similar to: How to get the length of an UTF-8 string

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

FYI, I tried installing ferret on my freebsd virtual server and got this: retango# gem install ferret --include-dependencies Attempting local installation of ''ferret'' Local gem file not found: ferret*.gem Attempting remote installation of ''ferret'' Updating Gem source index for: http://gems.rubyforge.org Building native extensions. This could take a while...

I suspect that this is simple, but thanks in advance for any advice... I have a dataframe, t2: V1 V2 aaa 3 aaaa 4 aaaaaa 6 a 1 aa 2 V2 is the length of the string in V1 using nchar(as.character(t1$V1)) I'd like to create a third column, that contains a string of the length of V2, but containing an alternate text, e.g. V1 V2 V3

Hi guys, I've got a question about the API mkchar(). I have met some difficulty in parsing utf-8 string to mkchar() in R-2.7.0. I was intending to parse an utf-8 string str_jan (some Japanese characters such as?, whose utf-8 code is E381B5) to R API SEXP mkChar(const char *name) , we only need to create the SEXP using the string that we parsed. Unfortunately, I found when parsing the

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

Dear R-helpers, I have two data frames, op and em4: > str(op) 'data.frame': 37 obs. of 5 variables: $ m : num 0.202 0.336 0.122 0.139 0.14 ... $ lln : num 0.798 0.643 0.863 0.835 0.823 ... $ rrn : num 0.789 0.702 0.894 0.895 0.923 ... $ asym2: num 0.177 0.304 0.108 0.187 0.274 ... $ asym3: num 0.0755 0.0975 0.0818 0.0651 0.13 ... > str(rownames(op)) chr

Dear all, I would like to know how can I compute the length of a string in a dataframe. Example: SEQUENCE ID TGCTCCCATCTCCACGG HR04FS000000645 ACTGAACTCCCATCTCCAAT HR00000595847847 I would like to know how to compute the length of each SEQUENCE. Best regards, João Fadista [[alternative HTML version deleted]]

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()

Dear R-users, I'd like to create filenames in a mask "file000.dat" numbered from 1 to e.g. 123. The last problem I'm dealing with is creating the sequence of numbers with equal length, i.e. 001, 002,.... 023, 024,.... 122, 123. The closest I got is by a repetition: Sequence <- c(1:123) for(i in c(1:length(Sequence))) { print( paste(rep("0",

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

I want to extract the last 3 letters of a string. So far, I've done this: > symbol = 'XYZ.VX" > substr(symbol,nchar(symbol)-2,nchar(symbol)) [1] ".VX" It works, but the code looks UGLY as hell. Am I missing something? Or is this the way it's supposed to be? Thanks, Sergio On 10/15/07, pintinho <diego at bpgomes.com> wrote: > > Hi everyone, >

Hello again, Let say I have following string: Vec <- c("sada", "asdsa", "sa") Now I want to make each element of this vector with equal length. Basically I want following vector: c("sada ", "asdsa", "sa ") Therefore we can get: > nchar(c("sada ", "asdsa", "sa ")) [1] 5 5 5 Is there any

Hi, i''m working on a project using Ferret for indexing it''s datas. I''m very happy with it but i need to code an extension to implement a .to_json method to TopDocs class, because ruby''s json implementation is really really slow... It''s my second (the first was the tutorial :/ ) ruby C extension, so i''m not really at ease with ruby C

Hi all, Can somebody explain why length("") returns 1 and not 0? How do I test if a given string is the empty string? Thanks, Steffen. [[alternative HTML version deleted]]

This is very basic but I have not been able to find an answer. Basically I want to find the length of a string. length("Text") returns 1 so I know that is not right. Thank you. Kevin

Hi all, I'm getting the following error from substring: > substr("<I>Jens Oehlschl\xe4gel-Akiyoshi", 1, 100) Error in substr("<I>Jens Oehlschl\xe4gel-Akiyoshi", 1, 100) : invalid multibyte string at '<e4>gel-A<6b>iyoshi' Is that normal / intended? I've tried setting the Encoding/locale to Latin-1/UTF-8 but that does not help. nchar

Hi all, I'm using R to find duplicates in a set of 6 files containing Part Number information. Before applying the intersect method to identify the duplicates I need to normalize the P/Ns. Converting the P/N to uppercase if alphanumerical and applying an 18 char long zero padding if numerical. When I apply the pn_formatting function (see code below) to "Part Number" column of the

Hi, ?rawToChar 'rawToChar' converts raw bytes either to a single character string or a character vector of single bytes. (Note that a single character string could contain embedded nuls.) Allowing embedded nuls in a string might be an interesting experiment but it seems to cause some troubles to most of the string manipulation functions. A string with an embedded 0:

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

Hi, I have a function in R that compares two very large strings for about 1 million records. The strings are very large URLs like:- http://query.nytimes.com/gst/sitesearch_selector.html?query=US+Visa+Laws&type=nyt&x=25&y=8. .. or of larger lengths. The data-frame looks like:- id url 1