This may be a silly question, but its doing my head in. I have always used gsub, but for some reason, it isnt working for this one:
Example of dataset
ColumnS
I 2,[3],4:i:-
I 2,[3],4:i:-
I 2,[3],4:b:-
Give
Derby
Panama
Kentucky
This is what I've been trying
dataset$ColumnS<-gsub("I 2,[3],4","2,[3],4", dataset$ColumnS)
What is wrong with it?
Square brackets are special characters when it comes to pattern recognition and if you want to match them, you have to use escape characters to inform R.
dataset$ColumnS <- gsub("I 2,\\[3\\],4","2,[3],4", dataset$ColumnS)
You can also use the argument fixed=TRUE which will take the pattern as a string.
dataset$ColumnS <- gsub("I 2,[3],4","2,[3],4", dataset$ColumnS, fixed=TRUE)
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 a column within a data frame with a series of identifiers in, a letter and 8 numbers, i.e. B15006788.
Is there a way to remove all instances of B15.... to make them empty cells (there’s thousands of variations of numbers within each category) but keep B16.... etc?
I know if there was just one thing I wanted to remove, like the B15, I could do;
sub(“B15”, ””, df$col)
But I’m not sure on the how to remove a set number of characters/numbers (or even all subsequent characters after B15).
Thanks in advance :)
Welcome to SO! This is a case of regex. You can use base R as I show here or look into the stringR package for handy tools that are easier to understand. You can also look for regex rules to help define what you want to look for. For what you ask you can use the following code example to help:
testStrings <- c("KEEPB15", "KEEPB15A", "KEEPB15ABCDE")
gsub("B15.{2}", "", testStrings)
gsub is the base R function to replace a pattern with something else in one or a series of inputs. To test our regex I created the testStrings vector for different examples.
Breaking down the regex code, "B15" is the pattern you're specifically looking for. The "." means any character and the "{2}" is saying what range of any character we want to grab after "B15". You can change it as you need. If you want to remove everything after "B15". replace the pattern with "B15.". the "" means everything till the end.
edit: If you want to specify that "B15" must be at the start of the string, you can add "^" to the start of the pattern as so: "^B15.{2}"
https://www.rstudio.com/wp-content/uploads/2016/09/RegExCheatsheet.pdf has a info on different regex's you can make to be more particular.
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)