CODEWITHSUNDEEP
pythonstringpython-3.x
Athena
Athena

Reputation: 37

How to get consonant repeated, but not vowel in Python

I have two functions, the first one is related to the second, and it codes for whether or not a letter is a vowel (True) or a consonant (False).

def vowel(c):
    """(str) -> bool
    Return whether the string c is a vowel.
    >>> vowel('e')
    True
    >>> vowel('t')
    False
    """
    for char in c:
        if char.lower() in 'aeiou':
            return True
        else:
            return False


def repeated(s, k):
    """(str) -> str
    Return a string where consonants in the string s is repeated k times.
    >>> repeated('', 24)
    ''
    >>> repeated('eoa', 2)
    'eoa'
    >>> repeated('m', 5)
    'mmmmm'
    >>> repeated('choice', 4)
    'cccchhhhoicccce'
    """
    result = ''

    for c in s:
        if c is not vowel(c):
            result = result + (c * k)
    return result

This is what I have for the function, but the examples fail and doesn't skip the vowels.

    repeat('eoa', 2)
Expected:
    'eoa'
Got:
    'eeooaa'

Thanks in advance!

Upvotes: 0

Views: 382

Answers (1)

Dimitris Fasarakis Hilliard
Dimitris Fasarakis Hilliard

Reputation: 160587

Two things. In the vowel function, there's no need for a loop. You're sending a single character so you simply need to check it:

def vowel(c):
    if c.lower() in 'aeiou':
        return True
    else:
        return False

Or:

def vowel(c):
    return True if c.lower() in 'aeiou' else False

Then, in repeated, don't use c is not vowel(c). That compares if the identity of c, the character, is or isn't equal to True/False. Just use the value returned from vowel directly and conditionally add to result:

def repeated(s, k):
    result = ''
    for c in s:
        if not vowel(c):
            result += (c * k)
        else:
            result += c
    return result

Upvotes: 0

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