Python 索引器：字符串索引超出递归函数的范围

因此，我正在学习python，并试图计算一个句子中元音的数量。我已经知道如何使用count（）函数和迭代来实现它，但现在我正尝试使用递归来实现它。当我尝试以下方法时，我得到一个错误“IndexError:字符串索引超出范围”。这是我的密码

sentence = input(": ")

def count_vowels_recursive(sentence):
    total = 0
    if sentence[0] == "a" or sentence[0] == "e" or sentence[0] == "i" or sentence[0] == "o" or sentence[0] == "u":
        total = total + 1 + count_vowels_recursive(sentence[1:])
    else:
        total = total + count_vowels_recursive(sentence[1:])   
    return the_sum

print(count_vowels_recursive(sentence))

这里是我前面的两个解决方案

def count_vowels(sentence):
    a = sentence.count("a")
    b = sentence.count("e")
    c = sentence.count("i")
    d = sentence.count("o")
    e = sentence.count("i")
    return (a+b+c+d+e)



def count_vowels_iterative(sentence):
    a_ = 0
    e_ = 0
    i_ = 0
    o_ = 0
    u_ = 0
    for i in range(len(sentence)):
        if "a" == sentence[i]:
            a_ = a_ + 1
        elif "e" == sentence[i]:
            e_ = e_ + 1
        elif "i" == sentence[i]:
            i_ = i_ + 1
        elif "o" == sentence[i]:
            o_ = o_ + 1
        elif "u" == sentence[i]:
            u_ = u_ + 1
        else:
            continue
    return (a_ + e_ + i_ + o_ + u_)

---

你没有基本的理由。函数将一直递归，直到

语句

为空，在这种情况下，第一个if语句将导致该索引错误

---

您应该首先检查句子是否为空，如果为空，则返回0

您可以大大缩短句子：

def count_vowels_recursive(sentence):
    # this base case is needed to stop the recursion
    if not sentence:  
        return 0
    # otherwise, sentence[0] will raise an exception for the empty string
    return (sentence[0] in "aeiou") + count_vowels_recursive(sentence[1:])
    # the boolean expression `sentence[0] in "aeiou"` is cast to an int for the addition

您可以尝试以下方法：

def count_vowels_recursive(s, count):
   if not s:
      return count
   else:
       new_count = count
       if s[0] in ["a", "e", "i", "o", "u"]:
          new_count += 1
       return count_vowels_recursive(s[1:], new_count)

---

提示：当递归到达字符串末尾时，您尝试测试

语句[0]

？您可以进一步缩短它：

return（在“aeiou”中的句子[0]）+count\u元音\u recursive（句子[1:]）

（Python将False计为0，将True计为1）。True，尽管这种强制对初学者来说可能更加模糊。