Regex - Substitute character in a matching substring - r

Let's say I have the following string:
input = "askl jmsp wiqp;THIS IS A MATCH; dlkasl das, fm"
I need to replace the white-spaces with underscores, but only in the substrings that match a pattern. (In this case the pattern would be a semi-colon before and after.)
The expected output should be:
output = "askl jmsp wiqp;THIS_IS_A_MATCH; dlkasl das, fm"
Any ideas how to achieve that, preferably using regular expressions, and without splitting the string?
I tried:
gsub("(.*);(.*);(.*)", "\\2", input) # Pattern matching and
gsub(" ", "_", input) # Naive gsub
Couldn't put them both together though.

Regarding the original question:
Substitute character in a matching substring
You may do it easily with gsubfn:
> library(gsubfn)
> input = "askl jmsp wiqp;THIS IS A MATCH; dlkasl das, fm"
> gsubfn(";([^;]+);", function(g1) paste0(";",gsub(" ", "-", g1, fixed=TRUE),";"), input)
[1] "askl jmsp wiqp;THIS-IS-A-MATCH; dlkasl das, fm"
The ;([^;]+); matches any string starting with ; and up to the next ; capturing the text in-between and then replacing the whitespaces with hyphens only inside the captured part.
Another approach is to use a PCRE regex with a \G based regex with gsub:
p = "(?:\\G(?!\\A)|;)(?=[^;]*;)[^;\\s]*\\K\\s"
> gsub(p, "-", input, perl=TRUE)
[1] "askl jmsp wiqp;THIS-IS-A-MATCH; dlkasl das, fm"
See the online regex demo
Pattern details:
(?:\\G(?!\\A)|;) - a custom boundary: either the end of the previous successful match (\\G(?!\\A)) or (|) a semicolon
(?=[^;]*;) - a lookahead check: there must be a ; after 0+ chars other than ;
[^;\\s]* - 0+ chars other than ; and whitespaces
\\K - omitting the text matched so far
\\s - 1 single whitespace character (if multiple whitespaces are to be replaced with 1 hyphen, add + after it).

Related

cleaning phonenumbers using regex

I have the following composition of phonenumbers where 33 is the area code:
+331234567
+3301234567
00331234567
003301234567
0331234567
033-123-456-7
0033.1234567
where Im expecting only 331234567
What I have tried to clean those numbers using R
R::tidyverse::str_replace_all(c("+331234567", "033-123-456-7", "0033.1234567"), pattern = "[^0-9.]", replacement = "") removing non-numeric characters
R::tidyverse::str_replace_all("0331234567", pattern = "^0", replacement = "") removing the leading 0
R::tidyverse::str_replace_all("00331234567", pattern = "^00", replacement = "") removing the leading 00
my question is how to remove the zeros in between: 3301234567 or 003301234567 or +3301234567 or 03301234567
Appreciate any help
You can use
gsub("^(?:00?|\\+)330?|\\W", "", x, perl=TRUE)
See the regex demo. See the R demo online.
If there can be more 0s after 33 before the number you need to extract, replace 0? with 0*.
Details
^ - start of string
(?:00?|\+) - 00, 0 or +
330? - 33 or 330
| - or
\W - any non-word char.
You can use ^\+?0*3*0*|[^\s\d]
Pattern explanation:
^ - match beginning of the string
\+? - match + literally, zero or one time.
0* - match zero or more 0
3* - match zero or more 3
| - alternation
[^\s\d] - negated character class - match any character other from whitespace and digit (you could remove \s if you handle one number at a time, it just prevents from matching newline in demo)
Regex demo
It will match unwanted parts separately. First part will clean beginning of a number if it starts with + or 0, second part will clean non-digits inside the number.

Extracting matches from strings with lookaround in R

I have textual data (storytellings) and my aim is to extract certain words that are defined by a co-occurrence pattern, namely that they occur immediately prior to overlap, which is indicated by square brackets. The data are like this:
who <- c("Sue:", NA, "Carl:", "Sue:", NA, NA, NA, "Carl:", "Sue:","Carl:", "Sue:","Carl:")
story <- c("That’s like your grand:ma. did that with::=erm ",
"with Ju:ne (.) once or [ twice.] ",
" [ Yeah. ] ",
"And June wanted to go out and yo- your granny said (0.8)",
"“make sure you're ba(hh)ck before midni(hh)ght.” ",
"[Mm.] ",
"[There] she was (.) a ma(h)rried woman with a(h)- ",
"She’s a right wally. ",
"mm [kids as well ] ",
" [They assume] an awful lot man¿ ",
"°°ye:ah,°° ",
"°°the elderly do.°° ")
CAt <- data.frame(who, story)
Now, defining the pattern:
pattern <- "\\w.*\\s\\[[^]].*]"
and using grep():
grep(pattern, CAt$story, value = T)
[1] "with Ju:ne (.) once or [ twice.] "
[2] "mm [kids as well ] "
I get the two strings that contain the target matches but what I'm really after are the target words only, in this case the words "or" and "mm". This, to me, seems to call for positive lookahead. So I redefined the pattern thus:
pattern <- "\\w.*(?=\\s\\[[^]].*])"
which says something along the lines: "match the word iff you see a space followed by square brackets with some content on the right of that word". Now to extract only the exact matches, I normally use this code, which works fine as long as no lookaround is involved, but here it throws an error:
unlist(regmatches(CAt$story, gregexpr(pattern, CAt$story)))
Error in gregexpr(pattern, CAt$story) :
invalid regular expression, reason 'Invalid regexp'
Why is this? And how can the exact matches be extracted?
In your code, you could add perl=TRUE to gregexpr.
In your pattern \w.* will match a single word char followed by matching any char 0+ times.
This part \[[^]].*] will match [, then 1 char which is not ] and then .* which will match any char 0+ times followed by ].
You could update your pattern to repeating the word char and the character class itself instead.
\w+(?=\s\[[^]]*])
Explanation
\w+ Match 1+ word chars
(?= Positive lookahead, assert what is directly to the right is
\s Match single whitespace char
\[[^]]*] Match from opening[ to closing ] using a negated character class
) Close positive lookahead
Regex demo
Using doubled backslashes:
\\w+(?=\\s\\[[^]]*])
As an alternative you could use a capturing group instead of using a lookahead
(\w+)\s\[[^]]*]
Regex demo

Extract substring between two colons / special characters

I want to extract "SUBSTRING" with sub() from the following string:
attribute <- "S4Q7b1_t1_r1: SUBSTRING: some explanation: some explanation - ..."
I used the following code, but unfortunately it didn't work:
sub(".*: (.*) : .*", "\\1", attribute)
Does anyone know an answer for that?
You may use
sub("^[^:]*: ([^:]*).*", "\\1", attribute)
See the regex demo
You need to rely on negated character classes, [^:] that matches any char but :, since .* matches greedily any 0 or more chars. Also, your pattern contains a space before : and it is missing in the string.
Details
^ - start of string
[^:]* - any 0+ chars other than :
: - a colon with a space
-([^:]*) - Capturing group 1 (\1 refers to this value): any 0+ chars other than :
.* - the rest of the string.
R Demo:
attribute <- "S4Q7b1_t1_r1: SUBSTRING: some explanation: some explanation - ..."
sub("^[^:]*: ([^:]*).*", "\\1", attribute)
## => [1] "SUBSTRING"

Extract digits after matching the certain string second time

I want to extract the digits after second occurance of under score _ from a pattern.
by following the similar posts here
Matching different digits after a lookahead
regex - return all before the second occurrence
I tried
library(stringr)
pattern <- c("1/2/3_500k/855kk_1400k/AVBB")
str_extract(pattern, "(^_){2}(\\d+\\.*\\d*)")
which outputs
[1] NA
instead of 1400. Could you help?
You may use a base R solution with regexpr/regmatches:
regmatches(x, regexpr("^(?:[^_]*_){2}[^_0-9]*\\K\\d*\\.?\\d+", x, perl=TRUE))
Or, with sub:
sub("^(?:[^_]*_){2}[^_0-9]*(\\d*\\.?\\d+).*", "\\1", x)
See the R demo online.
The regex is
^(?:[^_]*_){2}[^_0-9]*\K\d*\.?\d+
See the online regex demo.
Details
^ - start of string
(?:[^_]*_){2} - 2 repetitions of
[^_]* - any 0+ chars other than _
_ - an underscore
[^_0-9]* - any 0+ chars other than _ and digits
\K - match reset operator discarding all text matched so far
\d*\.?\d+ - a float or integer number pattern (0+ digits, an optional . and then 1+ digits).
In the sub regex variation, the \K is not necessary, the number pattern is captured into a capturing group and the rest of string is matched with .* pattern. The result is the contents of Group 1, referred to with the \1 placeholder.
One option could be as:
pattern <- c("1/2/3_500k/855kk_1400k/AVBB")
sub(".*_*_(\\d+).*","\\1", pattern, perl = TRUE)
[1] "1400"
The regex is:
".*_*_(\\d+).*"
Details:
.*_ anything before first _
.*_ anything after first _ and before 2nd _
\\d+ look for digits and take those as selection.
.* anything afterwards.
\\1 replaces matching strings with values found for 1st group.

Replace a specific character only between parenthesis

Lest's say I have a string:
test <- "(pop+corn)-bread+salt"
I want to replace the plus sign that is only between parenthesis by '|', so I get:
"(pop|corn)-bread+salt"
I tried:
gsub("([+])","\\|",test)
But it replaces all the plus signs of the string (obviously)
If you want to replace all + symbols that are inside parentheses (if there may be 1 or more), you can use any of the following solutions:
gsub("\\+(?=[^()]*\\))", "|", x, perl=TRUE)
See the regex demo. Here, the + is only matched when it is followed with any 0+ chars other than ( and ) (with [^()]*) and then a ). It is only good if the input is well-formed and there is no nested parentheses as it does not check if there was a starting (.
gsub("(?:\\G(?!^)|\\()[^()]*?\\K\\+", "|", x, perl=TRUE)
This is a safer solution since it starts matching + only if there was a starting (. See the regex demo. In this pattern, (?:\G(?!^)|\() matches the end of the previous match (\G(?!^)) or (|) a (, then [^()]*? matches any 0+ chars other than ( and ) chars, and then \K discards all the matched text and \+ matches a + that will be consumed and replaced. It still does not handle nested parentheses.
Also, see an online R demo for the above two solutions.
library(gsubfn)
s <- "(pop(+corn)+unicorn)-bread+salt+malt"
gsubfn("\\((?:[^()]++|(?R))*\\)", ~ gsub("+", "|", m, fixed=TRUE), s, perl=TRUE, backref=0)
## => [1] "(pop(|corn)|unicorn)-bread+salt+malt"
This solves the problem of matching nested parentheses, but requires the gsubfn package. See another regex demo. See this regex description here.
Note that in case you do not have to match nested parentheses, you may use "\\([^()]*\\)" regex with the gsubfn code above. \([^()]*\) regex matches (, then any zero or more chars other than ( and ) (replace with [^)]* to match )) and then a ).
We can try
sub("(\\([^+]+)\\+","\\1|", test)
#[1] "(pop|corn)-bread+salt"

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