I am finding sub string in a string repeating thrice consecutively and removing the obtained sub string from it using gregexpr. However, in my attempt to find sub strings I need to remove lookahead. For example, consider a string kajaaaaaaaaaaaa, here aaaa is outputting along with aaa, aa and a. Since the last three are included in aaaa how can I get rid of them? I have tried a lot but have been unable to do it. I want to capture a sub string repeating itself consecutively for atleast thrice in a string.
s <- 'kajaaaaaaaaaaaa'
m <- gregexpr(sprintf'(?=(.{2,})\\1{2,})',t) s, perl=TRUE)
unique(mapply(function(x, y) substr(s, x, x+y-1),
attr(m[[1]], 'capture.start'),
attr(m[[1]], 'capture.length')))
If I understand your regex correctly:
m <- gregexpr('(.)(?=(\1{3}))', s, perl=TRUE)
which will match anything repeating three times after the original
The result will be two match groups one for a and one for "aaa" use the latter, as you have to still have a match group to find repeats
I need to remove lookahead.
Just omit it, lookahead is not needed here:
> gregexpr('(..+)\\1{2,}', s, perl=TRUE) -> m
> mapply(function(x, y) substr(s, x, x+y-1), attr(m[[1]], 'capture.start')
+ , attr(m[[1]], 'capture.length'))
[1] "aaaa"
Related
I have been trying to extract a portion of string after the occurrence of a first ^ sign. For example, the string looks like abc^28092015^def^1234. I need to extract 28092015 sandwiched between the 1st two ^ signs.
So, I need to extract 8 characters from the occurrence of the 1st ^ sign. I have been trying to extract the position of the first ^ sign and then use it as an argument in the substr function.
I tried to use this:
x=abc^28092015^def^1234 `rev(gregexpr("\\^", x)[[1]])[1]`
Referring the answer discussed here.
But it continues to return the last position. Can anyone please help me out?
I would use sub.
x <- "^28092015^def^1234"
sub("^.*?\\^(.*?)\\^.*", "\\1", x)
# [1] "28092015"
Since ^ is a special char in regex, you need to escape that in-order to match literal ^ symbols.
or
Do splitting on ^ and get the value of second index.
strsplit(x,"^", fixed=-T)[[1]][2]
# [1] "28092015"
or
You may use gsub aslo.
gsub("^.*?\\^|\\^.*", "", x, perl=T)
# [1] "28092015"
Here's one option with base R:
x <- "abc^28092015^def^1234"
m <- regexpr("(?<=\\^)(.+?)(?=\\^)", x, perl = TRUE)
##
R> regmatches(x, m)
#[1] "28092015"
Another option is stri_extract_first from library(stringi)
library(stringi)
stri_extract_first_regex(str1, '(?<=\\^)\\d+(?=\\^)')
#[1] "28092015"
If it is any character between two ^
stri_extract(str1, regex='(?<=\\^)[^^]+')
#[1] "28092015"
data
str1 <- 'abc^28092015^def^1234'
x <- 'abc^28092015^def^1234'
library(qdapRegex)
unlist(rm_between(x, '^', '^', extract=TRUE))[1]
# [1] "28092015"
It would be better if you split it using ^. But if you still want the pattern, you can try this.
^\S+\^(\d+)(?=\^)
Then match group 1.
OUTPUT
28092015
See DEMO
I have a vector with strings like:
x <-c('kjsdf_class-X1(z)20_sample-318TT1X.3','kjjwer_class-Z3(z)29_sample-318TT2X.4')
I wanted to use regular expressions to get what is between substrings 'class-' and '_sample' (such as 'X1(z)20' and 'Z3(z)29' in x), and thought the lookaround regex ((?=...), (?!...),... and so) would do it. Cannot get it to work though!
Sorry if this is similar to other SO questions eg here or here).
This is a bit different then what you had in mind, but it will do the job.
gsub("(.*class-)|(.)|(_sample.*)", "\\2", x)
The logic is the following, you have 3 "sets" of strings:
1) characters .* ending in class-
2) characters .
3) Characters starting with _sample and characters afterwords .*
From those you want to keep the second "set" \\2.
Or another maybe easier to understand:
gsub("(.*class-)|(_sample.*)", "", x)
Take any number of characters that end in class- and the string _sample followed by any number of characters, and substitute them with the NULL character ""
We could use str_extract_all from library(stringr)
library(stringr)
unlist(str_extract_all(x, '(?<=class-)[^_]+(?=_sample)'))
#[1] "X1(z)20" "Z3(z)29"
This should also work if there are multiple instances of the pattern within a string
x1 <- paste(x, x)
str_extract_all(x1, '(?<=class-)[^_]+(?=_sample)')
#[[1]]
#[1] "X1(z)20" "X1(z)20"
#[[2]]
#[1] "Z3(z)29" "Z3(z)29"
Basically, we are matching the characters that are between the two lookarounds ((?<=class-) and (?=_sample)). We extract characters that is not a _ (based on the example) preceded by class- and succeded by _sample.
gsub('.*-([^-]+)_.*','\\1',x)
[1] "X1(z)20" "Z3(z)29"
Edit: Changing the whole question to make it clearer.
Can I remove a single character from one of the regular expression classes in R (such as [:alnum:])?
For example, match all punctuation ([:punct:]) except the _ character.
I am trying the replace underscores used in markdown for italicizing but the italicized substring may contain a single underscore which I would want to keep.
Edit: As another example, I want to capture everything between pairs of underscores (note one pair contains a single underscore that I want to keep between 1 and 10)
This is _a random_ string with _underscores: rate 1_10 please_
You won't believe it, but lazy matching achieved with a mere ? works as expected here:
str <- 'This is a _string with_ some _random underscores_ in it.'
gsub("_+([[:print:]]+?)_+", "\\1", str)
str <- 'This is a _random string with_ a scale of 1_10.'
gsub("_+([[:print:]]+?)_+", "\\1", str)
Result:
[1] "This is a string with some random underscores in it."
[1] "This is a random string with a scale of 1_10."
Here is the demo program
However, if you want to modify the [[:print:]] class, mind it is basically a [\x20-\x7E] range. The underscore being \x5F, you can easily exclude it from the range, and use [\x20-\x5E\x60-\x7E].
str <- 'This is a _string with_ some _random underscores_ in it.'
gsub("_+([\x20-\x5E\x60-\x7E]+)_+", "\\1", str)
Returns
[1] "This is a string with some random underscores in it."
Similar to #stribizhev:
x <- "This is _a random_ string with _underscores: rate 1_10 please_"
gsub("\\b_(.*?)_\\b", "\\1", x, perl=T)
produces:
[1] "This is a random string with underscores: rate 1_10 please"
Here we use word boundaries and lazy matching. Note that the default regexp engine has issues with lazy repetition and capture groups, so you may want to use perl=T
gsub('(?<=\\D)\\_(?=\\D|$)','',str,perl=T)
I need to match any 'r' that is preceded by two different vowels. For example, 'our' or 'pear' would be matching but 'bar' or 'aar' wouldn't. I did manage to match for the two different vowels, but I still can't make that the condition (...) of lookbehind for the ensuing 'r'. Neither (?<=...)r nor ...\\Kr yields any results. Any ideas?
x <- c('([aeiou])(?!\\1)(?=(?1))')
y <- c('our','pear','bar','aar')
y[grepl(paste0(x,collapse=''),y,perl=T)]
## [1] "our" "pear"`
These two solutions seem to work:
the why not way:
x <- '(?<=a[eiou]|e[aiou]|i[aeou]|o[aeiu]|u[aeio])r'
y[grepl(x, y, perl=T)]
the \K way:
x <- '([aeiou])(?!\\1)[aeiou]\\Kr'
y[grepl(x, y, perl=T)]
The why not way variant (may be more efficient because it searches the "r" before):
x <- 'r(?<=a[eiou]r|e[aiou]r|i[aeou]r|o[aeiu]r|u[aeio]r)'
or to quickly exclude "r" not preceded by two vowels (without to test the whole alternation)
x <- 'r(?<=[aeiou][aeiou]r)(?<=a[eiou]r|e[aiou]r|i[aeou]r|o[aeiu]r|u[aeio]r)'
As HamZa points out in the comments using skip and fail verbs is one way to do what we want. Basically we tell it to ignore cases where we have two identical vowels followed by "r"
# The following is the beginning of the regex and isn't just R code
# the ([aeiou]) captures the first vowel, the \\1 references what we captured
# so this gives us the same vowel two times in a row
# which we then follow with an "r"
# Then we tell it to skip/fail for this
([aeiou])\\1r(*SKIP)(*FAIL)
Now we told it to skip those cases so now we tell it "or cases where we have two vowels followed by an 'r'" and since we already eliminated the cases where those two vowels are the same this will get us what we want.
|[aeiou]{2}r
Putting it together we end up with
y <- c('our','pear','bar','aar', "aa", "ae", "are", "aeer", "ssseiras")
grep("([aeiou])\\1r(*SKIP)(*FAIL)|[aeiou]{2}r", y, perl = TRUE, value = TRUE)
#[1] "our" "pear" "sseiras"
Here is a less than elegant solution:
y[grepl("[aeiou]{2}r", y, perl=T) & !grepl("(.)\\1r", y, perl=T)]
Probably has some corner case failures where the first set matches at different location than the second set (will have to think about that), but something to get you started.
Another one through negative lookahead assertion.
> y <- c('our','pear','bar','aar', "aa", "ae", "are", "aeer", "ssseiras")
> grep("(?!(?:aa|ee|ii|oo|uu)r)[aeiou][aeiou]r", y, perl=TRUE, value=TRUE)
[1] "our" "pear" "ssseiras"
> grep("(?!aa|ee|ii|oo|uu)[aeiou][aeiou]r", y, perl=TRUE, value=TRUE)
[1] "our" "pear" "ssseiras"
(?!aa|ee|ii|oo|uu) asserts that the first two chars in the match won't be aa or ee or .... or uu. So this [aeiou][aeiou] would match any two vowels other but it wouldn't be repeated . That's why we set the condition at first. r matches the r which follows the vowels.
I am trying to extract all of the words in the string below contained within the brackets following the word 'tokens' only if the 'tokens' occurs after 'tag(noun)'.
For example, I have the string:
m<- "phrase('The New York State Department',[det([lexmatch(['THE']),
inputmatch(['The']),tag(det),tokens([the])]),mod([lexmatch(['New York State']),
inputmatch(['New','York','State']),tag(noun),tokens([new,york,state])]),
head([lexmatch([department]),inputmatch(['Department']),tag(noun),
tokens([department])])],0/29,[])."
I want to get a list of all of the words that occur within the brackets after the word 'tokens' only when the word tokens occurs after 'tag(noun)'.
Therefore, I want my output to be a vector of the following:
[1] new, york, state, department
How do I do this? I'm assuming I have to use a regular expression, but I'm lost on how to write this in R.
Thanks!
Remove newlines and then extract the portion matched to the part between parentheses in pattern pat. Then split apart such strings by commas and simplify into a character vector:
library(gsubfn)
pat <- "tag.noun.,tokens..(.*?)\\]"
strapply(gsub("\\n", "", m), pat, ~ unlist(strsplit(x, ",")), simplify = c)
giving:
[1] "new" "york" "state" "department"
Visualization: Here is the debuggex representation of the regular expression in pat. (Note that we need to double the backslash when put within R's double quotes):
tag.noun.,tokens..(.*?)\]
Debuggex Demo
Note that .*? means match the shortetst string of any characters such that the entire pattern matches - without the ? it would try to match the longest string.
How about something like this. Here i'll use the regcatputedmatches helper function to make it easier to extract the captured matches.
m<- "phrase('The New York State Department',[det([lexmatch(['THE']),inputmatch(['The']),tag(det),tokens([the])]),mod([lexmatch(['New York State']),inputmatch(['New','York','State']),tag(noun),tokens([new,york,state])]),head([lexmatch([department]),inputmatch(['Department']),tag(noun),tokens([department])])],0/29,[])."
rx <- gregexpr("tag\\(noun\\),tokens\\(\\[([^]]+)\\]\\)", m, perl=T)
lapply(regcapturedmatches(m,rx), function(x) {
unlist(strsplit(c(x),","))
})
# [[1]]
# [1] "new" "york" "state" "department"
The regular expression is a bit messy because your desired match contains many special regular expression symbols so we need to properly escape them.
Here is a one liner if you like:
paste(unlist(regmatches(m, gregexpr("(?<=tag\\(noun\\),tokens\\(\\[)[^\\]]*", m, perl=T))), collapse=",")
[1] "new,york,state,department"
Broken down:
# Get match indices
indices <- gregexpr("(?<=tag\\(noun\\),tokens\\(\\[)[^\\]]*", m, perl=T)
# Extract the matches
matches <- regmatches(m, indices)
# unlist and paste together
paste(unlist(matches), collapse=",")
[1] "new,york,state,department"