I'm looking for words starting with a hashtag: "#yolo"
My regex for this was very simple: /#\w+/
This worked fine until I hit words that ended with a question mark: "#yolo?".
I updated my regex to allow for words and any non whitespace character as well: /#[\w\S]*/.
The problem is I sometimes need to pull a match from a word starting with two '#' characters, up until whitespace, that may contain a special character in it or at the end of the word (which I need to capture).
Example:
"##yolo?"
And I would like to end up with:
"#yolo?"
Note: the regular expressions are for Ruby.
P.S. I'm testing these out here: http://rubular.com/
Maybe this would work
#(#?[\S]+)
What about
#[^#\s]+
\w is a subset of ^\s (i.e. \S) so you don't need both. Also, I assume you don't want any more #s in the match, so we use [^#\s] which negates both whitespace and # characters.
Related
So, there are a number of regular expression which matches a particular group like the following:
/./ - Any character except a newline.
/./m - Any character (the m modifier enables multiline mode)
/\w/ - A word character ([a-zA-Z0-9_])
/\s/ - Any whitespace character
And in ruby:
/[[:punct:]]/ - Punctuation character
/[[:space:]]/ - Whitespace character ([:blank:], newline, carriage return, etc.)
/[[:upper:]]/ - Uppercase alphabetical
So, here is my question: how do I get a regexp to match a group like this, but exempt a character out?
Examples:
match all punctuations apart from the question mark
match all whitespace characters apart from the new line
match all words apart from "go"... etc
Thanks.
You can use character class subtraction.
Rexegg:
The syntax […&&[…]] allows you to use a logical AND on several character classes to ensure that a character is present in them all. Intersecting with a negated character, as in […&&[^…]] allows you to subtract that class from the original class.
Consider this code:
s = "./?!"
res = s.scan(/[[:punct:]&&[^!]]/)
puts res
Output is only ., / and ? since ! is excluded.
Restricting with a lookahead (as sawa has written just now) is also possible, but is not required when you have this subtraction supported. When you need to restrict some longer values (more than 1 character) a lookahead is required.
In many cases, a lookahead must be anchored to a word boundary to return correct results. As an example of using a lookahead to restrict punctuation (single character matching generic pattern):
/(?:(?!!)[[:punct:]])+/
This will match 1 or more punctuation symbols but a !.
The puts "./?!".scan(/(?:(?!!)[[:punct:]])+/) code will output ./? (see demo)
Use character class subtraction whenever you need to restrict with single characters, it is more efficient than using lookaheads.
So, the 3rd scenario regex must look like:
/\b(?!go\b)\w+\b/
^^
If you write /(?!\bgo\b)\b\w+\b/, the regex engine will check each position in the input string. If you use a \b at the beginning, only word boundary positions will be checked, and the pattern will yield better performance. Also note that the ^^ \b is very important since it makes the regex engine check for the whole word go. If you remove it, it will only restrict to the words that do not start with go.
Put what you want to exclude inside a negative lookahead in front of the match. For example,
To match all punctuations apart from the question mark,
/(?!\?)[[:punct:]]/
To match all words apart from "go",
/(?!\bgo\b)\b\w+\b/
This is a general approach that is sometimes useful:
a = []
".?!,:;-".scan(/[[:punct:]]/) { |s| a << s unless s == '?' }
a #=> [".", "!", ",", ":", ";", "-"]
The content of the block is limited only by your imagination.
So I have a string that looks like this:
#jackie#test.com, #mike#test.com
What I want to do is before any email in this comma separated list, I want to remove the #. The issue I keep running into is that if I try to do a regular \A flag like so /[\A#]+/, it finds all the instances of # in that string...including the middle crucial #.
The same thing happens if I do /[\s#]+/. I can't figure out how to just look at the beginning of each string, where each string is a complete email address.
Edit 1
Note that all I need is the regex, I already have the rest of the stuff I need to do what I want. Specifically, I am achieving everything else like this:
str.gsub(/#/, '').split(',').map(&:strip)
Where str is my string.
All I am looking for is the regex portion for my gsub.
You may use the below negative lookbehind based regex.
str.gsub(/(?<!\S)#/, '').split(',').map(&:strip)
(?<!\S) Negative lookbehind asserts that the character or substring we are going to match would be preceeded by any but not of a non-space character. So this matches the # which exists at the start or the # which exists next to a space character.
Difference between my answer and hwnd's str.gsub(/\B#/, '') is, mine won't match the # which exists in :# but hwnd's answer does. \B matches between two word characters or two non-word characters.
Here is one solution
str = "#jackie#test.com, #mike#test.com"
p str.split(/,[ ]+/).map{ |i| i.gsub(/^#/, '')}
Output
["jackie#test.com", "mike#test.com"]
I am new to ruby and I'm trying to work with regex.
I have a text which looks something like:
HEADING
Some text which is always non capitalized. Headings are always capitalized, followed by a space or nothing more.
YOU CAN HAVE MULTIPLE WORDS IN HEADING
I'm using this regular expression to choose all headings:
^[A-Z]{2,}\s?([A-Z]{2,}\s?)*$
However, it matches all headings which does not contain chars as Č, Š, Ž(slovenian characters).
So I'm guessing [A-Z] only matches ASCII characters? How could I get utf8?
You are right in that when you define the ASCII range A-Z, the match is made literally only for those characters. This is to do with the history of characters on computers, more and more characters have been added over time, and they are not always structured in an encoding in ways that are easy to use.
You could make a larger character class that matches the slovenian characters you need, by listing them.
But there is a shortcut. Someone else has already added necessary data to the Unicode data so that you can write shorter matches for "all uppercase characters": /[[:upper:]]/. See http://ruby-doc.org//core-2.1.4/Regexp.html for more.
Altering your regular expression with just this adjustment:
^[[:upper:]]{2,}\s?([[:upper:]]{2,}\s?)*$
You may need to adjust it further, for instance it would not match the heading "I AM A HEADING" due to the match insisting each word is at least two letters long.
Without seeing all your examples, I would probably simplify the group matching and just allow spaces anywhere:
^[[:upper:]\s]+$
You can use unicode upper case letter:
\p{Lu}
Your regex:
\b\p{Lu}{2,}(?:\s*\p{Lu}{2,})\b
RegEx Demo
I want to extract #hashtags from a string, also those that have special characters such as #1+1.
Currently I'm using:
#hashtags ||= string.scan(/#\w+/)
But it doesn't work with those special characters. Also, I want it to be UTF-8 compatible.
How do I do this?
EDIT:
If the last character is a special character it should be removed, such as #hashtag, #hashtag. #hashtag! #hashtag? etc...
Also, the hash sign at the beginning should be removed.
The Solution
You probably want something like:
'#hash+tag'.encode('UTF-8').scan /\b(?<=#)[^#[:punct:]]+\b/
=> ["hash+tag"]
Note that the zero-width assertion at the beginning is required to avoid capturing the pound sign as part of the match.
References
String#encode
Ruby's POSIX Character Classes
This should work:
#hashtags = str.scan(/#([[:graph:]]*[[:alnum:]])/).flatten
Or if you don't want your hashtag to start with a special character:
#hashtags = str.scan(/#((?:[[:alnum:]][[:graph:]]*)?[[:alnum:]])/).flatten
How about this:
#hashtags ||=string.match(/(#[[:alpha:]]+)|#[\d\+-]+\d+/).to_s[1..-1]
Takes cares of #alphabets or #2323+2323 #2323-2323 #2323+65656-67676
Also removes # at beginning
Or if you want it in array form:
#hashtags ||=string.scan(/#[[:alpha:]]+|#[\d\+-]+\d+/).collect{|x| x[1..-1]}
Wow, this took so long but I still don't understand why scan(/#[[:alpha:]]+|#[\d\+-]+\d+/) works but not scan(/(#[[:alpha:]]+)|#[\d\+-]+\d+/) in my computer. The difference being the () on the 2nd scan statement. This has no effect as it should be when I use with match method.
Is there anyway to scan only if there is nothing before what I am scanning for.
For example I have a post and I am scanning for a forward slash and what follows it but I do not want to scan for a forward slash if it is not the beginning character.
I want to scan for /this but I do not want to scan for this/this or http://this.com.
The regular expression I am currently using is..
/\/(\w+)/
I am using this with gsub to link each /forwardslash.
I think what you are asking for is to only match words that begin with '/', not strings or lines beginning with '/'. If that is true, I believe the following regex will work: %r{(?:^|\s+)/(\w+)}:
For example:
"/foo /this this/that http://this".scan %r{(?:^|\s+)/(\w+)} # => [["foo"], ["this"]]
The caret (^) character means "beginning of string" -- a dollar sign ($) means "end of string."
So
/^\/(\w+)/
...will get you what you want -- only matching at the beginning of the string.
First thing, since you're using a regex with slashes change the delimiter to something else, then you won't have to escape the backslashes and it will be easier to read.
Secondly, if you want to replace the slash as well then include it in the capture.
On to the regex.
...if it is not the beginning
character...
...of a line:
!^(/\w+)!
if it is not the beginning
character...
...of a word:
!\s(/\w+)!
but that won't match if it's at the very beginning of a line. For that you'll need something a lot more complex, so I'd just run both the regexes here instead of creating that monster.