I'm trying to eliminate apostrophes from my a column of words in my dataset using
str_replace(tidy_posts$word, "'", "")
But the vector it returns still contains apostrophes. The vector's class is character, so I can't understand why this won't work. My only guess is that because the words came from data I got from the reddit API, the encoding is funky or something.
edit: "[:punct:]" also doesn't work.
When you are having some "special" punctuation such as . or ' a way to select them in regex is by using \\ before the punctuation of interest. This example may help you:
ch = c("The vector's class is character, so I can't understand why this won't work")
> stringr::str_replace_all(ch, "\\'","")
[1] "The vectors class is character, so I cant understand why this wont work"
Related
I'm working with the following code:
Y_Columns <- c("Y.1.1")
paste('{"ImportId":"', Y_Columns, '"}', sep = "")
The paste function produces the following output:
"{\"ImportId\":\"Y.1.1\"}"
How do I get the paste function to omit the \? Such that, the output is:
"{"ImportId":"Y.1.1"}"
Thank you for your help.
Note: I did do a search on SO to see if there were any Q's that asked "what is an escape character in R". But I didn't review all the 160 answers, only the first 20.
This is one way of demonstrating what I wrote in my comment:
out <- paste('{"ImportId":"', Y_Columns, '"}', sep = "")
out
#[1] "{\"ImportId\":\"Y.1.1\"}"
?print
print(out,quote=FALSE)
#[1] {"ImportId":"Y.1.1"}
Both R and regex patterns use escape characters to allow special characters to be displayed in print output or input. (And sometimes regex patterns need to have doubled escapes.) R has a few characters that need to be "escaped" in certain situation. You illustrated one such situation: including double-quote character inside a result that will be printed with surrounding double-quotes. If you were intending to include any single quotes inside a character value that was delimited by single quotes at the time of creation, they would have needed to be escaped as well.
out2 <- '\'quoted\''
nchar(out2)
#[1] 8 ... note that neither the surround single-quotes nor the backslashes get counted
> out2
[1] "'quoted'" ... and the default output quote-char is a double-quote.
Here's a good Q&A to review:How to replace '+' using gsub() function in R
It has two answers, both useful: one shows how to double escape a special character and the other shows how to use teh fixed argument to get around that requirement.
And another potentially useful Q&A on the topic of handling Windows paths:
File path issues in R using Windows ("Hex digits in character string" error)
And some further useful reading suggestions: Look at the series of help pages that start with capital letters. (Since I can never remember which one has which nugget of essential information, I tried ?Syntax first and it has a "See Also" list of essential reading: Arithmetic, Comparison, Control, Extract, Logic, NumericConstants, Paren, Quotes, Reserved. and I then realized what I wanted to refer you to was most likely ?Quotes where all the R-specific escape sequence letters should be listed.
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 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)
When doing some textual data cleaning in R, I can found some special characters. In order to get rid of them, I have to know their unicodes, for example € is \u20AC. I would like to know if it is possible "see" the unicodes with a function that take into account the string within the special character as an input?
Refering to Cath comment, iconv can do the job :
iconv("é", toRaw = TRUE)
Then, you may want to unlist and paste with \u00.
special_char <- "%"
Unicode::as.u_char(utf8ToInt(special_char))
I've been using gsub("toreplace","replacement", myvector) to clean out data in R. While this works for commas and the like, removing "$" has no effect. So if I do gsub("$","",myvector) all the dollar signs remain in place.
I think this is because $ is a special character in R. I tried escaping it "\$" but that yields the same result (no effect). And I couldn't find a resource on escaping special characters in R.
Obviously I should do this in preprocessing. But I was wondering if anyone out there knew how to either a) escape special characters in R b) get rid of pesky $ in R directly. For science.
You have to escape it twice, first for R, second for the regex.
gsub('\\$', '', c("a$a", "bb$"))
[1] "aa" "bb"
See ?Quotes for details on quoting and escaping.
Use fixed = TRUE:
gsub('$', '', c("a$a", "bb$"), fixed = TRUE)
Then you don't need to worry about any special characters. In stringr, this is implemented a little differently:
library(stringr)
str_replace_all(c("$100","ta$ty"), fixed("$"), "")
Thanks to DiggyF and James for the examples!
Escaping characters can be a pain some times, but just putting it in square brackets (make it a character class) helps with this:
> gsub("[$]","",c("$100","ta$ty"))
[1] "100" "taty"
if you have $ followed by number in set of data columns (e.g. $400,000) there is an easier way that worked like charm for me.
data%>%
mutate_at(5:6, parse_number)
where 5:6 are the data column numbers.