I've successfully split the data and removed the "," with the following code:
s = MSA_data$area_title
str_split(s, pattern = ",")
Result
[1] "Albany" " GA"
I need to trim this data, removing white space, however this places the comma back into the data which was initially removed.
"Albany, GA"
How can I successfully split and trim the data so that the result is:
[1] "Albany" "GA"
Thank you
An alternative is to use trimws function to trim the whitespace at the beginning and end of the string.
Result <- trimws(Result)
We just need to use zero or more spaces (\\s*) (the question OP asked) and this can be done in a single step
strsplit(MSA_data$area_title, pattern = ",\\s*")
If we are using the stringr, then make use of the str_trim
library(stringr)
str_trim(str_split("Albany, GA", ",")[[1]])
#[1] "Albany" "GA"
Related
I have
df<-c("That's","you're", "'am")
and I would like to remove the part of a word after and including the apostrophe which should return
c("That", "you", "")
tidyverse solution or a solution usable within a pipe |> structure preferable
Replace ' and whatever follows it, using str_replace in stringr.
library(stringr)
str_replace(df, "'.*", "")
#[1] "That" "you" ""
Using R base sub
> sub("'.*", "", df)
[1] "That" "you" ""
Your example data only has one word per string. If you also need it to work for strings containing multiple words then use:
gsub("'\\w*\\b","",df)
Using trimws in base R
trimws(df, whitespace = "'.*")
[1] "That" "you" ""
I have string, which should be split into parts from "random" locations. Split occurs always from next comma after colon.
My idea was to find colons with
stringr::str_locate_all(test, ":") %>%
unlist()
then find commas
stringr::str_locate_all(test, ",") %>%
unlist()
and from there to figure out position where it should be split up, but could not find suitable way to do it. Feels like there is always 6 characters after colon before the comma, but I can't be sure about that for whole data.
Here is example string:
dput(test)
"AA,KK,QQ,JJ,TT,99,88:0.5083,66,55:0.8303,AK,AQ,AJs,AJo:0.9037,ATs:0.0024,ATo:0.5678"
Here is what result should be
dput(result)
c("AA,KK,QQ,JJ,TT,99,88:0.5083", "66,55:0.8303", "AK,AQ,AJs,AJo:0.9037",
"ATs:0.0024", "ATo:0.5678")
Perehaps we can use regmatches like below
> regmatches(test, gregexpr("(\\w+,?)+:[0-9.]+", test))[[1]]
[1] "AA,KK,QQ,JJ,TT,99,88:0.5083" "66,55:0.8303"
[3] "AK,AQ,AJs,AJo:0.9037" "ATs:0.0024"
[5] "ATo:0.5678"
here is one option with strsplit - replace the , after the digit followed by the . and one or more digits (\\d+) with a new delimiter using gsub and then split with strsplit in base R
result1 <- strsplit(gsub("([0-9]\\.[0-9]+),", "\\1;", test), ";")[[1]]
-checking
> identical(result, result1)
[1] TRUE
If the number of characters are fixed, use a regex lookaround
result1 <- strsplit(test, "(?<=:.{6}),", perl = TRUE)[[1]]
I have a string that consists entirely of simple repeating patterns of a [:digit:]+[A-Z] for instance 12A432B4B.
I want to to use base::strsplit() to get:
[1] "12A" "432B" "4B"
I thought I could use lookahead to split by a LETTER and keep this pattern with unlist(strsplit("12A432B4B", "(?<=.)(?=[A-Z])", perl = TRUE)) but as can be seen I get the split wrongly:
[1] "12" "A432" "B4" "B"
Cant get my mind around a pattern that works with this strsplit strategy? Explanations would be really appreciated.
Bonus:
I also failed to use back reference in gsub (e.g. - pattern not working `gsub("([[:digit:]]+[A-Z])+", "\\1", "12A432B4B"), and can you retrieve more than \\1 to \\9 groups, say if [:digit:]+[A-Z] repeats for more than 9 times ?
We can use regex lookaround to split between an upper case letter and a digit
strsplit(str1, "(?<=[A-Z])(?=[0-9])", perl = TRUE)[[1]]
#[1] "12A" "432B" "4B"
data
str1 <- "12A432B4B"
The pattern mentioned in the post can be used as it is in str_extract_all :
str_extract_all(string, '[[:digit:]]+[A-Z]')[[1]]
#[1] "12A" "432B" "4B"
Or in base R :
regmatches(string, gregexpr('[[:digit:]]+[A-Z]', string))[[1]]
where string is :
string <- '12A432B4B'
I am setting up an automated data analysis procedure and, more or less at the end of the procedure, I would like to extract automatically the name of the file that has been analysed. I have a data frame with a column containing names, with the following style:
Baseline/Cell_Line_2_KB_1813_B_Baseline
Dose 0001/Cell_Line_3_KB1720_1_0001
Dose 0010/Cell_Line_1_KB1810 mat_0010
I would like to extract just the characters in bold: "KB_1813_B", "KB1720_1" and "KB1810 mat" in a separate column.
I used gsub with the following command:
df$column.with.names <- gsub(".*KB|_.*", "KB", df$column.with.new.names)
I could easily remove the first part of the problem, but I am stuck trying to remove the second part. Is there some command in gsub to remove everything, starting from the end of the name, until you encounter a special character ( "_" in my case)?
Thank you :)
We can use str_extract
library(stringr)
str_extract(df$column.with.new.names, "KB_*\\d+[_ ]*[^_]*")
#[1] "KB_1813_B" "KB1720_1" "KB1810 mat"
Or the same pattern can be captured as a group with sub
sub(".*(KB_*\\d+[_ ]*[^_]*).*", "\\1", df$column.with.new.names)
#[1] "KB_1813_B" "KB1720_1" "KB1810 mat"
data
df <- data.frame(column.with.new.names = c("Baseline/Cell_Line_2_KB_1813_B_Baseline",
"Dose 0001/Cell_Line_3_KB1720_1_0001",
"Dose 0010/Cell_Line_1_KB1810 mat_0010"), stringsAsFactors = FALSE)
The way to do this is using regex groups:
x <- c("Baseline/Cell_Line_2_KB_1813_B_Baseline",
"Dose 0001/Cell_Line_3_KB1720_1_0001",
"Dose 0010/Cell_Line_1_KB1810 mat_0010")
gsub("^.+Cell_Line_._(.+)_.+$", "\\1", x)
[1] "KB_1813_B" "KB1720_1" "KB1810 mat"
I find this problem >S
I scrap some data from the web and for instance I obtain this
"3.444.654" (As character)
If I use gsub("3.444.654", ".", "") in order to get 3444654...
R gives me
[1] ""
What could I do to get the integer!
> gsub(".", "", "3.444.654", fixed = TRUE)
[1] "3444654"
Maybe read the documentation for gsub for argument order etc. To then turn the string into a number, use as.numeric, as.integer etc.