search for: letters

list subselect by name ? ------------------------- I have this 'list of two elements of named elements.' > list(letters=letters, LETTERS=LETTERS)[c("letters","LETTERS")] $letters [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" &quo...

Dear useRs, I have written a very simple function to compute some probabilities on words (function is below). The function includes a which() statement applied to a vector of characters (word.split): sapply (word.split, function(x) which(letters==x)). This statement worked as expected when used outside the global function : > word <- "hello" > (word.split <- unlist(strsplit(word, split=c()))) [1] "h" "e" "l" "l" "o" > (used.letters <- sapply(word.split, fun...

...wing feature: - caller dials a given extension dedicated to a given language (german, english, ...) - Asterisk plays a welcome audio prompt - caller spells his or her first name letter by letter (for example, caller spell "A", "L", "I", ...) - Asterisk repeats spelled letters to caller. If possible, which module is needed ? Best regards -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.digium.com/pipermail/asterisk-users/attachments/20160323/31c38d2b/attachment.html>

Hi, I was wondering if anybody can help me. In the small data set below I would like to select the index which doesn't contain the letter 'N' in the variable 'letters'. How can I discard these rows when the letter has a different position everytime (but the same letter for the whole column)? index<-c(1,2,3) letters<-c("CCTTGGAA", "NNTTGGAAT", "AACCTTNN") z<-data.frame(index,letters) index letters 1 1 CCTTGGAA...

Hi, imagine the following matrix/data.frame Letter Number a 1 a 1 b 1 b 0 c 0 c 1 d 0 d 0 If the numbers for two identical letters are also identical then I want to remove either the first or the second row of that letter. If for a letter the numbers are 1 and 0 I want to remove the row with the 0. That means if the code works I would and up with the following matrix/data.frame Letter Number a 1 b 1 c 1 d 1 Many thanks, Sy...

...of the vector of strings", "second element" ) becomes: c( "This Is An Element Of The Vector Of Strings", "Second Element" ) My reason to try to do this is to get more readable abbreviations. (A suggestion would be to add an option to abbreviate() which changes letters after space to uppercase letters before executing the abbreviation algorithm.) Thanks - Wolfram

Hi, I've been watching my * Console and seems to be one call not well terminated or something: For 5 minutes at least my console is reporting this: ectory|general|ext-local|be: -- Playing 'letters/c' (language 'en') directory|general|ext-local|be: -- Playing 'letters/o' (language 'en') directory|general|ext-local|be: -- Playing 'dir-instr' (language 'en') directory|general|ext-local|be: -- Playing 'letters/a' (language 'en')...

Hi all If I have a string, say "ABCDA", and I want to convert this to the sum of the letter values, e.g. A -> 1 B -> 2 etc, so "ABCDA" = 1+2+3+4+1 = 11 Is there an elegant way to do this? Trying something like which(LETTERS %in% unlist(strsplit("ABCDA", ""))) is not quite correct, as it does not count repeated characters. I guess what I need is some kind of lookup table? Cheers Rory Rory Winston RBS Global Banking & Markets 280 Bishopsgate, London, EC2M 4RB Office: +44 20 7085 4476 ******...

...I have a multiline plot with each line labeled with a different letter. But I'm not able to make the legend display the same kind of pattern '-a-', instead the letter is overwritten by the line. A simpler legend with only the letter is not very visible and the pt.bg does nothing with letters. Any idea? plot(1:10,10:1,lty=1,type='b', lwd=2,pch='a') legend("left", legend=c("ds1","ds2"), bty='n', col=1:2, lty=2,lwd=4,pch=letters) Thanks for your help! mario -- Ing. Mario Valle Data Analysis and Visualization Group |...

Hi R users, I need a way to substitute the values 10:31 to the letters A:V (i.e 10=A, 11=B, ..., 31=V) in a data frame. For example: > y<-c(10,11,12,13) > z<-c(28,29,30,31) > df<-data.frame(y,z) > df y z 1 10 28 2 11 29 3 12 30 4 13 31 Then I would substitute it and obtain a data frame like this as a result of the function: > w<-c(&quo...

Hello, > X <- data.frame(letters=letters[1:3], flag=c(TRUE, FALSE, TRUE)) > X letters flag 1 a TRUE 2 b FALSE 3 c TRUE > apply(X, 1, as.list) [[1]] [[1]]$letters [1] "a" [[1]]$flag [1] " TRUE" [[2]] [[2]]$letters [1] "b" [[2]]$flag [1] "FALSE" [[3]] [[3]]$...

Dear helpers, how can I extract only the values from a row in a data frame? Using [X,] doesn't do the trick: > data.frame(letters[1:10],letters[11:20])->my.data > my.data[1,] letters.1.10. letters.11.20. 1 a k I would like to be able to extract only the values "a" and "k" without getting the row names and column names with them. I'm asking because I want to assign the...

Consider this: > letters[c(2, 3)] [1] "b" "c" > letters[c(2, NA)] [1] "b" NA > letters[c(NA, 3)] [1] NA "c" > letters[c(NA, NA)] [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA [26] NA The result is a 2-vector in each case until we get to c(...

Hi there, Is there a neat and easy way to implement pagination with letters rather than numbers eg : A B C D ... Z Rob

Hi All, I have strings made up of an unknown number of letters, digits, and spaces. Strings always start with one or two letters, and always end with one or two digits. A set of letters (one or two letters) is always followed by a set of digits (one or two digits), possibly with one or more spaces between the sets of letters and digits. A set of letters always...

8. You have join to this mail a text file with the few characters that don't = match. On Swat, it's easier. = I use Swat from my win98 box. When I write a share comment with some chara= cters, there are not translated properly inside smb.conf file. = When I read my smb.conf file, I noticed that for example a cp1252 characte= r like (0xC1) has been changed onto an ISO8859-15 character

Dear R Users, I am trying to read in a series of csv files which vary by the letter on the end of he file name. When I input what seems to be a logical for loop I get an error message that doesn't make sense to me. > for(i in 1:12){ paste("G&P", LETTERS[i],sep='') <-read.csv(paste("/Users/thomasjackson/Data/GEP&CO/GEP&CO",LETTERS[i],"/HPLC_",LETTERS[i],"12.csv",sep=''), header=T, sep=',')} Error in paste("G&P", LETTERS[i], sep = "") <- read.csv(paste(&quo...

I get the following error when I try to modify the names of the object returned by get(). Does anybody know how to do so? (I could use as.vector() to remove the names, but I'm interested in how to modify the object returned by get().) $ cat main_get_name_assign.R x=1:10 names(x)=letters[1:10] names(get('x'))=LETTERS[1:10] #names(x)=LETTERS[1:10] x $ Rscript main_get_name_assign.R > x=1:10 > names(x)=letters[1:10] > > names(get('x'))=LETTERS[1:10] Error in names(get("x")) = LETTERS[1:10] : target of assignment expands to non-language object...

Dear list, I have a vector of letter strings as follow: > LETTERS[c(1:7,9,15,18:25)]  [1] "A" "B" "C" "D" "E" "F" "G" "I" "O" "R" "S" "T" "U" "V" "W" "X" "Y"   I need to find sequences of length...

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 (...