How to implement a for loop in a method in Ruby - ruby

Question

I have been tasked to create a pig_latin method.
Pig Latin is a made-up children's language that's intended to be
confusing. It obeys a few simple rules (below) but when it's spoken
quickly it's really difficult for non-children (and non-native
speakers) to understand.
Rule 1: If a word begins with a vowel sound, add an "ay" sound to
the end of the word.
Rule 2: If a word begins with a consonant sound, move it to the end
of the word, and then add an "ay" sound to the end of the word.
(There are a few more rules for edge cases, and there are regional
variants too, but that should be enough to understand the tests.)
All my tests are passing save one, to translate many words.
This is my error:
#translate
translates a word beginning with a vowel
translates a word beginning with a consonant
translates a word beginning with two consonants
translates two words
translates a word beginning with three consonants
counts 'sch' as a single phoneme
counts 'qu' as a single phoneme
counts 'qu' as a consonant even when it's preceded by a consonant
translates many words (FAILED - 1)
Failures:
1) #translate translates many words
Failure/Error: expect(s).to eq("ethay ickquay ownbray oxfay")
expected: "ethay ickquay ownbray oxfay"
got: "ethay"
(compared using ==)
# ./spec/04_pig_latin_spec.rb:70:in `block (2 levels) in <top (required)>'
Finished in 0.00236 seconds (files took 0.10848 seconds to load)
9 examples, 1 failure
Failed examples:
rspec ./spec/04_pig_latin_spec.rb:68 # #translate translates many words
And this is my method:
def translate(str)
def add_ay(str)
return str + 'ay'
end
def word_begins_with_vowel(str)
if (!(str.match(' '))) && $vowels[str[0]]
return add_ay(str)
end
end
def begins_with_consonant(str)
if ((!$vowels[str[0]]) && (!$vowels[str[1]]) && (!$vowels[str[2]]))
first_three = str.split('').slice(0, 3).join('');
str = str.slice(3, str.length - 1)
return str + first_three + 'ay'
end
if ((!$vowels[str[0]]) && (!$vowels[str[1]]))
first_two = str.split('').slice(0, 2).join('');
str = str.slice(2, str.length - 1)
return str + first_two + 'ay'
end
if ((!$vowels[str[0]]))
first_char = str.split('').slice(0);
str = str.slice(1, str.length - 1)
return str + first_char +'ay'
end
end
def translates_two_words(str)
if (str.match(' '))
str = str.split(' ');
first_char = str[1].split('').slice(0);
str[1] = str[1].slice!(1, str[1].length - 1);
return str[0] + 'ay' + ' ' + str[1] + first_char + 'ay'
end
end
def translates_many_words(str)
str = str.split(' ');
if str.length > 2
str.each do |item|
return begins_with_consonant(item) || word_begins_with_vowel(item)
end
end
end
$vowels = {
'a' => add_ay(str),
'e' => add_ay(str),
'i' => add_ay(str),
'o' => add_ay(str),
'y' => add_ay(str)
}
return translates_many_words(str) || word_begins_with_vowel(str) || begins_with_consonant(str) || translates_two_words(str)
end
I would figure this would take care of many words:
def translates_many_words(str)
str = str.split(' ');
if str.length > 2
str.each do |item|
return begins_with_consonant(item) || word_begins_with_vowel(item)
end
end
end
but it's not.

Answer 1

As #theTinMan says, return - will decline next iteration and just return the first value in the first iteration, from my comment, I think, this should work for you(with minimum editing of your code):
def translates_many_words(str)
str = str.split(' ');
if str.length > 2
str.map do |item|
begins_with_consonant(item) || word_begins_with_vowel(item)
end.join(' ')
end
end
UPD
Also, I'll recommend you to refactor your code to make it more readable, it could help you in the future.
My variant of this method is:
def translates_many_words(str)
str = str.split
# line under - is a shortcut from `return nil if str.size <= 2`
# `#size` is more relative to this context if you will count elements of array
return unless str.size > 2
# Now, when we excluded possibility of work with array that have less then 2 elements,
# we can continue with our iteration
str.map do |item|
begins_with_consonant(item) || word_begins_with_vowel(item)
end.join(' ')
end