What would this mean in an expression?
(?m:.*?)
or this
(?m:\s*)
I mean, it appears to be something to do with whitespace but I'm unsure.
ADDITIONAL DETAILS:
The full expression I'm looking at is:
\A((?m:\s*)((\/\*(?m:.*?)\*\/)|(\#\#\# (?m:.*?)\#\#\#)|(\/\/ .* \n?)+|(\# .* \n?)+))+
(?...) is a way of applying modifiers to the regular expression inside the parentheses.
(?:...) allows you to treat the part between the parentheses as a group, without affecting the set of strings captured by the matching engine. But you can add option letters between the ? and the :, in which case the part of the regular expression between the parentheses behaves as if you had included those option letters when creating the regular expression. That is, /(?m:...)/ behaves the same as /.../m.
The m, in turn, enables "multiline" mode.
CORRECTED:
Here's where I got confused in the original answer, because this option has different meanings in different environments.
This question is tagged Ruby, in which "multiline mode" causes the dot character (.) to match newlines, whereas normally that's the one character it doesn't match:
irb(main):001:0> "a\nb" =~ /a.b/
=> nil
irb(main):002:0> "a\nb" =~ /a.b/m
=> 0
irb(main):003:0> "a\nb" =~ /(?m:a.b)/
=> 0
So your first regular expression, (?m:.*?) will match any number (including zero) of any characters (including newlines). Basically, it will match anything at all, including nothing.
In the second regular expression, (?m:\s*), the modifier has no effect at all because there are no dots in the contained expression to modify.
Back to the first expression. As Ωmega says, the ? after the * means that it is a non-greedy match. If that were the whole expression, or if there were no captures, it wouldn't matter. But when something follows that section and there are captures, you get different results. Without the ?, the longest possible match wins:
irb(main):001:0> /<(.*)>/.match("<a><b>")[1]
=> "a><b"
With the ?, you get the shortest one instead:
irb(main):002:0> /<(.*?)>/.match("<a><b>")[1]
=> "a"
Finally, about the above-mentioned /m confusion (though if you want to avoid becoming confused yourself, this might be a good place to stop reading):
In Perl 5 (which is the source of most regular expression extensions beyond the basic syntax), the behavior triggered by /m in Ruby is instead triggered by the /s option (which Ruby doesn't have, though if you put one on your regex it will silently ignore it). In Perl, /m, despite still being called "multiline mode", has a completely different effect: it causes the ^ and $ anchors to match at newlines within the string as well as at the beginning and end of the whole string respectively. But in Ruby, that behavior is the default, and there's not even an option to change it.
Pattern .*? will match any string, but as short string as possible, as there is a lazy operator ?.
Pattern \s* will match white-space characters (zero of more).
(?m) enables "multi-line mode". In this mode, the caret and dollar match before and after newlines in the subject string. To apply this mode to some sub-pattern only, sytax (?m:...) is used, where ... is a matching pattern.
For more information read http://www.regular-expressions.info/modifiers.html
Related
I want to match the path "/". I've tried the following alternatives, and the first two do match, but I don't know why the third doesn't:
/\A\/\z/.match("/") # <MatchData "/">
"/\A\/\z/".match("/") # <MatchData "/">
Regexp.new("/\A\/\z/").match("/") # nil
What's going on here? Why are they different?
The first snippet is the only correct one.
The second example is... misleading. That string literal "/\A\/\z/" is, obviously, not a regex. It's a string. Strings have #match method which converts its argument to a regexp (if not already one) and match against it. So, in this example, it's '/' that is the regular expression, and it matches a forward slash found in the other string.
The third line is completely broken: don't need the surrounding slashes there, they are part of regex literal, which you didn't use. Also use single quoted strings, not double quoted (which try to interpret escape sequences like \A)
Regexp.new('\A/\z').match("/") # => #<MatchData "/">
And, of course, none of the above is needed if you just want to check if a string consists of only one forward slash. Just use the equality check in this case.
s == '/'
I am trying to create a regex that matches a pattern in some part of a string, but not in another part of the string.
I am trying to match a substring that
(i) is surrounded by a balanced pair of one or more consecutive backticks `
(ii) and does not include as many consecutive backticks as in the surrounding patterns
(iii) where the surrounding patterns (sequence of backticks) are not adjacent to other backticks.
This is some variant of the syntax of inline code notation in Markdown syntax.
Examples of matches are as follows:
"xxx`foo`yyy" # => matches "foo"
"xxx``foo`bar`baz``yyy" # => matches "foo`bar`baz"
"xxx```foo``bar``baz```yyy" # => matches "foo``bar``baz"
One regex to achieve this is:
/(?<!`)(?<backticks>`+)(?<inline>.+?)\k<backticks>(?!`)/
which uses a non-greedy match.
I was wondering if I can get rid of the non-greedy match.
The idea comes from when the prohibited pattern is a single character. When I want to match a substring that is surrounded by a single quote ' that does not include a single quote in it, I can do either:
/'.+?'/
/'[^']+'/
The first one uses non-greedy match, and the second one uses an explicit non-matching pattern [^'].
I am wondering if it is possible to have something like the second form when the prohibited pattern is not a single character.
Going back to the original issue, there is negative lookahead syntax(?!), but I cannot restrict its effective scope. If I make my regex like this:
/(?<!`)(?<backticks>`+)(?<inline>(?!.*\k<backticks>).*)\k<backticks>(?!`)/
then the effect of (?!.*\k<backticks>) will not be limited to within (?<inline>...), but will extend to the whole string. And since that contradicts with the \k<backticks> at the end, the regex fails to match.
Is there a regex technique to ensure non-matching of a pattern (not-necessarily a single character) within a certain scope?
You can search for one or more characters which aren't the first character of a delimiter:
/(?<!`)(?<backticks>`+)(?<inline>(?:(?!\k<backticks>).)+)\k<backticks>(?!`)/
def test_period_is_a_shortcut_for_any_non_newline_character
assert_equal "abc123", "abc\n123"[/.*/m]
end
This test does not pass, as the expression [/.*/m] returns "abc\n123" instead of "abc123". What am I doing wrong? I am using . for non-newline characters, * to match all and m for multiline and yet it returns \n as well.
If you don't want to match the newline, then remove the /m from the test
From http://www.regular-expressions.info/ruby.html :
/m makes the dot match newlines. Ruby indeed uses /m, whereas Perl and many other programming languages use /s for "dot matches newlines".
Or indeed from the official documentation:
/./ - Any character except a newline.
/./m - Any character (the m modifier enables multiline mode)
As it stands, your test is an impossible one, you're trying to test that Ruby does the opposite of what it's designed to do (which, luckily, it doesn't).
Perhaps it would be a solution to go solve the problem from the opposite direction, and removing the particular string you don't want.
"abc\n123".gsub(/\n/, '')
Let's say I have a regexp with some arbitrary amount of capturing groups:
pattern = /(some)| ..a lot of masks combined.... |(other)/
Is there any way to determine a number of that groups?
If you can always find a string that matches the regex you are given, then it suffices to match it against the regex, and look at the match data length. However, determining whether a regexp has a string that it matches is np-hard[1]. This is only feasible if you know in advance what kind of regexes you'll be getting.
The next best best method in the Regexp class is Regexp#source or Regexp#to_s. However, we need to parse the regex if we do this.
I can't speak for the future, but as of Ruby 2.0, there is no better method in the Regexp core class.
A left parenthesis denotes a literal left parenthesis, if preceded by an unescaped backslash. A backslash is unescaped unless an unescaped backslash precedes. So, a character is escaped iff preceded by an odd number of backslashes.
An unescaped left parenthesis denotes a capturing group iff not followed by a question mark. With a question mark, it can mean various things: (?'name') and (?<name>) denote a named capturing group. Named and unnamed capturing groups cannot coexist in the same regex, however[2]. (?:) denote non-capturing groups. This is a special case of (?flags-flags:). (?>) denote atomic groups. (?=), (?!), (?<=) and (?<!) denote lookaround. (?#) denote comments.
Ruby regexp engine supports comments in regexes. Considering them in the main regex would be very difficult. We can try to strip them if we really want to support these, but supporting them fully will get messy due to the possibility of inline flags turning extended mode (and thus line comments) on and off in ways that a regular expression cannot capture. I will go ahead and not support unescaped parentheses in regex comments[3].
We want to count:
the number of left parentheses \(
that are not escaped by a backslash (?<!(?<!\\)(?:\\\\)*\\) (read: not preceded by an odd number of backslashes that are not preceded by yet another backslash) and
that are not followed by a question mark (?!\?)
Ruby doesn't support unbounded lookbehind, but if we reverse the source first, we can rewrite the first assertion slightly: (?!(?:\\\\)*(?!\\)). The second assertion becomes a lookbehind: (?<!\?).
the whole solution
def count_groups(regexp)
# named capture support:
# named_count = regexp.named_captures.count
# return named_count if named_count > 0
# main:
test = /(?!<\?)\((?!(?:\\\\)*(?!\\))/
regexp.source.scan(test).count
end
[1]: we can show the NP-hardness by converting the satisfiability problem to it:
AND: xy (x must be an assertion)
OR: x|y
NOT: (?!x)
atoms: (?=1), (?=.1), (?=..1), ..., (?!1), (?!.1)...
example(XOR): /^(?:(?=1)(?!.1)|(?!1)(?=.1))..$/
this extends to NP-completeness for any class of regexes that can be tested in polynomial time. This includes any regex with no nested repetition (or repeated backreferences to repetition or recursion) and with bounded nesting depth of optional matches.
[2]: /((?<name>..)..)../.match('abcdef').to_a returns ['abcdef', 'ab'], indicating that unnamed capturing groups are ignored when named capturing groups are present. Tested in Ruby 1.9.3
[3]: Inline comments start with (?# and end with ). They cannot contain an unescaped right parenthesis, but they can contain an unescaped left parenthesis. These can be stripped easily (even though we have to sprinkle the "unescaped" regex everywhere), are the lesser evil, but they're also less likely to contain anunescaped left parenthesis.
Line comments start with # and end with a newline. These are only treated as comments in the extended mode. Outside the extended mode, they match the literal # and newline. This is still easy, even if we have to consider escaping again. Determining if the regex has the extended flag set is not too difficult, but the flag modifier groups are a different beast entirely.
Even with Ruby's awesome recursive regexes, merely determining if a previously-open group modifying the extended mode is already closed would yield a very nasty regex (even if you replace one by one and don't have to skip comments, you have to account for escaping). It wouldn't be pretty (even with interpolation) and it wouldn't be fast.
Using Ruby, I am writing a regular expression, and I need to be a able to remove any colon that appears between parentheses. I understand that I can use
"This is a (string :)".sub!(/\([^\)]*:/, '')
to do this, but the problem is that this function will also remove the context along with it. Is there any way to specify that I only want it to remove the colon and not the entire matching expression?
So some regular expression engines support what are called look-ahead and look-behind matches that will match but not consume characters. Ruby does support look-ahead, but not look-behind (which is more difficult to do in a performant way), which means you could quite easily stick with sub and remove a colon that precedes a closing parenthesis, but only without ensuring it is after an opening parenthesis:
string = 'This is a (string :)'
string.sub /:(?=\))/, ''
# => 'This is a (string )'
The alternative would be to use subpattern capturing (which happens automatically when you use grouping in your regular expression) to rebuild the string without the undesirable portion, in this case the colon:
string.sub /(\([^:]+):\)/, '\1)'
The \1 is a back-reference to what is matched in the first group, which is delimited by the parentheses that are not escaped. You can see here I didn't bother capturing the closing parenthesis in a second group, opting instead simply to include it in the substitution. This works well in this case because it will not change, but if you don't know that the colon will appear at the end of the parentheses-enclosed content, you would need a second group:
string.sub /(\([^:]+):([^)]+\))/, '\1\2'
The prior answer will mostly work for deleting single colons within paren groups, but have trouble with multiples like '(thing:foo:bar)`. It would be nice to use lookbehind and lookahead to make the within parens assertion, but ruby (and most regexp engines) doesn't support non-deterministic length patterns in lookbehind.
irb> s = 'x (a:b:c) : (1:2:3) y'
=> "x (a:b:c) : (1:2:3) y"
irb> s.gsub /(?<=\([^\(]*):(?=[^\)]*\))/, ''
SyntaxError: (irb):10: invalid pattern in look-behind: /(?<=\([^\(]*):(?=[^\)]*\))/
from /Users/dbenhur/.rbenv/versions/1.9.2-wp/bin/irb:12:in `<main>'
You could instead use the block form of gsub to capture paren enclosed groups, then remove colons from each match:
irb> s.gsub(/\([^\)]*\)/) {|m| m.delete ':'}
=> "x (abc) : (123) y"
in regex in general, you can use '(\()(:)(\))', \1\3.
I'm not familiar with Ruby. Basically what you do is you have 3 groups, and from this three groups ( : and ) you get rid of the second one, the :.
I tested it in Notepad++ and it works.
I think this is called: regex backreference
Cheers.
If you can assume all parentheses will come in balanced pairs like they do in your example, this should be all you need:
"This is a (string :)".gsub!(/:(?=[^()]*\))/, '')
If the lookahead succeeds in finding a closing paren without seeing an opening paren first, the colon must be inside a (...) sequence. Notice how I excluded the opening paren as well as the closing paren; that's essential.