My goal is to replace all instances of the substring "test_" to "append.test_". This can occur multiple times in a sentence. For example:
I am test_a and test_b => I am append.test_a and append.test_b
The problem is that I can have something like this
I am test_a and I am test_b_test_c
I can't figure out the right regex because when I try to use
my_string.gsub('test_', "append.test_") it returns
I am append.test_a and I am append.test_b_append.test_c
My expected result:
I am append.test_a and I am append.test_b_test_c
Any ideas?
Assuming that you wish to avoid replacing in situations where nothing else precedes test_ in a string, you can use \b to check for a word boundary before test_.
s = "I am test_a and I am test_b_test_c"
s.gsub(/\btest_/, "append.test_")
# => "I am append.test_a and I am append.test_b_test_c"
Related
Given a sentence, I want to count all the duplicated words:
It is an exercice from Exercism.io Word count
For example for the input "olly olly in come free"
plain
olly: 2
in: 1
come: 1
free: 1
I have this test for exemple:
def test_with_quotations
phrase = Phrase.new("Joe can't tell between 'large' and large.")
counts = {"joe"=>1, "can't"=>1, "tell"=>1, "between"=>1, "large"=>2, "and"=>1}
assert_equal counts, phrase.word_count
end
this is my method
def word_count
phrase = #phrase.downcase.split(/\W+/)
counts = phrase.group_by{|word| word}.map {|k,v| [k, v.count]}
Hash[*counts.flatten]
end
For the test above I have this failure when I run it in the terminal:
2) Failure:
PhraseTest#test_with_apostrophes [word_count_test.rb:69]:
--- expected
+++ actual
## -1 +1 ##
-{"first"=>1, "don't"=>2, "laugh"=>1, "then"=>1, "cry"=>1}
+{"first"=>1, "don"=>2, "t"=>2, "laugh"=>1, "then"=>1, "cry"=>1}
My problem is to remove all chars except 'apostrophe...
the regex in the method almost works...
phrase = #phrase.downcase.split(/\W+/)
but it remove the apostrophes...
I don't want to keep the single quote around a word, 'Hello' => Hello
but Don't be cruel => Don't be cruel
Maybe something like:
string.scan(/\b[\w']+\b/i).each_with_object(Hash.new(0)){|a,(k,v)| k[a]+=1}
The regex employs word boundaries (\b).
The scan outputs an array of the found words and for each word in the array they are added to the hash, which has a default value of zero for each item which is then incremented.
Turns out my solution whilst finding all items and ignoring case will still leave the items in the case they were found in originally.
This would now be a decision for Nelly to either accept as is or to perform a downcase on the original string or the array item as it is added to the hash.
I'll leave that decision up to you :)
Given:
irb(main):015:0> phrase
=> "First: don't laugh. Then: don't cry."
Try:
irb(main):011:0> Hash[phrase.downcase.scan(/[a-z']+/)
.group_by{|word| word.downcase}
.map{|word, words|[word, words.size]}
]
=> {"first"=>1, "don't"=>2, "laugh"=>1, "then"=>1, "cry"=>1}
With your update, if you want to remove single quotes, do that first:
irb(main):038:0> p2
=> "Joe can't tell between 'large' and large."
irb(main):039:0> p2.gsub(/(?<!\w)'|'(?!\w)/,'')
=> "Joe can't tell between large and large."
Then use the same method.
But you say -- gsub(/(?<!\w)'|'(?!\w)/,'') will remove the apostrophe in 'Twas the night before. Which I reply you will eventually need to build a parser that can determine the distinction between an apostrophe and a single quote if /(?<!\w)'|'(?!\w)/ is not sufficient.
You can also use word boundaries:
irb(main):041:0> Hash[p2.downcase.scan(/\b[a-z']+\b/)
.group_by{|word| word.downcase}
.map{|word, words|[word, words.size]}
]
=> {"joe"=>1, "can't"=>1, "tell"=>1, "between"=>1, "large"=>2, "and"=>1}
But that does not solve 'Tis the night either.
Another way:
str = "First: don't 'laugh'. Then: 'don't cry'."
reg = /
[a-z] #single letter
[a-z']+ #one or more letters or apostrophe
[a-z] #single letter
'? #optional single apostrophe
/ix #case-insensitive and free-spacing regex
str.scan(reg).group_by(&:itself).transform_values(&:count)
#=> {"First"=>1, "don't"=>2, "laugh"=>1, "Then"=>1, "cry'"=>1}
This is my expected result.
Input a string and get three returned string.
I have no idea how to finish it with Regex in Ruby.
this is my roughly idea.
match(/(.*?)(_)(.*?)(\d+)/)
Input and expected output
# "R224_OO2003" => R224, OO, 2003
# "R2241_OOP2003" => R2244, OOP, 2003
If the example description I gave in my comment on the question is correct, you need a very straightforward regex:
r = /(.+)_(.+)(\d{4})/
Then:
"R224_OO2003".scan(r).flatten #=> ["R224", "OO", "2003"]
"R2241_OOP2003".scan(r).flatten #=> ["R2241", "OOP", "2003"]
Assuming that your three parts consist of (R and one or more digits), then an underbar, then (one or more non-whitespace characters), before finally (a 4-digit numeric date), then your regex could be something like this:
^(R\d+)_(\S+)(\d{4})$
The ^ indicates start of string, and the $ indicates end of string. \d+ indicates one or more digits, while \S+ says one or more non-whitespace characters. The \d{4} says exactly four digits.
To recover data from the matches, you could either use the pre-defined globals that line up with your groups, or you could could use named captures.
To use the match globals just use $1, $2, and $3. In general, you can figure out the number to use by counting the left parentheses of the specific group.
To use the named captures, include ? right after the left paren of a particular group. For example:
x = "R2241_OOP2003"
match_data = /^(?<first>R\d+)_(?<second>\S+)(?<third>\d{4})$/.match(x)
puts match_data['first'], match_data['second'], match_data['third']
yields
R2241
OOP
2003
as expected.
As long as your pattern covers all possibilities, then you just need to use the match object to return the 3 strings:
my_match = "R224_OO2003".match(/(.*?)(_)(.*?)(\d+)/)
#=> #<MatchData "R224_OO2003" 1:"R224" 2:"_" 3:"OO" 4:"2003">
puts my_match[0] #=> "R224_OO2003"
puts my_match[1] #=> "R224"
puts my_match[2] #=> "_"
puts my_match[3] #=> "00"
puts my_match[4] #=> "2003"
A MatchData object contains an array of each match group starting at index [1]. As you can see, index [0] returns the entire string. If you don't want the capture the "_" you can leave it's parentheses out.
Also, I'm not sure you are getting what you want with the part:
(.*?)
this basically says one or more of any single character followed by zero or one of any single character.
Say I build an array like this:
:001 > holder = "This.File.Q99P84.Is.Awesome"
=> "This.File.Q99P84.Is.Awesome"
:002 > name = holder.split(".")
=> ["This", "File", "Q99P84", "Is", "Awesome"]
Now, I can do:
name[2].include? 'Q99P84'
Instead of putting in 'Q99P84' I want to put in something like 'symbol for Q followed by
symbol for number, symbol for number, symbol for P, symbol for number, symbol for number
so the .include? function will be dynamic. So any file name I load that has Q##P## will return true.
I'm pretty sure this is possibly I just don't know exactly what to search. If you know the answer can you link me to the documentation.
What you're looking for is regular expression matching. The Ruby regexp object will help you. What you want is something like
/Q[\d+]P[\d+]/.match(name[2])
...which will return a truthy value if name[2] has a string which matches a character Q, one or more digits (0-9), a character P, then one or more digits. This is probably too flexible a match if the pattern you want has exactly two digits in those number spaces; for that you might try a more specific pattern:
/Q\d\dP\d\d/.match(name[2])
One way to do this is through Regular Expressions ('regex' for short). Two good sources of information is Regular Expression.info and Rubular which is more Ruby centric.
One way to use regex on a string is the String#match method:
names[2].match(/Q\d\dP\d\d/) # /d stands for digit
This will return the string if there is a match and it will return nil if there isn't.
"Q42P67".match(/Q\d\dP\d\d/) #=> "Q42P67"
"H33P55".match(/Q\d\dP\d\d/) #=> nil
This is helpful in an if condition since a returned string is 'truthy' and nil is 'falsely'.
names[2] = "Q42P67"
if names[2].match(/Q\d\dP\d\d/)
# Will execute code here
end
names[2] = "H33P55"
if names[2].match(/Q\d\dP\d\d/)
# Will not execute code here
end
I hope that helps until you dig further into your study of Regular Expressions.
EDIT:
Note that the Q and P in /Q\d\dP\d\d/ are capital letters. If case is not important, you can append an 'i' for case-insensitivity. /Q\d\dP\d\d/i will capture "Q42P67" and "q42P67"
I'm trying to write a regular expressions that will match a set of characters without regard to order. For example:
str = "act"
str.scan(/Insert expression here/)
would match:
cat
act
tca
atc
tac
cta
but would not match ca, ac or cata.
I read through a lot of similar questions and answers here on StackOverflow, but have not found one that matches my objectives exactly.
To clarify a bit, I'm using ruby and do not want to allow repeat characters.
Here is your solution
^(?:([act])(?!.*\1)){3}$
See it here on Regexr
^ # matches the start of the string
(?: # open a non capturing group
([act]) # The characters that are allowed and a capturing group
(?!.*\1) # That character is matched only if it does not occur once more, Lookahead assertion
){3} # Defines the amount of characters
$
The only special think is the lookahead assertion, to ensure the character is not repeated.
^ and $ are anchors to match the start and the end of the string.
[act]{3} or ^[act]{3}$ will do it in most regular expression dialects. If you can narrow down the system you're using, that will help you get a more specific answer.
Edit: as mentioned by #georgydyer in the comments below, it's unclear from your question whether or not repeated characters are allowed. If not, you can adapt the answer from this question and get:
^(?=[act]{3}$)(?!.*(.).*\1).*$
That is, a positive lookahead to check a match, and then a negative lookahead with a backreference to exclude repeated characters.
Here's how I'd go about it:
regex = /\b(?:#{ Regexp.union(str.split('').permutation.map{ |a| a.join }).source })\b/
# => /(?:act|atc|cat|cta|tac|tca)/
%w[
cat act tca atc tac cta
ca ac cata
].each do |w|
puts '"%s" %s' % [w, w[regex] ? 'matches' : "doesn't match"]
end
That outputs:
"cat" matches
"act" matches
"tca" matches
"atc" matches
"tac" matches
"cta" matches
"ca" doesn't match
"ac" doesn't match
"cata" doesn't match
I use the technique of passing an array into Regexp.union for a lot of things; I works especially well with the keys of a hash, and passing the hash into gsub for rapid search/replace on text templates. This is the example from the gsub documentation:
'hello'.gsub(/[eo]/, 'e' => 3, 'o' => '*') #=> "h3ll*"
Regexp.union creates a regex, and it's important to use source instead of to_s when extracting the actual pattern being generated:
puts regex.to_s
=> (?-mix:\b(?:act|atc|cat|cta|tac|tca)\b)
puts regex.source
=> \b(?:act|atc|cat|cta|tac|tca)\b
Notice how to_s embeds the pattern's flags inside the string. If you don't expect them you can accidentally embed that pattern into another, which won't behave as you expect. Been there, done that and have the dented helmet as proof.
If you really want to have fun, look into the Perl Regexp::Assemble module available on CPAN. Using that, plus List::Permutor, lets us generate more complex patterns. On a simple string like this it won't save much space, but on long strings or large arrays of desired hits it can make a huge difference. Unfortunately, Ruby has nothing like this, but it is possible to write a simple Perl script with the word or array of words, and have it generate the regex and pass it back:
use List::Permutor;
use Regexp::Assemble;
my $regex_assembler = Regexp::Assemble->new;
my $perm = new List::Permutor split('', 'act');
while (my #set = $perm->next) {
$regex_assembler->add(join('', #set));
}
print $regex_assembler->re, "\n";
(?-xism:(?:a(?:ct|tc)|c(?:at|ta)|t(?:ac|ca)))
See "Is there an efficient way to perform hundreds of text substitutions in Ruby?" for more information about using Regexp::Assemble with Ruby.
I will assume several things here:
- You are looking for permutations of given characters
- You are using ruby
str = "act"
permutations = str.split(//).permutation.map{|p| p.join("")}
# and for the actual test
permutations.include?("cat")
It is no regex though.
No doubt - the regex that uses positive/negative lookaheads and backreferences is slick, but if you're only dealing with three characters, I'd err on the side of verbosity by explicitly enumerating the character permutations like #scones suggested.
"act".split('').permutation.map(&:join)
=> ["act", "atc", "cat", "cta", "tac", "tca"]
And if you really need a regex out of it for scanning a larger string, you can always:
Regexp.union "act".split('').permutation.map(&:join)
=> /\b(act|atc|cat|cta|tac|tca)\b/
Obviously, this strategy doesn't scale if your search string grows, but it's much easier to observe the intent of code like this in my opinion.
EDIT: Added word boundaries for false positive on cata based on #theTinMan's feedback.
I want to do something like this
def get_count(string)
sentence.split(' ').count
end
I think there's might be a better way, string may have built-in method to do this.
I believe count is a function so you probably want to use length.
def get_count(string)
sentence.split(' ').length
end
Edit: If your string is really long creating an array from it with any splitting will need more memory so here's a faster way:
def get_count(string)
(0..(string.length-1)).inject(1){|m,e| m += string[e].chr == ' ' ? 1 : 0 }
end
If the only word boundary is a single space, just count them.
puts "this sentence has five words".count(' ')+1 # => 5
If there are spaces, line endings, tabs , comma's followed by a space etc. between the words, then scanning for word boundaries is a possibility:
puts "this, is./tfour words".scan(/\b/).size/2
I know this is an old question, but this might help someone stumbling here. Countring words is a complicated problem. What is a "word"? Do numbers and special characters count as words? Etc...
I wrote the words_counted gem for this purpose. It's a highly flexible, customizable string analyser. You can ask it to analyse any string for word count, word occurrences, and exclude words/characters using regexp, strings, and arrays.
counter = WordsCounted::Counter.new("Hello World!", exclude: "World")
counter.word_count #=> 1
counted.words #=> ["Hello"]
Etc...
The documentation and full source are on Github.
using regular expression will also cover multi spaces:
sentence.split(/\S+/).size
String doesn't have anything pre-built to do what you wanted. You can define a method in your class or extend the String class itself for what you want to do:
def word_count( string )
return 0 if string.empty?
string.split.size
end
Regex split on any non-word character:
string.split(/\W+/).size
...although it makes apostrophe use count as two words, so depending on how small the margin of error needs to be, you might want to build your own regex expression.
I recently found that String#count is faster than splitting up the string by over an order of magnitude.
Unfortunately, String#count only accepts a string, not a regular expression. Also, it would count two adjacent spaces as two things, rather than a single thing, and you'd have to handle other white space characters seperately.
p " some word\nother\tword.word|word".strip.split(/\s+/).size #=> 4
I'd rather check for word boundaries directly:
"Lorem Lorem Lorem".scan(/\w+/).size
=> 3
If you need to match rock-and-roll as one word, you could do like:
"Lorem Lorem Lorem rock-and-roll".scan(/[\w-]+/).size
=> 4