similar to: median filter

Dear R-users, I use unlist of POSIX dates to extract the year, hour etc. With that I can search for files in my database which are in the form 'yyyymmddhh_synops.txt' However, I get stucked during midnight where unlist just gives NA's. The script is given below, the problem accurs at acc.period[16] (midnight). However when I write out the character, unlist works well. But

Dear R-users, I'm experiencing a segmentation fault when using hdf5load(file,load=FALSE). Library(hdf5) loads without problems but when loading a file, R crashes. I compiled R under Unix (Solaris for Sun). There is nothing wrong with the files, as I can run the same script at another place where R runs under Linux. Is it possible it has something to do with the hdf5 libraries where

Dear R-users, For a data frame (say in this example X) I want to look up the corresponding value in a 'look-up data frame' (in this example Y). The for-loop works but is very time-consuming because 'X' in reality is very big. Therefore I would like to have a solution with apply. However, I do not succeed. Any suggestions? Thanks in advance, Hanneke

Does anyone know of an R-function or method to compare two C-statistics (Harrells's C - rcorr.cens) obtained from 2 different models in partially paired datasets (i.e. some similar and some different cases), with one continuous independent variable in each separate model? (in a survival analysis context)? I have noticed that the rcorrp.cens function can be used for paired data. Thanks

Hello, my test.R file contains two huge arrays (>3000 entries), from which R needs to calculate the Pearson Correlation, if I look at the file the numbers look correct. if I run R R < test.R --no-save I see things like this: 0.723, 0.838, 1.002, 0.364, 0.357, 0.227, 0.982+ , 0.963, 0.535, 1.214, 1.270, 0.832, 1.033, 0.632, 2.482, 1.239, 0.743, 1.077, 0.962, 1.052, 1.075, 1.427, 1.395,

Hey, Thanks for your great inputs. While "index = apply(mat == 0, MARGIN = 1, any)" gives you an idea of the rows containing zero(s), "index<-data[!apply(data==0,MARGIN=1,any),]" does the actual job of removing the rows with zeros. Kind regards Ogbos On 9 March 2010 15:12, Paul Hiemstra <p.hiemstra@geo.uu.nl> wrote: > Dimitris Rizopoulos wrote: > >>

Dear All, I try to use Sweave to make a document. But when I use the Sweave() command on it and build a pdf with pdflatex (3.141592-1.40.3) my apostrophes are replaced by some gibberish (an 'a' with a hat on it, a capital A with a arc pointing upwards on it and a capital Y with two points on it). If I manually replace the apostrophes using the keyboard, I get a different looking

We have a samba 2.2.3a-12.3 for Debian running, and something odd in printing. We use CUPs as linux print server. When we print from a w2k station, prints come out the printer, but the printer window reports a cycle of messages: - initializing printer - opening printer - failed to open, retrying what can be the cause? parts of the smb.conf: ============== [global] printcap

Hello everyone, I encountered the following confusing behavior of 'deparse' and 'substitute' while programming with S4 classes (see example below). It seems like the presence of '...' argument in the definition of the generic generates the problem. I have no clue why, can anyone explain that to me? Are there any "workarounds"? Thanks a lot for your time!

Hello dear useRs and wizaRds, I am currently developing a package that will enable to use administrative map of Poland in R plots. Among other things I wanted to include region names in proper Polish language so that they can be used in creating graphics etc. I am working on Windows and when I build the package it is complaining about non-ASCII characters R code files. I was wondering what would

Hello dear useRs and wizaRds, I encountered the following problem using the hist() method for the 'times' classes from package 'chron'. You should be able to recreate it using the code: library(chron) # pasted from chron help file (?chron) dts <- dates(c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92")) class(dts)

Dear useRs and wizaRds, I am currently developing a set of functions using S4 classes. On the way I encountered the problem exemplified with the code below. For some reason the 'validity' method does not seem to work, i.e. does not check for errors in the specification of the slots of the defined class. Any hints? My understanding of the whole S4 system was that validity checks are made

Hola a tod en s, Este es el código principal que utilizo para diseñar mi zona de estudio (cuenca): png("mapa_zona_estudio.png", units="in", width=9, height=8.5, res=300) # Vamos a pintar los distintos instrumentos de medida sobre el mapa (junto con éste) y añadimos leyenda ggmap(mapa, extend='device', legend="left", base_layer=ggplot(datos, aes(x=lon,

Dear, I am a newbie to R. Now I am learning to draw boxplot using graphics(). I want to highlight the median position with a round point and the value left (or on the top of)to the point. Please kindly share the command and thank you. Elaine [[alternative HTML version deleted]]

Hi, set.seed(45) test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data apply(test1,2,function(x) sprintf("%.1f",median(x))) #columnA columnB # "44.5"? "10.2" par(mfrow=c(1,2)) lapply(test1,function(x) {b<-

Dear List, I have worked through the examples given in the help on boxplot(). If I am reading it right, only the median is produced (as a default), which is represented by the horizontal line inside the box. What argument do I need to specify if I want to show the mean as well? Thanks, platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status

Dear R users,   I want to compute the 95% CI for difference in median survival times of two treatment groups. survfit function in R gives median and the corresponding CIs for each group. Could you please help me to get a CI for the difference in median times?   Thanks Ruwanthi [[alternative HTML version deleted]]

Hi, I am trying to calculated various summary statistics using the dcast function of reshape2. This works perfectly for getting the mean, sum, length, sd. But when I want to calculate the median I get an error. I tried it with and without removing NAs: my_median <- function(x) median(x, na.rm = FALSE) median_df <- dcast(patch_stats_dfm,formula=species~input+barriers,my_median) Error in

Does anyone can provide a code to implement least median squares regression in R (not using the lqs function or calling C functions)? Reason: teaching/learning purposes Thanks PM

Hi i have buy a Audiocode Median 2000 VoIP Gateway and connect it on : 1 E1 30 channels 1 Lan Port Anyone use this equipements with asterisk ? because i am search a config sample for AudioCode and for Asterisk (i am new in VoIP). I want that all calls arrives on the AudioCode are sent to the asterisk by SIP (trunk ?) and all outgoing call from Asterisk are sent to the AudioCode. I