豬拉丁文到英文轉換

Question

我們被賦予了將Pig-Latin轉換爲英文的任務。因此對於例如我必須編寫一個程序來轉換豬拉丁語句子：iway ovealay omputeracy ienceascay - 我愛計算機科學。我們也假設我們的程序爲piglatin.py並將其導入由它們提供的不同程序中。所以你可以看到這就是我所做的。

def toEnglish(s): 
    words = s.split(" ") ; # split based on space character 
    for word in words: 
    if word[-3:] == 'way': 
    qq = word[:-3] 
    return qq 

    elif word[-3:] != 'way': 
    x = word[:-2] 
    y = x[::-1] 
    z = y.index('a') 
    rrr = int(z) 
    a = y[:rrr] 
    d = a[::-1] 
    b = y[z+1:] 
    rr = b[::-1] 
    k = d+rr 
    return k 

我不得不導入到這個程序

import piglatin 

choice = input ("(E)nglish or (P)ig Latin?\n") 
action = choice[:1] 
if action == 'E': 
    s = input("Enter an English sentence:\n") 
    new_s = piglatin.toPigLatin(s) 
    print("Pig-Latin:") 
    print(new_s) 
elif action =='P': 
    s = input("Enter a Pig Latin sentence:\n") 
    new_s = piglatin.toEnglish(s) 
    print("English:") 
    print(new_s) 

#You can just ignore the middle part 'E' of the program. 

問題是，當我運行的代碼到這個程序這一點，它只是將第一字我打字。所以，如果我在atabay類型anamay，我只得到'蝙蝠'而不是'蝙蝠俠'。

好的，我得到了解決我的問題。

def toEnglish(s) 
    words = s.split(" ") ; # split based on space character 
    k = [] 
    for word in words: 
     if word[-3:] == 'way': 
      qq = word[:-3] 
      k.append(qq) 
     elif word[-3:] != 'way': 
      x = word[:-2] 
      y = x[::-1] 
      z = y.index('a') 
      rrr = int(z) 
      a = y[:rrr] 
      d = a[::-1] 
      b = y[z+1:] 
      rr = b[::-1] 
      l = d+rr 
      k.append(l) 
    return ' '.join(k) 

Answer 1

的return內置結束當場無論是後規定值的回報，永遠不會執行它之後會發生任何代碼的功能。您的代碼的問題是您的return在for循環內縮進。因此，您的循環將運行一次並返回您擁有的任何值。相反，您可能希望return在全部單詞被解密後返回該值。因此，您應該將您的return放在與for循環相同的縮進級別之後。下面是合法的代碼：

def toEnglish(s): 
    words = s.split(" ") ; # split based on space character 
    k = [] 
    for word in words: 
     if word[-3:] == 'way': 
      qq = word[:-3] 
      k.append(qq) 
     elif word[-3:] != 'way': 
      x = word[:-2] 
      y = x[::-1] 
      z = y.index('a') 
      rrr = int(z) 
      a = y[:rrr] 
      d = a[::-1] 
      b = y[z+1:] 
      rr = b[::-1] 
      l = d+rr 
      k.append(l) 
    return k 

>>> print ' '.join(toEnglish('atabay anamay')) 

蝙蝠男