I was wondering if it was possible to use negative matching on whole words, so that something like [^(<em>.*?<\/em>)] would match everything but text between (and including) <em>...</em>.
I was thinking about using negative lookahead, but I don't think this will work, as I need to check for the opening <em> as well.
Of course, I could just use the positive regex and then subtract the matches from the original text, but I'm looking for a more 'elegant' solution.
thx for any help
String#split works as negative match. It returns you an array of whatever part that does not match the regex.
'XXXXXXXX<em>YYYYYYY</em>ZZZZZZZZ'.split(%r|<em>.*?</em>|)
# => ['XXXXXXX', 'ZZZZZZZZ']
And if want it back into a string, just do join.
'XXXXXXXX<em>YYYYYYY</em>ZZZZZZZZ'.split(%r|<em>.*?</em>|).join
# => 'XXXXXXXZZZZZZZZ'
The whole thing with lookaround is that it doesn't consume any of the input. If you want to match everything but a pattern, it means that you want to match the prefix and the suffix of that pattern. To match the suffix, you probably want to consume --- and throw away -- the pattern that you don't want. But negative lookahead doesn't consume.
Related
I am reviewing regular expressions and cannot understand why a regular expression won't match a given string, specifically:
regex = /(ab*)+(bc)?/
mystring = "abbc"
The match matches "abb" but leaves the c off. I tested this using Rubular and in IRB and don't understand why the regex doesn't match the entire string. I thought that (ab*)+ would match "ab" and then (bc)? would match "bc".
Am I missing something in terms of precedence for regular expression operations?
Regular expressions try to match the first part of the regular expression as much as possible by default, and they do not backtrack to try to make larger sections match if they don't have to. Since you make (bc) optional, the (ab*) can match as much as it wants (the non-zero repetition after it doesn't have much to do) and doesn't try backtracking to try other matching alternatives.
If you want the whole string to be matched (which will force some backtracking in this case) make sure you anchor both ends of the string:
regex = /^(ab*)+(bc)?$/
The regex with parenthesis assumes you have two matches in your string.
The first one is abb because (ab*) means a and zero or more b. You have two b, so the match is abb. Then you have only c in your string, so it doesn't match the second condition which is bc.
I'm trying to match some text if it does not have another block of text in its vicinity. For example, I would like to match "bar" if "foo" does not precede it. I can match "bar" if "foo" does not immediately precede it using negative look behind in this regex:
/(?<!foo)bar/
but I also like to not match "foo 12345 bar". I tried:
/(?<!foo.{1,10})bar/
but using a wildcard + a range appears to be an invalid regex in Ruby. Am I thinking about the problem wrong?
You are thinking about it the right way. But unfortunately lookbehinds usually have be of fixed-length. The only major exception to that is .NET's regex engine, which allows repetition quantifiers inside lookbehinds. But since you only need a negative lookbehind and not a lookahead, too. There is a hack for you. Reverse the string, then try to match:
/rab(?!.{0,10}oof)/
Then reverse the result of the match or subtract the matching position from the string's length, if that's what you are after.
Now from the regex you have given, I suppose that this was only a simplified version of what you actually need. Of course, if bar is a complex pattern itself, some more thought needs to go into how to reverse it correctly.
Note that if your pattern required both variable-length lookbehinds and lookaheads, you would have a harder time solving this. Also, in your case, it would be possible to deconstruct your lookbehind into multiple variable length ones (because you use neither + nor *):
/(?<!foo)(?<!foo.)(?<!foo.{2})(?<!foo.{3})(?<!foo.{4})(?<!foo.{5})(?<!foo.{6})(?<!foo.{7})(?<!foo.{8})(?<!foo.{9})(?<!foo.{10})bar/
But that's not all that nice, is it?
As m.buettner already mentions, lookbehind in Ruby regex has to be of fixed length, and is described so in the document. So, you cannot put a quantifier within a lookbehind.
You don't need to check all in one step. Try doing multiple steps of regex matches to get what you want. Assuming that existence of foo in front of a single instance of bar breaks the condition regardless of whether there is another bar, then
string.match(/bar/) and !string.match(/foo.*bar/)
will give you what you want for the example.
If you rather want the match to succeed with bar foo bar, then you can do this
string.scan(/foo|bar/).first == "bar"
I have the following
address.gsub(/^\d*/, "").gsub(/\d*-?\d*$/, "").gsub(/\# ?\d*/,"")
Can this be done in one gsub? I would like to pass a list of patterns rather then just one pattern - they are all being replaced by the same thing.
You could combine them with an alternation operator (|):
address = '6 66-666 #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/, "")
# " 66-666 "
address = 'pancakes 6 66-666 # pancakes #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/,"")
# "pancakes 6 66-666 pancakes "
You might want to add little more whitespace cleanup. And you might want to switch to one of:
/\A\d*|\d*-?\d*\z|\# ?\d*/
/\A\d*|\d*-?\d*\Z|\# ?\d*/
depending on what your data really looks like and how you need to handle newlines.
Combining the regexes is a good idea--and relatively simple--but I'd like to recommend some additional changes. To wit:
address.gsub(/^\d+|\d+(?:-\d+)?$|\# *\d+/, "")
Of your original regexes, ^\d* and \d*-?\d*$ will always match, because they don't have to consume any characters. So you're guaranteed to perform two replacements on every line, even if that's just replacing empty strings with empty strings. Of my regexes, ^\d+ doesn't bother to match unless there's at least one digit at the beginning of the line, and \d+(?:-\d+)?$ matches what looks like an integer-or-range expression at the end of the line.
Your third regex, \# ?\d*, will match any # character, and if the # is followed by a space and some digits, it'll take those as well. Judging by your other regexes and my experience with other questions, I suspect you meant to match a # only if it's followed by one or more digits, with optional spaces intervening. That's what my third regex does.
If any of my guesses are wrong, please describe what you were trying to do, and I'll do my best to come up with the right regex. But I really don't think those first two regexes, at least, are what you want.
EDIT (in answer to the comment): When working with regexes, you should always be aware of the distinction between a regex the matches nothing and a regex that doesn't match. You say you're applying the regexes to street addresses. If an address doesn't happen to start with a house number, ^\d* will match nothing--that is, it will report a successful match, said match consisting of the empty string preceding the first character in the address.
That doesn't matter to you, you're just replacing it with another empty string anyway. But why bother doing the replacement at all? If you change the regex to ^\d+, it will report a failed match and no replacement will be performed. The result is the same either way, but the "matches noting" scenario (^\d*) results in a lot of extra work that the "doesn't match" scenario avoids. In a high-throughput situation, that could be a life-saver.
The other two regexes bring additional complications: \d*-?\d*$ could match a hyphen at the end of the string (e.g. "123-", or even "-"); and \# ?\d* could match a hash symbol anywhere in string, not just as part of an apartment/office number. You know your data, so you probably know neither of those problems will ever arise; I'm just making sure you're aware of them. My regex \d+(?:-\d+)?$ deals with the trailing-hyphen issue, and \# *\d+ at least makes sure there are digits after the hash symbol.
I think that if you combine them together in a single gsub() regex, as an alternation,
it changes the context of the starting search position.
Example, each of these lines start at the beginning of the result of the previous
regex substitution.
s/^\d*//g
s/\d*-?\d*$//g
s/\# ?\d*//g
and this
s/^\d*|\d*-?\d*$|\# ?\d*//g
resumes search/replace where the last match left off and could potentially produce a different overall output, especially since a lot of the subexpressions search for similar
if not the same characters, distinguished only by line anchors.
I think your regex's are unique enough in this case, and of course changing the order
changes the result.
I am getting completely different reults from string.scan and several regex testers...
I am just trying to grab the domain from the string, it is the last word.
The regex in question:
/([a-zA-Z0-9\-]*\.)*\w{1,4}$/
The string (1 single line, verified in Ruby's runtime btw)
str = 'Show more results from software.informer.com'
Work fine, but in ruby....
irb(main):050:0> str.scan /([a-zA-Z0-9\-]*\.)*\w{1,4}$/
=> [["informer."]]
I would think that I would get a match on software.informer.com ,which is my goal.
Your regex is correct, the result has to do with the way String#scan behaves. From the official documentation:
"If the pattern contains groups, each individual result is itself an array containing one entry per group."
Basically, if you put parentheses around the whole regex, the first element of each array in your results will be what you expect.
It does not look as if you expect more than one result (especially as the regex is anchored). In that case there is no reason to use scan.
'Show more results from software.informer.com'[ /([a-zA-Z0-9\-]*\.)*\w{1,4}$/ ]
#=> "software.informer.com"
If you do need to use scan (in which case you obviously need to remove the anchor), you can use (?:) to create non-capturing groups.
'foo.bar.baz lala software.informer.com'.scan( /(?:[a-zA-Z0-9\-]*\.)*\w{1,4}/ )
#=> ["foo.bar.baz", "lala", "software.informer.com"]
You are getting a match on software.informer.com. Check the value of $&. The return of scan is an array of the captured groups. Add capturing parentheses around the suffix, and you'll get the .com as part of the return value from scan as well.
The regex testers and Ruby are not disagreeing about the fundamental issue (the regex itself). Rather, their interfaces are differing in what they are emphasizing. When you run scan in irb, the first thing you'll see is the return value from scan (an Array of the captured subpatterns), which is not the same thing as the matched text. Regex testers are most likely oriented toward displaying the matched text.
How about doing this :
/([a-zA-Z0-9\-]*\.*\w{1,4})$/
This returns
informer.com
On your test string.
http://rubular.com/regexes/13670
I would like to replace only the group in parenthesis in this expression :
my_string.gsub(/<--MARKER_START-->(.)*<--MARKER_END-->/, 'replace_text')
so that I get : <--MARKER_START-->replace_text<--MARKER_END-->
I know I could repeat the whole MARKER_START and MARKER_END blocks in the substitution expression but I thought there should be a more simple way to do this.
You can do it with zero width look-ahead and look-behind assertions.
This regex should work in ruby 1.9 and in perl and many other places:
Note: ruby 1.8 only supports look-ahead assertions. You need both look-ahead and look-behind to do this properly.
s.gsub( /(?<=<--MARKER START-->).*?(?=<--MARKER END-->)/, 'replacement text' )
What happens in ruby 1.8 is the ?<= causes it to crash because it doesn't understand the look-behind assertion. For that part, you then have to fall back to using a backreference - like Greig Hewgill mentions
so what you get is
s.gsub( /(<--MARKER START-->).*?(?=<--MARKER END-->)/, '\1replacement text' )
EXPLANATION THE FIRST:
I've replaced the (.)* in the middle of your regex with .*? - this is non-greedy.
If you don't have non-greedy, then your regex will try and match as much as it can - if you have 2 markers on one line, it goes wrong. This is best illustrated by example:
"<b>One</b> Two <b>Three</b>".gsub( /<b>.*<\/b>/, 'BOLD' )
=> "BOLD"
What we actually want:
"<b>One</b> Two <b>Three</b>".gsub( /<b>.*?<\/b>/, 'BOLD' )
=> "BOLD Two BOLD"
EXPLANATION THE SECOND:
zero-width-look-ahead-assertion sounds like a giant pile of nerdly confusion.
What "look-ahead-assertion" actually means is "Only match, if the thing we are looking for, is followed by this other stuff.
For example, only match a digit, if it is followed by an F.
"123F" =~ /\d(?=F)/ # will match the 3, but not the 1 or the 2
What "zero width" actually means is "consider the 'followed by' in our search, but don't count it as part of the match when doing replacement or grouping or things like that.
Using the same example of 123F, If we didn't use the lookahead assertion, and instead just do this:
"123F" =~ /\dF/ # will match 3F, because F is considered part of the match
As you can see, this is ideal for checking for our <--MARKER END-->, but what we need for the <--MARKER START--> is the ability to say "Only match, if the thing we are looking for FOLLOWS this other stuff". That's called a look-behind assertion, which ruby 1.8 doesn't have for some strange reason..
Hope that makes sense :-)
PS: Why use lookahead assertions instead of just backreferences? If you use lookahead, you're not actually replacing the <--MARKER--> bits, only the contents. If you use backreferences, you are replacing the whole lot. I don't know if this incurs much of a performance hit, but from a programming point of view it seems like the right thing to do, as we don't actually want to be replacing the markers at all.
You could do something like this:
my_string.gsub(/(<--MARKER_START-->)(.*)(<--MARKER_END-->)/, '\1replace_text\3')