similar to: remove Punctuation characters

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Dear all, I wanted to remove all punctuations except commas from a string. I used: gsub("[[:punct:]]", "", string) but I don't know how to exclude the commas ",". Anyone would kindly answer my basic question? -- View this message in context: http://www.nabble.com/Remove-all-punctuations-except-commas-tp24845721p24845721.html Sent from the R help mailing list

Hello, I have a vector of dates and I would like to grep the year component from this vector (= all digits after the last punctuation character) dates <- c("28.7.08","28.7.2008","28/7/08", "28/7/2008", "28/07/2008", "28-07-2008", "28-07-08") the resulting vector should look like "08" "2008"

Buenos días, estoy realizando análisis de texto con Twitter y tengo un problema con unos carácteres que no logro quitar. Són cadenas de letras con forma similar a *xaexdfxdeaxoa*. Creo que surgen de la códificación de los emojis. Yo suelo utilizar, más o menos el siguiente codigo con gsub para limpiar texto, pero no me sirve # remove rt x = gsub("rt", "", x) # remove at x =

Hello, I have a file which contains a column with age, which is represented in the two following patterns 1. "007/A" or ''007/a" or ''7 /a" ..... In this case A or a means year and I would like to extract only the numeric values eg 7 in the above case if this pattern exits in a line of file. 2. "004/M" or "004/m" where M or m means month

Dear list, I was wondering how to best implement some sort of a "plugin" paradigm using R methods and the dispatcher: Say we have a function/method ('foo') that does something useful, but that should be open for extension in ONE specific area by OTHERS using my package. Of course they could go ahead and write a whole new 'foo' method including the features they'd

I am running the following code on a MacBook Pro 17" Unibody early 2009 with 8GB RAM, OS X 10.5.8, R 2.10.0 Patch from Nov. 2, 2009, in 64-bit mode. freq.stopwords <- numeric(0) freq.nonstopwords <- numeric(0) token.tables <- list(0) i.ss <- c(0) cat("Beginning at ", date(), ".\n") for (i.d in 1:length(tokens)) { tt <- list(0) for (i.s in

Hello does R support [:punct:] in regular expressions? I am trying to strip all regular expressions for a vector of strings. > x <- c("yoda-yoda","billy!") > gsub("/[:punct:]/","",x) [1] "yoda-yoda" "billy!" Thanks Dan -- ************************************************************** Daniel Brewer, Ph.D. Institute of Cancer

Dear List, I have a data frame of data taken every few seconds. I would like to subset the data to retain only the data taken on the quarter hour, and as close to the quarter hour as possible. So far I have figured out how to subset the data to the quarter hour, but not how to keep only the minimum time for each quarter hour. For example:

I tried making a wordcloud of Obama's State of the Union address using the tm package to process the text sotu <- scan(file="c:/R/data/sotu2012.txt", what="character") sotu <- tolower(sotu) corp <-Corpus(VectorSource(paste(sotu, collapse=" "))) corp <- tm_map(corp, removePunctuation) corp <- tm_map(corp, stemDocument) corp <- tm_map(corp,

http://bugzilla.mindrot.org/show_bug.cgi?id=550 Summary: Problems with .Ql in mdoc Product: Portable OpenSSH Version: -current Platform: All OS/Version: All Status: NEW Severity: minor Priority: P2 Component: Documentation AssignedTo: openssh-unix-dev at mindrot.org ReportedBy: cjwatson at

Buenas tardes, Estoy extrayendo un dato de una web y el separador de miles es el ".", que es el separador decimal en R. He intentado reemplazaro por el caracter, para convertirlo en ´numérico de R, vacío pero no me deja. Lo que he intentado es lo siguiente: > a<-"17.691.700" # valor que obtengo de la web > sub(".","",a) [1] "7.691.700"

I am sorry, I'd like to split my column ("names") such that all the beginning of a string ("X..") is gone and only the rest of the text is left. x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd")) x$names<-as.character(x$names) (x) str(x) Can't figure out how to apply strsplit in this situation - without using a

I just need to confirm something with pattern matching folks. I have a factor with the following levels in a very large data set: > levels(all$Classical.Statistic) [1] "" "AB;ABD" "CollapsedSteps" "CR_P" "CR_Prop;CR_P;AB" [6] "NMK"

Hello All, Can anyone tell help me understand why the function below doesn't work and how I can fix it? Below are some sample data, some code that works on individual rows of the data, and my attempt to translate that code into a function. My hope is to get the function working and then to apply it to the larger data frame using ddply() from the plyr package or possibly some other approach.

Hi I am searching for a way to do something like "ABC" -> c("A","B","C"). How can this be accomplished? I tried cut() and split(), but they do something else, it seems. The purpose for doing this is to find the number of common (and uncommon) characters, i.e. ultimately I want something like this: > foo("ABD","ADE") c(2,1) # 2

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Hi, I'm having a slight problem with samba permissions. Here goes my scenario: Red Hat Linux 8.0 with samba I got a samba share "public" Users can access the share and write there what they need, but if a user creates a directory, other users can't access it. How can I configure things so that everyone in the group accesses everything in the share?

Hi all, I am trying to parse a vector for caliculating minimum in that vector the vector having values like 1 Kontrolle 2 Placebo 3 125mg/kg 4 250mg/kg 5 500mg/kg 6 1000mg/kg hear i tries for comverting it into numeric with using "as.numaric()" function but i got values like 5 6 2 3 4 1 it gives 1000mg/kg is the least one but i have

Dear All, I've successfully import my synteny data to R by using scan command. Below show my results. My major problem with my data is how am i going to combine the column names with the data( splt) where i have tried on cbind but a warning message occur. I have realized that the splt data only have 5 column instead of 6. Please help me with this!! I want my data to be a numerical