similar to: How to use access results of gregexpr in data frames

Hi, I have a data frame with column names "1", "2", "3", ... and I'd like to extract a subset based on the values in the first column. None of the methods I tried worked (below). x <- subset(dframe, 1 = = "My Text") x <- subset(dframe, "1" = = "My Text") x <- subset(dframe, names(dframe)[1] = = "My Text") Q

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

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)

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, 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 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)

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 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: >

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

Hi, It seems that there's a bug in summary, in the max. output... but max() alone works fine. > hw04.dframe$area ... [41] 1790 1380 1296 2745 798 2306 438649 1481 1559 2450 ... > summary(hw04.dframe) area Min. : 798 1st Qu.: 1349 Median : 1690 Mean : 6962 3rd Qu.: 2306 Max. :438600 ### should read 438649 or, to the point,

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, Does anyone know if the alternatives in the 'switch' function can be specified as strings containing spaces?  Neither of the two approaches below works. switch(expr, "Choice 1"="My first choice", "Choice 2"="My 2nd choice", "Choice 3"="My 3rd choice") x <- c("Choice 1", "Choice 2", "Choice

> -----Original Message----- > From: William Dunlap > Sent: Tuesday, November 25, 2008 9:16 AM > To: 'johannes_graumann at web.de' > Subject: Re: [R] Efficient passing through big data.frame and > modifying select fields > > > Johannes Graumann johannes_graumann at web.de > > Tue Nov 25 15:16:01 CET 2008 > > > > Hi all, > > > >

Dear R friends: Can someone help me with the following problem? Many thanks in advance. # Problem Description: # I want to write functions which take a (character) vector of dataframe names as input argument. # For example, I want to extract the number of observations from a number of dataframes. # I tried the following: nobs.fun <- function (dframe.vec) { nobs.vec <-

Hello, After reading help file, various threads on this board, and other online tutorials, I've attempted to use gsub (using Perl-like syntax) to change a currency string into something that can be converted to numeric type using only one regular expression.  Can anybody point out my error?  Note that >  x <- "\"$ 1,200,300,400.50\"" Tried the following in an

Dear alltogether, I experience very strange behavior of imputation of NA's with the NORM library. I use R 2.2.0, win32. The code is below and the same dataset was also tried with MICE and aregImpute() from HMISC _without_ any problem. The problem is as follows: (1) using the whole dataset results in very strange imputations - values far beyond the maximum of the respective column, >