I have a string " hello world! How is it going?"
The output I need is " helloworld!Howisitgoing?"
So all the whitespaces after hello should be removed. I am trying to do this in ruby using regex.
I tried strip and delete(' ') methods but I didn't get what I wanted.
some_string = " hello world! How is it going?"
some_string.delete(' ') #deletes all spaces
some_string.strip #removes trailing and leading spaces only
Please help. Thanks in advance!
There are numerous ways this could be accomplished without without a regular expressions, but using them could be the "cleanest" looking approach without taking sub-strings, etc. The regular expression I believe you are looking for is /(?!^)(\s)/.
" hello world! How is it going?".gsub(/(?!^)(\s)/, '')
#=> " helloworld!Howisitgoing?"
The \s matched any whitespace character (including tabs, etc), and the ^ is an "anchor" meaning the beginning of the string. The ! indicates to reject a match with following criteria. Using those together to your goal can be accomplished.
If you are not familiar with gsub, it is very similar to replace, but takes a regular expression. It additionally has a gsub! counter-part to mutate the string in place without creating a new altered copy.
Note that strictly speaking, this isn't all whitespace "after a word" to quote the exact question, but I gathered from your examples that your intentions were "all whitespace except beginning of string", which this will do.
def remove_spaces_after_word(str, word)
i = str.index(/\b#{word}\b/i)
return str if i.nil?
i += word.size
str.gsub(/ /) { Regexp.last_match.begin(0) >= i ? '' : ' ' }
end
remove_spaces_after_word("Hey hello world! How is it going?", "hello")
#=> "Hey helloworld!Howisitgoing?"
Related
Based on "How to Delete Strings that Start with Certain Characters in Ruby", I know that the way to remove a string that starts with the character "#" is:
email = email.gsub( /(?:\s|^)#.*/ , "") #removes strings that start with "#"
I want to also remove strings that end in ".". Inspired by "Difference between \A \z and ^ $ in Ruby regular expressions" I came up with:
email = email.gsub( /(?:\s|$).*\./ , "")
Basically I used gsub to remove the dollar sign for the carrot and reversed the order of the part after the closing parentheses (making sure to escape the period). However, it is not doing the trick.
An example I'd like to match and remove is:
"a8&23q2aas."
You were so close.
email = email.gsub( /.*\.\s*$/ , "")
The difference lies in the fact that you didn't consider the relationship between string of reference and the regex tokens that describe the condition you wish to trigger. Here, you are trying to find a period (\.) which is followed only by whitespace (\s) or the end of the line ($). I would read the regex above as "Any characters of any length followed by a period, followed by any amount of whitespace, followed by the end of the line."
As commenters pointed out, though, there's a simpler way: String#end_with?.
I'd use:
words = %w[#a day in the life.]
# => ["#a", "day", "in", "the", "life."]
words.reject { |w| w.start_with?('#') || w.end_with?('.') }
# => ["day", "in", "the"]
Using a regex is overkill for this if you're only concerned with the starting or ending character, and, in fact, regular expressions will slow your code in comparison with using the built-in methods.
I would really like to stick to using gsub....
gsub is the wrong way to remove an element from an array. It could be used to turn the string into an empty string, but that won't remove that element from the array.
def replace_suffix(str,suffix)
str.end_with?(suffix)? str[0, str.length - suffix.length] : str
end
print "Enter your text here:"
user_text = gets.chomp
user_text_2 = user_text.gsub! "Damn", "Darn"
user_text_3 = user_text.gsub! "Shit", "Crap"
puts "Here is your edited text: #{user_text}"
I would like that code to also recognize when I use the lowercase versions of Shit and Damn and replace them with the substitute words. Right now it only recognizes when I type the words in with an uppercase first word. Is there any way to get it to recognize the lowercase words too, without adding more gsub! lines of code?
You can specify the i flag on your patten to ignore case:
user_text_2 = user_text.gsub! /Damn/i, "Darn"
Just a very short solution:
user_text.gsub!(/[Dd]amn/, 'Darn')
The more general approach, if this is what you want, is with an i which makes the regex case-insensitive.
user_text.gsub!(/damn/i, 'Darn')
You can do that with a Regexp and the i option, that makes the Regexp case insensitive:
foo = 'foo Foo'
foo.gsub(/foo/, 'bar')
#=> "bar Foo"
foo.gsub(/foo/i, 'bar') # with i
#=> "bar bar"
Use regular expressions instead of strings. So /damn/i instead of "damn". The 'i' at the end of a regular expression means ignore casing.
Say I have a string : "hEY "
I want to convert it to "Hey "
string.gsub!(/([a-z])([A-Z]+ )/, '\1'.upcase)
That is the idea I have, but it seems like the upcase method does nothing when I use it within the gsub method. Why is that?
EDIT: I came up with this method:
string.gsub!(/([a-z])([A-Z]+ )/) { |str| str.downcase!.capitalize! }
Is there a way to do this within the regex though? I don't really understand the '\1' '\2' thing. Is that backreferencing? How does that work
#sawa Has the simple answer, and you've edited your question with another mechanism. However, to answer two of your questions:
Is there a way to do this within the regex though?
No, Ruby's regex does not support a case-changing feature as some other regex flavors do. You can "prove" this to yourself by reviewing the official Ruby regex docs for 1.9 and 2.0 and searching for the word "case":
https://github.com/ruby/ruby/blob/ruby_1_9_3/doc/re.rdoc
https://github.com/ruby/ruby/blob/ruby_2_0_0/doc/re.rdoc
I don't really understand the '\1' '\2' thing. Is that backreferencing? How does that work?
Your use of \1 is a kind of backreference. A backreference can be when you use \1 and such in the search pattern. For example, the regular expression /f(.)\1/ will find the letter f, followed by any character, followed by that same character (e.g. "foo" or "f!!").
In this case, within a replacement string passed to a method like String#gsub, the backreference does refer to the previous capture. From the docs:
"If replacement is a String it will be substituted for the matched text. It may contain back-references to the pattern’s capture groups of the form \d, where d is a group number, or \k<n>, where n is a group name. If it is a double-quoted string, both back-references must be preceded by an additional backslash."
In practice, this means:
"hello world".gsub( /([aeiou])/, '_\1_' ) #=> "h_e_ll_o_ w_o_rld"
"hello world".gsub( /([aeiou])/, "_\1_" ) #=> "h_\u0001_ll_\u0001_ w_\u0001_rld"
"hello world".gsub( /([aeiou])/, "_\\1_" ) #=> "h_e_ll_o_ w_o_rld"
Now, you have to understand when code runs. In your original code…
string.gsub!(/([a-z])([A-Z]+ )/, '\1'.upcase)
…what you are doing is calling upcase on the string '\1' (which has no effect) and then calling the gsub! method, passing in a regex and a string as parameters.
Finally, another way to achieve this same goal is with the block form like so:
# Take your pick of which you prefer:
string.gsub!(/([a-z])([A-Z]+ )/){ $1.upcase << $2.downcase }
string.gsub!(/([a-z])([A-Z]+ )/){ [$1.upcase,$2.downcase].join }
string.gsub!(/([a-z])([A-Z]+ )/){ "#{$1.upcase}#{$2.downcase}" }
In the block form of gsub the captured patterns are set to the global variables $1, $2, etc. and you can use those to construct the replacement string.
I don't know why you are trying to do it in a complicated way, but the usual way is:
"hEY".capitalize # => "Hey"
If you insist in using a regex and upcase, then you would also need downcase:
"hEY".downcase.sub(/\w/){$&.upcase} # => "Hey"
If you really want to just swap the case of every letter in the string, you can avoid the complexity of regex entirely because There's A Method For That™.
"hEY".swapcase # => "Hey"
"HellO thERe".swapcase # => "hELLo THerE"
There's also swapcase! to do it destructively.
Let's say i have the following string:
"a test-eh'l"
I want to capitalize the start of each word. A word can be separated by a space, apostrophe, hyphen, a forward slash, a period, etc. So I want the string to turn out like this:
"A Test-Eh'L"
I'm not too worried about getting the first character capitalized from the gsub call, as that's easy to do after the fact. However, when I've been using IRB and match method, I only seem to be getting one result. When i use a scan, it collects the matches, but the problem is I cannot really do much with it, as i need to replace the contents of the original string.
Here's what i have so far:
"a test-eh'a".scan(/[\s|\-|\'][a-z]/)
=> [" t", "-e", "'a"]
"a test-eh'a".match(/[\s|\-|\'][a-z]/)
=> #<MatchData " t">
Then if i try the pattern using gsub:
"a test-eh'a".gsub(/[\s|\-|\'][a-z]/, $1)
TypeError: can't convert nil into String
In javascript, i would normally use parenthesis instead of square brackets on the front section. However, i wasn't getting correct results in the scan call when doing so.
"a test-eh'a".scan(/(\s|\-|\')[a-z]/)
=> [[" "], ["-"], ["'"]]
"a test-eh'a".gsub(/(\s|\-|\')[a-z]/, $1)
=> "a'est'h'"
Any help would be appreciated.
Try this:
"a test-eh'a".gsub(/(?:^|\s|-|')[a-z]/) { |r| r.upcase }
# => "A Test-Eh'A"
Suppose I said £ character as dangerous, and I want to be able to protect and to unprotect any string. And vice versa.
Example 1:
"Foobar £ foobar foobar foobar." # => dangerous string
"Foobar \£ foobar foobar foobar." # => protected string
Example 2:
"Foobar £ foobar £££££££foobar foobar." # => dangerous string
"Foobar \£ foobar \£\£\£\£\£\£\£foobar foobar." # => protected string
Example 3:
"Foobar \£ foobar \\£££££££foobar foobar." # => dangerous string
"Foobar \£ foobar \\\£\£\£\£\£\£\£foobar foobar." # => protected string
Is there an easy way, with Ruby, to escape (and unescape) a given character (such as £ in my example) from a string?
Edit: here is an explication about the behavior of this question.
First of all, thanks for your answers. I have a Rails app with a Tweet model having a content field. Example of tweet:
tweet = Tweet.create(content: "Hello #bob")
Inside the model, there's a serialization process that converte the string like this:
dump('Hello #bob') # => '["Hello £", 42]'
# ... where 42 is the id of bob username
Then, I'm able to deserialize and display its tweet like this:
load('["Hello £", 42]') # => 'Hello #bob'
In the same way, it's also possible to do so with more than one username:
dump('Hello #bob and #joe!') # => '["Hello £ and £!", 42, 185]'
load('["Hello £ and £!", 42, 185]') # => 'Hello #bob and #joe!'
That's the goal :)
But this find-and-replace could be hard to perform with something like:
tweet = Tweet.create(content: "£ Hello #bob")
'cause here we also have to escape £ char. And I think your solution is good for this. So the result become:
dump('£ Hello #bob') # => '["\£ Hello £", 42]'
load('["\£ Hello £", 42]') # => '£ Hello #bob'
Just perfect. <3 <3
Now, if there is this:
tweet = Tweet.create(content: "\£ Hello #bob")
I think we first should escape every \, and then escape every £, like:
dump('\£ Hello #bob') # => '["\\£ Hello £", 42]'
load('["\\£ Hello £", 42]') # => '£ Hello #bob'
However... how can we do in this case:
tweet = Tweet.create(content: "\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\£ Hello #bob")
...where tweet.content.gsub(/(?<!\\)(?=(?:\\\\)*£)/, "\\") seems not working.
Hopefully your version of ruby supports lookbehinds. If it doesn't my solution will not work for you.
Escape characters :
str = str.gsub(/(?<!\\)(?=(?:\\\\)*£)/, "\\")
Un-escape characters :
str = str.gsub(/(?<!\\)((?:\\\\)*)\\£/, "\1£")
Both regexes will work regardless of the amount of backslashes. They are complementing each other.
Escape explanation :
"
(?<! # Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind)
\\ # Match the character “\” literally
)
(?= # Assert that the regex below can be matched, starting at this position (positive lookahead)
(?: # Match the regular expression below
\\ # Match the character “\” literally
\\ # Match the character “\” literally
)* # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
£ # Match the character “£” literally
)
"
Not that I am matching a certain position. No text is consumed at all. When I pinpoint the position I want I insert a \.
Explanation of unescape :
"
(?<! # Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind)
\\ # Match the character “\” literally
)
( # Match the regular expression below and capture its match into backreference number 1
(?: # Match the regular expression below
\\ # Match the character “\” literally
\\ # Match the character “\” literally
)* # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
)
\\ # Match the character “\” literally
£ # Match the character “£” literally
"
Here I am saving all the backslashes minus one and and I replace this number of backslashes with the special character. Tricky stuff :)
If you are using Ruby 1.9, which has lookbehind, then FailedDev's answer should work quite well. If you are using Ruby 1.8, which does not have lookbehind (I think), a different approach may work. Give this a try:
text.gsub!(/(\\.)|£)/m) do
if ($1 != nil) # If escaped anything
"$1" # replace with self.
else # Otherwise escape the
"\\£" # unescaped £.
end
end
Note that I am not a Ruby programmer and this snippet is untested (in particular I'm not sure if the: if ($1 != nil) statement usage is correct - it may need to be: if ($1 != "") or if ($1)), but I do know that this general technique (using code in place of a simple replacement string) works. I recently used this same technique for my JavaScript solution to a similar question which was looking to find unescaped asterisks.
I'm not sure if this is what you want, but I think you can do a simple find-and-replace:
str = str.gsub("£", "\\£") # to escape
str = str.gsub("\\£", "£") # to unescape
Note that I changed \ to \\ because you have to escape the backslash in a double-quoted string.
Edit: I think what you want is a regex that matches an odd number of backslashes:
str = str.gsub(/(^|[^\\])((?:\\\\)*)\\£/, "\\1\\2£")
That does the following transformations
"£" #=> "£"
"\\£" #=> "£"
"\\\\£" #=> "\\\\£"
"\\\\\\£" #=> "\\\\£"