyr, faster than 'by' function
tsDaply - function(...)
{
library(plyr)
e1 - new.env(parent = baseenv()) #create a new env
res - daply(X, ID, buildTimeSeriesFromDataFrame,
env = e1)
return(get(xx, e1)) # return xx from env
}
##replicate 100 times
#Time03 - replicate(100,
#
Re: [R] Deriving formula with deriv
On Apr 4, 2011, at 6:35 AM, kitty wrote:
Dear list,
Hi,
I am trying to get the second derivative of a logistic formula, in R
summary
the model is given as :
###
$nls
Nonlinear regression model
model: data ~ logistic(time, A, mu, lambda, addpar)
data: parent.frame()
A mu lambda
0.53243 0.03741 6.94296
###
but I know the formula used is
# y~'A'/(1+exp((4*'mu'/'A')*('lambda'-'time'))+2))# from the
grofit (
package I am using to fit the model) documentation.
I have attempted to use the R function 'deriv' to get the
first derivative from which I can then reuse the deriv function to
get the
second derivative unfortunately this does not seem to work
###
express-expression(y~'A'/(1+exp((4*'mu'/'A')*('lambda'-'time'))+2))
express
expression(y ~ A/(1 + exp((4 * mu/A) * (lambda - time)) +
2))
d1-deriv(express)
Error in deriv.default(express) : element 2 is empty;
the part of the args list of '.Internal' being evaluated was:
(expr, namevec, function.arg, tag, hessian)
For one thing you are not specifying what variable you want to
differentiate with-respect-to: Assuming this to be `A` then:
express -
expression( A/(1+exp((4*mu/A)*(lambda-time))+ 2))
# The quotes looked wrong inside an expression, so I removed them
d1-deriv(express, A) # but the diff w.r.t variable needs to be
quoted.
d1
expression({
.expr1 - 4 * mu
.expr3 - lambda - time
.expr5 - exp(.expr1/A * .expr3)
.expr7 - 1 + .expr5 + 2
.value - A/.expr7
.grad - array(0, c(length(.value), 1L), list(NULL, c(A)))
.grad[, A] - 1/.expr7 + A * (.expr5 * (.expr1/A^2
* .expr3))/.expr7^2
attr(.value, gradient) - .grad
.value
})
All this should have been clear if you had looked at the examples in
help(deriv).
Why is this not working and how do I get the second derivative?
That , too, is clearly exemplified in the help page.
Thank you for reading my post, all help is appreciated,
Kitty
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] D'Agostino test
Juraj17 wrote: Do I have to write my own, or it exists yet? How name has it, or how can I use it. Try the R-function search. It return the function you are looking for as the first match. Dieter -- View this message in context: http://r.789695.n4.nabble.com/D-Agostino-test-tp3424952p3425833.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] How to speed up grouping time series, help please
On Mon, Apr 4, 2011 at 11:20 AM, Den Alpin den.al...@gmail.com wrote: I did some tests on Your and Gabor solutions, below my findings: - Your solution is fast as my solution in xts (below) but MUCH MORE READABLE, in particular I think your test should take into account xts creation from the data.frame (see below); - Gabor's solution with read.zoo is fast as xts but give an xts object that has some problems with time zones. read.zoo gives a zoo object, not an xts object. If you want an xts object try as.xts(z). If you were expecting an xts object that may be the source of your other problems. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace last 3 characters of string
Bert Jacobs-2 wrote: I would like to replace the last tree characters of the values of a certain column in a dataframe. Besides the mentioned standard method: I found the subset of string operations in Hadley Wickhams stringr package helpful. They have a much more consistent interface compared to the standard interface. str_sub: accepts negative positions, which are calculated from the left of the last character. Dieter -- View this message in context: http://r.789695.n4.nabble.com/replace-last-3-characters-of-string-tp3424354p3425855.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] gap.barplot doesn't support data arrays?
True - gapped stacked bar plots make no sense at all. I'm working my way up to a gapped bar plot with series next to each other (and error bars!), what you'd get if you put a gap in the y-axis of twogrp2-array(twogrp, dim=c(2,5)) barplot(twogrp2, beside=TRUE) I'm guessing I can do this if I spend enough time with the axis.break() function. But is there an easier way? --Drew -Original Message- From: Peter Ehlers [mailto:ehl...@ucalgary.ca] Sent: Monday, April 04, 2011 4:33 PM To: Andrew D. Steen Cc: r-help@r-project.org Subject: Re: [R] gap.barplot doesn't support data arrays? On 2011-04-04 06:39, Andrew D. Steen wrote: I am trying to make a barplot with a broken axis using gap.barplot (in the indispensable plotrix package). This works well when the data is a vector: twogrp-c(rnorm(10)+4,rnorm(10)+20) gap.barplot(twogrp,gap=c(8,16),xlab=Index,ytics=c(3,6,17,20),ylab=Gr oup values,main=Barplot with gap) But when the data is an array (for a bar plot with multiple series) I get an error and a strange plot with no y-tics and bars stretching downwards, as if all the values were negative: twogrp2-array(twogrp, dim=c(2,5)) gap.barplot(twogrp2,gap=c(8,16),xlab=Index,ytics=c(3,6,17,20),ylab=G roup values,main=Barplot with gap) Error in rect(xtics[bigones] - halfwidth, botgap, xtics[bigones] + halfwidth, : cannot mix zero-length and non-zero-length coordinates However, the main title and axis labels do appear correctly. Are data arrays unsupported for gap.barplot, or am I missing something? Looks like they're not supported, as you could easily see from the code. But do gapped stacked barplots even make sense? Not to me. Still, I think that the plotrix documentation is somewhat spotty. The help page for gap.barplot indicates that the input 'y' should be 'data values'; not overly informative. Peter Ehlers Thanks, Drew Steen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] help
Hi R Group Thanks for some suggestions. I have finally figured it out.. The following script called from within R Session does what i want..(to attach files called test1.pdf and Document-1.pdf into the file test2.pdf and then save the new file with attachments as test3.pdf at the given path) options(useFancyQuotes=FALSE) system(noquote(paste(pdftools -S,dQuote(paste(attachfiles=C:\\test1.pdf|C:\\Document-1.pdf)), paste(-i C:\\test2.pdf -o C:\\test3.pdf Regards Vijayan Padmanabhan On Sun, Apr 3, 2011 at 5:21 PM, Mike Marchywka marchy...@hotmail.comwrote: Date: Sun, 3 Apr 2011 01:35:16 +0530 From: nandan.a...@gmail.com To: padmanabhan.vija...@gmail.com CC: r-help@r-project.org Subject: Re: [R] help One way that u might have thought of is to create plot in PDF in R and the use pdftools. Additionally one can also think of running R script using R CMD and then using pdftools in a .sh script file if u r in linux. I am not aware of pdftools capability in R. On 2 April 2011 23:01, Vijayan Padmanabhan wrote: Dear R Help group I need to run a command line script from within R session. I am not clear how i can acheive this. I tried shell and system function, but i am missing something critical.can someone provide help? My intention is to create a pdf file of a plot in R and then attach existing files from my system as attachment into the newly created pdf file. Any help would be greatly appreciated.. Here is the command line script i want to execute from within R. pdftools -S attachfiles=C:\test1.pdf -i C:\test2.pdf -o C:\test4.pdf Regards Vijayan Padmanabhan I just tried system(pdftk --help) and it appeared to work as I have pdftk from cygwin.I routinely do this the other way however and invoke R from a bash script and then use external tools like this from the bash script after R is done. If I'm generating various pieces, it seems to make sense to get them all first and release any resources R has accumulated as pdf manipulation itself can often require lots of memory etc. [[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] merging data list in to single data frame
Dear R community members I did find a good way to merge my 200 text data files in to a single data file with one column added will show indicator for that file. filelist = list.files(pattern = K*cd.txt) # the file names are K1cd.txt .to K200cd.txt data_list -lapply(filelist, read.table, header=T, comment=;, fill=T) This will create list, but this is not what I want. I want a single dataframe (all separate dataframes have same variable headings) with additional row for example ; just for example, two small datasets are created by my component datasets are huge, need automation ;read from file K1cd.txt var1 var2 var3var4 1 6 0.3 8 3 4 0.4 9 2 3 0.4 6 1 0.40.9 3 ;read from file K2cd.txt var1 var2 var3var4 1 16 0.67 3 14 0.4 6 2 1 3 0.4 5 1 0.60.9 2 the output dataframe should look like Fileno var1 var2 var3var4 1 1 6 0.38 1 3 4 0.4 9 1 2 3 0.4 6 1 1 0.4 0.93 2 1 16 0.67 2 3 14 0.46 2 2 1 3 0.45 2 1 0.6 0.9 2 Please note that new file no column is added Thank you for the help. Umesh 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] Adjusting p values of a matrix
Dear all, I have an n x n matrix of p-values. The matrix is symmetrical, as it describes the each against each p values of correlation coefficients. How can I best correct the p values of the matrix? Notably, the total number of the tests performed is n(n-1)/2, since I do not test the correlation of each variable with itself. That means, I only want to correct one half of the matrix, not including the diagonal. Therefore, simply writing pmat - p.adjust( pmat, method= fdr ) # where pmat is an n x n matrix ...doesn't cut it. Of course, I can turn the matrix in to a three column data frame with n(n-1)/2 rows, but that is slow and not elegant. regards, j. -- Dr. January Weiner 3 -- Max Planck Institute for Infection Biology Charitéplatz 1 D-10117 Berlin, Germany Web : www.mpiib-berlin.mpg.de Tel : +49-30-28460514 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Creating multiple vector/list names-novice
Hi Thanks ,however I would need something different still... I would need to return a vector so if as to choose cc[[3]] [2] would return vector/list as in terms of c(b,d,e) Thanks -- View this message in context: http://r.789695.n4.nabble.com/Creating-multiple-vector-list-names-novice-tp3425616p3425759.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] Multithreading of Geneland
Hi all, I would like to multithread that script, to detect structure from multilocus genetic data : library(Geneland) geno = read.table(cot966gen_test.txt) #the file is show after MCMC(geno.dip.codom = geno, varnpop=T, npopmax=20, spatial = F, nit=10, thinnin=100, path.mcmc=./) PostProcessChain(path.mcmc=./, nxdom=100, nydom=100, burnin=200) I have an 8 cores computer, and since I have to compute that bit on a thousand of line like that : 209 209 217 217 180 180 154 154 181 181 192 192 -9 -9 211 211 -9 -9 160 160 -9 -9 -9 -9 -9 -9 254 254 140 140 181 181 -9 -9 -9 -9 211 211 158 158 160 160 158 146 209 209 223 223 348 348 186 186 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 154 154 146 146 212 212 291 291 218 218 170 170 187 187 194 194 190 190 214 208 152 152 158 158 155 155 209 209 -9 -9 180 180 150 150 185 185 -9 -9 -9 -9 211 211 152 152 -9 -9 155 155 212 212 209 209 234 234 166 166 187 187 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 205 205 236 236 164 164 187 187 194 194 183 183 214 214 163 163 160 160 155 155 212 212 203 203 238 236 164 164 187 187 -9 -9 179 179 -9 -9 -9 -9 -9 -9 -9 -9 212 212 203 203 236 236 164 164 187 187 194 194 -9 -9 214 214 165 165 160 160 155 155 212 212 209 209 236 234 164 164 187 187 194 194 179 179 -9 -9 -9 -9 -9 -9 -9 -9 212 212 205 205 236 236 166 166 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 -9 -9 180 180 147 147 181 181 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 -9 -9 180 180 147 147 181 181 198 198 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 209 209 234 234 166 166 187 187 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 209 209 234 234 166 166 187 187 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 209 209 236 236 166 166 187 187 194 194 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 207 207 236 236 164 164 -9 -9 194 194 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 212 205 205 238 238 164 164 187 187 -9 -9 -9 -9 -9 -9 -9 -9 160 160 155 155 212 212 209 207 234 234 164 164 187 187 194 194 183 183 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 166 166 -9 -9 194 194 183 183 -9 -9 -9 -9 -9 -9 -9 -9 212 212 209 205 234 234 164 164 187 187 194 194 179 179 214 214 165 165 160 160 155 155 212 212 205 205 234 234 164 164 187 187 194 194 183 183 214 214 163 163 160 160 155 155 212 212 209 209 236 236 166 166 187 187 194 194 -9 -9 -9 -9 -9 -9 -9 -9 -9 -9 212 209 -9 -9 180 180 150 147 187 187 -9 -9 -9 -9 211 211 158 154 -9 -9 155 155 209 209 207 207 328 328 184 184 181 181 -9 -9 -9 -9 211 211 165 165 -9 -9 155 155 212 212 253 253 212 212 168 150 187 181 -9 -9 -9 -9 208 208 160 160 -9 -9 155 146 212 212 -9 -9
Re: [R] Adjusting p values of a matrix
1. This is not an R question, AFAICS. 2. Sounds like a research topic. I don't think there's a meaningful simple answer. I suspect it strongly depends on the model and context. -- Bert On Mon, Apr 4, 2011 at 8:02 AM, January Weiner january.wei...@mpiib-berlin.mpg.de wrote: Dear all, I have an n x n matrix of p-values. The matrix is symmetrical, as it describes the each against each p values of correlation coefficients. How can I best correct the p values of the matrix? Notably, the total number of the tests performed is n(n-1)/2, since I do not test the correlation of each variable with itself. That means, I only want to correct one half of the matrix, not including the diagonal. Therefore, simply writing pmat - p.adjust( pmat, method= fdr ) # where pmat is an n x n matrix ...doesn't cut it. Of course, I can turn the matrix in to a three column data frame with n(n-1)/2 rows, but that is slow and not elegant. regards, j. -- Dr. January Weiner 3 -- Max Planck Institute for Infection Biology Charitéplatz 1 D-10117 Berlin, Germany Web : www.mpiib-berlin.mpg.de Tel : +49-30-28460514 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] power of 2 way ANOVA with interaction
You can use simulation: 1. decide what you think your data will look like 2. decide how you plan to analyze your data 3. write a function that simulates a dataset (common arguments include sample size(s) and effect sizes) then analyzes the data in your planned manner and returns the p-value(s) or other statistic(s) of interest 4. run the function from 3 a bunch (1000 or more) times, the replicate function is useful for this (progress bars can also be useful) 5. the proportion of times that the results are significant is your estimate of power -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Timothy Spier Sent: Sunday, April 03, 2011 5:11 PM To: R Help Subject: [R] power of 2 way ANOVA with interaction I've been searching for an answer to this for a while but no joy. I have a simple 2-way ANOVA with an interaction. I'd like to determine the power of this test for each factor (factor A, factor B, and the A*B interaction). How can I do this in R? I used to do this with proc Glmpower in SAS, but I can find no analogue 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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Adjusting p values of a matrix
On Apr 4, 2011, at 17:02 , January Weiner wrote: Dear all, I have an n x n matrix of p-values. The matrix is symmetrical, as it describes the each against each p values of correlation coefficients. How can I best correct the p values of the matrix? Notably, the total number of the tests performed is n(n-1)/2, since I do not test the correlation of each variable with itself. That means, I only want to correct one half of the matrix, not including the diagonal. Therefore, simply writing pmat - p.adjust( pmat, method= fdr ) # where pmat is an n x n matrix ...doesn't cut it. Of course, I can turn the matrix in to a three column data frame with n(n-1)/2 rows, but that is slow and not elegant. I don't think there's a really elegant way (have a look inside pairwise.table if you care). If you start one step further back, you could just use pairwise.table with a suitably defined comparison function. Otherwise, how about ltri - lower.tri(pmat) utri - upper.tri(pmat) pmat[ltri] - p.adjust(pmat[ltri], method = fdr) pmat[utri] - t(pmat)[utri] -- 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.
Re: [R] Adjusting p values of a matrix
There are also the multcomp and multcompView packages that might provide something of interest in this regard. multcomp has a companion book, Multiple Comparisons Using R (Bretz, Hothorn, Westfall, 2010, CRC Press), which I believe provides an excellent overview of the state of the art in multiple comparisons. The simple rule is Bonferroni, which involves multiplying the p-values by n or n(n-1)/2. Note, also, that one of the most important innovations in statistical methods of the past quarter century is the development of false discovery rate, which estimates the false alarm rate among the cases that the user actually sees, which is a mixture of true and false hypotheses. The p value, by contrast, is the probability of a decision error only among hypotheses that are true. For more info, see the Wikipedia entries on Bonferroni or false discovery rate -- or the book by Bretz, Hothorn and Westfall or the vignettes accompanying the multcomp package. Hope this helps. Spencer On 4/4/2011 8:54 AM, Bert Gunter wrote: 1. This is not an R question, AFAICS. 2. Sounds like a research topic. I don't think there's a meaningful simple answer. I suspect it strongly depends on the model and context. -- Bert On Mon, Apr 4, 2011 at 8:02 AM, January Weiner january.wei...@mpiib-berlin.mpg.de wrote: Dear all, I have an n x n matrix of p-values. The matrix is symmetrical, as it describes the each against each p values of correlation coefficients. How can I best correct the p values of the matrix? Notably, the total number of the tests performed is n(n-1)/2, since I do not test the correlation of each variable with itself. That means, I only want to correct one half of the matrix, not including the diagonal. Therefore, simply writing pmat- p.adjust( pmat, method= fdr ) # where pmat is an n x n matrix ...doesn't cut it. Of course, I can turn the matrix in to a three column data frame with n(n-1)/2 rows, but that is slow and not elegant. regards, j. -- Dr. January Weiner 3 -- Max Planck Institute for Infection Biology Charitéplatz 1 D-10117 Berlin, Germany Web : www.mpiib-berlin.mpg.de Tel : +49-30-28460514 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Spencer Graves, PE, PhD President and Chief Operating Officer Structure Inspection and Monitoring, Inc. 751 Emerson Ct. San José, CA 95126 ph: 408-655-4567 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] add zero in front of numbers
On 04.04.2011 12:35, Yan Jiao wrote: Dear R users, I need to add 0 in front of a series of numbers, e.g. 1-001, 19-019, Is there a fast way of doing that? formatC(c(1, 19), flag=0, width=3) Uwe Ligges Many thanks yan [[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] Adjusting p values of a matrix
How about as.matrix(p.adjust(as.dist(pmat))) Benno On 4.Apr.2011, at 17:02, January Weiner wrote: Dear all, I have an n x n matrix of p-values. The matrix is symmetrical, as it describes the each against each p values of correlation coefficients. How can I best correct the p values of the matrix? Notably, the total number of the tests performed is n(n-1)/2, since I do not test the correlation of each variable with itself. That means, I only want to correct one half of the matrix, not including the diagonal. Therefore, simply writing pmat - p.adjust( pmat, method= fdr ) # where pmat is an n x n matrix ...doesn't cut it. Of course, I can turn the matrix in to a three column data frame with n(n-1)/2 rows, but that is slow and not elegant. regards, j. -- Dr. January Weiner 3 -- Max Planck Institute for Infection Biology Charitéplatz 1 D-10117 Berlin, Germany Web : www.mpiib-berlin.mpg.de Tel : +49-30-28460514 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging data list in to single data frame
On 04.04.2011 16:41, Umesh Rosyara wrote: Dear R community members I did find a good way to merge my 200 text data files in to a single data file with one column added will show indicator for that file. filelist = list.files(pattern = K*cd.txt) I doubt you meant K*cd.txt but ^K[[:digit:]]*cd\\.txt$. # the file names are K1cd.txt .to K200cd.txt data_list-lapply(filelist, read.table, header=T, comment=;, fill=T) Replace by: data_list - lapply(filelist, function(x) cbind(Filename = x, read.table(x, header=T, comment=;, fill=TRUE)) And then: result - do.call(rbind, data_list) Uwe Ligges This will create list, but this is not what I want. I want a single dataframe (all separate dataframes have same variable headings) with additional row for example ; just for example, two small datasets are created by my component datasets are huge, need automation ;read from file K1cd.txt var1 var2 var3var4 1 6 0.3 8 3 4 0.4 9 2 3 0.4 6 1 0.40.9 3 ;read from file K2cd.txt var1 var2 var3var4 1 16 0.67 3 14 0.4 6 2 1 3 0.4 5 1 0.60.9 2 the output dataframe should look like Fileno var1 var2 var3var4 1 1 6 0.38 1 3 4 0.4 9 1 2 3 0.4 6 1 1 0.4 0.93 2 1 16 0.67 2 3 14 0.46 2 2 1 3 0.45 2 1 0.6 0.9 2 Please note that new file no column is added Thank you for the help. Umesh 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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 'RQuantLib for 2.12 version
Hi Mauricio, A Windows binary is now available on CRAN: http://dirk.eddelbuettel.com/blog/2011/04/04/#rquantlib_0.3.7 Best, -- Joshua Ulrich | FOSS Trading: www.fosstrading.com On Tue, Mar 29, 2011 at 10:38 AM, Mauricio Romero mauricio.rom...@quantil.com.co wrote: Dear R users, I have been trying to use RQuantLib in the 2.12.2 R version. I downloaded the .zip file from http://sourceforge.net/projects/quantlib/files/QuantLib/ http://sourceforge.net/projects/quantlib/files/QuantLib/ which didn't work, so I I downloaded the package source from http://sourceforge.net/projects/quantlib/files/QuantLib/1.0.1/ http://sourceforge.net/projects/quantlib/files/QuantLib/1.0.1/ Then I run: install.packages(RQuantLib_0.3.6.tar.gz, type=source, repos=NULL) and get the following error: Installing package(s) into 'C:\Users\Mauricio\Documents/R/win-library/2.12' (as 'lib' is unspecified) * installing *source* package 'RQuantLib' ... ** WARNING: this package has a configure script It probably needs manual configuration ** ** libs *** arch - i386 cygwin warning: MS-DOS style path detected: C:/PROGRA~1/R/R-212~1.2/etc/i386/Makeconf Preferred POSIX equivalent is: /cygdrive/c/PROGRA~1/R/R-212~1.2/etc/i386/Makeconf CYGWIN environment variable option nodosfilewarning turns off this warning. Consult the user's guide for more details about POSIX paths: http://cygwin.com/cygwin-ug-net/using.html#using-pathnames g++ -IC:/PROGRA~1/R/R-212~1.2/include -IC:/Users/Mauricio/Documents/R/win-library/2.12/Rcpp/include -I -I. -O2 -Wall -c asian.cpp -o asian.o asian.cpp:26:23: fatal error: rquantlib.h: No such file or directory compilation terminated. make: *** [asian.o] Error 1 ERROR: compilation failed for package 'RQuantLib' * removing 'C:/Users/Mauricio/Documents/R/win-library/2.12/RQuantLib' Warning messages: 1: running command 'C:\PROGRA~1\R\R-212~1.2/bin/i386/R CMD INSTALL -l C:\Users\Mauricio\Documents/R/win-library/2.12 RQuantLib_0.3.6.tar.gz' had status 1 2: In install.packages(RQuantLib_0.3.6.tar.gz, type = source, repos = NULL) : installation of package 'RQuantLib_0.3.6.tar.gz' had non-zero exit status Any ideas? Thanks, Mauricio Romero [[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] lattice: how to center a subtitle?
Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html). A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] converting affybatch object to matrix
Use exprs on the output from RMA (or another method you like) library(affy) myData -ReadAffy() myRMA - rma(myData) e = exprs(myRMA) Also, check out the Bioconductor mailing list where Bioconductor-related topics are discussed. On Fri, Apr 1, 2011 at 9:54 AM, Landes, Ezekiel ezekiel_lan...@hms.harvard.edu wrote: I have an Affybatch object called batch : batch AffyBatch object size of arrays=1050x1050 features (196 kb) cdf=HuGene-1_0-st-v1 (32321 affyids) number of samples=384 number of genes=32321 annotation=hugene10stv1 notes= Is there a way of converting a portion of this data into a matrix? More specifically, a matrix where the 384 samples are columns and the 32321 genes are rows? The exprs function returns a matrix that has 384 columns but for some reason there are 1050^2 rows. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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-squared for object timeseries
Hi, i am new in this forum. I hope someone can help me or correct me, if this is the false subforum to write this. I have to choose the best arima model from different possibilities of a timeseries. I know the AIC; BIC and similar. But now i would like to check the value called r-squared or adjusted r-squared in a linear model (in german called: Bestimmtheitsmaß). How can i get it in R? The function summary is not availiable for the type timeseries. If there is a routine or function that gets it, perfect, if not i am able to calculate this value for linear models. Is the equivalent thing for timeseries given, if i simply change the variable y from a linear model by the time? Thanks for any help! -- View this message in context: http://r.789695.n4.nabble.com/r-squared-for-object-timeseries-tp3425984p3425984.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] Adjusting p values of a matrix
There are, however, the multcomp and multcompView packages that might provide something of interest in this regard. multcomp has a companion book, Multiple Comparisons Using R (Bretz, Hothorn, Westfall, 2010, CRC Press), which I believe provides an excellent overview of the state of the art in multiple comparisons. The simple rule is Bonferroni, which involves multiplying the p-values by n or n(n-1)/2. Note, also, that one of the most important innovations in statistical methods of the past quarter century is the development of false discovery rate, which estimates the false alarm rate among the cases that the user actually sees, which is a mixture of true and false hypotheses. The p value, by contrast, is the probability of a decision error only among hypotheses that are true. For more info, see the Wikipedia entries on Bonferroni or false discovery rate -- or the book by Bretz, Hothorn and Westfall or the vignettes accompanying the multcomp package. Hope this helps. Spencer On 4/4/2011 8:54 AM, Bert Gunter wrote: 1. This is not an R question, AFAICS. 2. Sounds like a research topic. I don't think there's a meaningful simple answer. I suspect it strongly depends on the model and context. -- Bert On Mon, Apr 4, 2011 at 8:02 AM, January Weiner january.wei...@mpiib-berlin.mpg.de wrote: Dear all, I have an n x n matrix of p-values. The matrix is symmetrical, as it describes the each against each p values of correlation coefficients. How can I best correct the p values of the matrix? Notably, the total number of the tests performed is n(n-1)/2, since I do not test the correlation of each variable with itself. That means, I only want to correct one half of the matrix, not including the diagonal. Therefore, simply writing pmat- p.adjust( pmat, method= fdr ) # where pmat is an n x n matrix ...doesn't cut it. Of course, I can turn the matrix in to a three column data frame with n(n-1)/2 rows, but that is slow and not elegant. regards, j. -- Dr. January Weiner 3 -- Max Planck Institute for Infection Biology Charitéplatz 1 D-10117 Berlin, Germany Web : www.mpiib-berlin.mpg.de Tel : +49-30-28460514 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Spencer Graves, PE, PhD President and Chief Operating Officer Structure Inspection and Monitoring, Inc. 751 Emerson Ct. San José, CA 95126 ph: 408-655-4567 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 remaining: define any character as na.string RE: merging data list in to single data frame
Dear Uwe and R community members Thank you Uwe for the help. I have still a question remaining, I am trying to find answer from long time. While exporting my data, I have some characters mixed into it. I want to define any characters as na.string? Is it possible to do so? Thanks; Umesh -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: Monday, April 04, 2011 12:22 PM To: Umesh Rosyara Cc: r-help@r-project.org; rosyar...@gmail.com Subject: Re: [R] merging data list in to single data frame On 04.04.2011 16:41, Umesh Rosyara wrote: Dear R community members I did find a good way to merge my 200 text data files in to a single data file with one column added will show indicator for that file. filelist = list.files(pattern = K*cd.txt) I doubt you meant K*cd.txt but ^K[[:digit:]]*cd\\.txt$. # the file names are K1cd.txt .to K200cd.txt data_list-lapply(filelist, read.table, header=T, comment=;, fill=T) Replace by: data_list - lapply(filelist, function(x) cbind(Filename = x, read.table(x, header=T, comment=;, fill=TRUE)) And then: result - do.call(rbind, data_list) Uwe Ligges This will create list, but this is not what I want. I want a single dataframe (all separate dataframes have same variable headings) with additional row for example ; just for example, two small datasets are created by my component datasets are huge, need automation ;read from file K1cd.txt var1 var2 var3var4 1 6 0.3 8 3 4 0.4 9 2 3 0.4 6 1 0.40.9 3 ;read from file K2cd.txt var1 var2 var3var4 1 16 0.67 3 14 0.4 6 2 1 3 0.4 5 1 0.60.9 2 the output dataframe should look like Fileno var1 var2 var3var4 1 1 6 0.38 1 3 4 0.4 9 1 2 3 0.4 6 1 1 0.4 0.93 2 1 16 0.67 2 3 14 0.46 2 2 1 3 0.45 2 1 0.6 0.9 2 Please note that new file no column is added Thank you for the help. Umesh 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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
On Apr 4, 2011, at 12:45 PM, Marius Hofert wrote: Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 And if you wanted different fontface (underline, italic or bold) then you could use plotmath expressions: xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = expression( atop(but~subtitles2~are~underline(now)~centered, bold(bla) )), scales = list(alternating = c(1,1), tck = c(1,0))) -- David. Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html) . A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] merging 2 frames while keeping all the entries from the reference frame
Hello! I have my data frame mydata (below) and data frame reference - that contains all the dates I would like to be present in the final data frame. I am trying to merge them so that the the result data frame contains all 8 dates in both subgroups (i.e., Group1 should have 8 rows and Group2 too). But when I merge it it's not coming out this way. Any hint would be greatly appreciated! Dimitri mydata-data.frame(mydate=rep(seq(as.Date(2008-12-29), length = 8, by = week),2), group=c(rep(Group1,8),rep(Group2,8)),values=rnorm(16,1,1)) (reference);(mydata) set.seed(1234) out-sample(1:16,5,replace=F) mydata-mydata[-out,]; dim(mydata) (mydata) # reference contains the dates I want to be present in the final data frame: reference-data.frame(mydate=seq(as.Date(2008-12-29), length = 8, by = week)) # Merging: new.data-merge(mydata,reference,by=mydate,all.x=T,all.y=T) new.data-new.data[order(new.data$group,new.data$mydate),] (new.data) # my new.data contains only 7 rows in Group 1 and 4 rows in Group 2 -- Dimitri Liakhovitski Ninah Consulting __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Questions remaining: define any character as na.string RE: merging data list in to single data frame
On Apr 4, 2011, at 12:37 PM, Umesh Rosyara wrote: Dear Uwe and R community members Thank you Uwe for the help. I have still a question remaining, I am trying to find answer from long time. While exporting my data, I have some characters mixed into it. I want to define any characters as na.string? Is it possible to do so? Option 1: do it in an editor that is regex aware. Option 2: input the file with readLines, use gsub to remove the unwanted characters, read.table(textConnection(obj)) on the resulting object. [There are many worked examples in the archives. Search on read.table(textConnection( .] -- David. Thanks; Umesh -Original Message- From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de] Sent: Monday, April 04, 2011 12:22 PM To: Umesh Rosyara Cc: r-help@r-project.org; rosyar...@gmail.com Subject: Re: [R] merging data list in to single data frame On 04.04.2011 16:41, Umesh Rosyara wrote: Dear R community members I did find a good way to merge my 200 text data files in to a single data file with one column added will show indicator for that file. filelist = list.files(pattern = K*cd.txt) I doubt you meant K*cd.txt but ^K[[:digit:]]*cd\\.txt$. # the file names are K1cd.txt .to K200cd.txt data_list-lapply(filelist, read.table, header=T, comment=;, fill=T) Replace by: data_list - lapply(filelist, function(x) cbind(Filename = x, read.table(x, header=T, comment=;, fill=TRUE)) And then: result - do.call(rbind, data_list) Uwe Ligges This will create list, but this is not what I want. I want a single dataframe (all separate dataframes have same variable headings) with additional row for example ; just for example, two small datasets are created by my component datasets are huge, need automation ;read from file K1cd.txt var1 var2 var3var4 1 6 0.3 8 3 4 0.4 9 2 3 0.4 6 1 0.40.9 3 ;read from file K2cd.txt var1 var2 var3var4 1 16 0.67 3 14 0.4 6 2 1 3 0.4 5 1 0.60.9 2 the output dataframe should look like Fileno var1 var2 var3var4 1 1 6 0.38 1 3 4 0.4 9 1 2 3 0.4 6 1 1 0.4 0.93 2 1 16 0.67 2 3 14 0.46 2 2 1 3 0.45 2 1 0.6 0.9 2 Please note that new file no column is added Thank you for the help. Umesh 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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging 2 frames while keeping all the entries from the reference frame
To clarify just in case, here is the result I am trying to get: mydate group values 12/29/2008 Group1 0.453466522 1/5/2009Group1 NA 1/12/2009 Group1 0.416548943 1/19/2009 Group1 2.066275155 1/26/2009 Group1 2.037729638 2/2/2009Group1 -0.598040483 2/9/2009Group1 1.658999227 2/16/2009 Group1 -0.869325211 12/29/2008 Group2 NA 1/5/2009Group2 NA 1/12/2009 Group2 NA 1/19/2009 Group2 0.375284194 1/26/2009 Group2 0.706785401 2/2/2009Group2 NA 2/9/2009Group2 2.104937151 2/16/2009 Group2 2.880393978 On Mon, Apr 4, 2011 at 1:09 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Hello! I have my data frame mydata (below) and data frame reference - that contains all the dates I would like to be present in the final data frame. I am trying to merge them so that the the result data frame contains all 8 dates in both subgroups (i.e., Group1 should have 8 rows and Group2 too). But when I merge it it's not coming out this way. Any hint would be greatly appreciated! Dimitri mydata-data.frame(mydate=rep(seq(as.Date(2008-12-29), length = 8, by = week),2), group=c(rep(Group1,8),rep(Group2,8)),values=rnorm(16,1,1)) (reference);(mydata) set.seed(1234) out-sample(1:16,5,replace=F) mydata-mydata[-out,]; dim(mydata) (mydata) # reference contains the dates I want to be present in the final data frame: reference-data.frame(mydate=seq(as.Date(2008-12-29), length = 8, by = week)) # Merging: new.data-merge(mydata,reference,by=mydate,all.x=T,all.y=T) new.data-new.data[order(new.data$group,new.data$mydate),] (new.data) # my new.data contains only 7 rows in Group 1 and 4 rows in Group 2 -- Dimitri Liakhovitski Ninah Consulting -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
Dear David, do you know how to get plotmath-like symbols in both rows? I tried s.th. like: lab - expression(paste(alpha==1, , , beta==2, sep=)) xlab - substitute(expression( atop(lab==lab., bold(foo)) ), list(lab.=lab)) xyplot(0 ~ 0, xlab = xlab) Cheers, Marius On 2011-04-04, at 18:59 , David Winsemius wrote: On Apr 4, 2011, at 12:45 PM, Marius Hofert wrote: Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 And if you wanted different fontface (underline, italic or bold) then you could use plotmath expressions: xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = expression( atop(but~subtitles2~are~underline(now)~centered, bold(bla) )), scales = list(alternating = c(1,1), tck = c(1,0))) -- David. Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html). A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
On Apr 4, 2011, at 1:27 PM, Marius Hofert wrote: Dear David, do you know how to get plotmath-like symbols in both rows? I tried s.th. like: lab - expression(paste(alpha==1, , , beta==2, sep=)) xlab - substitute(expression( atop(lab==lab., bold(foo)) ), list(lab.=lab)) xyplot(0 ~ 0, xlab = xlab) I _did_ have plotmath functions in both rows: But here is your solution: xyplot(0 ~ 0, xlab = expression( atop(paste(alpha==1,, beta==2), bold(bla) )) ) ) Note that `paste` in plotmath is different than `paste` in regular R. It has no `sep` argument. I did try both substitute and bquote on you externally expression, but lattice seems to be doing some non- standard evaluation and I never got it to work. Using what I thought _should_ work, does work with `plot`: x=1;y=2 plot(0 ~ 0, xlab = bquote( atop(alpha==.(x)*,~beta==.(y), bold(foo) ) ) + ) But the same expression throws an error with xyplot: x=1;y=2 xyplot(0 ~ 0, xlab = bquote( atop(alpha==.(x)*,~beta==.(y), bold(foo) ) ) + ) Error in trellis.skeleton(formula = 0 ~ 0, cond = list(1L), aspect = fill, : could not find function atop -- David. Cheers, Marius On 2011-04-04, at 18:59 , David Winsemius wrote: On Apr 4, 2011, at 12:45 PM, Marius Hofert wrote: Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 And if you wanted different fontface (underline, italic or bold) then you could use plotmath expressions: xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = expression( atop(but~subtitles2~are~underline(now)~centered, bold(bla) )), scales = list(alternating = c(1,1), tck = c(1,0))) -- David. Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html) . A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x- axis label. David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
On 2011-04-04 10:27, Marius Hofert wrote: Dear David, do you know how to get plotmath-like symbols in both rows? I tried s.th. like: lab- expression(paste(alpha==1, , , beta==2, sep=)) xlab- substitute(expression( atop(lab==lab., bold(foo)) ), list(lab.=lab)) xyplot(0 ~ 0, xlab = xlab) Marius, I always find paste a bit tricky with plotmath. Maybe this will do what you want: mylab - expression( atop(lab==list(alpha==1, beta==2), bold(foo)) ) xyplot(0 ~ 0, xlab = mylab) Peter Ehlers Cheers, Marius On 2011-04-04, at 18:59 , David Winsemius wrote: On Apr 4, 2011, at 12:45 PM, Marius Hofert wrote: Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 And if you wanted different fontface (underline, italic or bold) then you could use plotmath expressions: xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = expression( atop(but~subtitles2~are~underline(now)~centered, bold(bla) )), scales = list(alternating = c(1,1), tck = c(1,0))) -- David. Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html). A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
Maybe: xyplot(0 ~ 0, xlab = bquote(expression(atop(alpha==.(x)*,~beta==.(y), bold(foo) )) )) On Mon, Apr 4, 2011 at 2:58 PM, David Winsemius dwinsem...@comcast.net wrote: On Apr 4, 2011, at 1:27 PM, Marius Hofert wrote: Dear David, do you know how to get plotmath-like symbols in both rows? I tried s.th. like: lab - expression(paste(alpha==1, , , beta==2, sep=)) xlab - substitute(expression( atop(lab==lab., bold(foo)) ), list(lab.=lab)) xyplot(0 ~ 0, xlab = xlab) I _did_ have plotmath functions in both rows: But here is your solution: xyplot(0 ~ 0, xlab = expression( atop(paste(alpha==1, , beta==2), bold(bla) )) ) ) Note that `paste` in plotmath is different than `paste` in regular R. It has no `sep` argument. I did try both substitute and bquote on you externally expression, but lattice seems to be doing some non-standard evaluation and I never got it to work. Using what I thought _should_ work, does work with `plot`: x=1;y=2 plot(0 ~ 0, xlab = bquote( atop(alpha==.(x)*,~beta==.(y), bold(foo) ) ) + ) But the same expression throws an error with xyplot: x=1;y=2 xyplot(0 ~ 0, xlab = bquote( atop(alpha==.(x)*,~beta==.(y), bold(foo) ) ) + ) Error in trellis.skeleton(formula = 0 ~ 0, cond = list(1L), aspect = fill, : could not find function atop -- David. Cheers, Marius On 2011-04-04, at 18:59 , David Winsemius wrote: On Apr 4, 2011, at 12:45 PM, Marius Hofert wrote: Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 And if you wanted different fontface (underline, italic or bold) then you could use plotmath expressions: xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = expression( atop(but~subtitles2~are~underline(now)~centered, bold(bla) )), scales = list(alternating = c(1,1), tck = c(1,0))) -- David. Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html). A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Granger Causality in a VAR Model
Dear Community, I am new to R and have a question concerning the causality () test in the vars package. I need to test whether, say, the variable y Granger causes the variable x, given z as a control variable. I estimated the VAR model as follows: model-VAR(cbind(x,y,z),p=2) Then I did the following: causality(model, cause=y). I thing this tests the Granger causality of y on the vector (x,z), though. How can I implement the test for y causing x controlled for z? Thus, the F-test comparing the two models M1:x~lagged(x)+lagged(z) and M2:x~lagged(x)+lagged(y)+lagged(z)? Thank you in advance. Best Regards __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: how to center a subtitle?
Dear all, many thanks, that helped a lot! Cheers, Marius On 2011-04-04, at 19:58 , David Winsemius wrote: On Apr 4, 2011, at 1:27 PM, Marius Hofert wrote: Dear David, do you know how to get plotmath-like symbols in both rows? I tried s.th. like: lab - expression(paste(alpha==1, , , beta==2, sep=)) xlab - substitute(expression( atop(lab==lab., bold(foo)) ), list(lab.=lab)) xyplot(0 ~ 0, xlab = xlab) I _did_ have plotmath functions in both rows: But here is your solution: xyplot(0 ~ 0, xlab = expression( atop(paste(alpha==1,, beta==2), bold(bla) )) ) ) Note that `paste` in plotmath is different than `paste` in regular R. It has no `sep` argument. I did try both substitute and bquote on you externally expression, but lattice seems to be doing some non-standard evaluation and I never got it to work. Using what I thought _should_ work, does work with `plot`: x=1;y=2 plot(0 ~ 0, xlab = bquote( atop(alpha==.(x)*,~beta==.(y), bold(foo) ) ) + ) But the same expression throws an error with xyplot: x=1;y=2 xyplot(0 ~ 0, xlab = bquote( atop(alpha==.(x)*,~beta==.(y), bold(foo) ) ) + ) Error in trellis.skeleton(formula = 0 ~ 0, cond = list(1L), aspect = fill, : could not find function atop -- David. Cheers, Marius On 2011-04-04, at 18:59 , David Winsemius wrote: On Apr 4, 2011, at 12:45 PM, Marius Hofert wrote: Dear David, I intended to use another x-label. But your suggestion brings me to the idea of just using a two-line xlab, so s.th. like print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered\nbla, scales = list(alternating = c(1,1), tck = c(1,0 And if you wanted different fontface (underline, italic or bold) then you could use plotmath expressions: xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = expression( atop(but~subtitles2~are~underline(now)~centered, bold(bla) )), scales = list(alternating = c(1,1), tck = c(1,0))) -- David. Thanks! Cheers, Marius On 2011-04-04, at 16:47 , David Winsemius wrote: On Apr 4, 2011, at 7:39 AM, Marius Hofert wrote: Dear expeRts, I recently asked for a real centered title (see, e.g., http://tolstoy.newcastle.edu.au/R/e13/help/11/01/0135.html). A nice solution (from Deepayan Sarkar) is to use xlab.top instead of main: library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, sub = but subtitles are not centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() library(lattice) trellis.device(pdf) print(xyplot(0 ~ 0, xlab.top = This title is now 'centered' for the human's eye, xlab = but subtitles are _now_ centered, scales = list(alternating = c(1,1), tck = c(1,0 dev.off() (I realize that those are not really subtitles by a 'lab', but that appears acceptable in your current test case.) My question is whether there is something similar for *sub*titles [so something like xlab.bottom]? As you can see from the plot, the subtitle does not seem to be centered for the human's eye. I would like to center it according to the x-axis label. David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] automating regression or correlations for many variables
Dear All, I have a large data frame with 10 rows and 82 columns. I want to apply the same function to all of the columns with a single command. e.g. zl - lm (snp$a_109909 ~ snp$lat) will fit a linear model to the values in lat and a_109909. What I want to do is fit linear models for the values in each column against lat. I tried doing zl - (snp[,2:82] ~ snp$lat[,1]) but got the following error message Error in model.frame.default(formula = snp[, 2:82] ~ snp[, 1], drop.unused.levels = TRUE) : invalid type (list) for variable 'snp[, 2:82]'. Does this mean I cannot use a data frame to do this? Should I have my data in a matrix instead? Thanks -- View this message in context: http://r.789695.n4.nabble.com/automating-regression-or-correlations-for-many-variables-tp3426091p3426091.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] Structural equation modeling in R(lavaan,sem)
Thanks you for your response For lavaan package can i have more information about this example you have applied in the section 7 the meanings of The variables (c1,c2,c3,c4, i ,s ,x1,x2) I think i have need more information to learn more about how able to apply growth model in my data (longitudianl data) Thanks a lot Antra EL MOUSSELLY Date: Mon, 4 Apr 2011 02:16:42 -0500 From: ml-node+3424797-1256108745-225...@n4.nabble.com To: antr...@hotmail.com Subject: Re: Structural equation modeling in R(lavaan,sem) On 04/03/2011 09:38 PM, jouba wrote: Daer all, I have a question concerning longitudinal data: When we have a longitudinal data and we have to do sem analysis there is in the package lavaan some functions,options in this package that help to do this or we can treat these data like non longitudinal data The function 'growth' (in the lavaan package) can be used for (standard) growth modeling. Good material about growth modeling (using Mplus) can be found here: http://statistics.ats.ucla.edu/stat/mplus/seminars/gm/default.htm Next, you can read how to do growth modeling with lavaan by reading section 7 in the lavaan intro, which you can download from the documentation section on the lavaan website (http://lavaan.org). Yves. -- Yves Rosseel -- http://www.da.ugent.be Department of Data Analysis, Ghent University Henri Dunantlaan 1, B-9000 Gent, Belgium __ [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/Structural-equation-modeling-in-R-lavaan-sem-tp3409642p3424797.html To unsubscribe from Structural equation modeling in R(lavaan,sem), click here. -- View this message in context: http://r.789695.n4.nabble.com/Structural-equation-modeling-in-R-lavaan-sem-tp3409642p3426097.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.
Re: [R] merging 2 frames while keeping all the entries from the reference frame
On Mon, Apr 4, 2011 at 1:09 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Hello! I have my data frame mydata (below) and data frame reference - that contains all the dates I would like to be present in the final data frame. I am trying to merge them so that the the result data frame contains all 8 dates in both subgroups (i.e., Group1 should have 8 rows and Group2 too). But when I merge it it's not coming out this way. Any hint would be greatly appreciated! Dimitri mydata-data.frame(mydate=rep(seq(as.Date(2008-12-29), length = 8, by = week),2), group=c(rep(Group1,8),rep(Group2,8)),values=rnorm(16,1,1)) (reference);(mydata) set.seed(1234) out-sample(1:16,5,replace=F) mydata-mydata[-out,]; dim(mydata) (mydata) # reference contains the dates I want to be present in the final data frame: reference-data.frame(mydate=seq(as.Date(2008-12-29), length = 8, by = week)) # Merging: new.data-merge(mydata,reference,by=mydate,all.x=T,all.y=T) new.data-new.data[order(new.data$group,new.data$mydate),] (new.data) # my new.data contains only 7 rows in Group 1 and 4 rows in Group 2 It might make more sense to put each group into its own column since then the object is a multivariate time series: library(zoo) z - merge(read.zoo(mydata, split = 2), zoo(, reference[[1]]), all = c(FALSE, TRUE)) z Group1Group2 2008-12-29 2.0266215NA 2009-01-05NANA 2009-01-12 1.0255344NA 2009-01-19 1.3880938 0.8135788 2009-01-26 1.4380978 1.6068682 2009-02-02 1.1764965NA 2009-02-09 1.1578531 1.4484447 2009-02-16 0.6673568 1.4760864 although if you really need to you could string them out like this: library(reshape2) melt(data.frame(time(z), coredata(z)), id = 1) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging 2 frames while keeping all the entries from the reference frame
Try this: merge(mydata, cbind(reference, group = rep(unique(mydata$group), each = nrow(reference))), all = TRUE) On Mon, Apr 4, 2011 at 2:24 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: To clarify just in case, here is the result I am trying to get: mydate group values 12/29/2008 Group1 0.453466522 1/5/2009 Group1 NA 1/12/2009 Group1 0.416548943 1/19/2009 Group1 2.066275155 1/26/2009 Group1 2.037729638 2/2/2009 Group1 -0.598040483 2/9/2009 Group1 1.658999227 2/16/2009 Group1 -0.869325211 12/29/2008 Group2 NA 1/5/2009 Group2 NA 1/12/2009 Group2 NA 1/19/2009 Group2 0.375284194 1/26/2009 Group2 0.706785401 2/2/2009 Group2 NA 2/9/2009 Group2 2.104937151 2/16/2009 Group2 2.880393978 On Mon, Apr 4, 2011 at 1:09 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Hello! I have my data frame mydata (below) and data frame reference - that contains all the dates I would like to be present in the final data frame. I am trying to merge them so that the the result data frame contains all 8 dates in both subgroups (i.e., Group1 should have 8 rows and Group2 too). But when I merge it it's not coming out this way. Any hint would be greatly appreciated! Dimitri mydata-data.frame(mydate=rep(seq(as.Date(2008-12-29), length = 8, by = week),2), group=c(rep(Group1,8),rep(Group2,8)),values=rnorm(16,1,1)) (reference);(mydata) set.seed(1234) out-sample(1:16,5,replace=F) mydata-mydata[-out,]; dim(mydata) (mydata) # reference contains the dates I want to be present in the final data frame: reference-data.frame(mydate=seq(as.Date(2008-12-29), length = 8, by = week)) # Merging: new.data-merge(mydata,reference,by=mydate,all.x=T,all.y=T) new.data-new.data[order(new.data$group,new.data$mydate),] (new.data) # my new.data contains only 7 rows in Group 1 and 4 rows in Group 2 -- Dimitri Liakhovitski Ninah Consulting -- Dimitri Liakhovitski Ninah Consulting www.ninah.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merging data list in to single data frame
Hi: Here's an alternative using ldply() from the plyr package. The idea is to read the data frames into a list, name them accordingly and then call ldply(). # Read in the test data frames (you want to use list.files() instead to input the data per Uwe's guidelines) df1 - read.table(textConnection( + var1 var2 var3var4 + 1 6 0.3 8 + 3 4 0.4 9 + 2 3 0.4 6 + 1 0.40.9 3), header = TRUE) df2 - read.table(textConnection( + var1 var2 var3 var4 + 1 16 0.6 7 + 3 14 0.4 6 + 2 13 0.4 5 + 1 0.6 0.9 2), header = TRUE) closeAllConnections() # generate the list dl - list(df1, df2) # Name the list components by number and then call ldply(): names(dl) - 1:2 # more generally, names(dl) - 1:length(dl) library(plyr) ldply(dl, rbind) .id var1 var2 var3 var4 1 11 6.0 0.38 2 13 4.0 0.49 3 12 3.0 0.46 4 11 0.4 0.93 5 21 16.0 0.67 6 23 14.0 0.46 7 22 13.0 0.45 8 21 0.6 0.92 You can always change .id to fileno afterwards. HTH, Dennis On Mon, Apr 4, 2011 at 7:41 AM, Umesh Rosyara rosy...@msu.edu wrote: Dear R community members I did find a good way to merge my 200 text data files in to a single data file with one column added will show indicator for that file. filelist = list.files(pattern = K*cd.txt) # the file names are K1cd.txt .to K200cd.txt data_list -lapply(filelist, read.table, header=T, comment=;, fill=T) This will create list, but this is not what I want. I want a single dataframe (all separate dataframes have same variable headings) with additional row for example ; just for example, two small datasets are created by my component datasets are huge, need automation ;read from file K1cd.txt var1 var2 var3var4 1 6 0.3 8 3 4 0.4 9 2 3 0.4 6 1 0.40.9 3 ;read from file K2cd.txt var1 var2 var3var4 1 16 0.67 3 14 0.4 6 2 1 3 0.4 5 1 0.60.9 2 the output dataframe should look like Fileno var1 var2 var3var4 1 1 6 0.38 1 3 4 0.4 9 1 2 3 0.4 6 1 1 0.4 0.93 2 1 16 0.67 2 3 14 0.46 2 2 1 3 0.45 2 1 0.6 0.9 2 Please note that new file no column is added Thank you for the help. Umesh 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. [[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] automating regression or correlations for many variables
Hi: Here's a small example: df - data.frame(y1 = rnorm(10), y2 = rnorm(10), y3 = rnorm(10), lat = rnorm(10)) m - lm(cbind(y1, y2, y3) ~ lat, data = df) summary(m) snip...provides summaries for each response The LHS of the model formula needs to be a matrix. In your case, something like m - lm(as.matrix(snp[, -1]) ~ lat, data = snp) ought to work. HTH, Dennis On Mon, Apr 4, 2011 at 10:13 AM, geral geral...@mail.ubc.ca wrote: Dear All, I have a large data frame with 10 rows and 82 columns. I want to apply the same function to all of the columns with a single command. e.g. zl - lm (snp$a_109909 ~ snp$lat) will fit a linear model to the values in lat and a_109909. What I want to do is fit linear models for the values in each column against lat. I tried doing zl - (snp[,2:82] ~ snp$lat[,1]) but got the following error message Error in model.frame.default(formula = snp[, 2:82] ~ snp[, 1], drop.unused.levels = TRUE) : invalid type (list) for variable 'snp[, 2:82]'. Does this mean I cannot use a data frame to do this? Should I have my data in a matrix instead? Thanks -- View this message in context: http://r.789695.n4.nabble.com/automating-regression-or-correlations-for-many-variables-tp3426091p3426091.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] zoo:rollapply by multiple grouping factors
On Mon, Apr 4, 2011 at 3:40 PM, Mark Novak mnov...@ucsc.edu wrote: Thank you very much Gabor! It looks like that's gonna work wonderfully. I didn't even know 'ave' existed. For others out there: I only needed to add a comma: dat[,c(Site, Plot, Sp)] Actually, if dd is a data frame dd[, ix] and dd[ix] give the same result. e.g. dd - data.frame(a = 1:3, b = 11:13, c = 21:23) identical(dd[, c(b, c)], dd[c(b, c)]) [1] TRUE Small follow up Q: Is there any reason to use 'aggregate' vs. 'ave' in general? aggregate reduces the data to fewer rows. ave adds a potentially additional column to the original data. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Linear Model with curve fitting parameter?
-Original Message- From: stephen sefick [mailto:ssef...@gmail.com] Sent: April-03-11 5:35 PM To: Steven McKinney Cc: R help Subject: Re: [R] Linear Model with curve fitting parameter? Steven: You are exactly right sorry I was confused. ### so log(y-intercept)+log(K) is a constant called b0 (is this right?) Doesn't look right to me based on the information you've provided. I don't see anything labeled y in your previous emails, so I'm not clear on what y is and how it relates to the original model you described I have a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. If the model is Q=K*A*(R^r)*(S^s) then log(Q) = log(K) + log(A) + r*log(R) + s*log(S) Rearranging yields log(Q) - log(A) = log(K) + r*log(R) + s*log(S) Let Z = log(Q) - log(A) = log(Q/A) so Z = log(K) + r*log(R) + s*log(S) and a linear model fit of Z ~ log(R) + log(S) will yield parameter estimates for the linear equation E(Z) = B0 + B1*log(R) + B2*log(S) (E(Z) = expected value of Z) so B0 estimate is an estimate of log(K) B1 estimate is an estimate of r B2 estimate is an estimate of s and these are the only parameters you described in the original model. lm(log(Q)~log(A)+log(R)+log(S)-1) is fitting the model log(Q)=a*log(A)+r*log(R)+s*log(S) (no beta 0) and lm(log(Q)~log(A)+log(R)+log(S)) is fitting the model log(Q)=b0+a*log(A)+r*log(R)+s*log(S) K has disappeared from these equations so these model fits do not correspond to the model originally described. Now a b0 appears, and is used in models below. I think changing notation is also adding confusion. What are y and intercept you discuss above, in relation to your original notation? ## These are the models I am trying to fit and if I have reasoned correctly above then I should be able to fit the below models similarly. You will be able to fit models appropriately once you have a clearly defined system of notation that allows you to map between the proposed data model, the parameters in that model, and the corresponding regression equations. Once you have consistent notation, you will be able to see if you can express your model as a linear regression, or if not, what kind of non-linear regression you will need to do to get estimates for the parameters in your model. Best Steve McKinney manning log(Q)=log(b0)+log(K)+log(A)+r*log(R)+s*log(S) dingman log(Q)=log(b0)+log(K)+a*log(A)+r*log(R)+s*(log(S))^2 bjerklie log(Q)=log(b0)+log(K)+a*log(A)+r*log(R)+s*log(S) ### Thank you for all of your help! Stephen On Fri, Apr 1, 2011 at 2:44 PM, Steven McKinney smckin...@bccrc.ca wrote: -Original Message- From: stephen sefick [mailto:ssef...@gmail.com] Sent: April-01-11 5:44 AM To: Steven McKinney Cc: R help Subject: Re: [R] Linear Model with curve fitting parameter? Setting Z=Q-A would be the incorrect dimensions. I could Z=Q/A. I suspect this is confusion about what Q is. I was presuming that the Q in this following formula was log(Q) with Q from the original data. I have taken the log of the data that I have and this is the model formula without the K part lm(Q~offset(A)+R+S, data=x) If the model is Q=K*A*(R^r)*(S^s) then log(Q) = log(K) + log(A) + r*log(R) + s*log(S) Rearranging yields log(Q) - log(A) = log(K) + r*log(R) + s*log(S) so what I labeled 'Z' below is Z = log(Q) - log(A) = log(Q/A) so Z = log(K) + r*log(R) + s*log(S) and a linear model fit of Z ~ log(R) + log(S) will yield parameter estimates for the linear equation E(Z) = B0 + B1*log(R) + B2*log(S) (E(Z) = expected value of Z) so B0 estimate is an estimate of log(K) B1 estimate is an estimate of r B2 estimate is an estimate of s More details and careful notation will eventually lead to a reasonable description and analysis strategy. Best Steve McKinney Is fitting a nls model the same as fitting an ols? These data are hydraulic data from ~47 sites. To access predictive ability I am removing one site fitting a new model and then accessing the fit with a myriad of model assessment criteria. I should get the same answer with ols vs nls? Thank you for all of your help. Stephen On Thu, Mar 31, 2011 at 8:34 PM, Steven McKinney smckin...@bccrc.ca wrote: -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of stephen sefick Sent: March-31-11 3:38 PM To: R help Subject: [R] Linear Model with curve fitting parameter? I have a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. I have
Re: [R] zoo:rollapply by multiple grouping factors
Thank you very much Gabor! It looks like that's gonna work
wonderfully. I didn't even know 'ave' existed.
For others out there: I only needed to add a comma: dat[,c(Site,
Plot, Sp)]
Small follow up Q: Is there any reason to use 'aggregate' vs. 'ave' in
general?
-mark
On 4/3/11 3:27 PM, Gabor Grothendieck wrote:
Try ave:
dat$cv- ave(dat$Count, dat[c(Site, Plot, Sp)], FUN =
function(x) rollapply(zoo(x), 2, cv, na.pad = TRUE, align = right))
On Sun, Apr 3, 2011 at 11:58 AM, Mark Novakmnov...@ucsc.edu wrote:
# Hi there,
# I am trying to apply a function over a moving-window for a large number of
multivariate time-series that are grouped in a nested set of factors. I
have spent a few days searching for solutions with no luck, so any
suggestions are much appreciated.
# The data I have are for the abundance dynamics of multiple species
observed in multiple fixed plots at multiple sites. (I total I have 7
sites, ~3-5 plots/site, ~150 species/plot, for 60 time-steps each.) So my
data look something like this:
dat-data.frame(Site=rep(1), Plot=rep(c(rep(1,8),rep(2,8),rep(3,8)),1),
Time=rep(c(1,1,2,2,3,3,4,4)), Sp=rep(1:2), Count=sample(24))
dat
# Let the function I want to apply over a right-aligned window of w=2 time
steps be:
cv-function(x){sd(x)/mean(x)}
w-2
# The final output I want would look something like this:
Out-data.frame(dat,CV=round(c(NA,NA,runif(6,0,1),c(NA,NA,runif(6,0,1))),2))
# I could reshape and apply zoo:rollapply() to a given plot at a given site,
and reshape again as follows:
library(zoo)
a-subset(dat,Site==1Plot==1)
b-reshape(a[-c(1,2)],v.names='Count',idvar='Time',timevar='Sp',direction='wide')
d-zoo(b[,-1],b[,1])
d
out-rollapply(d, w, cv, na.pad=T, align='right')
out
# I would thereby have to loop through all my sites and plots which,
although it deals with all species at once, still seems exceedingly
inefficient.
# So the question is, how do I use something like aggregate.zoo or tapply or
even lapply to apply rollapply on each species' time series.
# The closest I've come is the following two approaches:
# First let:
datx-list(Site=dat$Site,Plot=dat$Plot,Sp=dat$Sp)
daty-dat$Count
# Method 1.
out1-tapply(seq(along=daty),datx,function(i,x=daty){ rollapply(zoo(x[i]),
w, cv, na.pad=T, align='right') })
out1
out1[,,1]
# Which works in that it gives me the right answers, but in a format from
which I can't figure out how to get back into the format I want.
# Method 2.
fun-function(x){y-zoo(x);coredata(rollapply(y, w,
cv,na.pad=T,align='right'))}
out2-aggregate(daty,by=datx,fun)
out2
# Which superficially works better, but again only in a format I can't
figure out how to use because the output seems to be a mix of data.frame and
lists.
out2[1,4]
out2[1,5]
is.data.frame(out2)
is.list(out2)
# The situation is made more problematic by the fact that the time point of
first survey can differ between plots (e.g., site1-plot3 may only start at
time-point 3). As in...
dat2-dat
dat2-dat2[-which(dat2$Plot==3 dat2$Time3),]
dat2
# I must therefore ensure that I'm keeping track of the true time associated
with each value, not just the order of their occurences. This information
is (seemingly) lost by both methods.
datx-list(Site=dat2$Site,Plot=dat2$Plot,Sp=dat2$Sp)
daty-dat2$Count
# Method 1.
out3-tapply(seq(along=daty),datx,function(i,x=daty){ rollapply(zoo(x[i]),
w, cv, na.pad=T, align='right') })
out3
out3[1,3,1]
time(out3[1,3,1])
# Method 2
out4-aggregate(daty,by=datx,fun)
out4
time(out4[3,4])
# Am I going about this all wrong? Is there a different package to try?
Any thoughts and suggestions are much appreciated!
# R 2.12.2 GUI 1.36 Leopard build 32-bit (5691); zoo 1.6-4
# Thanks!
# -mark
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] General binary search?
Is there a generic binary search routine in a standard library which
a) works for character vectors
b) runs in O(log(N)) time?
I'm aware of findInterval(x,vec), but it is restricted to numeric vectors.
I'm also aware of various hashing solutions (e.g. new.env(hash=TRUE) and
fastmatch), but I need the greatest-lower-bound match in my application.
findInterval is also slow for large N=length(vec) because of the O(N)
checking it does, as Duncan Murdoch has pointed
outhttps://stat.ethz.ch/pipermail/r-help/2008-September/174584.html:
though
its documentation says it runs in O(n * log(N)), it actually runs in O(n *
log(N) + N), which is quite noticeable for largish N. But that is easy
enough to work around by writing a variant of findInterval which calls
find_interv_vec without checking.
-s
PS Yes, binary search is a one-liner in R, but I always prefer to use
standard, fast native libraries when possible
binarysearch - function(val,tab,L,H) {while (H=L) { M=L+(H-L) %/% 2; if
(tab[M]val) H-M-1 else if (tab[M]val) L-M+1 else return(M)};
return(L-1)}
[[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] merging data list in to single data frame
Thank you Dennis for the solution. It is a step ahead..However I need to read all 200 files as dataframes one-by-one. Can we automate this process. I used the following step to read all file at once however the data_list ended as list. filelist = list.files(pattern = K*cd.txt) # the file names are K1cd.txt .to K200cd.txt data_list -lapply(filelist, read.table, header=T, comment=;, fill=T) names(filelist) - 1:length(filelist) library(plyr) ldply(data_list, rbind) I tried to use your approach to list, is not successful to have the var .id (otherwise it is binding the component dataframes !), probably this is applicable to component data frames not list with many data frames. Do you any suggestion on using fuctions that can read the files (as I did above) and save as new dataframe (for example DF1.DF2) not a list of 200 data frames? If we can do that then we will able to use this approach. Thank you so much, Umesh R From: Dennis Murphy [mailto:djmu...@gmail.com] Sent: Monday, April 04, 2011 3:25 PM To: Umesh Rosyara Cc: r-help@r-project.org; rosyar...@gmail.com Subject: Re: [R] merging data list in to single data frame Hi: Here's an alternative using ldply() from the plyr package. The idea is to read the data frames into a list, name them accordingly and then call ldply(). # Read in the test data frames (you want to use list.files() instead to input the data per Uwe's guidelines) df1 - read.table(textConnection( + var1 var2 var3var4 + 1 6 0.3 8 + 3 4 0.4 9 + 2 3 0.4 6 + 1 0.40.9 3), header = TRUE) df2 - read.table(textConnection( + var1 var2 var3 var4 + 1 16 0.6 7 + 3 14 0.4 6 + 2 13 0.4 5 + 1 0.6 0.9 2), header = TRUE) closeAllConnections() # generate the list dl - list(df1, df2) # Name the list components by number and then call ldply(): names(dl) - 1:2 # more generally, names(dl) - 1:length(dl) library(plyr) ldply(dl, rbind) .id var1 var2 var3 var4 1 11 6.0 0.38 2 13 4.0 0.49 3 12 3.0 0.46 4 11 0.4 0.93 5 21 16.0 0.67 6 23 14.0 0.46 7 22 13.0 0.45 8 21 0.6 0.92 You can always change .id to fileno afterwards. HTH, Dennis On Mon, Apr 4, 2011 at 7:41 AM, Umesh Rosyara rosy...@msu.edu wrote: Dear R community members I did find a good way to merge my 200 text data files in to a single data file with one column added will show indicator for that file. filelist = list.files(pattern = K*cd.txt) # the file names are K1cd.txt .to K200cd.txt data_list -lapply(filelist, read.table, header=T, comment=;, fill=T) This will create list, but this is not what I want. I want a single dataframe (all separate dataframes have same variable headings) with additional row for example ; just for example, two small datasets are created by my component datasets are huge, need automation ;read from file K1cd.txt var1 var2 var3var4 1 6 0.3 8 3 4 0.4 9 2 3 0.4 6 1 0.40.9 3 ;read from file K2cd.txt var1 var2 var3var4 1 16 0.67 3 14 0.4 6 2 1 3 0.4 5 1 0.60.9 2 the output dataframe should look like Fileno var1 var2 var3var4 1 1 6 0.38 1 3 4 0.4 9 1 2 3 0.4 6 1 1 0.4 0.93 2 1 16 0.67 2 3 14 0.46 2 2 1 3 0.45 2 1 0.6 0.9 2 Please note that new file no column is added Thank you for the help. Umesh 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. [[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] automating regression or correlations for many variables
Thanks! I must confess I am just a beginner, but I followed your suggestion and did 'm - lm(as.matrix(snp[, -1]) ~ lat, data = snp) ' and it worked perfectly. I would like to understand what is being done here. as.matrix I understand makes my data frame be a matrix, but I don't understand the part snp[,-1]. I think I am saying use all rows of... but I don't understand the -1, does that mean, all columns but column 1? I really appreciate the help! It was really really valuable. A related question is, what is m? If I do is.object(m) I get true. so I guess it is an object. I now need to use each of the fitted models for more things, such as calculate anova, etc... How do I do that? When I do this column by column I just do anova(m), but it does not seem to work now... Thanks! AG -- View this message in context: http://r.789695.n4.nabble.com/automating-regression-or-correlations-for-many-variables-tp3426091p3426519.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] General binary search?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Stavros Macrakis
Sent: Monday, April 04, 2011 1:15 PM
To: r-help
Subject: [R] General binary search?
Is there a generic binary search routine in a standard library which
a) works for character vectors
b) runs in O(log(N)) time?
I'm aware of findInterval(x,vec), but it is restricted to
numeric vectors.
xtfrm(x) will convert a character (or other) vector to
a numeric vector with the same ordering. findInterval
can work on that. E.g.,
f0 - function(x, vec) {
tmp - xtfrm(c(x, vec))
findInterval(tmp[seq_along(x)], tmp[-seq_along(x)])
}
f0(c(Baby, Aunt, Dog), LETTERS)
[1] 2 1 4
I've never looked at its speed.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
I'm also aware of various hashing solutions (e.g.
new.env(hash=TRUE) and
fastmatch), but I need the greatest-lower-bound match in my
application.
findInterval is also slow for large N=length(vec) because of the O(N)
checking it does, as Duncan Murdoch has pointed
outhttps://stat.ethz.ch/pipermail/r-help/2008-September/174584.html:
though
its documentation says it runs in O(n * log(N)), it actually
runs in O(n *
log(N) + N), which is quite noticeable for largish N. But
that is easy
enough to work around by writing a variant of findInterval which calls
find_interv_vec without checking.
-s
PS Yes, binary search is a one-liner in R, but I always prefer to use
standard, fast native libraries when possible
binarysearch - function(val,tab,L,H) {while (H=L) {
M=L+(H-L) %/% 2; if
(tab[M]val) H-M-1 else if (tab[M]val) L-M+1 else return(M)};
return(L-1)}
[[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] moving mean and moving variance functions
Thanks, I had recently seen reference to caTools but had forgotten about it.
Much appreciated.
friedman.st...@gmail.com
517-648-6290
-Original message-
From: Gabor Grothendieck ggrothendi...@gmail.com
To: Steve Friedman skfgla...@gmail.com
Cc: r-help@r-project.org
Sent: Mon, Apr 4, 2011 13:17:36 GMT+00:00
Subject: Re: [R] moving mean and moving variance functions
On Mon, Apr 4, 2011 at 8:30 AM, Steve Friedman skfgla...@gmail.com wrote:
Hello
Lets say as an example I have a dataframe with the following attributes:
rownum(1:405), colnum(1:287), year(2000:2009), daily(rownum x colnum x year)
and foragePotential (0:1, by 0.01). Â The data is actually stored in a netcdf
file and I'm trying to provide a conceptual version of the data.
Ok. I need to calculate a moving mean and a moving variance for each cell on
the following temporal
windows - 7 day, 14 day, and 28 day. So far I have code for the moving
average.
ma - function(x , n) {
     filter(x, rep(1/n, n), sides = 1)
   }  # note that when the function is used, n is defined for the
temporal period (7, 14, and 28), and x is the input variable.
ma7 - Â ma(dat, 7) Â # where dat is accessing the foraging potential of the
birds.
ma14 - ma(dat, 14)
ma28 - ma(dat, 28)
This works fine. Â What I don't have is the code for a moving variance.
filter in the function above is included in the stats package and conducts a
linear filtering on a Time Series.
Is there comparable code some place in R for a moving variance?
See rollmean and rollapply in the zoo package and runmean and runsd in
the caTools package.
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] how to handle no lines in input with pipe()
This has to do with using pipe() and grep and read.csv()
I have a .csv file that I grep using pipe() and read.csv() as follows:
read.csv(pipe('grep foo bar.csv'))
However, is there a way to have this command run when for example,
there is no foo text in the bar.csv file? I get an error message
(appropriately):
Error in read.table(file = file, header = header, sep = sep, quote = quote, :
no lines available in input
Is there a way to inspect the output of pipe before passing it on to
read.csv()?
Thanks,
Andrew
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
