similar to: vector of character with unequal width

I am trying to upgrade to R 2.14 from R 2.13.1 I have compied all the libraries from the 'library' directory in my existing installation (2.13.1) to the installed R 2.14. Now I want to uninstall the old installation (R 2.13.1) and I get the error: Internal Error: Cannot find utCompiledCode record for this version of the uninstaller. Any ideas? Kevin [[alternative HTML

Hello, I am currently running R on ubuntu and everything is working perfectly fine. However, I would like to connect to R via Windows using Eclipse StatEt plugin. Is this possible to do? or do I have to have a version of R running on Windows also? I prefer to have Linux do the heavy lifting and Windows Eclipse to be a sort of GUI. Any thoughts?

Hi, i have a vector filled with names: [1] Alvaro Adela ... [25] Beatriz Berta ... ... [100000] ... I would like to drop last character in every name. I use the next program: for (i in 1:100000) { ? ? ? ? ? ? ? ? ? ? ? ? ? largo <- nchar(names[i]-1) ? ? ? ? ? ? ? ? ? ? ? ? ? names[i] <- substring (names[i],1,largo] ? ? ? ? ? ? ? ? ? ? ? ? ?} Is another and faster way of do it? Thanks,

Dear list, If I include the zero-width non-breaking space (\ufeff) in a string, nchar seems to compute the wrong number of columns used by 'cat'. > x <- "f\ufeffoo" > x [1] "f?oo" > nchar(x,type="width") [1] 2 I would expect "3" here. Going through the documentation of 'Encoding' and 'encodeString', I don't think

Hi -- I'm looking for alternatives to regex for a fairly simply 'reformatting' problem. Alternatives only because a lot of folks have trouble parsing/interpreting regex expressions, and I'm looking for suggestions for something more 'transparent'. Here is an example of what I'm trying to do. Take the following string, which I call x, and for each character in the

Hi, (First, apology for my earlier incorrectly addressed "subscribe" post.) Can somebody tell me what exactly is going on below. Basically, I am running into some kind of "string truncation" problem when I try to get a substring starting past the 8192nd character (see sample session below). There doesn't appear to be any problem creating the string, and nchar()

Hello Arnaud, You posted this question a long long time ago, however I found your answer so I decided to post it anyway in case somebody else have the same problem as you and me. You were actually very close in finding your solution. The function DoWritedbf is an internal function from the foreign package. To access it outside of the package just do: foreign:::DoWritedbf so in your line:

Hi, Is nchar function knowingly slow in R? I'm doing some string formatting that requires multiple call to nchar, and nchar seems to be very slow. Experiment 1, pass nchar inside sprintf, and it takes 0.7 seconds > system.time(for (i in 1:10000) + str = sprintf('0005%020d', nchar(op)) + )[3] [1] 0.7 Experiment 2, get the length of op separately using nchar, and then pass

Dear R users, I created a matrix that tells me the first day of use of a category by id. #Calculate time difference test$tdiff<-as.numeric(difftime(as.Date("2002-09-01"), test$ftime, units = "days")) # obtain the index date per person and dcategory index.date.test<-tapply(test$tdiff, list(test$id, test$rcat), max) Nonetheless, at the moment I think will be

Malcolm, I tested the code on a clean R 3.2.0 session. Not even in RStudio, just to rule that out. > sessionInfo() R version 3.2.0 (2015-04-16) Platform: x86_64-w64-mingw32/x64 (64-bit) Running under: Windows 8 x64 (build 9200) locale: [1] LC_COLLATE=English_United Kingdom.1252 [2] LC_CTYPE=English_United Kingdom.1252 [3] LC_MONETARY=English_United Kingdom.1252 [4] LC_NUMERIC=C [5]

In R 2.9.1 Windows: > nchar(factor(paste('sdf',1:10))) [1] 1 1 1 1 1 1 1 1 2 1 so it appears that nchar is counting the number of characters in the numeric representation, just like: > nchar(as.numeric(factor(paste('sdf',1:10)))) [1] 1 1 1 1 1 1 1 1 2 1 but ?nchar says explicitly: x: character vector, or a vector to be coerced to a character vector.

On 05/10/2015 8:25 PM, Matt Dowle wrote: > > On Mon, Oct 5, 2015 at 4:57 PM, Duncan Murdoch <murdoch.duncan at gmail.com > <mailto:murdoch.duncan at gmail.com>> wrote: > > On 05/10/2015 7:24 PM, Matt Dowle wrote: > > Joris Meys <jorismeys <at> gmail.com <http://gmail.com>> writes: > > > >> > >> Hi all,

Hi all, I have a puzzling problem related to nchar. In R 3.2.1, the internal nchar gained an extra argument (see https://stat.ethz.ch/pipermail/r-announce/2015/000586.html) I've been testing code using the package copula, and at home I'm still running R 3.2.0 (I know, I know...). When trying the following code, I got an error: > library(copula) > fgmCopula(0.8) Error in

Dear all, There is something I'm missing in order to understand the following behavior: > aa <- c("test", "name") > ifelse(any(nchar(aa) < 3), aa[-which(nchar(aa) < 3)], aa) [1] "test" > any(nchar(aa) < 3) [1] FALSE Shouldn't the ifelse function return the whole aa vector? Using if and else separately, I get the correct result... >

Help-Rs,   I'm doing some string manipulation in a file where I converted a string date in mm/dd/yyyy format and returned the date yyyy.   I've used regexpr (hat tip to Gabor G for a very nice earlier post on this function) in steps (I've un-nested the code and provided it and an example of what I did below.  My question is: is there a more efficient way to do this.  Specifically is

Hi I am not sure if it is more readable > paste(paste(unlist(strsplit(x,"")),".", sep=""), collapse="") [1] "1.0.1.1.0.1.1.1." If you did not want last dot, it is a bit shorter. > paste(unlist(strsplit(x,"")),collapse=".") [1] "1.0.1.1.0.1.1.1" > Cheers Petr Tento e-mail a jak?koliv k n?mu p?ipojen?

Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? #Works T13702 <- TRACKDATA[["13702.xls"]][["data"]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])))) 16553 #Works d<-2 assign(paste("T",substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d]

All, I am having trouble with a "read.table()" function that is inside of another function. But if I call the function by itself, it works fine. Moreover, if I run the script on a Mac OS X (with the default Mac OS X version of R installed, rev 2.8), it works fine. But it does not work if I run it on windows vista (also default Windows version of R, rev. 2.8). Again, both

On 05/10/2015 7:24 PM, Matt Dowle wrote: > Joris Meys <jorismeys <at> gmail.com> writes: > >> >> Hi all, >> >> I have a puzzling problem related to nchar. In R 3.2.1, the internal > nchar >> gained an extra argument (see >> https://stat.ethz.ch/pipermail/r-announce/2015/000586.html) >> >> I've been testing code using the

this function is supposed to canonicalize the language: --8<---------------cut here---------------start------------->8--- canonicalize.language <- function (s) { s <- tolower(s) long <- nchar(s) == 5 s[long] <- sub("^([a-z]{2})[-_][a-z]{2}$","\\1",s[long]) s[nchar(s) != 2 & s != "c"] <- "unknown" s }