similar to: regular expression change in R version 2.3.0?

Hi all I have a question regarding differences in the way gregpexr works in R 2.3.0 and R 2.4.0. In R 2.3.0, this is what happens: > gregexpr(" [a-z] [a-z] ", " a b c d e f ", perl=T) [[1]] [1] 1 3 5 7 9 attr(,"match.length") [1] 5 5 5 5 5 ... while in R 2.4.0, this is what happens: > gregexpr(" [a-z] [a-z] ", " a b c d e f ", perl=T)

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

\b is word boundary. But, unexpectedly, strsplit("dia ma", "\\b") splits character by character. > strsplit("dia ma", "\\b") [[1]] [1] "d" "i" "a" " " "m" "a" > strsplit("dia ma", "\\b", perl=TRUE) [[1]] [1] "d" "i" "a" " "

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)

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

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)

completion is semi-broken in today's r-devel, and the reason seems to be some regular expression changes: > sessionInfo() R version 2.6.0 Under development (unstable) (2007-05-22 r41673) i686-pc-linux-gnu locale: [...] attached base packages: [1] "stats" "graphics" "grDevices" "utils" "datasets" "methods" [7]

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

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 am getting stuck over an apparently simple problem in the use of regular expressions : To collect together the first letters of the words from the Perl motto, ?There is more than one way to do it? in the following form ? TIMTOWTDI. I tried the following code : ? ##### A regex problem with the Perl motto astr<-"There is more than one way to do it" b1<-grep("\\<",

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

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])

Hello, Is there any way I can use the gregexpr functions (or a different function) in a manner that will also return overlapping (i.e. non disjoint) regular expressions? For instance, when running gregexpr("AAA","AAAAAA"), I get two matches, one at position 1 and one at position 4. I'd like to receive 4 matches at positions 1, 2, 3 and 4. Thanks, Schraga

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

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

I have a two part question Part 1) I am trying to remove characters in a string based on the position of a key character in another string.? I have a solution that works but it requires a for-loop.? A vectorized way of doing this has alluded me.? CleanRead<-function(x,y) { ? if (!is.character(x)) ??? x <- as.character(x) ? if (!is.character(y)) ??? y <- as.character(y) ?

Last one for you guys: The command: length(gregexpr('cus','hocus pocus')[[1]]) [1] 2 returns the number of times the substring 'cus' appears in 'hocus pocus' (which is two) It's returning the number of **disjoint** matches. So: length(gregexpr('aa','aaa')[[1]]) [1] 1 returns 1. **What I want to do:** I'm looking for a way to count

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: >

Hi, In R 3.4.0, the "Pattern Matching and Replacement" documentation that describes regexec(), gregexpr(), etc. states that the "text" argument to regexec is a character vector, "or an object which can be coerced by as.character to a character vector": regexec(pattern, text, ignore.case = FALSE, perl = FALSE, fixed = FALSE, useBytes = FALSE)