这个功能是如何工作的？

Question

ordered_vowel_words方法和ordered_vowel_word?辅助方法接受一个单词并返回单词，如果单词的元音符合（a，e，i，o，u）的顺序。

我无法理解逻辑。特别是帮助方法中的最后一个块(0...(vowels_arr.length - 1)).all? do...如何工作。

有人可以解释这是如何工作的？我不明白如何在range上调用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

我添加了一些意见:)

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 

希望这有助于！

编辑我可能会补充说，最后一块并不觉得Ruby-ish。我建议这个替代实现：

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?块基本上遍历vowels_arr阵列，比较每个值与它的下一个。如果所有的比较返回true那么all?也将返回true，这意味着数组是有序的。如果其中一个迭代返回false，返回值all?也将是false，这意味着该集合是无序的。

您可以在Rangehttp呼吁all?：//www.ruby-doc.org/core-1.9.3/Range.html对象，因为Range混合在Enumerablehttp：//www.ruby-doc.org/core -1.9.3/Enumerable.html模块，它定义了all?。

您可以尝试在IRB以下验证这一点：

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

Answer 3

1. 第一部分(0...(vowels_arr.length - 1))创造多少元音在词的范围 0。
2. all?在该范围内迭代并返回true如果全部为 范围元素的某些条件为真否则为false。
3. do |i|引入了i作为表示 步骤1中创建
4. 最后的范围内的每个元素，条件是在范围内的每个索引变量的块，现在由i表示的，它会检查是否vowels_arr[i] <= vowels_arr[i+1]为真。