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.
Related
I'm writing a simple method to detect and strip tags from text strings. Given this input string:
{{foobar}}
The function has to return
foobar
I thought I could just chain multiple chomp! methods, like so:
"{{foobar}}".chomp!("{{").chomp!("}}")
but this won't work, because the first chomp! returns a NilClass. I can do it with regular chomp statements, but I'm really looking for a one-line solution.
The String class documentation says that chomp! returns a Str if modifications have been made - therefore, the second chomp! should work. It doesn't, however. I'm at a loss at what's happening here.
For the purposes of this question, you can assume that the input string is always a tag which begins and ends with double curly braces.
You can definitely chain multiple chomp statements (the non-bang version), still having a one-line solution as you wanted:
"{{foobar}}".chomp("{{").chomp("}}")
However, it will not work as expected because both chomp! and chomp removes the separator only from the end of the string, not from the beginning.
You can use sub
"{{foobar}}".sub(/{{(.+)}}/, '\1')
# => "foobar"
"alfa {{foobar}} beta".sub(/{{(.+)}}/, '\1')
# => "alfa foobar beta"
# more restrictive
"{{foobar}}".sub(/^{{(.+)}}$/, '\1')
# => "foobar"
Testing this out, it's clear that chomp! will return nil if the separator it's provided as an argument is not present at the end of the string.
So "{{text}}".chomp!("}}") returns a string, but "{{text}}".chomp!("{{") reurns nil.
See here for an answer of how to chomp at the beginning of a string. But recognize that chomp only looks at the end of the string. So you can call str.reverse.chomp!("{{").reverse to remove the opening brackets.
You could also use a regex:
string = "{{text}}"
puts [/^\{\{(.+)\}\}$/, 1]
# => "text"
Try tr:
'{{foobar}}'.tr('{{', '').tr('}}', '')
You can also use gsub or sub but if the replacement is not needed as pattern, then tr should be faster.
If there are always curly braces, then you can just slice the string:
'{{foobar}}'[2...-2]
If you plan to make a method which returns the string without curly braces then DO NOT use bang versions. Modifying the input parameter of a method will be suprising!
def strip(string)
string.tr!('{{', '').tr!('}}', '')
end
a = '{{foobar}}'
b = strip(a)
puts b #=> foobar
puts a #=> foobar
I want to replace the content (or delete it) that does not match with my filter.
I think the perfect description would be an opposite sub. I cannot find anything similar in the docs, and I'm not sure how to invert the regex, but I think a method would probably be the more convenient.
An example of how it would work (I've just changed the words to make it more clear)
"bird.cats.dogs".opposite_sub(/(dogs|cats)\.(dogs|cats)/, '')
#"cats.dogs"
I hope it's easy enough to understand.
Thanks in advance.
String#[] can take a regular expression as its parameter:
▶ "bird.cats.dogs"[/(dogs|cats)\.(dogs|cats)/]
#⇒ "cats.dogs"
For multiple matches one can use String#scan:
▶ "bird.cats.dogs.bird.cats.dogs".scan /(?:dogs|cats)\.(?:dogs|cats)/
#⇒ ["cats.dogs", "cats.dogs"]
So you want to extract the part that matches your regex?
You can use String#slice, for example:
"bird.cats.dogs".slice(/(dogs|cats)\.(dogs|cats)/)
#=> "cats.dogs"
And String#[] does the same.
"bird.cats.dogs"[/(dogs|cats)\.(dogs|cats)/]
#=> "cats.dogs"
You cannot have a single replacement string because the part of the string that matches the regex might not be at the beginning or end of the string, in which case it's not clear whether the replacement string should precede or follow the matching string. I've therefore written the following with two replacement strings, one for pre-match, the other for post_match. I've made this a method of the String class as that's what you've asked for (though I've given the method a less-perfect name :-) )
class String
def replace_non_matching(regex, replace_before, replace_after)
first, match, last = partition(regex)
replace_before + match + replace_after
end
end
r = /(dogs|cats)\.(dogs|cats)/
"birds.cats.dogs.pigs".replace_non_matching(r, "", "")
#=> "cats.dogs"
"birds.cats.dogs".replace_non_matching(r, "snakes.", ".hens")
#=> "snakes.cats.dogs.hens"
"birds.cats.dogs.mice.cats.dogs.bats".replace_non_matching(r, "snakes.", ".hens")
#=> "snakes.cats.dogs.hens"
Regarding the last example, the method could be modified to replace "birds.", ".mice." and ".bats", but in that case three replacement strings would be needed. In general, determining in advance the number of replacement strings needed could be problematic.
If I wanted to remove things like:
.!,'"^-# from an array of strings, how would I go about this while retaining all alphabetical and numeric characters.
Allowed alphabetical characters should also include letters with diacritical marks including à or ç.
You should use a regex with the correct character property. In this case, you can invert the Alnum class (Alphabetic and numeric character):
"◊¡ Marc-André !◊".gsub(/\p{^Alnum}/, '') # => "MarcAndré"
For more complex cases, say you wanted also punctuation, you can also build a set of acceptable characters like:
"◊¡ Marc-André !◊".gsub(/[^\p{Alnum}\p{Punct}]/, '') # => "¡MarcAndré!"
For all character properties, you can refer to the doc.
string.gsub(/[^[:alnum:]]/, "")
The following will work for an array:
z = ['asfdå', 'b12398!', 'c98347']
z.each { |s| s.gsub! /[^[:alnum:]]/, '' }
puts z.inspect
I borrowed Jeremy's suggested regex.
You might consider a regular expression.
http://www.regular-expressions.info/ruby.html
I'm assuming that you're using ruby since you tagged that in your post. You could go through the array, put it through a test using a regexp, and if it passes remove/keep it based on the regexp you use.
A regexp you might use might go something like this:
[^.!,^-#]
That will tell you if its not one of the characters inside the brackets. However, I suggest that you look up regular expressions, you might find a better solution once you know their syntax and usage.
If you truly have an array (as you state) and it is an array of strings (I'm guessing), e.g.
foo = [ "hello", "42 cats!", "yöwza" ]
then I can imagine that you either want to update each string in the array with a new value, or that you want a modified array that only contains certain strings.
If the former (you want to 'clean' every string the array) you could do one of the following:
foo.each{ |s| s.gsub! /\p{^Alnum}/, '' } # Change every string in place…
bar = foo.map{ |s| s.gsub /\p{^Alnum}/, '' } # …or make an array of new strings
#=> [ "hello", "42cats", "yöwza" ]
If the latter (you want to select a subset of the strings where each matches your criteria of holding only alphanumerics) you could use one of these:
# Select only those strings that contain ONLY alphanumerics
bar = foo.select{ |s| s =~ /\A\p{Alnum}+\z/ }
#=> [ "hello", "yöwza" ]
# Shorthand method for the same thing
bar = foo.grep /\A\p{Alnum}+\z/
#=> [ "hello", "yöwza" ]
In Ruby, regular expressions of the form /\A………\z/ require the entire string to match, as \A anchors the regular expression to the start of the string and \z anchors to the end.
I had to convert a series of sentences into camel-cased method names. I ended writing something for it. I am still curious if there's something simpler for it.
Given the string a = "This is a test." output thisIsATest
I used for following:
a.downcase.gsub(/\s\w/){|b| b[-1,1].upcase }
Not sure it's better as your solution but it should do the trick:
>> "This is a test.".titleize.split(" ").join.camelize(:lower)
=> "thisIsATest."
titleize: uppercase every first letter of each word
split(" ").join: create an array with each word and join to squeeze the spaces out
camelize(:lower): make the first letter lowercase
You can find some more fun functions in the Rails docs: http://api.rubyonrails.org/classes/ActiveSupport/CoreExtensions/String/Inflections.html
"active_record".camelize(:lower)
output : "activeRecord"
use these
"Some string for you".gsub(/\s+/,'_').camelize(:lower) #=> "someStringForYou"
gsub: Replace spaces by underscores
camelize: java-like method camelcase
You might try using the 'English' gem, available at http://english.rubyforge.org/
require 'english/case'
a = "This is a test."
a.camelcase().uncapitalize() # => 'thisIsATest
How can I remove the very first "1" from any string if that string starts with a "1"?
"1hello world" => "hello world"
"112345" => "12345"
I'm thinking of doing
string.sub!('1', '') if string =~ /^1/
but I' wondering there's a better way. Thanks!
Why not just include the regex in the sub! method?
string.sub!(/^1/, '')
As of Ruby 2.5 you can use delete_prefix or delete_prefix! to achieve this in a readable manner.
In this case "1hello world".delete_prefix("1").
More info here:
https://blog.jetbrains.com/ruby/2017/10/10-new-features-in-ruby-2-5/
https://bugs.ruby-lang.org/issues/12694
'invisible'.delete_prefix('in') #=> "visible"
'pink'.delete_prefix('in') #=> "pink"
N.B. you can also use this to remove items from the end of a string with delete_suffix and delete_suffix!
'worked'.delete_suffix('ed') #=> "work"
'medical'.delete_suffix('ed') #=> "medical"
https://bugs.ruby-lang.org/issues/13665
I've answered in a little more detail (with benchmarks) here: What is the easiest way to remove the first character from a string?
if you're going to use regex for the match, you may as well use it for the replacement
string.sub!(%r{^1},"")
BTW, the %r{} is just an alternate syntax for regular expressions. You can use %r followed by any character e.g. %r!^1!.
Careful using sub!(/^1/,'') ! In case the string doesn't match /^1/ it will return nil. You should probably use sub (without the bang).
This answer might be more optimised: What is the easiest way to remove the first character from a string?
string[0] = '' if string[0] == '1'
I'd like to post a tiny improvement to the otherwise excellent answer by Zach. The ^ matches the beginning of every line in Ruby regex. This means there can be multiple matches per string. Kenji asked about the beginning of the string which means they have to use this regex instead:
string.sub!(/\A1/, '')
Compare this - multiple matches with this - one match.