similar to: Read txt File contains character ""

I am lousy at simple regex and I have not found a solution to a simple problem. I have a vector with some character values that I want to split. Sample data dd1 <- c( "XXY (mat harry)","XXY (jim bob)", "CAMP (joe blow)", "ALP (max jack)") Desired result dd2 <- data.frame( xx = c("XXY", "XXY", "CAMP",

?s 06:47 de 08/04/2024, Dave Dixon escreveu: > Greetings, > > I have a csv file of 76 fields and about 4 million records. I know that > some of the records have errors - unmatched quotes, specifically. > Reading the file with readLines and parsing the lines with read.csv(text > = ...) is really slow. I know that the first 2459465 records are good. > So I try this: >

?s 16:21 de 07/12/2023, Sorkin, John escreveu: > Colleagues, > > I have a matrix of character data that represents date and time. The format of each element of the matrix is > "2020-09-17_00:00:00" > How can I convert the elements into a valid R date-time constant? > > Thank you, > John > > > > John David Sorkin M.D., Ph.D. > Professor of

I thought this is a common question but rseek/google searches don't yield any relevant hit. I have a matrix of character strings, which are time stamps, > time.m[1:5,1:5] [,1] [,2] [,3] [,4] [,5] [1,] "08:00:20.799" "08:00:20.799" "08:00:20.799" "08:00:20.799" "08:00:20.799" [2,]

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

Dear all, I want to manipulate a character string such as ex<-"cbind(data$response1,data$response2)" in R in two ways: 1) extracting the "response1" portion of ex 2) replacing "$" with "." I am wondering that is it possible efficiently doing these in R? Best Ozgur -- View this message in context:

Isn't this like trying to tie up the horse after it has left the barn? Why not figure all this out _before_ converting to xts? On April 12, 2023 12:29:49 PM PDT, akshay kulkarni <akshay_e4 at hotmail.com> wrote: >Dear Rui, > Not working. I have entirely removed the column containing % but am still bootless: > >> head(coredata(INFYTX)) > INFY

Hello, > > The problem is that I get a the following error bacause anything after the > # is ignored. > > Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, > : > line 6 did not have 500 elements > > R thinks that line 6 has only 2 elements because of the #. > Use 'readLines' instead, followed by 'strsplit'. In the

?s 19:57 de 12/04/2023, akshay kulkarni escreveu: > Dear members, > I have an xts object: > >> head(INFYTX) > INFY Historical Data INFY Historical Data.1 INFY Historical Data.2 > 2003-04-16 "47.26" "44.28" "47.56" > 2003-04-17 "46.30" "44.92"

Dear R users, I want to aggregate all *d *data from all combination of n *plots* taken by k. Thank very much! My data is like that: plot d 1 14 1 13 1 12 1 14 1 18 1 20 1 21 1 43 1 108 1 43 2 41 2 61 2 83 2 61 2 84 2 45 2 21 2 12 2 11 ... 100 10 100 12 -- --- Catalin-Constantin ROIBU Forestry engineer, PhD Forestry Faculty of Suceava

Hi there, I just noticed that "crt" is a graphical parameter, which is stated as follows in the help page of "par": A numerical value specifying (in degrees) how single characters should be rotated. It is unwise to expect values other than multiples of 90 to work... However, I did not find any code example to demo the usage of "crt", and even know how to present a

R 2.15.1 OS X Colleagues, I am reading a 1 GB file into R using read.table. The file consists of 100 tables, each of which is headed by two lines of characters. The first of these lines is: TABLE NO. 1 The second is a list of column headers. For example: TABLE NO. 1 COL1 COL2 COL3 COL4 COL5 COL6 COL7 COL8 COL9 COL10

Dear Rui, Not working. I have entirely removed the column containing % but am still bootless: > head(coredata(INFYTX)) INFY Historical Data INFY Historical Data INFY Historical Data INFY Historical Data [1,] "47.26" "44.28" "47.56" "44.28" [2,] "46.30" "44.92"

Thanks. This is exactly the doubt I had. Rui Barradas ?s 05:26 de 05/08/2018, Kenny Bell escreveu: > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)]

Short question: Is it possible to use statistical tests, like the Augmented Dickey-Fuller test, in functions with for-loops? If not, are there any alternative ways to scale measures? Detailed explanation: I am working with time-series, and I want to flag curves that are not stationary and which display pulses, trends, or level shifts. >df DATE ID VALUE2012-03-06 1

?s 15:51 de 04/08/2018, I?aki ?car escreveu: > El s?b., 4 ago. 2018 a las 15:32, Rui Barradas > (<ruipbarradas at sapo.pt>) escribi?: >> >> Hello, >> >> Maybe I am not understanding how negative indexing works but >> >> 1) This is right. >> >> (1:10)[-1] >> #[1] 2 3 4 5 6 7 8 9 10 >> >> 2) Are these right? They

Hi, I have a small problem with the function split() and would appreciate your help. I have a table called ?XXX? with 2 columns and 49 rows. The 49 rows belong to 8 different levels (electrode1, ...,electrode8). I want to split the table always at the row where ?electrode1? starts again so that I can export 7 individual dataframes (numbered ?dataframe1? to ?dataframe7?) which contain always

sir i could not find the plant breeding libraray in Rgui3.0.0 [[alternative HTML version deleted]]

Hi R users, I am a new user, still learning basics of R. Is there anyway to extract y (or x) value for a known x (or y) value from ecdf (empirical cumulative distribution function) curve? Thanks in advance. Mano. [[alternative HTML version deleted]]

convertpdf2txt <- function(dirpath){ files <- list.files(dirpath, pattern = "Consoli.*\\.pdf$", full.names = TRUE) files <- chartr("\\", "/", files) x <- lapply(files, function(x){ pdftools::pdf_text(x) %>% paste0(collapse = " ") %>% stringr::str_squish() }) new_names <-