I want to replace DM*13:01:01:02 with DM*13:01:01:01. However my script also changes DM*11:01:01:01, DM*03:01:01:01, DM*01:01:01:01 to DM*13:01:01:01. I do not want these to be changed
The script I use:
> papST$DM_c1 <-gsub("[DM*]\\d[13][:]\\d[01][:]\\d[01][:]\\d[02]", "*13:01:01:01", papST$DM_o1, perl = TRUE)
Based on the examples you have given, you don't really need to use any fancy regex features to do the specific replacement you have mentioned. The only thing you need to include in your pattern is a backslash so that * doesn't get treated as a special character:
x = c("DM*13:01:01:02", "DM*11:01:01:01", "DM*03:01:01:01", "DM*01:01:01:01")
gsub("DM\\*13:01:01:02", "DM*13:01:01:01", x)
If there are more values that need replacing, like you want to replace all values ending in 02, then you may need to bring in some of the "pattern matching" features in regular expressions, but it's important not to overcomplicate things.
For reference, to replace all 02s at the end of your strings, you could use a simple regex that uses $, which matches at the end of a string:
gsub("02$", "01", x)
Related
I have column names in the following format:
col= c('UserLanguage','Q48','Q21...20','Q22...21',"Q22_4_TEXT...202")
I would like to get the column names without everything that is after ...
[1] "UserLanguage" "Q48" "Q21" "Q22" "Q22_4_TEXT"
I am not sure how to code it. I found this post here but I am not sure how to specify the pattern in my case.
You can use gsub.
gsub("\\...*","",col)
#[1] "UserLanguage" "Q48" "Q21" "Q22" "Q22_4_TEXT"
Or you can use stringr
library(stringr)
str_remove(col, "\\...*")
Since . matches any character, we need to "escape" (\) to specify exactly what we want to match in the regular expression (and not use the special behavior of the .). So, to match a period, we would need \.. However, the backslash (\) is used to escape special behavior (e.g., escape symbol in strings) in regexps. So, to create the regular expression, we need an additional backslash, \\. In this case, we want to match additional periods, so we can add those here, hence \\.... Then, * specifies that the previous expression (everything the three periods) may occur 0 or more times.
You could sub and capture the first word in each column:
col <- c("UserLanguage", "Q48", "Q21...20", "Q22...21", "Q22_4_TEXT...202")
sub("^(\\w+).*$", "\\1", col)
[1] "UserLanguage" "Q48" "Q21" "Q22" "Q22_4_TEXT"
The regex pattern used here says to match:
^ from the start of the input
(\w+) match AND capture the first word
.* then consume the rest
$ end of the input
Then, using sub we replace with \1 to retain just the first word.
Trying to code up a Regex in R to match everything before the first occurrence of a colon.
Let's say I have:
time = "12:05:41"
I'm trying to extract just the 12. My strategy was to do something like this:
grep(".+?(?=:)", time, value = TRUE)
But I'm getting the error that it's an invalid Regex. Thoughts?
Your regex seems fine in my opinion, I don't think you should use grep, also you are missing perl=TRUE that is why you are getting the error.
I would recommend using :
stringr::str_extract( time, "\\d+?(?=:)")
grep is little different than it is being used here, its good for matching separate values and filtering out those which has similar pattern, but you can't pluck out values within a string using grep.
If you want to use Base R you can also go for sub:
sub("^(\\d+?)(?=:)(.*)$","\\1",time, perl=TRUE)
Also, you may split the string using strsplit and filter out the first string like below:
strsplit(time, ":")[[1]][1]
I have the following string
"31032017"
and I want to use regular expressions in R to get
"31.03.2017"
What is the best function to do it?
And a general question, how can I repeat the matched part, like as in sed in bash? There, we use \1 to repeat the first matched part.
You need to put the single parts in round brackets like this:
sub("([0-9]{2})([0-9]{2})([0-9]{4})", "\\1.\\2.\\3", "31032017")
You can then use \\1 to access the part matched by the first group, \\2 for the second and so on.
Note that if your string is a date, there are better ways to parse / reformat it than directly using regex.
date_vector = c("31032017","28052017","04052022")
as.character(format(as.Date(date_vector, format = "%d%m%Y"), format = "%d.%m.%Y"))
#[1] "31.03.2017" "28.05.2017" "04.05.2022"
If you want to work/do math with dates, omit as.character.
I would like to split strings like the following:
x <- "abc-1230-xyz-[def-ghu-jkl---]-[adsasa7asda12]-s-[klas-bst-asdas foo]"
by dash (-) on the condition that those dashes must not be contained inside a pair of []. The expected result would be
c("abc", "1230", "xyz", "[def-ghu-jkl---]", "[adsasa7asda12]", "s",
"[klas-bst-asdas foo]")
Notes:
There is no nesting of square brackets inside each other.
The square brackets can contain any characters / numbers / symbols except square brackets.
The other parts of the string are also variable so that we can only assume that we split by - whenever it's not inside [].
There's a similar question for python (How to split a string by commas positioned outside of parenthesis?) but I haven't yet been able to accurately adjust that to my scenario.
You could use look ahead to verify that there is no ] following sooner than a [:
-(?![^[]*\])
So in R:
strsplit(x, "-(?![^[]*\\])", perl=TRUE)
Explanation:
-: match the hyphen
(?! ): negative look ahead: if that part is found after the previously matched hyphen, it invalidates the match of the hyphen.
[^[]: match any character that is not a [
*: match any number of the previous
\]: match a literal ]. If this matches, it means we found a ] before finding a [. As all this happens in a negative look ahead, a match here means the hyphen is not a match. Note that a ] is a special character in regular expressions, so it must be escaped with a backslash (although it does work without escape, as the engine knows there is no matching [ preceding it -- but I prefer to be clear about it being a literal). And as backslashes have a special meaning in string literals (they also denote an escape), that backslash itself must be escaped again in this string, so it appears as \\].
Instead of splitting, extract the parts:
library(stringr)
str_extract_all(x, "(\\[[^\\[]*\\]|[^-])+")
I am not familiar with r language, but I believe it can do regex based search and replace. Instead of struggling with one single regex split function, I would go in 3 steps:
replace - in all [....] parts by a invisible char, like \x99
split by -
for each element in the above split result(array/list), replace \x99 back to -
For the first step, you can find the parts by \[[^]]
I need to find attribute values in an ASPX file using regular expressions.
That means you don't need to worry about malformed HTML or any HTML related issues.
I need to find the value of a particular attribute (LocText). I want to get what's inside the quotes.
Any ASPX tags such as <%=, <%#, <%$ etc. inside the value don't make sense for this attribute therefore are considered as part of it.
The regex I began with looks like this:
LocText="([^"]+)"
This works great, the first group, which is the result text, gets everything except the double quotes, which are not allowed there (" ; must be used instead)
But the ASPX file allows using of single quotes - second regular expression must be applied then.
LocText='([^']+)'
I could use these two regular expressions but I'm looking for a way to connect them.
LocText=("([^"]+)"|'([^']+)')
This also works but doesn't seem very efficient as it's creating unnecessary number of groups. I think this could be somehow done by using backreferences, but I can't get it to work.
LocText=(["']{1})([^\1]+)\1
I thought that by this, I save the single/double quote to the first group and then I tell it to read anything that is NOT the char found in the first group. This is enclosed again by the quote from the first group. Obviously, I'm wrong and it's not working like that.
Is there any way, how to connect the first two expressions together creating just a minimum amount of groups with one group being the value of the attribute I want to get? Is it possible using a backreference for the single/double quote value, or have I completely misunderstood the meaning of them?
I'd say your solution with alternation isn't that bad, but you could use named captures so the result will always be found in the same group's value:
Regex regexObj = new Regex(#"LocText=(?:""(?<attr>[^""]+)""|'(?<attr>[^']+)')");
resultString = regexObj.Match(subjectString).Groups["attr"].Value;
Explanation:
LocText= # Match LocText=
(?: # Either match
"(?<attr>[^"]+)" # "...", capture in named group <attr>
| # or match
'(?<attr>[^']+)' # '...', also capture in named group <attr>
) # End of alternation
Another option would be to use lookahead assertions ([^\1] isn't working because you can't place backreferences inside a character class, but you can use them in lookarounds):
Regex regexObj = new Regex(#"LocText=([""'])((?:(?!\1).)*)\1");
resultString = regexObj.Match(subjectString).Groups[2].Value;
Explanation:
LocText= # Match LocText=
(["']) # Match and capture (group 1) " or '
( # Match and capture (group 2)...
(?: # Try to match...
(?!\1) # (unless it's the quote character we matched before)
. # any character
)* # repeat any number of times
) # End of capturing group 2
\1 # Match the previous quote character