Ruby：拉丁語翻譯器中的字符串範圍

Question

我目前正在做Test First的rspec教程，並且有一個與Pig_Latin問題有關的問題。

具體我想知道關於字符串範圍。這裏是我的代碼段：

if phonemes.include?(word[0]) && phonemes.include?(word[1]) && phonemes.include?(word[2]) 
<do something> 
end 

相反，我嘗試了上面的：

if phonemes.include?(word[0..2]) # i added that character to the list of phonemes 
    <do something>     # e.g. if the word is school i added "sch" to 
end        # the array called phonemes 

然而，即使"sch"是phonemes不工作，word[0..2] == "sch"

我的問題是爲什麼我可以不使用字符串範圍來操縱結果。 （我會後我完整的代碼底部的情況下，這是不清楚）

代碼（工作正在進行中）：

def translate(string) 
array = string.split(" ") 
alphabet = ("a".."z").to_a 
vowels = ["a", "e", "i", "o", "u"] 
phonemes = alphabet - vowels 
phonemes << ["qu", "sch", "thr"] 
result = [] 
array.each do |word| 
    if vowels.include?(word[0]) 
     result << (word + "ay") 
    elsif phonemes.include?(word[0..1]) 
     result << "do something" 
    elsif phonemes.include?(word[0]) && phonemes.include?(word[1]) && phonemes.include?(word[2]) 
     result << (word[3..-1] + (word[0..2] + "ay")) 
    elsif phonemes.include?(word[0]) && phonemes.include?(word[1]) 
     result << (word[2..-1] + (word[0..1] + "ay")) 
    elsif phonemes.include?(word[0..1]) 
     result << "do something else" 
    elsif phonemes.include?(word[0]) 
     result << (word[1..-1] + (word[0]+ "ay")) 
    end 
end 
return result.join(" ") 
end 

一如既往的提示，使代碼更高效，將不勝感激（但最對我來說重要的是要理解爲什麼字符串範圍不工作）。 謝謝。

Answer 1

您的語句phonemes << ["qu", "sch", "thr"]將該數組添加爲phonemes的最後一個元素，這就是include?失敗的原因。 <<運算符用於將單個元素添加到陣列。如果要將該數組中的所有元素添加到phonemes，則可以使用+=運算符。

Answer 2

這不是您的主要問題的答案，但您需要提示以改善您的代碼。我建議你考慮使用一個案例陳述，你有很長的if-else。它使其更具可讀性並減少重複。事情是這樣的：

result << case 
    when vowels.include?(word[0]) 
    word + "ay" 
    when phonemes.include?(word[0..1]) 
    "do something" 
    when phonemes.include?(word[0]) && phonemes.include?(word[1]) 
    if phonemes.include?(word[2]) 
     word[3..-1] + word[0..2] + "ay" 
    else 
     word[2..-1] + word[0..1] + "ay" 
    end 
    when phonemes.include?(word[0..1]) 
    "do something else" 
    when phonemes.include?(word[0]) 
    word[1..-1] + word[0]+ "ay" 
    else 
    "do something else or raise an error if you reach this point." 
end 

我沒有在你的代碼仔細一看，但我沒有，你有兩次phonemes.include?(word[0..1])通知，所以第二個永遠不會被執行。