This is a silly question, but I can't seem to find a solution in R online. I am trying to remove an isolated number from a long string. For example, I would like to remove the number 27198 from the sentence below.
x <- "hello3 my name 27198 is 5joey"
I tried the following:
gsub("[0-9]","",x)
Which results in:
"hello my name is joey"
But I want:
"hello3 my name is 5joey"
This seems really simple, but I am not well versed with regular expressions. Thanks for your help!
We can specify word boundary (\\b) at the end of one or more digits ([0-9]+)
gsub("\\b[0-9]+\\b", "", x)
#[1] "hello3 my name is 5joey"
Related
I'm trying to grep strings that end in a dash in R, but having trouble. I've worked out how to grep strings ending in any punctuation mark, maybe not the best way but this worked:
grep("\\#[[:print:]]+[[:punct:]]$",c)
Can't for the life of me work out how to grep strings that end specifically in a dash
for example these strings:
- # (piano) - not this.
- # hello hello - not this either.
I'd like to sub all the stuff between the dashes (and including the dashes) with nothing "" and leave the text to the right of the second dash, which end in full stops. So, I would like the output to be (for example, based on the example above):
not this.
and
not this either.
Any help would be appreciated.
Thank you!
Maro
UPDATE:
Hi again everyone,
I'm just updating my original question again:
So what I had in my original data was these three examples (I tried to simplify in my original post above, but I think it might be helpful for you all to see what I was actually dealing with):
- # (Piano) - no, and neither can you.
- # (Piano) - uh-huh.
- # Many dreams ago - Try it again.
(numbers 1-3 are for the purposes of making things clearer, they are not part of the strings)
I was trying to find a way to delete all the stuff between and including the two dashes, and leave all the stuff after the second dash, so I wanted my output to be:
no, and neither can you.
uh-huh.
Try it again.
I ended up using this:
gsub(("-[[:blank:]]#[[:blank:]]\\(?[A-Z][a-z]*\\)?[[:blank:]]-", "", c)
which helped me get 1. and 2. in one go. But this didn't help with 3 - I thought by including the question mark after the open and close bracket (which I thought meant 'optional') this would help me get all three targets, but for some reason it didn't. To then get 3, I just ended up targeting that specific string i.e. - # Many dreams ago -, by using:
gsub(("- # Many dreams ago -"), "", c)
I'm new to this, so not the best solution I'm sure.
In my original post (this has been edited a couple of times) I included square brackets around the three strings, which explains some of the answers I originally received from members of the community. Apologies for the confusion!
Thanks everyone - if there's anything that doesn't make sense, please let me know, and I'll try to clarify.
Maro
If you want to stay in between the square brackets you can start the match at #, then use a negated character class [^][]* matching optional chars other than an opening or closing square bracket, and match the last -
Replace the match with an empty string.
c <- "[- # (piano) - not this.]"
sub("#[^][]*-", "", c)
Output
[1] "[- not this.]"
For a more specific match of that string format, you can match the whole line including the square brackets, the # and the string ending on a full stop, and capture what you want to keep.
In the replacement use the capture group value.
c <- c("[- # (piano) - not this.]", "[- # hello hello - not this either.]")
sub("\\[[^][#]*#[^][]*-\\s*([^][]*\\.)]", "\\1", c)
Output
[1] "not this." "not this either."
There is a vector of strings that looks like the following (text with two or more substrings in quotation marks):
vec <- 'ab"cd"efghi"j"kl"m"'
The text within the first pair of quotation marks (cd) contains a useful identifier (cd is the desired output). I have been studying how to use regular expressions but I haven't learned how to find the first and second occurrences of something like quotation marks.
Here's how I have been getting cd:
tmp <- strsplit(vec,split="")[[1]]
paste(tmp[(which(tmp=='\"')[1]+1):(which(tmp=='\"')[2]-1)],collapse="")
"cd"
My question is, is there another way to find "cd" using regular expressions? in order to learn more how to use them. I prefer base R solutions but will accept an answer using packages if that's the only way. Thanks for your help.
Match everything except " then capture everything upto next " and replace captured group by itself.
gsub( '[^"]*"([^"]*).*', '\\1', vec)
[1] "cd"
For detailed explanation of regex you can see this demo
I have been mucking around with regex strings and strsplit but can't figure out how to solve my problem.
I have a collection of html documents that will always contain the phrase "people own these". I want to extract the number immediately preceding this phrase. i.e. '732,234 people own these' - I'm hoping to capture the number 732,234 (including the comma, though I don't care if it's removed).
The number and phrase are always surrounded by a . I tried using Xpath but that seemed even harder than a regex expression. Any help or advice is greatly appreciated!
example string: >742,811 people own these<
-> 742,811
Could you please try following.
val <- "742,811 people own these"
gsub(' [a-zA-Z]+',"",val)
Output will be as follows.
[1] "742,811"
Explanation: using gsub(global substitution) function of R here. Putting condition here where it should replace all occurrences of space with small or capital alphabets with NULL for variable val.
Try using str_extract_all from the stringr library:
str_extract_all(data, "\\d{1,3}(?:,\\d{3})*(?:\\.\\d+)?(?= people own these)")
I have this vector Target <- c( "tes_1123_SS1G_340T01", "tes_23_SS2G_340T021". I want to remove anything before SS and anything after T0 (including T0).
Result I want in one line of code:
SS1G_340 SS2G_340
Code I have tried:
gsub("^.*?SS|\\T0", "", Target)
We can use str_extract
library(stringr)
str_extract(Target, "SS[^T]*")
#[1] "SS1G_340" "SS2G_340"
Try this:
gsub(".*(SS.*)T0.*","\\1",Target)
[1] "SS1G_340" "SS2G_340"
Why it works:
With regex, we can choose to keep a pattern and remove everything outside of that pattern with a two-step process. Step 1 is to put the pattern we'd like to keep in parentheses. Step 2 is to reference the number of the parentheses-bound pattern we'd like to keep, as sometimes we might have multiple parentheses-bound elements. See the example below for example:
gsub(".*(SS.*)+(T0.*)","\\1",Target)
[1] "SS1G_340" "SS2G_340"
Note that I've put the T0.* in parentheses this time, but we still get the correct answer because I've told gsub to return the first of the two parentheses-bound patterns. But now see what happens if I use \\2 instead:
gsub(".*(SS.*)+(T0.*)","\\2",Target)
[1] "T01" "T021"
The .* are wild cards by the way. If you'd like to learn more about using regex in R, here's a reference that can get you started.
I looked around both here and elsewhere, I found many similar questions but none which exactly answer mine. I need to clean up naming conventions, specifically replace/remove certain words and phrases from a specific column/variable, not the entire dataset. I am migrating from SPSS to R, I have an example of the code to do this in SPSS below, but I am not sure how to do it in R.
EG:
"Acadia Parish" --> "Acadia" (removes Parish and space before Parish)
"Fifth District" --> "Fifth" (removes District and space before District)
SPSS syntax:
COMPUTE county=REPLACE(county,' Parish','').
There are only a few instances of this issue in the column with 32,000 cases, and what needs replacing/removing varies and the cases can repeat (there are dozens of instances of a phrase containing 'Parish'), meaning it's much faster to code what needs to be removed/replaced, it's not as simple or clean as a regular expression to remove all spaces, all characters after a specific word or character, all special characters, etc. And it must include leading spaces.
I have looked at the replace() gsub() and other similar commands in R, but they all involve creating vectors, or it seems like they do. What I'd like is syntax that looks for characters I specify, which can include leading or trailing spaces, and replaces them with something I specify, which can include nothing at all, and if it does not find the specific characters, the case is unchanged.
Yes, I will end up repeating the same syntax many times, it's probably easier to create a vector but if possible I'd like to get the syntax I described, as there are other similar operations I need to do as well.
Thank you for looking.
> x <- c("Acadia Parish", "Fifth District")
> x2 <- gsub("^(\\w*).*$", "\\1", x)
> x2
[1] "Acadia" "Fifth"
Legend:
^ Start of pattern.
() Group (or token).
\w* One or more occurrences of word character more than 1 times.
.* one or more occurrences of any character except new line \n.
$ end of pattern.
\1 Returns group from regexp
Maybe I'm missing something but I don't see why you can't simply use conditionals in your regex expression, then trim out the annoying white space.
string <- c("Arcadia Parish", "Fifth District")
bad_words <- c("Parish", "District") # Write all the words you want removed here!
bad_regex <- paste(bad_words, collapse = "|")
trimws( sub(bad_regex, "", string) )
# [1] "Arcadia" "Fifth"
dataframename$varname <- gsub(" Parish","", dataframename$varname)