I have a string with exclamation marks. I want to remove the exclamation marks at the end of the word, not the ones before a word. Assume there is no exclamation mark by itself/ not accompanied by a word. By word I mean [a..z], can be uppercased.
For example:
exclamation("Hello world!!!")
#=> ("Hello world")
exclamation("!!Hello !world!")
#=> ("!!Hello !world")
I have read How do I remove substring after a certain character in a string using Ruby? ; these two are close, but different.
def exclamation(s)
s.slice(0..(s.index(/\w/)))
end
# exclamation("Hola!") returns "Hol"
I have also tried s.gsub(/\w!+/, ''). Although it retains the '!' before word, it removes both the last letter and exclamation mark. exclamation("!Hola!!") #=> "!Hol".
How can I remove only the exclamation marks at the end?
If you don't want to use regex that sometimes difficult to understand use this:
def exclamation(sentence)
words = sentence.split
words_wo_exclams = words.map do |word|
word.split('').reverse.drop_while { |c| c == '!' }.reverse.join
end
words_wo_exclams.join(' ')
end
Although you haven't given a lot of test data, here's an example of something that might work:
def exclamation(string)
string.gsub(/(\w+)\!(?=\s|\z)/, '\1')
end
The \s|\z part means either a space or the end of the string, and (?=...) means to just peek ahead in the string but not actually match against it.
Note that this won't work in the case of things like "I'm mad!" where the exclamation mark is not adjacent to a space, but you could always add that as another potential end-of-word match.
"!!Hello !world!, world!! I say".gsub(r, '')
#=> "!!Hello !world, world! I say"
where
r = /
(?<=[[:alpha:]]) # match an uppercase or lowercase letter in a positive lookbehind
! # match an exclamation mark
/x # free-spacing regex definition mode
or
r = /
[[:alpha:]] # match an uppercase or lowercase letter
\K # discard match so far
! # match an exclamation mark
/x # free-spacing regex definition mode
If the above example should return "!!Hello !world, world I say", change ! to !+ in the regexes.
Related
I have a string
a="Tamilnadu is far away from Kashmir"
If I split this string using "Tamilnadu", then I don't find Tamilnadu as a part of the array, I find empty string there, If I split the string "away" then away is not present in the result array, it's having empty string in the place of away. What should I do include it instead of having empty string.
Example
a="Tamilnadu is far away from Kashmir"
p a.split("Tamilnadu")
then Output is
["", " is far away from Kashmir"]
But I want
["Tamilnadu", " is far away from Kashmir"]
From docs:
If pattern is a Regexp, str is divided where the pattern matches. Whenever the pattern matches a zero-length string, str is split into individual characters. If pattern contains groups, the respective matches will be returned in the array as well.
So... to split by "Tamilnadu" and keep it in the list, make it a capture group:
"Tamilnadu is far away from Kashmir".split(/(Tamilnadu)/)
# => ["", "Tamilnadu", " is far away from Kashmir"]
or, if you want to split after "Tamilnadu", make a zero-width match after it using lookbehind:
"Tamilnadu is far away from Kashmir".split(/(?<=Tamilnadu)/)
# => ["Tamilnadu", " is far away from Kashmir"]
If you don't know where "Tamilnadu" is in the string but you want to split the string before and after it, and not have any empty strings in the resulting array, you can use String#scan:
def split_it(str, substring)
str.scan(/\A.+(?= #{substring}\b)|\b#{substring}\b|(?<=\b#{substring} ).+/)
end
substring = "Tamilnadu"
split_it("Tamilnadu is far away from Kashmir", substring)
#=> ["Tamilnadu", "is far away from Kashmir"]
split_it("Far away is Tamilnadu from Kashmir", substring)
#=> ["Far away is", "Tamilnadu", "from Kashmir"]
split_it("Far away from Kashmir is Tamilnadu", substring)
#=> ["Far away from Kashmir is", "Tamilnadu"]
split_it("Far away is Daluth from Kashmir", substring)
#=> []
split_it("Far away is Tamilnaduland from Kashmir", substring)
#=> []
I've assumed that substring appears at most once in the string.
The regular expression can be written in free-spacing mode to make it self-documenting:
substring = "Tamilnadu"
/
\A.+ # match the beginning of the string followed by > 0 characters
(?=\ #{substring}\b) # match the value of substring preceded by a space and
# followed by a word break, in a positive lookahead
| # or
\b#{substring}\b # match the value of substring with a word break before and after
| # or
(?<=\b#{substring}\ ) # match the value of substring preceded by a word break
# and followed by a space, in a positive lookbehind
.+ # match > 0 characters
/x # free-spacing regex definition mode
#=>
/
\A.+ # ...
(?=\ Tamilnadu\b) # ...
| # ...
\bTamilnadu\b # ...
| # ...
(?<=\bTamilnadu\ ) # ...
.+ # ...
/x
Free-spacing mode removes all spaces before the regex is parsed, including spaces that may be intended to be part of the expression. It was for that reason that I escaped the two spaces. I could alternatively put each in a character class ([ ]) or use \s, [[:space:]] or \p{Space}, though they match whitespace, which is not quite the same.
I am trying to come up with a Ruby Regex that will match the following string:
MAINT: Refactor something
STRY-1: Add something
STRY-2: Update something
But should not match the following:
MAINT: Refactored something
STRY-1: Added something
STRY-2: Updated something
MAINT: Refactoring something
STRY-3: Adding something
STRY-4: Updating something
Basically, the first word after : should not end with either ed or ing
This is what I have currently:
^(MAINT|(STRY|PRB)-\d+):\s([A-Z][a-z]+)\s([a-zA-Z0-9._\-].*)
I have tried [^ed] and [^ing] but they would not work here since I am targeting more than single character.
I am not able to come up with a proper solution to achieve this.
You could use
^[-\w]+:\s*(?:(?!(?:ed|ing)\b)\w)+\b.+
See a demo on regex101.com.
Broken down this says:
^ # start of the line/string
[-\w]+:\s* # match - and word characters, 1+ then :
(?: # non-capturing group
(?!(?:ed|ing)\b) # neg. lookahead: no ed or ing followed by a word boundary
\w # match a word character
)+\b # as long as possible, followed by a boundary
.* # match the rest of the string, if any
I have no experience in Ruby but I guess you could alternatively do a split and check if the second word ends with ed or ing. The latter approach might be easier to handle for future programmers/colleagues.
r = /
\A # match beginning of string
(?: # begin a non-capture group
MAINT # match 'MAINT'
| # or
STRY\-\d+ # match 'STRY-' followed by one or more digits
) # end non-capture group
:[ ] # match a colon followed by a space
[[:alpha:]]+ # match one or more letters
(?<! # begin a negative lookbehind
ed # match 'ed'
| # or
ing # match 'ing'
) # end negative lookbehind
[ ] # match a space
/x # free-spacing regex definition mode
"MAINT: Refactor something".match?(r) #=> true
"STRY-1: Add something".match?(r) #=> true
"STRY-2: Update something".match?(r) #=> true
"MAINT: Refactored something".match?(r) #=> false
"STRY-1: Added something".match?(r) #=> false
"STRY-2: Updated something".match?(r) #=> false
"A MAINT: Refactor something".match?(r) #=> false
"STRY-1A: Add something".match?(r) #=> false
This regular expression is conventionally written as follows.
r = /\A(?:MAINT|STRY\-\d+): [[:alpha:]]+(?<!ed|ing) /
Expressed this way the two spaces can each be represented a space character. In free-spacing mode, however, all spaces outside character classes are removed, which is why I needed to enclose each space in a character class.
(Posted on behalf of the question author).
This is what I ended up using:
^(MAINT|(STRY|PRB)-\d+):\s(?:(?!(?:ed|ing)\b)[A-Za-z])+\s([a-zA-Z0-9._\-].*)
I have a list of users grabbed by the Etc Ruby library:
Thomas_J_Perkins
Jennifer_Scanner
Amanda_K_Loso
Aaron_Cole
Mark_L_Lamb
What I need to do is grab the full first name, skip the middle name (if given), and grab the first character of the last name. The output should look like this:
Thomas P
Jennifer S
Amanda L
Aaron C
Mark L
I'm not sure how to do this, I've tried grabbing all of the characters: /\w+/ but that will grab everything.
You don't always need regular expressions.
Some people, when confronted with a problem, think "I know, I'll use
regular expressions." Now they have two problems. Jamie Zawinski
You can do it with some simple Ruby code
string = "Mark_L_Lamb"
string.split('_').first + ' ' + string.split('_').last[0]
=> "Mark L"
I think its simpler without regex:
array = "Thomas_J_Perkins".split("_") # split at _
array.first + " " + array.last[0] # .first prints first name .last[0] prints first char of last name
#=> "Thomas P"
You can use
^([^\W_]+)(?:_[^\W_]+)*_([^\W_])[^\W_]*$
And replace with \1_\2. See the regex demo
The [^\W_] matches a letter or a digit. If you want to only match letters, replace [^\W_] with \p{L}.
^(\p{L}+)(?:_\p{L}+)*_(\p{L})\p{L}*$
See updated demo
The point is to match and capture the first chunk of letters up to the first _ (with (\p{L}+)), then match 0+ sequences of _ + letters inside (with (?:_\p{L}+)*_) and then match and capture the last word first letter (with (\p{L})) and then match the rest of the string (with \p{L}*).
NOTE: replace ^ with \A and $ with \z if you have independent strings (as in Ruby ^ matches the start of a line and $ matches the end of the line).
Ruby code:
s.sub(/^(\p{L}+)(?:_\p{L}+)*_(\p{L})\p{L}*$/, "\\1_\\2")
I'm in the don't-use-a-regex-for-this camp.
str1 = "Alexander_Graham_Bell"
str2 = "Sylvester_Grisby"
"#{str1[0...str1.index('_')]} #{str1[str1.rindex('_')+1]}"
#=> "Alexander B"
"#{str2[0...str2.index('_')]} #{str2[str2.rindex('_')+1]}"
#=> "Sylvester G"
or
first, last = str1.split(/_.+_|_/)
#=> ["Alexander", "Bell"]
first+' '+last[0]
#=> "Alexander B"
first, last = str2.split(/_.+_|_/)
#=> ["Sylvester", "Grisby"]
first+' '+last[0]
#=> "Sylvester G"
but if you insist...
r = /
(.+?) # match any characters non-greedily in capture group 1
(?=_) # match an underscore in a positive lookahead
(?:.*) # match any characters greedily in a non-capture group
(?:_) # match an underscore in a non-capture group
(.) # match any character in capture group 2
/x # free-spacing regex definition mode
str1 =~ r
$1+' '+$2
#=> "Alexander B"
str2 =~ r
$1+' '+$2
#=> "Sylvester G"
You can of course write
r = /(.+?)(?=_)(?:.*)(?:_)(.)/
This is my attempt:
/([a-zA-Z]+)_([a-zA-Z]+_)?([a-zA-Z])/
See demo
Let's see if this works:
/^([^_]+)(?:_\w)?_(\w)/
And then you'll have to combine the first and second matches into the format you want. I don't know Ruby, so I can't help you there.
And another attempt using a replacement method:
result = subject.gsub(/^([^_]+)(?:_[^_])?_([^_])[^_]+$/, '\1 \2')
We capture the entire string, with the relevant parts in capturing groups. Then just return the two captured groups
using the split method is much better
full_names.map do |full_name|
parts = full_name.split('_').values_at(0,-1)
parts.last.slice!(1..-1)
parts.join(' ')
end
/^[A-Za-z]{5,15}\s[A-Za-z]{1}]$/i
This will have the following criteria:
5-15 characters for first name then a whitespace and finally a single character for last name.
So I've got a string that's an improperly formatted name. Let's say, "Jean-paul Bertaud-alain".
I want to use a regex in Ruby to find the first character after every dash and make it uppercase. So, in this case, I want to apply a method that would yield: "Jean-Paul Bertaud-Alain".
Any help?
String#gsub can take a block argument, so this is as simple as:
str = "Jean-paul Bertaud-alain"
str.gsub(/-[a-z]/) {|s| s.upcase }
# => "Jean-Paul Bertaud-Alain"
Or, more succinctly:
str.gsub(/-[a-z]/, &:upcase)
Note that the regular expression /-[a-z]/ will only match letters in the a-z range, meaning it won't match e.g. à. This is because String#upcase does not attempt to capitalize characters with diacritics anyway, because capitalization is language-dependent (e.g. i is capitalized differently in Turkish than in English). Read this answer for more information: https://stackoverflow.com/a/4418681
"Jean-paul Bertaud-alain".gsub(/(?<=-)\w/, &:upcase)
# => "Jean-Paul Bertaud-Alain"
I suggest you make the test more demanding by requiring the letter to be upcased: 1) be preceded by a capitalized word followed by a hypen and 2) be followed by lowercase letters followed by a word break.
r = /
\b # Match a word break
[A-Z] # Match an upper-case letter
[a-z]+ # Match >= 1 lower-case letters
\- # Match hypen
\K # Forget everything matched so far
[a-z] # Match a lower-case letter
(?= # Begin a positive lookahead
[a-z]+ # Match >= 1 lower-case letters
\b # Match a word break
) # End positive lookahead
/x # Free-spacing regex definition mode
"Jean-paul Bertaud-alain".gsub(r) { |s| s.upcase }
#=> "Jean-Paul Bertaud-Alain"
"Jean de-paul Bertaud-alainM".gsub(r) { |s| s.upcase }
#=> "Jean de-paul Bertaud-alainM"
def get_hashtags(post)
tags = []
post.scan(/(?<![0-9a-zA-Z])(#+)([a-zA-Z]+)/){|x,y| tags << y}
tags
end
Test.assert_equals(get_hashtags("two hashs##in middle of word#"), [])
#Expected: [], instead got: ["in"]
Should it not look behind to see if the match doesnt begin with a word or number? Why is it still accepting 'in' as a valid match?
You should use \K rather than a negative lookbehind. That allows you to simplify your regex considerably: no need for a pre-defined array, capture groups or a block.
\K means "discard everything matched so far". The key here is that variable-length matches can precede \K, whereas (in Ruby and most other languages) variable-length matches are not permitted in (negative or positive) lookbehinds.
r = /
[^0-9a-zA-Z#] # do not match any character in the character class
\#+ # match one or more pound signs
\K # discard everything matched so far
[a-zA-Z]+ # match one or more letters
/x # extended mode
Note # in \#+ need not be escaped if I weren't writing the regex in extended mode.
"two hashs##in middle of word#".scan r
#=> []
"two hashs&#in middle of word#".scan r
#=> ["in"]
"two hashs#in middle of word&#abc of another word.###def ".scan r
#=> ["abc", "def"]