similar to: gregexpr in R 2.3.0 != gregexpr in R 2.4.0

The interpretation of regular expressions with repetition quantifiers in the 'gregexpr' function seems to have changed between R Version 2.2.0 and 2.3.0. The 'gsub' function, however, gives the same results in R Versions 2.2.0 and 2.3.0. Below is an example that demonstrates the version differences of the 'gregexpr' function. I am not sure whether this new behavior is

Hi I have a question concerning how to match word boundaries which I bet has a very simple answer, but I haven't found it with trial and error nor by searching the help archives for the terms in the subject line. The problem is this: I have a vector of two character strings. text<-c("This is a first example sentence.", "And this is a second example sentence.") If I

Full_Name: Stefan Th. Gries Version: 2.2.1 OS: Windows XP (Home and Professional) Submission from: (NULL) (68.6.34.104) The problem is this: I have a vector of two character strings. > text<-c("This is a first example sentence.", "And this is a second example sentence.") If I now look for word boundaries with regexpr, this is what I get: >

Full_Name: Peter Dolan Version: 2.5.1 OS: Windows Submission from: (NULL) (128.193.227.43) gregexpr does not find all matching substrings if the substrings overlap: > gregexpr("abab","ababab") [[1]] [1] 1 attr(,"match.length") [1] 4 It does work correctly in Version 2.3.1 under linux.

Hi, I have a silly question regarding the usage of two commands: read.table and gregexpr： For read.table, if I read a matrix and set header = T, I found that all the dash ("-") becomes dots (".") A = read.table("Matrix.txt", sep = "\t", header = F) A[1,1] # "A-B-C-D". A = read.table("Matrix.txt", sep = "\t", header = T)

Dear list, I am trying to use gregexpr to see if entries in a dataframe have either of two possible values for a string. here's an example text<-c("fat", "rat", "cat", "dog", "log", "fish") If I just wanted to find if any one of the elements in text match the pattern "at" I would do gregexpr("\\at", text)

Gregexpr - extract results with lapply Hello, I need to extract sequences of three upper case letters in a string. In other words, in this string: str <-c("ABC", "this WOUld be gOOD") The result I'm looking for is ABC WOU OOD. With gregexpr, I can get the position and length of the sequences gregexpr('[A-Z]{3}',str,perl=TRUE) [[1]] [1] 1

Full_Name: Reid Thompson Version: 2.8.0 RC (2008-10-12 r46696) OS: darwin9.5.0 Submission from: (NULL) (129.98.107.177) the gregexpr() function does NOT return a complete list of global matches as it should. this occurs when a pattern matches two overlapping portions of a string, only the first match is returned. the following function call demonstrates this error (although this is not how I

Full_Name: Reid Thompson Version: 2.8.0 RC (2008-10-12 r46696) OS: darwin9.5.0 Submission from: (NULL) (129.98.107.177) the gregexpr() function does NOT return a complete list of global matches as it should. this occurs when a pattern matches two overlapping portions of a string, only the first match is returned. the following function call demonstrates this error (although this is not how I

Hi all, Several people have noticed that gregexpr is very slow for large subject strings when perl=TRUE is specified. - https://stackoverflow.com/questions/31216299/r-faster-gregexpr-for-very-large-strings - http://r.789695.n4.nabble.com/strsplit-perl-TRUE-gregexpr-perl-TRUE-very-slow-for-long-strings-td4727902.html - https://stat.ethz.ch/pipermail/r-help/2008-October/178451.html I figured out

Hi Folks, I'm trying to extract just the backreferences from a regex. > temp = "abcd1234abcd1234" > regmatches(temp, gregexpr("(?:abcd)(1234)", temp)) [[1]] [1] "abcd1234" "abcd1234" What I would like is: [1] "1234" "1234" Note: I know I can just match 1234 here, but the actual example is complicated enough that I have to

Hello, I'm trying to figure out how to find the index of the second occurrence of "/" in a string (which happens to represent a date) within a data frame column. I've used the following code successfully to find the first instance of "/". dframe <- data.frame(date=c("5/14/2011", "4/7/2011")) dframe$x1 <- regexpr("/", dframe[, 1])

Dear All, I have a long string and need to search for regular expressions in there. However it becomes horribly slow as the string length increases. Below is an example: when "i" increases by 5, the time spent increases by more! (my string is 11,000,000 letters long!) I also noticed that - the search time increases dramatically with the number of matches found. - the perl=T option

Hi, Tried with R 2.6 and R 2.7: > gregexpr("", "abc", fixed=TRUE) *** caught segfault *** address 0x1c09000, cause 'memory not mapped' Traceback: 1: gregexpr("", "abc", fixed = TRUE) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace

I have a question regarding accessing the names of list components when applying a function with lapply. Here is an example that demonstrates what I'd like to do. I have a list like this one: mylist <- list(a=letters[1:10], b=letters[10:1], c=letters[1:3]) Now I would like to append the names of the list components to their corresponding vectors with the c() function. I thought this

First, my site will be down December 27-28 because of a network upgrade at Penn. It will also be down at least one day before that, while I upgrade the operating system. (And another day some time in January because of a planned power outage.) Second, I have replaced the search engine in my R site: http://finzi.psych.upenn.edu/ I am now using Namazu instead of HtDig. The direct link to the

The CRAN host in Tampa, FL (cran.hostingzero.net) isn't responding, and hasn't responded in quite a while -- at least problems were reported more than a year ago (Oct 2007) http://finzi.psych.upenn.edu/R/Rhelp02a/archive/116706.html although someone apparently succeeded in July 2008 http://finzi.psych.upenn.edu/R/Rhelp02a/archive/137451.html Perhaps it's just flaky, and not

Hi All, Sorry that I didn't provide enough information. I've been trying to import SAS xport files that contain multiple files using package foreign's read.xport. I first attempted this back in 2005 and had problems. Some of files that were present in the SAS xport file weren't being created in R. I submitted my problem to the community:

How can I get prediction intervals from a mixed-effects model? Consider the following example: library(nlme) fm3 <- lme(distance ~ age*Sex, data = Orthodont, random = ~ 1) df3.1 <- with(Orthodont, data.frame(age=seq(5, 20, 5), Subject=rep(Subject[1], 4), Sex=rep(Sex[1], 4))) predict(fm3, df3.1, interval='prediction') # M01 M01

Dear All, I have the following function tstpar <- function(n = 200, want.pdf = FALSE, pdfFileName = NULL){ oldpar <- par(no.readonly = TRUE) on.exit(par(oldpar)) steps <- seq(from = 1, to = 8, by = 1) h <- 10; w <- 6 if(want.pdf){pdf(file = pdfFileName, onefile = TRUE, paper = "letter", width = w, height = h)} par(mfrow = c(4,2)) for(i