similar to: Pattern match

Hi, I have a simple question. Suppose I have a string "x$Expensive". I want to find the position of the $ in this string; i.e., I want a function that returns 2. I tried grep, regexpr, etc with no luck, unless I'm just using them incorrectly. Any suggestions? Thanks, Walt ________________________ Walter R. Paczkowski, Ph.D. Data Analytics Corp. 44 Hamilton Lane Plainsboro,

Patrick, ## Run the following script an notice the different values of the dataframe "data" in each instance. # I understand you have done something like the following: data <- data.frame(COL1 = c("6/1/14", "7/1/14"), COL2 = c("5/1/15", "5/1/15"), stringsAsFactors = FALSE) data$Date_Flag <- ifelse(data$COL2 >

A small example data set that illustrates your question will be of great value to those trying to help. This appears to be a transformation that you are wanting to do (timestamp to units of time) so a data representing what you have (dput() is handy for this) and one representing what you want to have with any guidance regarding how to use the other columns in you data set (e.g., the event(0/1)).

Hello, I want to perform a latent class analysis using poLCA package. My formula is: substances <- cbind(subs1, subs2, subs3, subs4, subs5, subs6) ~ gender+age+education+income+occupation+urban+dbehavior+incarceration+treatment+depression+alcriteria I want to include sample weights in the model, I have read that poLCA does not take into account weights, but when I introduce them, it seems

Mark, Below is the sampled simulated granular data format for pumps for trial period of 3 months that I need to transform for survival analysis: 3 months = (60*24*90) minutes i.e 129600 minutes pump_id timings events vibration temprature flow pump1 01-07-2017 00:00 0 3.443 69.6 139.806 pump1 01-07-2017 00:10 1 0.501 45.27 140.028

Hi Iago, This is not a bug. It is expected. Patterns may not overlap. However, there is a way to get the result you want using perl: ```R gsub("([aeiouAEIOU])(?=[aeiouAEIOU])", "\\1_", "aerioue", perl = TRUE) ``` The specific change I made is called a positive lookahead, you can read more about it here: https://www.regular-expressions.info/lookaround.html

Hi Iris, Thank you. Further, very nice solution. Best, Iago On 01/03/2024 12:49, Iris Simmons wrote: > Hi Iago, > > > This is not a bug. It is expected. Patterns may not overlap. However, there > is a way to get the result you want using perl: > > ```R > gsub("([aeiouAEIOU])(?=[aeiouAEIOU])", "\\1_", "aerioue", perl = TRUE) > ``` >

Hi all, I tested next command: gsub("([aeiouAEIOU])([aeiouAEIOU])", "\\1_\\2", "aerioue") with the following output: [1] "a_eri_ou_e" So, there are two consecutive vowels where an underscore is not added. May it be a bug? Is it expected (bug or not)? Is there any chance to get what I want (an underscore between each pair of consecutive vowels)?

Here's another *incorrect* way to do it -- incorrect because it will not always work, unlike Iris's correct solution. But it does not require PERL type matching. The idea: separate the two vowels in the regex by a character that you know cannot appear (if there is such) and match it optionally, e.g. with '*" repetition specifier. I used "?" for the optional character

Hello, I am trying to build a regex to extract vowels and consonants from a string. So far, I am able to extract the basic a-e-i-o-u sequence using the following extension to the String class: class String def vowels scan(/[aeiou]/i) end def consonants scan(/[^aeiou]/i) end end examples: >> "Mary had a little lamb".vowels => aaaiea >> "Mary had a

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 have this problem where I need to find if there is any numbers in a string, this is no problem if theres only one number per string. I would then simply use the regexpr() funtion togheter with the substring function to extract the number. But regexpr only picks one number per string either from the beginning or the end, but not multiple. Can this be done? And how for example My string <-

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]

Hello, What about an `invert` argument in grep, to return elements that are *not* matching a regular expression : R> grep("pink", colors(), invert = TRUE, value = TRUE) would essentially return the same as : R> colors() [ - grep("pink", colors()) ] I'm attaching the files that I modified (against today's tarball) for that purpose. Cheers, Romain --

Hi, I am looking for a way to search a file for position of some expression, from within R. My current code: sha1Pos = gregexpr("<sha1>", readChar(filename, file.info(filename)$size))[[1]] Works fine for small files, but text files I will be working with might get up to Gb range, so I was trying to accomplish the same without loading the whole file into R. I realize this is

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: >

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

Dear R fellows, I created a new column Date_flag to compare the dates of COL1 and COL2 using the code below. But it showed that 5/1/15 is greater than 6/1/2014 and 5/1/2015 greater than 7/1/2014 despite the year is greater. How do I fix that? I did try to format as %y/%m/%d but it does not fix that. data$Date_Flag <- ifelse(data$COL2 > data$COL1, 0,1) COL1 COL2 6/1/14