similar to: sequence of equal-length numbers (for filenames)

Hi everybody, given a fresh rgui.exe load on winxp OS, I enter (a minimal exaple) n <- 12.357531 Then the following command: n <- (n - floor(n))*10; n gives the following outputs: [1] 3.57531 [1] 5.7531 [1] 7.531 [1] 5.31 [1] 3.1 [1] 1 === still as expected [1] 10 === not expected, count with me: 1 - floor(1) is zero, times 10 gives 0, not 10!!!! [1] 10 === should

Hi Everyone, I'm new to programming R and have accomplished my goal, but feel that there is probably a more efficient way of coding this. I'd appreciate any guidance that a more advanced programmer can provide. My goal -- I would like to find the length of the longest word in a string containing many words separated by spaces. How I did it -- I was able to find the length of the

Dear R-users, in a minimal example exists() gives FALSE on an object which obviously does exist. How can I check on that list object anyway else, please? > SmoothData <- list(exists=TRUE, span=0.001) > SmoothData $exists [1] TRUE $span [1] 0.001 > exists("SmoothData") TRUE > exists("SmoothData$span") FALSE > exists("SmoothData[[2]]") FALSE

I would propose to add " See also: `nchar' for counting the number of character in character vectors. " to the helpfile of length(), because it is rather difficult to find nchar() if one has only search terms as "length", "len", "strlen" in mind. Sincerly Wolfram Fischer

Hi there, I am intending to get the length of an UTF-8 string which contains some Japanese characters (let's say, rstr) in R language. I try to use the nchar(rstr) to get its length, however, it returns the "NA" for it contains some multi-byte characters. Is there any alternatives to return the length of this rstr? Any suggestion is appreciated. Long

Full_Name: Volkmar Klatt Version: 2.6.2 OS: linux Submission from: (NULL) (84.147.0.178) Hello, the documentation article ?Quotes from package:base could be improved by discussing the meaning of "\0" escape sequence. In R (currently used: R-2.6.2), a "\0" in a string will silently terminate that string, just as it would in C code. There is no warning about an unknown escape

Hello r-devel, The attached Code page 437-encoded file contains 245 characters (including the final newline), but readLines only reads 242 of them: > test_text <- readLines(file('437__characters.txt', encoding='437')) Warning message: In readLines(file("437__characters.txt", : incomplete final line found on '437__characters.txt' > test_text [1]

Hi I'm using kronecker() with a matrix and a vector. I'm interested in the column names that kronecker() returns: > a <- matrix(1:9,3,3) > rownames(a) <- letters[1:3] > colnames(a) <- LETTERS[1:3] > b <- c(x=1,y=2) > kronecker(a,b,make.dimnames=TRUE) A: B: C: a:x 1 4 7 a:y 2 8 14 b:x 2 5 8 b:y 4 10 16 c:x 3 6 9 c:y 6 12 18 > The

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'd like to turn a vector so it starts with it's end. For better understanding, this set of commands will do what I need: i <- seq(1:10) i_turned <- i for (j in 1:length(i)) i_turned[j] <- i[length(i)-j+1] now, i_turned is what I call turned. Is there a function which would make a script lighter? Thank you upfront for any hint. Best regards, Zroutik

Hi, I'm looking for a manual (we based or pdf) which would explain in detail with graphical examples what all the option can do in plot and par. Does anybody now anything like this? A couple of manuals to R I went through do have plot parameters mentioned, but sometimes it is hard to understand or imagine what the parameter can do (the same with the related help page) e.g. I met "From

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... >

Dear R users, The best in this new year 2011. I am dealing with a character vector (xx) whose nchar are not the same. Ex. nchar(xx) [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [38] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 4 4 4 4 4 4 4 4 [75] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 ....... 9 I need xx to be nchar = 9 My

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]

I was wondering if anyone has a way out of the limitation that you must use equal length x,y, and z sequence lengths. For instance, x<-seq(1,100) y<-seq(1,100) z<-rnorm(100) scatterplot3d(z,x,y) works fine. However, if I get some results that has a different y subset length, such as x<-seq(1,100) y<-seq(1,300) z<-rnorm(100) scatterplot3d(z,x,y) I get the following error:

Dear R users, I plot data with points(my_data[x]~x, col = x, type = "o", lwd="4") where x is an integer running from 1 to 10, I get points drawn at the plot. When want to do a legend to this I try legend(leg.txt[x], col = x, text.col = 1, pch = 1, bty = "n") where leg.txt contains the names of the variable and x behaves the same. The difference between the