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" ""
Related
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"
The string is as shown below:
s <- "12N10-3A 12N10-3A-1 12N10-3A-2 YB10L-A2"
I can get the strings except from second one.
gsub("\\s.*","",s) #12N10-3A
gsub(".*\\s","",s) #YB10L-A2
gsub(".*\\s.*\\s(.*).*\\s(.*)","\\1",s) #12N10-3A-2
How to get the second string from s and what's short approach for each code line? I tried what I learnt on regex101.com
We can use stri_extract_last from stringi
library(stringi)
stri_extract_last(s, regex = '\\S+')
#[1] "YB10L-A2"
Or use word from stringr
library(stringr)
word(s, 4)
#[1] "YB10L-A2"
Just use strsplit:
items <- strsplit(s, "\\s+")[[1]]
If you want to access the last item, then just use:
items[4]
[1] "YB10L-A2"
If you really wanted to isolate the last term using sub, then here is one way:
sub(".*\\s+", "", s)
Looking for some guidance on how to replace a curly apostrophe with a straight apostrophe in an R list of character vectors.
The reason I'm replacing the curly apostrophes - later in the script, I check each list item, to see if it's found in a dictionary (using qdapDictionary) to ensure it's a real word and not garbage. The dictionary uses straight apostrophes, so words with the curly apostrophes are being "rejected."
A sample of the code I have currently follows. In my test list, item #6 contains a curly apostrophe, and item #2 has a straight apostrophe.
Example:
list_TestWords <- as.list(c("this", "isn't", "ideal", "but", "we", "can’t", "fix", "it"))
func_ReplaceTypographicApostrophes <- function(x) {
gsub("’", "'", x, ignore.case = TRUE)
}
list_TestWords_Fixed <- lapply(list_TestWords, func_ReplaceTypographicApostrophes)
The result: No change. Item 6 still using curly apostrophe. See output below.
list_TestWords_Fixed
[[1]]
[1] "this"
[[2]]
[1] "isn't"
[[3]]
[1] "ideal"
[[4]]
[1] "but"
[[5]]
[1] "we"
[[6]]
[1] "can’t"
[[7]]
[1] "fix"
[[8]]
[1] "it"
Any help you can offer will be most appreciated!
This might work: gsub("[\u2018\u2019\u201A\u201B\u2032\u2035]", "'", x)
I found it over here: http://axonflux.com/handy-regexes-for-smart-quotes
You might be running up against a bug in R on Windows. Try using utf8::as_utf8 on your input. Alternatively, this also works:
library(utf8)
list_TestWords <- as.list(c("this", "isn't", "ideal", "but", "we", "can’t", "fix", "it"))
lapply(list_TestWords, utf8_normalize, map_quote = TRUE)
This will replace the following characters with ASCII apostrophe:
U+055A ARMENIAN APOSTROPHE
U+2018 LEFT SINGLE QUOTATION MARK
U+2019 RIGHT SINGLE QUOTATION MARK
U+201B SINGLE HIGH-REVERSED-9 QUOTATION MARK
U+FF07 FULLWIDTH APOSTROPHE
It will also convert your text to composed normal form (NFC).
I see a problem in your call to gsub:
gsub("/’", "/'", x, ignore.case = TRUE)
You are prefixing the curly single quote with a forward slash. I don't know why you are doing this. I could speculate that you are trying to escape the quote characters, but this is having the side effect that your pattern is now trying to match a forward slash followed by a quote. As this never occurs in your text, no replacements are being made. You should be doing this:
gsub("’", "'", x, ignore.case = TRUE)
Follow the link below for a demo which shows that using the above gsub calls works as you expect.
Demo
Was about to say the same thing.
Try using str_replace from stringr package, will not need to use slashes
I was facing similar problem. Somehow non of the solutions worked for me. So I devised an indirect way of doing it by identifying apostrophe and replacing it with the required format.
gsub("(\\w)(\\W)(\\w\\s)", "\\1'\\3","sid’s bicycle")
[1] "sid's bicycle"
Hope it helps someone.
I have a basic problem in R, everything I'm working with is familiar to me (data, functions) but for some reason I can't get the strsplit or the gsub function to work as expected. I also tried the stringr package. I'm not going to bother putting up code using that package because I know this problem is simple and can be done with the two functions mentioned above. Personally, I feel like putting up a page for this isn't even necessary but my patience is pretty thin at this point.
I am trying to remove the "." and the number followed by the '.' in an Ensemble Gene ID. Simple, I know.
id <- "ENSG00000223972.5"
gsub(".*", "", id)
strsplit(id, ".")
The asterisk symbol was meant to catch anything after the '.' and remove it but I don't know for sure if that's what it does. The strsplit should definitely output a list of two items, the first being everything before the '.' and the second being the one digit after. All it returns is a list with 17 "" symbols, for no space and one for each character in the string. I think it's an obvious thing that I'm missing but I haven't been able to figure it out. Thanks in advance.
Read the help file for ?strsplit, you cannot use "."
id <- "ENSG00000223972.5"
gsub("[.]", "", id)
strsplit(id, split = "[.]")
Output:
> gsub("[.]", "", id)
[1] "ENSG000002239725"
> strsplit(id, split = "[.]")
[[1]]
[1] "ENSG00000223972" "5"
Help:
unlist(strsplit("a.b.c", "."))
## [1] "" "" "" "" ""
## Note that 'split' is a regexp!
## If you really want to split on '.', use
unlist(strsplit("a.b.c", "[.]"))
## [1] "a" "b" "c"
## or
unlist(strsplit("a.b.c", ".", fixed = TRUE))
I have a string, say
fruit <- "()goodapple"
I want to remove the brackets in the string. I decide to use stringr package because it usually can handle this kind of issues. I use :
str_replace(fruit,"()","")
But nothing is replaced, and the following is replaced:
[1] "()good"
If I only want to replace the right half bracket, it works:
str_replace(fruit,")","")
[1] "(good"
However, the left half bracket does not work:
str_replace(fruit,"(","")
and the following error is shown:
Error in sub("(", "", "()good", fixed = FALSE, ignore.case = FALSE, perl = FALSE) :
invalid regular expression '(', reason 'Missing ')''
Anyone has ideas why this happens? How can I remove the "()" in the string, then?
Escaping the parentheses does it...
str_replace(fruit,"\\(\\)","")
# [1] "goodapple"
You may also want to consider exploring the "stringi" package, which has a similar approach to "stringr" but has more flexible functions. For instance, there is stri_replace_all_fixed, which would be useful here since your search string is a fixed pattern, not a regex pattern:
library(stringi)
stri_replace_all_fixed(fruit, "()", "")
# [1] "goodapple"
Of course, basic gsub handles this just fine too:
gsub("()", "", fruit, fixed=TRUE)
# [1] "goodapple"
The accepted answer works for your exact problem, but not for the more general problem:
my_fruits <- c("()goodapple", "(bad)apple", "(funnyapple")
str_replace(my_fruits,"\\(\\)","")
## "goodapple" "(bad)apple", "(funnyapple"
This is because the regex exactly matches a "(" followed by a ")".
Assuming you care only about bracket pairs, this is a stronger solution:
str_replace(my_fruits, "\\([^()]{0,}\\)", "")
## "goodapple" "apple" "(funnyapple"
Building off of MJH's answer, this removes all ( or ):
my_fruits <- c("()goodapple", "(bad)apple", "(funnyapple")
str_replace_all(my_fruits, "[//(//)]", "")
[1] "goodapple" "badapple" "funnyapple"