Question

我最初写了一个方法，用一个词来判断它的元音是否按字母顺序排列。我是通过使用以下代码完成的：

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

但是，我决定尝试使用all来改变它？方法。我尝试使用以下代码执行此操作：

def ordered_vowel_word?(word)
  vowels = ["a","e", "i", "o", "u"]
  splitted_word = word.split("")
  vowels_in_word = []
  vowels_in_word = splitted_word.select {|word| vowels.include?(word)}

  vowels_in_word.all? {|x| vowels_in_word[x]<= vowels_in_word[x+1]}


end


ordered_vowel_word?("word")

任何人都有任何想法为什么它不起作用？我原以为这会起作用。此外，如果有人有更好的解决方案，请随时发布。谢谢！

例如：

it "does not return a word that is not in order" do
  ordered_vowel_words("complicated").should == ""
end

it "handle double vowels" do
  ordered_vowel_words("afoot").should == "afoot"
end

it "handles a word with a single vowel" do
  ordered_vowel_words("ham").should == "ham"
end

it "handles a word with a single letter" do
  ordered_vowel_words("o").should == "o"
end

it "ignores the letter y" do
  ordered_vowel_words("tamely").should == "tamely"
end

Answer 1

我将如何做到这一点：

#!/usr/bin/ruby

def ordered?(word)
  vowels = %w(a e i o u)
  check = word.each_char.select { |x| vowels.include?(x) }
  # Another option thanks to @Michael Papile
  # check = word.scan(/[aeiou]/)
  puts check.sort == check
end

ordered?("afoot")
ordered?("outaorder")

输出是：

true
false

在原始示例中，您使用数组值（String）作为数组索引，当all?方法触发时，它应该是整数。

Answer 2

def ordered_vowel_word?(word)
    vowels = ["a","e", "i", "o", "u"]
    splitted_word = word.split("")
    vowels_in_word = []
    vowels_in_word = splitted_word.select {|word| vowels.include?(word)}
    p vowels_in_word #=> ["o"]
    vowels_in_word.all? {|x| vowels_in_word[x]<= vowels_in_word[x+1]}
end

p ordered_vowel_word?("word")
#=> `[]': no implicit conversion of String into Integer (TypeError)

vowels_in_word仅包含'o'，而在vowels_in_word.all? {|x| vowels_in_word[x]<= vowels_in_word[x+1]}内，表达式vowels_in_word[x]表示vowels_in_word["o"]，因为索引永远不会{{1} }}

为什么我的全部？功能不起作用？我的语法有什么问题？

2 个答案: