Why does strsplit use positive lookahead and lookbehind assertion matches differently? - regex

Common sense and a sanity-check using gregexpr() indicate that the look-behind and look-ahead assertions below should each match at exactly one location in testString:
testString <- "text XX text"
BB <- "(?<= XX )"
FF <- "(?= XX )"
as.vector(gregexpr(BB, testString, perl=TRUE)[[1]])
# [1] 9
as.vector(gregexpr(FF, testString, perl=TRUE)[[1]][1])
# [1] 5
strsplit(), however, uses those match locations differently, splitting testString at one location when using the lookbehind assertion, but at two locations -- the second of which seems incorrect -- when using the lookahead assertion.
strsplit(testString, BB, perl=TRUE)
# [[1]]
# [1] "text XX " "text"
strsplit(testString, FF, perl=TRUE)
# [[1]]
# [1] "text" " " "XX text"
I have two questions: (Q1) What's going on here? And (Q2) how can one get strsplit() to be better behaved?
Update: Theodore Lytras' excellent answer explains what's going on, and so addresses (Q1). My answer builds on his to identify a remedy, addressing (Q2).

I am not sure whether this qualifies as a bug, because I believe this is expected behaviour based on the R documentation. From ?strsplit:
The algorithm applied to each input string is
repeat {
if the string is empty
break.
if there is a match
add the string to the left of the match to the output.
remove the match and all to the left of it.
else
add the string to the output.
break.
}
Note that this means that if there is a match at the beginning of
a (non-empty) string, the first element of the output is ‘""’, but
if there is a match at the end of the string, the output is the
same as with the match removed.
The problem is that lookahead (and lookbehind) assertions are zero-length. So for example in this case:
FF <- "(?=funky)"
testString <- "take me to funky town"
gregexpr(FF,testString,perl=TRUE)
# [[1]]
# [1] 12
# attr(,"match.length")
# [1] 0
# attr(,"useBytes")
# [1] TRUE
strsplit(testString,FF,perl=TRUE)
# [[1]]
# [1] "take me to " "f" "unky town"
What happens is that the lonely lookahead (?=funky) matches at position 12. So the first split includes the string up to position 11 (left of the match), and it is removed from the string, together with the match, which -however- has zero length.
Now the remaining string is funky town, and the lookahead matches at position 1. However there's nothing to remove, because there's nothing at the left of the match, and the match itself has zero length. So the algorithm is stuck in an infinite loop. Apparently R resolves this by splitting a single character, which incidentally is the documented behaviour when strspliting with an empty regex (when argument split=""). After this the remaining string is unky town, which is returned as the last split since there's no match.
Lookbehinds are no problem, because each match is split and removed from the remaining string, so the algorithm is never stuck.
Admittedly this behaviour looks weird at first glance. Behaving otherwise however would violate the assumption of zero length for lookaheads. Given that the strsplit algorithm is documented, I belive this does not meet the definition of a bug.

Based on Theodore Lytras' careful explication of substr()'s behavior, a reasonably clean workaround is to prefix the to-be-matched lookahead assertion with a positive lookbehind assertion that matches any single character:
testString <- "take me to funky town"
FF2 <- "(?<=.)(?=funky)"
strsplit(testString, FF2, perl=TRUE)
# [[1]]
# [1] "take me to " "funky town"

Looks like a bug to me. This doesn't appear to just be related to spaces, specifically, but rather any lonely lookahead (positive or negative):
FF <- "(?=funky)"
testString <- "take me to funky town"
strsplit(testString,FF,perl=TRUE)
# [[1]]
# [1] "take me to " "f" "unky town"
FF <- "(?=funky)"
testString <- "funky take me to funky funky town"
strsplit(testString,FF,perl=TRUE)
# [[1]]
# [1] "f" "unky take me to " "f" "unky "
# [5] "f" "unky town"
FF <- "(?!y)"
testString <- "xxxyxxxxxxx"
strsplit(testString,FF,perl=TRUE)
# [[1]]
# [1] "xxx" "y" "xxxxxxx"
Seems to work fine if given something to capture along with the zero-width assertion, such as:
FF <- " (?=XX )"
testString <- "text XX text"
strsplit(testString,FF,perl=TRUE)
# [[1]]
# [1] "text" "XX text"
FF <- "(?= XX ) "
testString <- "text XX text"
strsplit(testString,FF,perl=TRUE)
# [[1]]
# [1] "text" "XX text"
Perhaps something like that might function as a workaround.

Related

strsplit by parentheses [duplicate]

This question already has answers here:
Regular Expression to get a string between parentheses in Javascript
(10 answers)
Closed 7 years ago.
Suppose I have a string like "A B C (123-456-789)", I'm wondering what's the best way to retrieve "123-456-789" from it.
strsplit("A B C (123-456-789)", "\\(")
[[1]]
[1] "A B C" "123-456-789)"
If we want to extract the digits with - between the braces, one option is str_extract. If there are multiple patterns within a string, use str_extract_all
library(stringr)
str_extract(str1, '(?<=\\()[0-9-]+(?=\\))')
#[1] "123-456-789"
str_extract_all(str2, '(?<=\\()[0-9-]+(?=\\))')
In the above codes, we are using regex lookarounds to extract the numbers and the -. The positive lookbehind (?<=\\()[0-9-]+ matches numbers along with - ([0-9-]+) in (123-456-789 and not in 123-456-789. Similarly the lookahead ('[0-9-]+(?=\)') matches numbers along with - in 123-456-789) and not in 123-456-798. Taken together it matches all the cases that satisfy both the conditions (123-456-789) and extract those in between the lookarounds and not with cases like (123-456-789 or 123-456-789)
With strsplit you can specify the split as [()]. We keep the () inside the square brackets to [] to treat it as characters or else we have to escape the parentheses ('\\(|\\)').
strsplit(str1, '[()]')[[1]][2]
#[1] "123-456-789"
If there are multiple substrings to extract from a string, we could loop with lapply and extract the numeric split parts with grep
lapply(strsplit(str2, '[()]'), function(x) grep('\\d', x, value=TRUE))
Or we can use stri_split from stringi which has the option to remove the empty strings as well (omit_empty=TRUE).
library(stringi)
stri_split_regex(str1, '[()A-Z ]', omit_empty=TRUE)[[1]]
#[1] "123-456-789"
stri_split_regex(str2, '[()A-Z ]', omit_empty=TRUE)
Another option is rm_round from qdapRegex if we are interested in extracting the contents inside the brackets.
library(qdapRegex)
rm_round(str1, extract=TRUE)[[1]]
#[1] "123-456-789"
rm_round(str2, extract=TRUE)
data
str1 <- "A B C (123-456-789)"
str2 <- c("A B C (123-425-478) A", "ABC(123-423-428)",
"(123-423-498) ABCDD",
"(123-432-423)", "ABC (123-423-389) GR (124-233-848) AK")
or with sub from base R:
sub("[^(]+\\(([^)]+)\\).*", "\\1", "A B C (123-456-789)")
#[1] "123-456-789"
Explanation:
[^(]+ : matches anything except an opening bracket
\\( : matches an opening bracket, which is just before what you want
([^)]+) : matches the pattern you want to capture (which is then retrieved in replacement="\\1"), which is anything except a closing bracket
\\).* matches a closing bracket followed by anything, 0 or more times
Another option with look-ahead and look-behind
sub(".*(?<=\\()(.+)(?=\\)).*", "\\1", "A B C (123-456-789)", perl=TRUE)
#[1] "123-456-789"
The capture groups in sub will target your desired output:
sub('.*\\((.*)\\).*', '\\1', str1)
[1] "123-456-789"
Extra check to make sure I pass #akrun's extended example:
sub('.*\\((.*)\\).*', '\\1', str2)
[1] "123-425-478" "123-423-428" "123-423-498" "123-432-423" "124-233-848"
You may try these gsub functions.
> gsub("[^\\d-]", "", x, perl=T)
[1] "123-456-789"
> gsub(".*\\(|\\)", "", x)
[1] "123-456-789"
> gsub("[^0-9-]", "", x)
[1] "123-456-789"
Few more...
> gsub("[0-9-](*SKIP)(*F)|.", "", x, perl=T)
[1] "123-456-789"
> gsub("(?:(?![0-9-]).)*", "", x, perl=T)
[1] "123-456-789"
Try this also:
k<-"A B C (123-456-789)"
regmatches(k,gregexpr("*.(\\d+).*",k))[[1]]
[1] "(123-456-789)"
With suggestion from #Arun:
regmatches(k, gregexpr('(?<=\\()[^A-Z ]+(?=\\))', k, perl=TRUE))[[1]]
With suggestion from #akrun:
regmatches(k, gregexpr('[0-9-]+', k))[[1]]

Unable to replace string with back reference using gsub in R

I am trying to replace some text in a character vector using regex in R where, if there is a set of letters inside a bracket, the bracket content is to erplace the whole thing. So, given the input:
tst <- c("85", "86 (TBA)", "87 (LAST)")
my desired output would be equivalent to c("85", "TBA", "LAST")
I tried gsub("\\(([[:alpha:]])\\)", "\\1", tst) but it didn't replace anything. What do I need to correct in my regular expression here?
I think you want
gsub(".*\\(([[:alpha:]]+)\\)", "\\1", tst)
# [1] "85" "TBA" "LAST"
Your first expression was trying to match exactly one alpha character rather than one-or-more. I also added the ".*" to capture the beginning part of the string so it gets replaced as well, otherwise, it would be left untouched.
gsub("(?=.*\\([^)]*\\)).*\\(([^)]*)\\)", "\\1", tst, perl=TRUE)
## [1] "85" "TBA" "LAST"
You can try this.See demo.Replace by \1.
https://regex101.com/r/sH8aR8/38
The following would work. Note that white-spaces within the brackets may be problematic
A<-sapply(strsplit(tst," "),tail,1)
B<-gsub("\\(|\\)", "", A)
I like the purely regex answers better. I'm showing a solution using the qdapRegex package that I maintain as the result is pretty speedy and easy to remember and generalize. It pulls out the strings that are in parenthesis and then replaces any NA (no bracket) with the original value. Note that the result is a list and you'd need to use unlist to match your desired output.
library(qdpRegex)
m <- rm_round(tst, extract=TRUE)
m[is.na(m)] <- tst[is.na(m)]
## [[1]]
## [1] "85"
##
## [[2]]
## [1] "TBA"
##
## [[3]]
## [1] "LAST"

Split on first/nth occurrence of delimiter

I am trying something I thought would be easy. I'm looking for a single regex solution (though others are welcomed for completeness). I want to split on n occurrences of a delimiter.
Here is some data:
x <- "I like_to see_how_too"
pat <- "_"
Desired outcome
Say I want to split on first occurrence of _:
[1] "I like" "to see_how_too"
Say I want to split on second occurrence of _:
[1] "I like_to see" "how_too"
Ideally, if the solution is a regex one liner generalizable to nth occurrence; the solution will use strsplit with a single regex.
Here's a solution that doesn't fit my parameters of single regex that works with strsplit
x <- "I like_to see_how_too"
y <- "_"
n <- 1
loc <- gregexpr("_", x)[[1]][n]
c(substr(x, 1, loc-1), substr(x, loc + 1, nchar(x)))
Here is another solution using the gsubfn package and some regex-fu. To change the nth occurrence of the delimiter, you can simply swap the number that is placed inside of the range quantifier — {n}.
library(gsubfn)
x <- 'I like_to see_how_too'
strapply(x, '((?:[^_]*_){1})(.*)', c, simplify =~ sub('_$', '', x))
# [1] "I like" "to see_how_too"
If you would like the nth occurrence to be user defined, you could use the following:
n <- 2
re <- paste0('((?:[^_]*_){',n,'})(.*)')
strapply(x, re, c, simplify =~ sub('_$', '', x))
# [1] "I like_to see" "how_too"
Non-Solution
Since R is using PCRE, you can use \K to remove everything that matches the pattern before \K from the main match result.
Below is the regex to split the string at the 3rd _
^[^_]*(?:_[^_]*){2}\K_
If you want to split at the nth occurrence of _, just change 2 to (n - 1).
Demo on regex101
That was the plan. However, strsplit seems to think differently.
Actual execution
Demo on ideone.com
x <- "I like_to see_how_too but_it_seems to_be_impossible"
strsplit(x, "^[^_]*(?:_[^_]*)\\K_", perl=TRUE)
strsplit(x, "^[^_]*(?:_[^_]*){1}\\K_", perl=TRUE)
strsplit(x, "^[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# strsplit(x, "^[^_]*(?:_[^_]*)\\K_", perl=TRUE)
# [[1]]
# [1] "I like_to see" "how_too but" "it_seems to" "be_impossible"
# strsplit(x, "^[^_]*(?:_[^_]*){1}\\K_", perl=TRUE)
# [[1]]
# [1] "I like_to see" "how_too but" "it_seems to" "be_impossible"
# strsplit(x, "^[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# [[1]]
# [1] "I like" "to see" "how" "too but" "it"
# [6] "seems to" "be" "impossible"
It still fails to work on a stronger assertion \A
strsplit(x, "\\A[^_]*(?:_[^_]*){0}\\K_", perl=TRUE)
# [[1]]
# [1] "I like" "to see" "how" "too but" "it"
# [6] "seems to" "be" "impossible"
Explanation?
This behavior hints at the fact that strsplit find the first match, do a substring to extract the first token and the remainder part, and find the next match in the remainder part.
This removes all the states from the previous matches, and leaves us with a clean state when it tries to match the regex on the remainder. This makes the task of stopping the strsplit function at first match and achieving the task at the same time impossible. There is not even a parameter in strsplit to limit the number of splits.
Rather than split you do match to get your split strings.
Try this regex:
^((?:[^_]*_){1}[^_]*)_(.*)$
Replace 1 by n-1 where you're trying to get split on nth occurrence of underscore.
RegEx Demo
Update: It seems R also supports PCRE and in that case you can do split as well using this PCRE regex:
^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_
Replace 1 by n-1 where you're trying to get split on nth occurrence of underscore.
(*FAIL) behaves like a failing negative assertion and is a synonym for (?!)
(*SKIP) defines a point beyond which the regex engine is not allowed to backtrack when the subpattern fails later
(*SKIP)(*FAIL) together provide a nice alternative of restriction that you cannot have a variable length lookbehind in above regex.
RegEx Demo2
x <- "I like_to see_how_too"
strsplit(x, "^((?:[^_]*_){0}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
strsplit(x, "^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## > strsplit(x, "^((?:[^_]*_){0}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## [[1]]
## [1] "I like" "to see" "how" "too"
## > strsplit(x, "^((?:[^_]*_){1}[^_]*)(*SKIP)(*F)|_", perl=TRUE)
## [[1]]
## [1] "I like_to see" "how_too"
This uses gsubfn to to preprocess the input string so that strsplit can handle it. The main advantage is that one can specify a vector of numbers, k, indicating which underscores to split on.
It replaces the occurrences of underscore defined by k by a double underscore and then splits on double underscore. In this example we split at the 2nd and 4th underscore:
library(gsubfn)
k <- c(2, 4) # split at 2nd and 4th _
p <- proto(fun = function(., x) if (count %in% k) "__" else "_")
strsplit(gsubfn("_", p, "aa_bb_cc_dd_ee_ff"), "__")
giving:
[[1]]
[1] "aa_bb" "cc_dd" "ee_ff"
If empty fields are allowed then use any other character sequence not in the string, e.g. "\01" in place of the double underscore.
See section 4 of the gusbfn vignette for more info on using gusbfn with proto objects to retain state between matches.

remove leading zeroes from timestamp %j%Y %H:%M

My timestamp is in the form
0992006 09:00
I need to remove the leading zeros to get this form:
992006 9:00
Here's the code I'm using now, which doesn't remove leading zeros:
prediction$TIMESTAMP <- as.character(format(prediction$TIMESTAMP, '%j%Y %H:%M'))
Simplest way is to create your own boundary that asserts either the start of the string or a space precedes.
gsub('(^| )0+', '\\1', '0992006 09:00')
# [1] "992006 9:00"
You could do the same making the replacement exempt using a trick. \K resets the starting point of the reported match and any previously consumed characters are no longer included.
gsub('(^| )\\K0+', '', '0992006 09:00', perl=T)
# [1] "992006 9:00"
Or you could use sub and match until the second set of leading zeros.
sub('^0+([0-9]+ )0+', '\\1', '0992006 09:00')
# [1] "992006 9:00"
And to cover all possibilities, if you know that you will ever have a format like 0992006 00:00, simply remove the + quantifier from zero in the regular expression so it only removes the first leading zero.
str1 <- "0992006 09:00"
gsub("(?<=^| )0+", "", str1, perl=TRUE)
#[1] "992006 9:00"
For situations like below, it could be:
str2 <- "0992006 00:00"
gsub("(?<=^| )0", "", str2, perl=TRUE)
#[1] "992006 0:00"
Explanation
Here the idea is to use look behind (?<=^| )0+ to match 0s
if it occurs either at the beginning of the string
(?<=^
or |
if it follows after a space )0+
and replace those matched 0s by "" in the second part of the gsub argument.
In the second string, the hour and minutes are all 0's. So, using the first code would result in:
gsub("(?<=^| )0+", "", str2, perl=TRUE)
#[1] "992006 :00"
Here, it is unclear what the OP would accept as a result. So, I thought, instead of removing the whole 0s before the :, it would be better if one 0 was left. So, I replaced the multiple 0+ code to just one 0 and replace that by "".
Here's another option using a lookbehind
gsub("(^0)|(?<=\\s)0", "", "0992006 09:00", perl = TRUE)
## [1] "992006 9:00"
With sub:
sub("^[0]+", "", prediction$TIMESTAMP)
[1] "992006 09:00"
You can also use stringr without a regular expression, by using the substrings.
> library(stringr)
> str_c(str_sub(word(x, 1:2), 2), collapse = " ")
# [1] "992006 9:00"
Some more Perl regexes,
> gsub("(?<!:)\\b0+", "", "0992006 09:00", perl=T)
[1] "992006 9:00"
> gsub("(?<![\\d:])0+", "", "0992006 09:00", perl=T)
[1] "992006 9:00"

Regular Expression to anonymize emails

i use in R the regular expression
regexp <- "(^|[^([:alnum:]|.|_)])abc#abc.de($|[^[:alnum:]])"
to find the email-adress abc#abc.de in an spefic text and replace it by an anonym-mail-adress.
tmp <- c("aaaaabc#abc.debbbb", ## <- should not be matched
"aaaa abc#abc.de bbbb", ## <- should be matched
"abc#abc.de", ## <- should be matched
"aaa.abc#abc.de", ## <- should not be matched
"aaaa_abc#abc.de", ## <- should not be matched
"(abc#abc.de)", ## <- should be matched
"aaaa (abc#abc.de) bbbb") ## <- should be matched
replacement <- paste("\\1", "anonym#anonym.de", "\\2", sep="")
gsub(regexp, replacement, tmp, ignore.case=TRUE)
as result I get
> gsub(regexp, replacement, tmp, ignore.case=TRUE)
[1] "aaaaabc#abc.debbbb" "aaaa anonym#anonym.de bbbb"
[3] "anonym#anonym.de" "aaa.abc#abc.de"
[5] "aaaa_abc#abc.de" "(abc#abc.de)"
[7] "aaaa (abc.abc.de) bbbb"
I don't know why the last two elements of the array are not matched?
Thank you and best regards.
How about this?
gsub("^(abc#abc)|(?<=[ (])(abc#abc)", "anonym#anonym", tmp, perl=T)
The pattern before |: ^(abc#abc) checks for beginning with abc#abc, of course.
The pattern after | uses positive lookbehind and searches for abc#abc preceded by space or ( (left paranthesis) and if found, replaces with anonym#anonym.
This is what I get: (Note: I replaced abc.abc in the last string with abc#abc)
[1] "aaaaabc#abc.debbbb" "aaaa anonym#anonym.de bbbb"
[3] "anonym#anonym.de" "aaa.abc#abc.de"
[5] "aaaa_abc#abc.de" "(anonym#anonym.de)"
[7] "aaaa (anonym#anonym.de) bbbb"
Edit: To explain the problem with your regexp, it seems like a problem with the part:
[^([:alnum:]|.|_)]
I think the negation has to be present in every | statement. Also, you should use [.] instead of . as the latter implies any character. Alternatively, instead of using negation in for every character you're checking, we can condense this part by removing all unncessary | as:
[^.[:alpha:]_] # not a . or _ or any alphanumeric
# using gsub on it:
gsub("(^|[^.[:alpha:]_])abc#abc", " anonym#anonym", tmp)
# [1] "aaaaabc#abc.debbbb" "aaaa anonym#anonym.de bbbb"
# [3] " anonym#anonym.de" "aaa.abc#abc.de"
# [5] "aaaa_abc#abc.de" " anonym#anonym.de)"
# [7] "aaaa anonym#anonym.de) bbbb"
You get every abc#abc replaced. But, you'll lose the character before abc#abc everytime because you're checking for it in the pattern as well. So, you'll have to use the capture group. That is, if you wrap a regular expression with () then you can refer to that "capture" using special variables such as \\1, \\2 etc... Here, we have captured (^|[^.[:alpha:]_]), i.e., the part before abc#abc. Since it is the first capture, we'll refer to it as \\1 to use it to recover the missing character in the previous result:
gsub("(^|[^.[:alpha:]_])abc#abc", "\\1anonym#anonym", tmp)
# [1] "aaaaabc#abc.debbbb" "aaaa anonym#anonym.de bbbb"
# [3] "anonym#anonym.de" "aaa.abc#abc.de"
# [5] "aaaa_abc#abc.de" "(anonym#anonym.de)"
# [7] "aaaa (anonym#anonym.de) bbbb"
This is the result you needed. And this is the same as my initial answer using positive look-behind. In that case, since it just checks if it is preceded by something, you don't have to capture anything special. Only the abc#abc part got replaced. Hope this helps.