How do I remove an interpunct (aka interpoint, middle dot, middot) from a string? I am looking for something like trimws, but trimws doesn't work on the interpunct.
Cheers
I believe this is what you're looking for.
string <- c("· interpunct", "interpunct · interpunct", "interpunct · ")
#[1] "· interpunct" "interpunct · interpunct" "interpunct · "
sub("(?:\\s?)+·(?:\\s?)+", "", string)
#[1] "interpunct" "interpunctinterpunct" "interpunct"
Related
I have then following pattern Set(?:Value)? in R as follows:
grepl('Set(?:Value)?', 'Set(Value)', perl=T)
this pattern is macthed by
1- Set
2- Set Value
3- Set(Value)
But I want to match only for two first cases and for for third case. Can anybody help me?
Thank you
You can use
grepl('^Set(?:\\s+Value)?$', x)
grepl('\\bSet(?!\\(Value\\))(?:\\s+Value)?\\b', x, perl=TRUE)
See regex demo #1 and regex demo #2.
Details:
^Set(?:\\s+Value)?$ - start of string, Set, an optional sequence of one or more whitespaces (\s+) and a Value and then end of string
\bSet(?!\(Value\))(?:\s+Value)?\b:
\b - word boundary
Set - Set string
(?!\(Value\)) - no (Value) string allowed at this very location
(?:\s+Value)? - an optional sequence of one or more whitespaces (\s+) and a Value
\b - word boundary
See an R demo:
x <- c("Set", "Set Value", "Set(Value)")
grep('^Set(?:\\s+Value)?$', x, value=TRUE)
## => [1] "Set" "Set Value"
grep('\\bSet(?!\\(Value\\))(?:\\s+Value)?\\b', x, perl=TRUE, value=TRUE)
## => [1] "Set" "Set Value"
My string patterns looks like this:
UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870 and I am trying to extract everything after the second last +, i.e. 180101:0050+10870.
Thus far, I managed to address the second last block 180101:0050 with this expression (?<=\+)[^\+]+(?=\+[^\+]*$) but fail to include the last block including the last +. Here is my sample: regex101
The expression is meant for R and I still need to escape the characters later on. This format it just for testing purposes in Regex101.
We could capture group based on the occurrence of + from the end ($) of the string.
sub(".*\\+([^+]+\\+[^+]+$)", "\\1", str1)
#[1] "180101:0050+10870"
data
str1 <- "UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870"
You may use
\+\K[^+]+\+[^+]*$
Or, if you would like to use it with stringr::str_extract:
(?<=\+)[^+]+\+[^+]*$
See the regex demo. Details:
\+ - a + char
\K - match reset operator
(?<=\+) - location right after a + symbol
[^+]+ - one or more chars other than +
\+ - a +
[^+]+ - one or more chars other than +
$ - end of string.
See R demo online:
x <- "UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870"
regmatches(x, regexpr("\\+\\K[^+]+\\+[^+]*$", x, perl=TRUE))
## => [1] "180101:0050+10870"
library(stringr)
str_extract(x, "(?<=\\+)[^+]+\\+[^+]*$")
## => [1] "180101:0050+10870"
Another way you can do in this case:
library(stringr)
str_extract("UNB+UNOC:3+4399945681577+_GLN_Company__+180101:0050+10870", "\\d+:\\d+\\+\\d+")
#"180101:0050+10870"
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
I want to combine words in one string having spaces in between, which are similar to words in another string without spaces in between them (In R).
eg
s1 = 'this is an example of an undivided string case here'
s2 = 'Please note th is is an un di vid ed case right he r e for you!'
s2 needs to be converted into
s2 = 'Please note this is an undivided case right here for you!'
based on combined words in s1 which are same as non combined successive/continuous words in s2(with spaces in between)
I am new to R and tried with gsub, and different combinations of '\s', but not able to get the desired result.
You may achieve what you need by
removing all whitespaces from the string you want to search for (s1) (with gsub("\\s+", "", x)), then
insert whitespace patterns (\s*) in between each char (use something like sapply(strsplit(unspace(s1), ""), paste, collapse="\\s*")), and then
replace all the matches with the replacement with gsub(pattern, s1, s2).
See the R demo:
s2 = 'Please note th is is an un di vid ed case right he r e for you!'
s1 = 'this is an undivided case right here'
unspace <- function(x) { gsub("\\s+", "", x) }
pattern <- sapply(strsplit(unspace(s1), ""), paste, collapse="\\s*")
gsub(pattern, s1, s2)
## => [1] "Please note this is an undivided case right here for you!"
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"