similar to: strsplit on comma, with a trailing comma in input

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,

Full_Name: Charles Dupont Version: 2.2.0 OS: linux Submission from: (NULL) (160.129.129.136) when strsplit("", " ") returns character(0) where as strsplit("a", " ") returns "a". these return values are not constiant with each other. Charles Dupont

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),"\\.") >

Hello R Gurus: I would like to take a character string and split at the $ sign. I thought that strsplit would do it, but here are the results: > vv [1] "whine$ts1" > vv [1] "whine$ts1" > strsplit(vv,"$") [[1]] [1] "whine$ts1" Does anyone have any suggestions, please? Thanks, Edna Bell

Dear R People: I am trying to split a character vector into a set of individual letters: Ideal: x3 <- c("dog") "d" "o" "g" I tried the following: > strsplit(x3) Error in strsplit(x3) : argument "split" is missing, with no default > strsplit(x3,1) [[1]] [1] "dog" I know that this is incredibly simple, but what am I doing

Hi, As show in the code below, strsplit can be applied to a matrix but not a data.frame. I don't understand why R is designed in this way. Can somebody help me understand it? How to split all the strings in x$y? x=data.frame(x=1:10,y=rep("abc",10)) strsplit(x$y,'b') #Error in strsplit(x$y, "b") : non-character argument y=cbind(1:10,rep("abc",10))

Dear Researchers, I have several files as this example: Myfile_MyArea1_sample1.txt i wish to split in "Myfile", "MyArea1", "sample1", and "txt", becasue i need to use "sample1" label. I try to use "strsplit" but I am able just to split as "Myfile_MyArea1_sample1" and "txt" OR "Myfile", "MyArea1",

Hi, I tried to use strsplit for separating a string with || like strsplit(string,"\\||") but it returned each single character was separated. For example: strsplit("a||bc","\\||") [[1]] [1] "a" "" "" "b" "c" where I want the result to be "a" and "bc". Any ideas? Thanks! Best,

Hi All, is there a way of using strsplit with a forward slash '/' as the splitting point? For data such as: 1 T/T C/C 16/33 2 T/T C/C 33/36 3 T/T C/C 16/34 4 T/T C/C 16/31 5 C/C C/C 28/29 6 T/T C/C 16/34 strsplit(my.data[1,1], "/") # and any variation thereof Error in strsplit(apoe[1, 1], "/") : non-character

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]

strsplit() is a convenient way to get a list of items from a string when you have a regular expression for what is not an item. E.g., > strsplit("1.2, 34, 1.7e-2", split="[ ,] *") [[1]]: [1] "1.2" "34" "1.7e-2" However, sometimes is it more convenient to give a pattern for the items you do want. E.g., suppose you want to pull

Hi all, Came across an error in ?strsplit in the details section: Details: Arguments 'x' and 'split' will be coerced to character, so you will see uses with 'split = NULL' to mean 'split = character(0)', including in the examples below. I caught myself trying to use strsplit() on a factor and the coercion on 'x' is not done. According

\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 all, > diff_operator <- "\\(" > strsplit(cond, diff_operator) [[1]] [1] "andsin" "log_angle_1_4)" > diff_operator <- "\\sin(" > strsplit(cond, diff_operator) Error in strsplit(cond, diff_operator) : invalid regular expression '\sin(', reason 'Missing ')'' When I am going to split with "("

Hello, I would like to know what can I do if I use strplit with a string and I want to use the middle left,I mean I have this: strsplit("bA531F16-rep","\\-") [[1]] [1] "bA531F16" "rep" I would like to work just with bA531F16 in another variable, what could I do?, Thank you --------------------------------- Sé un Mejor Amante

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:

Hi all, In the help of strsplit one can read split character vector (or object which can be coerced to such) containing regular expression<http://127.0.0.1:39783/help/library/base/help/regular%20expression>(s) (unless fixed = TRUE) to use for splitting. If empty matches occur, in particular if split has length 0, x is split into single characters. Ifsplit has length greater than 1, it is

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]

Colleagues, I have some text: TEXT <- c("a", "bb;ccc", "dddd;eeeee;ffffff") I want to retrieve the portion of each element before the first semicolon. I can split each element using strsplit: SPLIT <- strsplit(TEXT, ";") This yields: > SPLIT [[1]] [1] "a" [[2]] [1] "bb" "ccc" [[3]] [1] "dddd"

Hello, I am attempting to manipulate strings in which there are differing amounts of whitespace before and after each element taht I want to keep (any word, letter, or number). However, after strsplit and unlist, I know how to select specific elements with "[ ]", but I want to select instead all elements that are not missing. That is, do not select any elements that are simply