IndexError: String Index out of range for recursive function

Question

So I am learning python and am trying to count the number of vowels in a sentence. I figured out how to do it both using the count() function and an iteration but now I am trying to do it using recursion. When I try the following method I get an error "IndexError: string index out of range". Here is my code.

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))

Here are my previous two solutions.

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_)

Hint: what happens when your recursion reaches the end of the string and you try to test sentence[0]? — Hugh Bothwell
– Hugh Bothwell, Commented Sep 15, 2017 at 21:33

Daniel Roseman · Accepted Answer · 2017-09-15 21:35:57Z

2

You have no base case. The function will keep recursing until sentence is empty, in which case your first if statement will cause that index error.

You should first of all check if sentence is empty, and if so return 0

answered Sep 15, 2017 at 21:35

Daniel Roseman

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Comments

user2390182 · Accepted Answer · 2017-09-15 21:38:28Z

1

You can shorten things up quite a bit:

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

edited Sep 15, 2017 at 21:38

answered Sep 15, 2017 at 21:36

user2390182

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2 Comments

Hugh Bothwell Over a year ago

You can shorten it even more: return (sentence[0] in "aeiou") + count_vowels_recursive(sentence[1:]) (Python counts False as 0 and True as 1).

user2390182 Over a year ago

True, even though this coercion is probably even more obscure to the beginner.

Ajax1234 · Accepted Answer · 2017-09-15 21:52:10Z

1

You can try this:

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)

answered Sep 15, 2017 at 21:52

Ajax1234

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