Why does the output from my map/regex block not capitalize? - ruby

I'm working through the Test First Ruby Master problems. My code for 08/book_titles is this:
class Book
attr_accessor :title
def title
if #title.include?(' ')
correct = #title.split.each_with_index.map {|x, index| ((x =~ /^a|an|of|the|or|in|and$/) && index != 0) ? x : x.capitalize}
correct.join(' ')
# this is throwing a weird error, the code looks right but isn't capitalizing last word (returns 'To Kill a mockingbird')
else #title.capitalize
end
end
end
I tested the map portion separately, and it works fine. But in the entirety of the problem, it does not capitalize as it should be. It throws an rspec error:
1) Book title should capitalize every word except... articles a
Failure/Error: expect(#book.title).to eq("To Kill a Mockingbird")
expected: "To Kill a Mockingbird"
got: "To Kill a mockingbird"
Anyone know why?
I originally didn't include ^/$ in the regex. I got the same error with a different title, and adding those anchors fixed it for that case. But then the error showed up again with the title.

Because mockingbird contains in
('mockingbird' =~ /^a|an|of|the|or|in|and$/) => 4
I think you want this regex:
/^a$|^an$|^of$|^the$|^or$|^in$|^and$/

It is not necessary to break the string into words, modify the words and join them back into a string. In fact, doing that has the disadvantage that spacing between words may be altered. Here's one way of operating on the string directly.
wee_words = ["a", "an", "of", "the", "or", "in", "and"]
str = "a dAy in the life of waltEr mITTY"
str.capitalize.gsub(/\w+/) { |s| wee_words.include?(s) ? s : s.capitalize }
#=> "A Day in the Life of Walter Mitty"
str.capitalize upcases the first letter of the string and downcases all subsequent letters. As a result, the first word will never be treated as a wee_word, since it is capitalized (e.g., wee_words.include?("The") #=> false).

The regex is slightly incorrect. The way to read it as it is can be done this way:
Match any string that
starts with 'a'
or contains 'an'
or contains 'of'
or contains 'the'
or contains 'or'
or contains 'in'
or ends in 'and'
What you really seem to want is something that reads like this:
Match any string that
only contains any of 'a', 'an', 'of', 'the', 'or', 'in', 'and'
To get this, you want your regex to be written like this:
/^(a|an|of|the|or|in|and)$/
Note the parenthesis around the alternation. (Alternation is the formal term for multiple choices in a regex, where choices are separated by '|').
If you're comparing against book or movie titles, this is much closer to the type of match you'd expect. It will match correctly for titles such as "Chariots of Fire" and "Benny and Joon", but not against falsely the 'in' of "To Kill a Mockingbird", which is a significant improvement.
However, it still won't quite work yet on something like "Benny AND Joon", because 'AND' is uppercase in this title (assuming that incoming titles may be arbitrarily mixed case). One last change will do it:
/^(a|an|of|the|or|in|and)$/i
That last letter 'i' at the end of the regex says to 'ignore case', so that matches can occur regardless of whether the 'AND' is uppercase, lowercase, or mixed case.
This should get you close to what you're trying to achieve and handle a few bumpy use cases in the process.

Related

Regular expression returns only one match

I have a set of keywords. Any keyword can contain a space symbol ['one', 'one two']. I generate a regexp from these kyewords like this /\b(?i:one|one\ two|three)\b/. Full example below:
keywords = ['one', 'one two', 'three']
re = /\b(?i:#{ Regexp.union(keywords).source })\b/
text = 'Some word one and one two other word'
text.downcase.scan(re)
the result of this code is
=> ["one", "one"]
How to find match of the second keyword one two and get result like this?
=> ["one", "one two"]
Regexes are eager to match. Once they find a match, they don't try to find another possibly longer one (with one important exception).
/\b(?i:one|one\ two|three)\b/ is never going to match one two because it will always match one first. You'd need /\b(?i:one two|one|three)\b/ so it tries one two first. Probably the simplest way to automate this is to sort by the longest keywords first.
keywords = ['one', 'one two', 'three']
re = Regexp.union(keywords.sort { |a,b| b.length <=> a.length }).source
re = /\b#{re}\b/i;
text = 'Some word one and one two other word'
puts text.scan(re)
Note that I set the whole regex to be case-insensitive, easier to read than (?:...), and that downcasing the string is redundant.
The exception is repetition like +, * and friends. They are greedy by default. .+ is going to match as many characters as it can. That's greedy. You can make it lazy, to match the first thing it sees, with a ?. .+? will match a single character.
"A foot of fools".match(/(.*foo)/); # matches "A foot of foo"
"A foot of fools".match(/(.*?foo)/); # matches "A foo"
The point is that \bone\b matches one in one two and since this branch appears before one two branch, it "wins" (see Remember That The Regex Engine Is Eager).
You need to sort the keyword array in a descending order before building a regex. It will then look like
(?-mix:\b(?i:three|one\ two|one)\b)
This way the longer one two will be before the shorter one and will get matched.
See the Ruby demo:
keywords = ['one', 'one two', 'three']
keywords = keywords.dup.sort.reverse
re = /\b(?i:#{ Regexp.union(keywords).source })\b/
text = 'Some word one and one two other word'
puts text.downcase.scan(re)
# => [ one, one two ]
I tried your example by moving the first element to the second position of the array and it works (e.g. http://rubular.com/r/4F2Hc46wHT).
In fact, it looks like the first keyword "overlaps" the second.
This response may be unhelpful if you can't change keywords order.

Ruby regular expressions and bracket. What do the brackets do?

I am going through the Peter Cooper book "Beginning Ruby" and I have some questions regarding some of the string methods and regular expression usage. I think I'm clear on what a regular expression is: "a string that describes a pattern for matching elements in other strings."
So:
"This is a test".scan(/\w\w/) {|x| puts x}
Output:
Th
is
is
te
st
=> "This is a test"
So it prints two characters at a time. I didn't realize it also returns the original string. Why is this?
Also,
"This is a test".scan(/[aeiou]/) { |x| puts x }
What do the brackets do? I think they are called character classes, but I am not sure exactly what they do. The explanation in Cooper's book isn't totally verbose and clear.
Explanation of character classes:
"The last important aspect of regular expressions you need to understand at this stage is
character classes. These allow you to match against a specific set of characters. For example, you can scan through all the vowels in a string:"
Yes, it is called a character class.
A character class defines a set of characters. Saying, "match one character specified by the class". The two implementations of a character class are considered a positive class [ ] and a negative class [^ ]. The positive character class allows you to define a list of characters, any one of which may appear in a string for a match to occur while the negative class allows you to define a list of characters that must NOT appear in a string for a match to occur.
Explanation of your character class:
[aeiou] # any character of: 'a', 'e', 'i', 'o', 'u'
The scan method usually returns an array with the matches, but it optionally accepts a block, which is equivalent to do an each of the resulting array.
Here is the documentation: http://www.ruby-doc.org/core-2.1.3/String.html#method-i-scan
To the second question, #hwnd already gave you a clear answer. The best way to learn this is to experiment, regex101.com is the online tool I usually use. It lists explanations for all your matching elements, so it's a wonderful learning resource too.
Some things you might like to try:
123abab12ab1234 with pattern [123]
123abab12ab1234 with pattern [ab]+
123abab12ab1234 with pattern b[1|a]
One thing to remember is that a character class matches ONE character, for example:
str = 'XXXaeiouXXX'
puts str
str.sub!(/[aeiou]/, '.')
puts str
--output:--
XXXaeiouXXX
XXX.eiouXXX
A character class says, "Match this character OR this character OR this character...ONE TIME ".
Also check out rubular:
http://rubular.com/
I didn't realize it also returns the original string. Why is this?
So that you can chain methods together:
my_str.scan(...).downcase.capitalize.each_char {|char| puts char}.upcase.chomp

Take an array and a letter as arguments and return a new array with words that contain that letter

I can run a search and find the element I want and can return those words with that letter. But when I start to put arguments in, it doesn't work. I tried select with include? and it throws an error saying, private method. This is my code, which returns what I am expecting:
my_array = ["wants", "need", 3, "the", "wait", "only", "share", 2]
def finding_method(source)
words_found = source.grep(/t/) #I just pick random letter
print words_found
end
puts finding_method(my_array)
# => ["wants", "the", "wait"]
I need to add the second argument, but it breaks:
def finding_method(source, x)
words_found = source.grep(/x/)
print words_found
end
puts finding_method(my_array, "t")
This doesn't work, (it returns an empty array because there isn't an 'x' in the array) so I don't know how to pass an argument. Maybe I'm using the wrong method to do what I'm after. I have to define 'x', but I'm not sure how to do that. Any help would be great.
Regular expressions support string interpolation just like strings.
/x/
looks for the character x.
/#{x}/
will first interpolate the value of the variable and produce /t/, which does what you want. Mostly.
Note that if you are trying to search for any text that might have any meaning in regular expression syntax (like . or *), you should escape it:
/#{Regexp.quote(x)}/
That's the correct answer for any situation where you are including literal strings in regular expression that you haven't built yourself specifically for the purpose of being a regular expression, i.e. 99% of cases where you're interpolating variables into regexps.

How the Anchor \z and \G works in Ruby?

I am using Ruby1.9.3. I am newbie to this platform.
From the doc I just got familiared with two anchor which are \z and \G. Now I little bit played with \z to see how it works, as the definition(End or End of String) made me confused, I can't understand what it meant say - by End. So I tried the below small snippets. But still unable to catch.
CODE
irb(main):011:0> str = "Hit him on the head me 2\n" + "Hit him on the head wit>
=> "Hit him on the head me 2\nHit him on the head with a 24\n"
irb(main):012:0> str =~ /\d\z/
=> nil
irb(main):013:0> str = "Hit him on the head me 24 2\n" + "Hit him on the head >
=> "Hit him on the head me 24 2\nHit him on the head with a 24\n"
irb(main):014:0> str =~ /\d\z/
=> nil
irb(main):018:0> str = "Hit1 him on the head me 24 2\n" + "Hit him on the head>
=> "Hit1 him on the head me 24 2\nHit him on the head with a11 11 24\n"
irb(main):019:0> str =~ /\d\z/
=> nil
irb(main):020:0>
Every time I got nil as the output. So how the calculation is going on for \z ? what does End mean? - I think my concept took anything wrong with the End word in the doc. So anyone could help me out to understand the reason what is happening with the out why so happening?
And also i didn't find any example for the anchor \G . Any example please from you people to make visualize how \G used in real time programming?
EDIT
irb(main):029:0>
irb(main):030:0* ("{123}{45}{6789}").scan(/\G(?!^)\{\d+\}/)
=> []
irb(main):031:0> ('{123}{45}{6789}').scan(/\G(?!^)\{\d+\}/)
=> []
irb(main):032:0>
Thanks
\z matches the end of the input. You are trying to find a match where 4 occurs at the end of the input. Problem is, there is a newline at the end of the input, so you don't find a match. \Z matches either the end of the input or a newline at the end of the input.
So:
/\d\z/
matches the "4" in:
"24"
and:
/\d\Z/
matches the "4" in the above example and the "4" in:
"24\n"
Check out this question for example of using \G:
Examples of regex matcher \G (The end of the previous match) in Java would be nice
UPDATE: Real-World uses for \G
I came up with a more real world example. Say you have a list of words that are separated by arbitrary characters that cannot be well predicted (or there's too many possibilities to list). You'd like to match these words where each word is its own match up until a particular word, after which you don't want to match any more words. For example:
foo,bar.baz:buz'fuzz*hoo-har/haz|fil^bil!bak
You want to match each word until 'har'. You don't want to match 'har' or any of the words that follow. You can do this relatively easily using the following pattern:
/(?<=^|\G\W)\w+\b(?<!har)/
rubular
The first attempt will match the beginning of the input followed by zero non-word character followed by 3 word characters ('foo') followed by a word boundary. Finally, a negative lookbehind assures that the word which has just been matched is not 'har'.
On the second attempt, matching picks back up at the end of the last match. 1 non-word character is matched (',' - though it is not captured due to the lookbehind, which is a zero-width assertion), followed by 3 characters ('bar').
This continues until 'har' is matched, at which point the negative lookbehind is triggered and the match fails. Because all matches are supposed to be "attached" to the last successful match, no additional words will be matched.
The result is:
foo
bar
baz
buz
fuzz
hoo
If you want to reverse it and have all words after 'har' (but, again, not including 'har'), you can use an expression like this:
/(?!^)(?<=har\W|\G\W)\w+\b/
rubular
This will match either a word which is immediately preceeded by 'har' or the end of the last match (except we have to make sure not to match the beginning of the input). The list of matches is:
haz
fil
bil
bak
If you do want to match 'har' and all following words, you could use this:
/\bhar\b|(?!^)(?<=\G\W)\w+\b/
rubular
This produces the following matches:
har
haz
fil
bil
bak
Sounds like you want to know how Regex works? Or do you want to know how Regex works with ruby?
Check these out.
Regexp Class description
The Regex Coach - Great for testing regex matching
Regex cheat sheet
I understand \G to be a boundary match character. So it would tell the next match to start at the end of the last match. Perhaps since you haven't made a match yet you cant have a second.
Here is the best example I can find. Its not in ruby but the concept should be the same.
I take it back this might be more useful

How do I write a regular expression that will match characters in any order?

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.

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