CODEWITHSUNDEEP
pythonstringrecursion
b0b0
b0b0

Reputation: 38

python recursion vowel count

def count_vowels(s):
    if not s:
        return 0
    elif s[0] in VOWELS:
        return 1 + count_vowels(s[1:])
    else:
        return 0 + count_vowels(s[1:])

I see that this code works perfectly fine for finding the number of vowels in a string. I also understand recursion always calls for a base case. I guess my main problem with this code is what does the first if statement actually mean? What is it checking?

if not s:
    return 0

if what not s? Is there any way to write the code without that portion?

Upvotes: 1

Views: 1175

Answers (2)

Salvador Dali
Salvador Dali

Reputation: 222641

This is your exit from recursion. Your recursion has to stop at some point of time (otherwise it will run forever).

It means that if the string is empty - return 0 (there are no vowels in empty string) and stop.

Upvotes: 2

FogleBird
FogleBird

Reputation: 76792

if not s checks if the string is empty. Empty strings are "false-y" objects in Python. If the string is empty, it must not have any vowels.

I realize you are probably required to use recursion here, but this would be a better way to do it in practice:

>>> VOWELS = set('aeiou')
>>> def count_vowels(s):
...     return sum(x in VOWELS for x in s)
... 
>>> count_vowels('hello world')
3

Upvotes: 1

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