similar to: regexp bug in very recent r-devel

Dear R core developers, One feature from Python that I have been wanting in R is the ability to capture groups in regular expressions using names. Consider the following example in R. > notables <- c(" Ben Franklin and Jefferson Davis","\tMillard Fillmore") > name.rex <- "(?<first>[A-Z][a-z]+) (?<last>[A-Z][a-z]+)" > (parsed <-

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

\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" " "

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)

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)

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

While doing some speed testing I noticed that in R-3.2.3 the perl=TRUE variants of strsplit() and gregexpr() took time proportional to the square of the number of pattern matches in their input strings. E.g., the attached test function times gsub, strsplit, and gregexpr, with perl TRUE (PCRE) and FALSE (TRE), when the input string contains 'n' matches to the given pattern. Notice the

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, 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)

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])

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 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

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: 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.

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

there seems to be something wrong with r's regexing. consider the following example: gregexpr('a*|b', 'ab') # positions: 1 2 # lengths: 1 1 gsub('a*|b', '.', 'ab') # .. where the pattern matches any number of 'a's or one b, and replaces the match with a dot, globally. the answer is correct (assuming a dfa engine).

Full_Name: Wacek Kusnierczyk Version: 2.10.0 r48181 OS: Ubuntu 8.04 Linux 32bit Submission from: (NULL) (129.241.199.135) there seems to be something wrong with r's regexing. consider the following example: gregexpr('a*|b', 'ab') # positions: 1 2 # lengths: 1 1 gsub('a*|b', '.', 'ab') # .. where the pattern matches any number of

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

That's basically what I did 1. Get text lines using readLines 2. use tryCatch to parse each line using read.csv(text=...) 3. in the catch, use?gregexpr to find any quotes not adjacent to a comma (gregexpr("[^,]\"[^,]",...) 4. escape any quotes found by adding a second quote (using str_sub from stringr) 6. parse the patched text using read.csv(text=...) 7. write out the parsed