search for: letter

Displaying 20 results from an estimated 7980 matches for "letter".

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2017 Sep 09
1
list subselect by name ?
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" &qu...
2006 Apr 11
1
error in which(): recursive default argument reference
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, fu...
2016 Mar 23
5
How to recognize a name spelled letter by letter ?
Hello, I'm wonddering if it is possible, with Asterisk and any third party module or service, to build the following 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/piper...
2009 Feb 13
2
select on letter
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,lett...
2011 Oct 13
8
Remove specific rows in a matrix/data.frame
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 wit...
2005 Apr 11
6
How to change letters after space into capital letters
What is the easiest way to change within vector of strings each letter after a space into a capital letter? E.g.: c( "this is an element 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 abbreviatio...
2006 Apr 10
1
Directory App() is running for a while, like blocked/freeze? in the same name...
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')...
2008 Nov 24
4
Calculating sum of letter values
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 som...
2010 Apr 26
5
How to make legend with line+ character
Dear all, 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=&...
2012 Nov 09
3
General function to substitute values in a data frame
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(&qu...
2009 Nov 04
3
inconsistent behavior for logical vectors when using apply (" TRUE")
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]]...
2011 Jul 11
2
Extract only the values from a row
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...
2010 Dec 17
3
Surprising behavior of letters[c(NA, NA)]
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...
2006 Apr 10
9
Pagination with letter (A B C D ... Z)
Hi there, Is there a neat and easy way to implement pagination with letters rather than numbers eg : A B C D ... Z Rob
2007 Jul 09
2
parsing strings
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 alway...
2003 Dec 01
0
No subject
...y hand the samba configuration file, I can read the right ch= aracter with window explorer. Of course, characters match only if 8859-15 = and cp1252 characters match ... Is it possible to set Samba and Swat with Unicode, to specify unicode=5Fma= p for client and unicode=5Fmap for server and to base letter transcription= s between client and server trought Unicode ? Or is it possible to force value Client -> Server and Server -> Client ? Thank you Jacques note: when characters match between cp1252 and ISO8859-15 files, Unicode e= ntries match too. --=====================_1042820539=_ Content...
2009 Nov 19
3
Reading in a series of files using a for loop
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",LETTE...
2011 Feb 15
2
Error when modifying names of the object returned by get()
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 objec...
2010 Jun 30
1
All possible permutations of letter A with other letters
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...
2009 Oct 09
2
Problems with code containing a for loop
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...