這個功能是如何工作的？

Question

ordered_vowel_words方法和ordered_vowel_word?輔助方法接受一個單詞並返回單詞，如果單詞的元音符合（a，e，i，o，u）的順序。

我無法理解邏輯。特別是幫助方法中的最後一個塊(0...(vowels_arr.length - 1)).all? do...如何工作。

有人可以解釋這是如何工作的？我不明白如何在range上調用all?。

def ordered_vowel_words(str) 
    words = str.split(" ") 

    ordered_vowel_words = words.select do |word| 
    ordered_vowel_word?(word) 
    end 

    ordered_vowel_words.join(" ") 
end 

def ordered_vowel_word?(word) 
    vowels = ["a", "e", "i", "o", "u"] 

    letters_arr = word.split("") 
    vowels_arr = letters_arr.select { |l| vowels.include?(l) } 

    (0...(vowels_arr.length - 1)).all? do |i| 
    vowels_arr[i] <= vowels_arr[i + 1] 
    end 
end 

Answer 1

我添加了一些意見:)

def ordered_vowel_words(str) 
    # words is a string with words separated by a whitespace. 
    # split generates an array of words from a string 
    words = str.split(" ") 

    # select only the ordered vowel words from the previous array 
    ordered_vowel_words = words.select do |word| 
    ordered_vowel_word?(word) 
    end 

    # join the ordered vowel words in a single string 
    ordered_vowel_words.join(" ") 
end 

def ordered_vowel_word?(word) 
    # THESE ARE THE VOWELS YOU FOOL 
    vowels = ["a", "e", "i", "o", "u"] 

    # transform the word in an array of characters 
    letters_arr = word.split("") 

    # select only the vowels in this array 
    vowels_arr = letters_arr.select { |l| vowels.include?(l) } 

    # generate a range from 0 to the length of the vowels array minus 2: 
    # there is this weird range because we want to iterate from the first vowel 
    # to the second to last; all? when called on a range returns true if... 
    (0...(vowels_arr.length - 1)).all? do |i| 
    # for each number in the range, the current vowel is smaller that the next vowel 
    vowels_arr[i] <= vowels_arr[i + 1] 
    end 
end 

希望這有助於！

編輯我可能會補充說，最後一塊並不覺得Ruby-ish。我建議這個替代實現：

def ordered_vowel_word?(word) 
    vowels = ["a", "e", "i", "o", "u"] 

    # transform the word in an array of characters 
    letters_arr = word.split("") 

    # select only the vowels in this array 
    vowels_arr = letters_arr.select { |l| vowels.include?(l) } 

    # from this array generate each possible consecutive couple of characters 
    vowels_arr.each_cons(2).all? do |first, second| 
    first <= second 
    end 
end 

require 'rspec/autorun' 

describe "#ordered_vowel_word?" do 
    it "tells if word is ordered" do 
    expect(ordered_vowel_word?("aero")).to be_true 
    end 

    it "or not" do 
    expect(ordered_vowel_word?("rolling")).to be_false 
    end 
end 

Answer 2

的all?塊基本上遍歷vowels_arr陣列，比較每個值與它的下一個。如果所有的比較返回true那麼all?也將返回true，這意味着數組是有序的。如果其中一個迭代返回false，返回值all?也將是false，這意味着該集合是無序的。

您可以在Rangehttp呼籲all?：//www.ruby-doc.org/core-1.9.3/Range.html對象，因爲Range混合在Enumerablehttp：//www.ruby-doc.org/core -1.9.3/Enumerable.html模塊，它定義了all?。

您可以嘗試在IRB以下驗證這一點：

Range.included_modules # => => [Enumerable, Kernel] 

Answer 3

1. 第一部分(0...(vowels_arr.length - 1))創造多少元音在詞的範圍 0。
2. all?在該範圍內迭代並返回true如果全部爲 範圍元素的某些條件爲真否則爲false。
3. do |i|引入了i作爲表示 步驟1中創建
4. 最後的範圍內的每個元素，條件是在範圍內的每個索引變量的塊，現在由i表示的，它會檢查是否vowels_arr[i] <= vowels_arr[i+1]爲真。