strsplit

I have a table called LOAN containing column named RATE in which the observations are given in percentage for example 14.49% how can i format the table so that all value in rate are edited and % is removed from the entries so that i can use plot function on it .I tried using strsplit. strsplit(LOAN$RATE,"%") but got error non character argument Items that appear to be character when printed but for which R thinks otherwise are generally factor classes objects. I'm also guessing htat you are not going to be happy with the list output that strsplit will return Try: gsub( "%", "", as.character

I have the following regex that splits on any space or punctuation. How can I exclude 1 or more punctuation characters from :punct: ? Let's say I'd like to exclude apostrophes and commas. I know I could explicitly use [all punctuation marks in here] instead of [[:punct:]] but I'm hoping for an exclusion method. X <- "I'm not that good at regex yet, but am getting better!" strsplit(X, "[[:space:]]|(?=[[:punct:]])", perl=TRUE) [1] "I" "'" "m" "not" "that" "good" "at" "regex" "yet" [10] "," "" "but" "am" "getting" "better" "!" Joshua Ulrich It's not clear to me what you want the result to be, but

I have a data table containing 20000+ rows and one column. The string in each column has different number of words. I want to split the words and put each of them in a new column. I know how I can do it word by word: Data [ , Word1 := as.character(lapply(strsplit(as.character(Data$complaint), split=" "), "[", 1))] ( Data is my data table and complaint is the name of the column) Obviously, this is not efficient because each cell in each row has different number of words. Could you please tell me about a more efficient way to do this? Check out cSplit from my "splitstackshape" package. It works

I have the vector length # [1] 15,34, 12,24, 225, # Levels: 12,24, 15,34, 225, and I want to separate them by the comma to eventually make a list of these values Tried: strsplit(length, ",") but keep getting the error message Error in strsplit(length, ",") : non-character argument Your "length" object is a factor : As the error message indicates, strsplit expects a character vector as the input. Try: strsplit(as.character(length), ",") Demo x <- factor(c("1,2", "3,4", "5,6")) strsplit(x, ",") # Error in strsplit(x, ",") : non-character argument strsplit(as.character(x), ",") # [[1]] # [1] "1"

I have the following string: [1] "10012 ---- ---- ---- ---- CAB UNCH CAB" I want to split this string by the gaps, but the gaps have a variable number of spaces. Is there a way to use strsplit() function to split this string and return a vector of 8 elements that has removed all of the gaps? One line of code is preferred. Just use strsplit with \\s+ to split on: x <- "10012 ---- ---- ---- ---- CAB UNCH CAB" x # [1] "10012 ---- ---- ---- ---- CAB UNCH CAB" strsplit(x, "\\s+")[[1]] # [1] "10012" "----" "----" "----" "----" "CAB" "UNCH" "CAB" length(.Last.value) # [1] 8 Or, in this case, scan

I could solve this using loops, but I am trying think in vectors so my code will be more R-esque. I have a list of names. The format is firstname_lastname. I want to get out of this list a separate list with only the first names. I can't seem to get my mind around how to do this. Here's some example data: t <- c("bob_smith","mary_jane","jose_chung","michael_marx","charlie_ivan") tsplit <- strsplit(t,"_") which looks like this: > tsplit [[1]] [1] "bob" "smith" [[2]] [1] "mary" "jane" [[3]] [1] "jose" "chung" [[4]] [1] "michael" "marx" [[5]] [1] "charlie" "ivan" I could get out what I want using