How to remove double single quotes but not the single quote in a string in Ruby? For instance from That's 'large', to That's large.
Try this regex:
\B'((?:(?!'\B)[\s\S])*)'
Replace each match with \1
Click for Demo
Code(Result):
re = /\B'((?:(?!'\B)[\s\S])*)'/m
str = 'That\'s \'large\'
The 69\'ers\' drummer won\'t like this.
He said, \'it\'s clear this does not work\'. It does not fit the \'contractual obligations\''
subst = '\\1'
result = str.gsub(re, subst)
# Print the result of the substitution
puts result
Explanation:
\B - matches a non-word boundary
((?:(?!'\B)[\s\S])*) - matches 0+ occurrences of any character [\s\S] which (does not start with ' followed by a non-word boundary). This is captured in Group 1.
' - matches a '
This is one of those quagmires ike parsing XML or HTML that can't be done with a regex, but you can sorta pretend like it's mostly going to work. You can tweak it forever and not get right.
You could look for balanced quotes, that is only quotes in pairs, but this doesn't help. Is That's 'large' to be stripped as Thats large' or That's large?
Instead you need to give it an understanding of English grammar and when a ' is an apostrophe versus a quote. Something simple that knows the basics of contractions and possessives. Contractions: don't, won't, I'll. Possessives: Joe's and s'. And maybe you can knock up a regex to skip those.
But it rapidly gets complicated. KO'd. Or what if you wish to indicate a particular pronunciation: fo'c's'le. Or someone's name O'Doole.
What you might be able to get away with stripping a pair of quotes that start at the beginning of a word and the end of a word. It's clear he said, 'this isn't a contraction'. Matching the quote in front of this and the quote at the end of contraction is probably maybe safe.
# Use negative look behind and ahead to look for quotes which are
# not after and before a word character.
# Use a non-greedy match to catch multiple pairs of quotes.
re = /(?<!\w)'(.*?)'(?!\w)/
sentence.gsub(re, '\1')
This works in a lot of cases.
That's 'large' -> That's large
Eat at Joe's -> Eat at Joe's
I'll be Jane's -> I'll be Jane's
Jones' three cats' toys. -> Jones' three cats' toys.
It's clear he said, 'this isn't a contraction'. -> It's clear he said, this isn't a contraction.
'scare quotes' -> scare quotes
The 69'ers' drummer -> The 69'ers' drummer
Was She's success greater, or King Solomon's Mines's? -> Was She's success greater, or King Solomon's Mines's?
The 69'er's drummer and their 'contractual obligations'. -> The 69'er's drummer and their contractual obligations.
He said, 'it's clear this doesn't work'. -> He said, it's clear this doesn't work.
But not always.
His 'n' Hers's first track is called 'Joyriders'. -> His n Hers's first track is called Joyriders.
Like I said, this is one of those problems that looks simple but is extremely complicated and you can never get quite right. It can suck down a lot of time. I'd recommend ditching the requirement if possible.
A slight variation — if the single quotes only occur around word characters, that is a character from a-z, A-Z, 0-9 or the _ (underscore) character. you can use this:
phrase = "That's 'large' and not 'small', but it's still 'amazing'."
phrase.gsub(/'(\w*)'/, '\1')
=> "That's large and not small, but it's still amazing."
But as Schwern says, if you're trying to do anything other than a bit of simple text manipulation, you'll soon find yourself bogged down by edge cases.
Related
I am using python to parse a string that is passed in by the optparse module.
I want to split the string on certain delimiters but not in between quote marks.
A sample string is:
--state-basedir /dir/dir/dir/ --cmd=\"param load $v2param\" --master=/dev/ttyUSB0 --console --map --out=udp:192.168.1.1:14550
This string is passed in as a single optparse argument, I am then going to pass it to another process.
I have been trying various things at http://pythex.org/
The closest I have gotten is:
`(?<!")[\s=](?![\s0-9a-zA-Z\$\\]*")`
The issue is that the = sign after --cmd and the space before --master are not matched.
In plain English, this is how I am reading my regex:
match either a space character or an equal character as long as it is not preceded by a quotation mark and as long as it is not proceeded by a combination of any other letter,numbers,punctuation and another quotation mark
I had a feeling that there was something else I was missing, like greediness, so I tried adding ? after my look-ahead and look-behind terms. If I put a ? after my look-behind one I can get the space before --master but if I put the ? after my look-ahead term I get the spaces in the quotation marks now, which I don't want.
The idea here is that I am going to use re.split to handle things.
Thanks for any explanations as to what I am doing wrong.
This is not a regex answer and it's also not pretty, but it is one line.
sum([[x] if '"' in x else re.split(' |=',x) for x in re.split('=(\".+?\" )',a)],[])
output:
['--state-basedir', '/dir/dir/dir/', '--cmd', '"param load $v2param" ', '--master', '/dev/ttyUSB0', '--console', '--map', '--out', 'udp:192.168.1.1:14550']
Starting from the re.split('=(\".+?\" )',a)] this splits out text surrounded by quotes (more specifically ="something another thing"). The split pieces are then split further with re.split(' |=',x) if they do not have a " in them, or are just returned as is [x] if they do. The last step is collapsing the resulting 2d list by overloading sum with sum(two_d_list,[]).
I hope this answer helps but I understand if it isn't what you're looking for
I'm writing a Python chatbot. No matter what the technique is(Levenshtein, LCS, regex, etc.), I want a pattern like My name is [ A ]. smart enough to match strings like:
My name is Tslmy. #Distance should = 0, and groupdict()['a'] outputs "Tslmy"
My name is Tesla Tahomana. #Distance should = 0(!), and groupdict()['a'] outputs "Tesla Tahomana"
my naem ist tslmy . #With a little typo, the distance = 5, and groupdict()['a'] outputs "tslmy "
Allow me to use groupdict()['a'] to refer to what the [ A ] thing (actually (?P<identifier>match)) has captured, please.
In other way, I'm looking for a "Levenshtein" with omits/skippings/blanks/neglects, and pick out what has been skipped as well.
In another way, I'm looking for a fuzzy(a.k.a. approximate) regex that can be less strict with the pattern, still provides the good old groupdict(), as well as a "fuzziness" value (or "edit distance", required to determine "the best matched pattern to the string" later).
This is the preferred solution, since it provides "sufficient" groupdict() if well managed.
However, The TRE library and the REGEX library, which is found to be the closest solution, don't seem to provide a "fuzziness" value. If this can be solved, then so much the better!
Is that possible? Thanks for paying attention.
Update:
I decided to use the powerful regex module in the end, but still unable to get the "fuzziness value".
Since the question on this page is theoratically solved, appending too further will be dishonorable. So I put forward another question about this new issue, and hopes you could solve it!
You could use a RegEx for the basic match:
r"My name is (\w+){1,2}."
And then use the TRE library to allow for variations.
DAT REGEX O_O
(?i)(?:(?:my|ym).?|.?(?:my|ym))\s+(?:.?(?:..me|n..e|na..)|(?:..me|n..e|na..).?)\s+(?:(?:is|si).?|.?(?:is|si))\s+(\w[\w\s])\s
Let's split it up:
(?i) : set the i modifier to match case insensitive
(?:(?:my|ym).?|.?(?:my|ym)) : this will match my, ym, My, Ym, may, amy etc...
\s+ : match white space one or more times
(?:.?(?:..am|n..e|na..)|(?:..am|n..e|na..).?) : match name, naao, tame, lame, n99e, names, Naats etc...
\s+ : match white space one or more times
(?:(?:is|si).?|.?(?:is|si)) : Match is, si, ist, sit, siR etc...
\s+ : match white space one or more times
(\w[\w\s]*) : match words and spaces one or more times and group it (it must start with a word \w)
\s* : match white spaces zero or more times
Online demo
I have a text input field for titles of various things and to help minimize false negatives on search results(internal search is not the best), I need to have a REGEX pattern which looks at the first four characters of the input string and removes the word(and space after the word) _the _ if it is there at the beginning only.
For example if we are talking about the names of bands, and someone enters The Rolling Stones , what i need is for the entry to say only Rolling Stones
Can a regex be used to automatically strip these 4characters?
Applying the regex
^(?:\s*the\s*)?(.*)$
will match any string, and capture it in backreference no. 1, unless it starts with the (optionally surrounded by whitespace), in which case backref no. 1 will contain whatever follows.
You need to set the case-insensitive option in your regex engine for this to work.
You can use the ^ identifier to match a pattern at the beginning of a line, however for what you are using this for, it can be considered overkill.
A lot of languages support string manipulations, which is a more suitable choice. I can provide an example to demonstrate in Python,
>>> def func(n):
n = n[4:len(n)] if n[0:4] == "The " else n
return n
>>> func("The Rolling Stones")
'Rolling Stones'
>>> func("They Might Be Giants")
'They Might Be Giants'
As you don't clarify with language, here is a solution in Perl :
my $str = "The Rolling Stones";
$str =~ s/^the //i;
say $str; # Rolling Stones
How do I get the substring " It's big \"problem " using a regular expression?
s = ' function(){ return " It\'s big \"problem "; }';
/"(?:[^"\\]|\\.)*"/
Works in The Regex Coach and PCRE Workbench.
Example of test in JavaScript:
var s = ' function(){ return " Is big \\"problem\\", \\no? "; }';
var m = s.match(/"(?:[^"\\]|\\.)*"/);
if (m != null)
alert(m);
This one comes from nanorc.sample available in many linux distros. It is used for syntax highlighting of C style strings
\"(\\.|[^\"])*\"
As provided by ePharaoh, the answer is
/"([^"\\]*(\\.[^"\\]*)*)"/
To have the above apply to either single quoted or double quoted strings, use
/"([^"\\]*(\\.[^"\\]*)*)"|\'([^\'\\]*(\\.[^\'\\]*)*)\'/
Most of the solutions provided here use alternative repetition paths i.e. (A|B)*.
You may encounter stack overflows on large inputs since some pattern compiler implements this using recursion.
Java for instance: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6337993
Something like this:
"(?:[^"\\]*(?:\\.)?)*", or the one provided by Guy Bedford will reduce the amount of parsing steps avoiding most stack overflows.
/(["\']).*?(?<!\\)(\\\\)*\1/is
should work with any quoted string
"(?:\\"|.)*?"
Alternating the \" and the . passes over escaped quotes while the lazy quantifier *? ensures that you don't go past the end of the quoted string. Works with .NET Framework RE classes
/"(?:[^"\\]++|\\.)*+"/
Taken straight from man perlre on a Linux system with Perl 5.22.0 installed.
As an optimization, this regex uses the 'posessive' form of both + and * to prevent backtracking, for it is known beforehand that a string without a closing quote wouldn't match in any case.
This one works perfect on PCRE and does not fall with StackOverflow.
"(.*?[^\\])??((\\\\)+)?+"
Explanation:
Every quoted string starts with Char: " ;
It may contain any number of any characters: .*? {Lazy match}; ending with non escape character [^\\];
Statement (2) is Lazy(!) optional because string can be empty(""). So: (.*?[^\\])??
Finally, every quoted string ends with Char("), but it can be preceded with even number of escape sign pairs (\\\\)+; and it is Greedy(!) optional: ((\\\\)+)?+ {Greedy matching}, bacause string can be empty or without ending pairs!
An option that has not been touched on before is:
Reverse the string.
Perform the matching on the reversed string.
Re-reverse the matched strings.
This has the added bonus of being able to correctly match escaped open tags.
Lets say you had the following string; String \"this "should" NOT match\" and "this \"should\" match"
Here, \"this "should" NOT match\" should not be matched and "should" should be.
On top of that this \"should\" match should be matched and \"should\" should not.
First an example.
// The input string.
const myString = 'String \\"this "should" NOT match\\" and "this \\"should\\" match"';
// The RegExp.
const regExp = new RegExp(
// Match close
'([\'"])(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))' +
'((?:' +
// Match escaped close quote
'(?:\\1(?=(?:[\\\\]{2})*[\\\\](?![\\\\])))|' +
// Match everything thats not the close quote
'(?:(?!\\1).)' +
'){0,})' +
// Match open
'(\\1)(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))',
'g'
);
// Reverse the matched strings.
matches = myString
// Reverse the string.
.split('').reverse().join('')
// '"hctam "\dluohs"\ siht" dna "\hctam TON "dluohs" siht"\ gnirtS'
// Match the quoted
.match(regExp)
// ['"hctam "\dluohs"\ siht"', '"dluohs"']
// Reverse the matches
.map(x => x.split('').reverse().join(''))
// ['"this \"should\" match"', '"should"']
// Re order the matches
.reverse();
// ['"should"', '"this \"should\" match"']
Okay, now to explain the RegExp.
This is the regexp can be easily broken into three pieces. As follows:
# Part 1
(['"]) # Match a closing quotation mark " or '
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
# Part 2
((?: # Match inside the quotes
(?: # Match option 1:
\1 # Match the closing quote
(?= # As long as it's followed by
(?:\\\\)* # A pair of escape characters
\\ #
(?![\\]) # As long as that's not followed by an escape
) # and a single escape
)| # OR
(?: # Match option 2:
(?!\1). # Any character that isn't the closing quote
)
)*) # Match the group 0 or more times
# Part 3
(\1) # Match an open quotation mark that is the same as the closing one
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
This is probably a lot clearer in image form: generated using Jex's Regulex
Image on github (JavaScript Regular Expression Visualizer.)
Sorry, I don't have a high enough reputation to include images, so, it's just a link for now.
Here is a gist of an example function using this concept that's a little more advanced: https://gist.github.com/scagood/bd99371c072d49a4fee29d193252f5fc#file-matchquotes-js
here is one that work with both " and ' and you easily add others at the start.
("|')(?:\\\1|[^\1])*?\1
it uses the backreference (\1) match exactley what is in the first group (" or ').
http://www.regular-expressions.info/backref.html
One has to remember that regexps aren't a silver bullet for everything string-y. Some stuff are simpler to do with a cursor and linear, manual, seeking. A CFL would do the trick pretty trivially, but there aren't many CFL implementations (afaik).
A more extensive version of https://stackoverflow.com/a/10786066/1794894
/"([^"\\]{50,}(\\.[^"\\]*)*)"|\'[^\'\\]{50,}(\\.[^\'\\]*)*\'|“[^”\\]{50,}(\\.[^“\\]*)*”/
This version also contains
Minimum quote length of 50
Extra type of quotes (open “ and close ”)
If it is searched from the beginning, maybe this can work?
\"((\\\")|[^\\])*\"
I faced a similar problem trying to remove quoted strings that may interfere with parsing of some files.
I ended up with a two-step solution that beats any convoluted regex you can come up with:
line = line.replace("\\\"","\'"); // Replace escaped quotes with something easier to handle
line = line.replaceAll("\"([^\"]*)\"","\"x\""); // Simple is beautiful
Easier to read and probably more efficient.
If your IDE is IntelliJ Idea, you can forget all these headaches and store your regex into a String variable and as you copy-paste it inside the double-quote it will automatically change to a regex acceptable format.
example in Java:
String s = "\"en_usa\":[^\\,\\}]+";
now you can use this variable in your regexp or anywhere.
(?<="|')(?:[^"\\]|\\.)*(?="|')
" It\'s big \"problem "
match result:
It\'s big \"problem
("|')(?:[^"\\]|\\.)*("|')
" It\'s big \"problem "
match result:
" It\'s big \"problem "
Messed around at regexpal and ended up with this regex: (Don't ask me how it works, I barely understand even tho I wrote it lol)
"(([^"\\]?(\\\\)?)|(\\")+)+"
I have a regular expression to match a persons name.
So far I have ^([a-zA-Z\'\s]+)$ but id like to add a check to allow for a maximum of 4 spaces. How do I amend it to do this?
Edit: what i meant was 4 spaces anywhere in the string
Don't attempt to regex validate a name. People are allowed to call themselves what ever they like. This can include ANY character. Just because you live somewhere that only uses English doesn't mean that all the people who use your system will have English names. We have even had to make the name field in our system Unicode. It is the only Unicode type in the database.
If you care, we actually split the name at " " and store each name part as a separate record, but we have some very specific requirements that mean this is a good idea.
PS. My step mum has 5 spaces in her name.
^ # Start of string
(?!\S*(?:\s\S*){5}) # Negative look-ahead for five spaces.
([a-zA-Z\'\s]+)$ # Original regex
Or in one line:
^(?!(?:\S*\s){5})([a-zA-Z\'\s]+)$
If there are five or more spaces in the string, five will be matched by the negative lookahead, and the whole match will fail. If there are four or less, the original regex will be matched.
Screw the regex.
Using a regex here seems to be creating a problem for a solution instead of just solving a problem.
This task should be 'easy' for even a novice programmer, and the novel idea of regex has polluted our minds!.
1: Get Input
2: Trim White Space
3: If this makes sence, trim out any 'bad' characters.
4: Use the "split" utility provided by your language to break it into words
5: Return the first 5 Words.
ROCKET SCIENCE.
replies
what do you mean screw the regex? your obviously a VB programmer.
Regex is the most efficient way to work with strings. Learn them.
No. Php, toyed a bit with ruby, now going manically into perl.
There are some thing ( like this case ) where the regex based alternative is computationally and logically exponentially overly complex for the task.
I've parse entire php source files with regex, I'm not exactly a novice in their use.
But there are many cases, such as this, where you're employing a logging company to prune your rose bush.
I could do all steps 2 to 5 with regex of course, but they would be simple and atomic regex, with no weird backtracking syntax or potential for recursive searching.
The steps 1 to 5 I list above have a known scope, known range of input, and there's no ambiguity to how it functions. As to your regex, the fact you have to get contributions of others to write something so simple is proving the point.
I see somebody marked my post as offensive, I am somewhat unhappy I can't mark this fact as offensive to me. ;)
Proof Of Pudding:
sub getNames{
my #args = #_;
my $text = shift #args;
my $num = shift #args;
# Trim Whitespace from Head/End
$text =~ s/^\s*//;
$text =~ s/\s*$//;
# Trim Bad Characters (??)
$text =~ s/[^a-zA-Z\'\s]//g;
# Tokenise By Space
my #words = split( /\s+/, $text );
#return 0..n
return #words[ 0 .. $num - 1 ];
} ## end sub getNames
print join ",", getNames " Hello world this is a good test", 5;
>> Hello,world,this,is,a
If there is anything ambiguous to anybody how that works, I'll be glad to explain it to them. Noted that I'm still doing it with regexps. Other languages I would have used their native "trim" functions provided where possible.
Bollocks -->
I first tried this approach. This is your brain on regex. Kids, don't do regex.
This might be a good start
/([^\s]+
(\s[^\s]+
(\s[^\s]+
(\s[^\s]+
(\s[^\s]+|)
|)
|)
|)
)/
( Linebroken for clarity )
/([^\s]+(\s[^\s]+(\s[^\s]+(\s[^\s]+|)|)|))/
( Actual )
I've used [^\s]+ here instead of your A-Z combo for succintness, but the point is here the nested optional groups
ie:
(Hello( this( is( example))))
(Hello( this( is( example( two)))))
(Hello( this( is( better( example))))) three
(Hello( this( is()))))
(Hello( this()))
(Hello())
( Note: this, while being convoluted, has the benefit that it will match each name into its own group )
If you want readable code:
$word = '[^\s]+';
$regex = "/($word(\s$word(\s$word(\s$word(\s$word|)|)|)|)|)/";
( it anchors around the (capture|) mantra of "get this, or get nothing" )
#Sir Psycho : Be careful about your assumptions here. What about hyphenated names? Dotted names (e.g. Brian R. Bondy) and so on?
Here's the answer that you're most likely looking for:
^[a-zA-Z']+(\s[a-zA-Z']+){0,4}$
That says (in English): "From start to finish, match one or more letters, there can also be a space followed by another 'name' up to four times."
BTW: Why do you want them to have apostrophes anywhere in the name?
^([a-zA-Z']+\s){0,4}[a-zA-Z']+$
This assumes you want 4 spaces inside this string (i.e. you have trimmed it)
Edit: If you want 4 spaces anywhere I'd recommend not using regex - you'd be better off using a substr_count (or the equivalent in your language).
I also agree with pipTheGeek that there are so many different ways of writing names that you're probably best off trusting the user to get their name right (although I have found that a lot of people don't bother using capital letters on ecommerce checkouts).
Match multiple whitespace followed by two characters at the end of the line.
Related problem ----
From a string, remove trailing 2 characters preceded by multiple white spaces... For example, if the column contains this string -
" 'This is a long string with 2 chars at the end AB "
then, AB should be removed while retaining the sentence.
Solution ----
select 'This is a long string with 2 chars at the end AB' as "C1",
regexp_replace('This is a long string with 2 chars at the end AB',
'[[[:space:]][a-zA-Z][a-zA-Z]]*$') as "C2" from dual;
Output ----
C1
This is a long string with 2 chars at the end AB
C2
This is a long string with 2 chars at the end
Analysis ----
regular expression specifies - match and replace zero or more occurences (*) of a space ([:space:]) followed by combination of two characters ([a-zA-Z][a-zA-Z]) at the end of the line.
Hope this is useful.