similar to: To hell with OpenSuse, ditch it and go to Ubuntu

Dear List, I have the following data: >> test <- data.frame(date = as.Date(c('2007-01-01','2008-03-24','2003-03-02','2008-05-03','2002-05-23','2001-06-30','2005-12-04')), nr = c(2000,2000,2000,2001,2002,2003,2003)) test date nr 1 2007-01-01 2000 2 2008-03-24 2000 3 2003-03-02 2000 4 2008-05-03 2001 5 2002-05-23 2002 6

My uncle recently bought a new computer and he wants me to help him install Guitar Pro. The thing is, he doesn't have Windows, and I'm thinking of installing Ubuntu with wine on the comp. Quiz time! Can you use wine, with GP, and if you can, deos it perform poorly in any ways?

hi ,everyone, I have a script shown as below: > bj2=bjerrdata$tyerr[bjyearnum[2]+1:bjyearnum[3]] > length(bj2) [1] 448 while > b=bjyearnum[3] > a=bjyearnum[2]+1 > bj1=bjerrdata$tyerr[a:b] > length(bj1) [1] 169 it is different with bj2 and bj1 . and the array bjyearnum is [1] 0 279 448 633 1021 1365 1813 2237 2839 3314 3798 4157 12 why bj1 is 169 while bj2 is 448

Installing R on Ubuntu 8.10, ( using sudo apt-get install r-base , and using one of the cran sites (cran.cnr.berkeley.edu)) the installation process says something about not having some gpg public key and "are you sure you want to download non-authenticated stuff [y/n]" (to which I answered yes). I'm assuming this warning can be ignored? Also: even though the Ubuntu install and

With virtio_gpu_object_create() being called earlier ttm_bo_init() will handle the virtio_gpu_object_attach() (via backend binding), so we can drop the extra calls now. Signed-off-by: Gerd Hoffmann <kraxel at redhat.com> --- drivers/gpu/drm/virtio/virtgpu_gem.c | 9 --------- drivers/gpu/drm/virtio/virtgpu_ioctl.c | 2 -- 2 files changed, 11 deletions(-) diff --git

The following code isn't working and we can't figure out why.. letters = c("A","B","C","D","E","F","G","H","I","J") numbers = 1:3 for(i in 1:6){ #6 letters for (j in 1:3) { #3 numbers for (k in -1:1) {

installing on Ubuntu, how to do it and have people found it to be glitchy? which is easier , binary install or from source ? With the source install, are you less likely to have a dependencies issue ? ( Ubuntu does the GCC install seamlessly, but has no mention of R )

I am trying to reshape a data set. Could someone please help me with the reshape, cast, and melt functions? I am new to R and I have tried reading up on how to use the reshape package, but I am very confused. Here is an example of what I am trying to do: subject coder score time [1,] 1 1 20 5 [2,] 1 2 30 4 [3,] 2 3 10 10 [4,] 2 2

Hi all, I use "write.csv" and "write.table" to write a data frame in a file like following: write.csv2(allRandomTestCase_XDroped, "allRandomTestCase.csv") But in the created file "allRandomTestCase.csv" an additional column with consecutive numbers is automatically added to the column of the data frame "allRandomTestCase_XDroped". That is why my

Dear I am trying to find a 90% credible interval. I am using the following code. fit<-bugs( model.file=BUGScode, data=data, inits = list(geninits1,geninits2), parameters.to.save=keepers, n.chains=nchains, n.iter=runs, n.burnin=burn, n.thin=nthin, DIC= TRUE, bugs.directory="C:/Program Files/WINBUGS.14", \ ) But this is only giving 95%

Am I asking for too much: for any object that a stat proc returns ( y <- lm( y~x) , etc ) ) , is there a super convenient function like give_all_extractors( y ) that lists all extractor functions , the datatype returned , and a text descriptor field ("pairwisepval" "lsmean" etc) That would just be so convenient. What are my options for querying an object so that I can

I have a bit of a quandy. I'm working with a data set for which I have sampled sites at a variety of dates. I want to use this data, and get a running average of the sampled values for the current and previous date. I originally thought something like ddply would be ideal for this, however, I cannot break up my data by date, and then apply a function that requires information

I am trying to calculate Bayesian Credible Intervals for a proportion (disease prevalence values to be more specific) and am having trouble using R to do this. I am working with ncredint() function but have not had success with it. Please help! Example: Positive samples = 3 Total sampled = 10 Prevalence = 0.3 pvec <- seq(1,10,by=1) npost = dbinom(pvec,10,prob=0.3, log=FALSE) ncredint(pvec,

Hi all, I have the data from the posterior distribution for some parameter. I want to find the 95% credible interval. I think "t.test(data)" is only for the confidence interval. I did not fine function for the Bayesian credible interval. Could some one suggest me? Thanks [[alternative HTML version deleted]]

I have a simple 1 way anova coded like summary(ANOVA1way <- aov(Value ~ WellID, data = welldata)) How can I use the BY function to do this ANOVA for each group using another variable in the dataset?? I tried coding it like this, but it doesn't seem to work. summary(ANOVA1way <- by(welldata, Analyte, function(x) aov(Value ~ WellID, data = welldata))) In SAS I would code it like this:

Dear all, I would like to plot credible interval for a function estimate in R. I would like to plot the credible intervals as shaded region using polygon function. Does anyone ever used that? I tried several times but I could not obtain the right figure. xis=sort(xi,decreasing=TRUE) plot(xi,fm) polygon(c(xi,xis),c(f05m,f95m)) The above piece of code produces something else. Many thanks in

Hi all, I'm new to the list and am hoping to get some advice. I have a set of multivariate data and would like to find the densest part of the data cloud containing 95% of the data, like a 95% HPD credible volume. Is there any R code available to compute that? Thank you very much! Your help and patience are much appreciated. G.S. [[alternative HTML version deleted]]

I have a function "eval.delta" which does what I want but isn't very elegant. I have consulted the R documents, MASS, and S Programming. Is there a practical way to optimize the code? Thank you very much. Peter B. Function: eval.delta <- function(delta){ cat("VALUES\n") vlr <- NULL k <- 0 for(j in 1:length(delta)) if(delta[j] <= 2){

This is a question of clarification. IN 2009 Higgins, Thompson and Spiegelhalter (J R Statist Soc A 172:137-159) gave WinBUGs code to get credible intervals from random effects meta analysis for the prediction interval of a new study. It appears that the predict.rma function creates approximate credible intervals (pending a function revision by the author) for that purpose. Is my assumption

Dear R Users, I'm trying to use the 'cast' function in the 'reshape' package to convert column-format data to gridded-format data. A sample of my dataset is as follows: head(finalframe) Latitude Longitude Temperature OrigLat p-value Blaney 1 -90 -38.75 NA -87.75 17.10167 NA 2 -90 135.75 NA -87.75 17.10167 NA 3 -90 80.25