similar to: extending strsplit(): supply pattern to keep, not to split by

Hi all, Does anyone have a version of strsplit that keeps the string that is split by. e.g. from x <- "A: 123 B: 456 C: 678" I'd like to get c("A:", "123 ", "B: ", "456 ", "C: ", 678) but strsplit(x, "[A-Z]+:") gives me c("", " 123 ", " 456 ", " 678") Any ideas? Thanks,

\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, if I have this string: "a.b.c.d" and I use this function: unlist(strsplit("a.b.c.d", "\\.")), I get this as my output: "a", "b", "c", and "d". Is there a way to just split on the first period so I obtain only two pieces like: "a" and "b.c.d"? Anyways, thanks in advance! -- View this message in context:

Hi there, I have a problem about lapply, strsplit, and accessing list elements, which I don't understand or cannot solve: I have e.g. a character vector with three elements: x = c("349/077,349/074,349/100,349/117", "340/384.2,340/513,367/139,455/128,D13/168", "600/437,128/903,128/904") The task I want to perform, is to generate a list,

On Wed, 12 Apr 2023 08:29:50 +0000 Emily Bakker <emilybakker at outlook.com> wrote: > Some example data: > ?leucocyten + gramnegatieve staven +++ grampositieve staven ++? > ?leucocyten ? grampositieve coccen +? > ? > I want to split the strings such that I get the following result: > c(?leucocyten +?, ??gramnegatieve staven +++?, > ??grampositieve staven ++?) >

Dear list, I am trying to split a string using regexp: x <- "2 Value 34 a-c 45 t" strsplit(x, "[0-9]") [[1]] [1] "" " Value " "" " a-c " "" " t" But I don't want to lose the digits (pattern), the result should be: [[1]] [1] "2" " Value " "34" " a-c "

Hi All, Is there a means to extract the "10" from "23:10:34" in one pass using strsplit (or something else)? tst <- "23:10:34" For example my attempt strsplit(as.character(tst),"^[0-9]*:") gives [[1]] [1] "" "" "34" Obviously it is matching the first two instances of [0-9]. Note that there may be only one digit

Dear R People: I have the following set of data > Block[1:5] [1] "5600-5699" "6100-6199" "9700-9799" "9400-9499" "8300-8399" and I want to split at the - > strsplit(Block[1:5],"-") [[1]] [1] "5600" "5699" [[2]] [1] "6100" "6199" [[3]] [1] "9700" "9799" [[4]] [1]

I am sure there is an obvious answer to this that I'm missing but I can't find it. I'm parsing headers of Emails and most have a date like this: "Wed, 16 Nov 2005 05:28:00 -0800" and I can parse that using: tmp.dat.data <- matrix(unlist(strsplit(headers$Date.line,",")), ncol = 2, byrow = TRUE) before going on to look at the day and date/time data.

Dear list, I'm seeking advice to extract some numeric values from a log file created by an external program. Consider the following example, input <- readLines(textConnection( "some text <ax> = 1.3770E-03 <bx> = 3.4644E-07 <ay> = 1.9412E-04 <by> = 4.8840E-08 other text <aax> = 1.3770E-03 <bbx> = 3.4644E-07

Dear R People: Here is a toy example: > x <- c("2E","5W","12H") > substr(x,2,2) [1] "E" "W" "2" > Sometimes x has 3 elements, sometimes 2. I want to extract the last element, and then extract the other 1 or 2 elements. How can I do this, please? TIA, Sincerely, Erin -- Erin Hodgess Associate Professor Department of

Greetings, I have a vector of values that are a word followed by a number, e.g., x = c("Apple12","HP42","Dell91"). The goal is to split this vector into two vectors such that the first vector contains just the words and the second contains just the numbers. I cannot use strsplit (or at least I do not know how) as there is no obvious separator. I can use sub to

Hi, I find this behavior unexpected: --8<---------------cut here---------------start------------->8--- > strsplit(c("a,b;c","d;e,f"),c(",",";")) [[1]] [1] "a" "b;c" [[2]] [1] "d" "e,f" --8<---------------cut here---------------end--------------->8--- I thought that it should be identical to this:

Any ideas what is wrong? > strsplit("a.b", ".") # generates empty strings with split="." [[1]] [1] "" "" "" > strsplit("a b", " ") # seems to work fine with split=" ", and other characters... [[1]] [1] "a" "b" > > R.Version() $platform [1]

Dear all, I'm having a problem understanding why a split does not occur with in the 2nd use of the function strsplit below: # text strings > txt <- c("sales to 23 August 2008 published 29 August", + "sales to 6 September 2008 published?11 September") # first use > strsplit(txt, 'published', fixed=TRUE) [[1]] [1] "sales to 23 August 2008 "

Hello, Anyone knows how can I do this in a cleaner way? mynumber = 1001 as.numeric(unlist(strsplit(as.character(mynumber),""))) [1] 1 0 0 1 Thanks in advance, Gustavo

Dear fellow R-help members, I hope to seek your advice on how to parse/manage a dataset with hundreds of columns. Two examples of these columns, 'cancer.problems', and 'neuro.problems' are depicted below. Essentially, I need to parse this into a useful dataset, and unfortunately, I am not familiar with perl or any such language. data <- data.frame(id=c(1:10))

Hi, I have a vector like this: DF <- c("Aetna, Inc.", "Alexander's Inc.", "Allegheny Energy, Inc") For each element in the vector I would like to remove the "incorporated" info, so that my vector looks like this: DF <- c("Aetna", "Alexander's", "Allegheny Energy") That means that I have to strip: strip <-

Colleagues Assume that I have a vector containing some text strings, some of which contain a particular character. I could like to apply "toupper" to the text before the character. For example (in this case, "|" is the particular character): ORIGINAL: TEXT <- c("aaaa", "bbb|cc", "|ddd") AFTER APPLICATION OF toupper: TEXT <-

Dear all, I want to use string split to parse column names, however, I am having some errors that I don't understand. I see a problem when I try to rbind the output from strsplit. please let me know if I'm missing something obvious, thanks, alison here are my commands: >strsplit<-strsplit(as.character(Rumino_Reps_agreeWalign$geneid),"\\.") >