Delete repeated vowels in string

Question

Here is my code:

def del_rep_vow(s):
    '''
    >>> del_rep_vow('adaptation')
    'adpttion'
    >>> del_rep_vow('repetitions')
    'reptitons'
    >>> del_rep_vow('kingkong')
    'kingkong'
    >>> del_rep_vow('aeiouaeiou')
    'aeiou'
    '''
    final_list = []
    for i in s:
        if i not in final_list:
            final_list.append(i)
    return ''.join(final_list)

if __name__ == "__main__":
    import doctest
    doctest.testmod(verbose=True)

I don’t know how to make the restriction that only repeated vowels have to be deleted once they are in the list. Any help will be very welcome.

Output of ‘adaptation’ must be ‘adpttion’, for example.

Asked By: Alba

||

Source

Answer 1

You should maintain a Python set of vowel characters already seen. For each new encountered letter as you walk down the string, only append a vowel if it is not in the set.

def del_rep_vow(s):
    vowels_seen = set()
    final_list = []
    for i in s:
        if i in ['a', 'e', 'i', 'o', 'u']:
            if i not in vowels_seen:
                final_list.append(i)
                vowels_seen.add(i)
        else:
            final_list.append(i)

    return ''.join(final_list)

print(del_rep_vow('adaptation'))  # 'adpttion'
print(del_rep_vow('repetitions')) # 'reptitons'
print(del_rep_vow('kingkong'))    # 'kingkong'
print(del_rep_vow('aeiouaeiou'))  # 'aeiou'

Answered By: Tim Biegeleisen

Answer 2

You can maintain 2 lists of letters: one for the vowels you have seen so far, and other for the remaining letters in the string.

vowels = []
remaining = []
for c in s:
    if c in 'aeiou':
        if c in vowels:
            continue
        vowels.append(c)
    remaining.append(c)

Answered By: Rodrigo Rodrigues

Answer 3

Here’s another approach

def del_repts(x):
    vowels = 'aeiou'
    foundVowels = ''
    rslt = ''
    for c in x:
        if c in vowels:
            if c not in foundVowels:
                rslt += c
                foundVowels += c
        else:
            rslt += c
    return rslt  

inl = ['adaptation', 'repetitions', 'kingkong', 'aeiouaeiou']
for itm in inl:
    print(f'{itm}t->t{del_repts(itm)}' )

Yields:

adaptation  ->  adpttion
repetitions ->  reptitons
kingkong    ->  kingkong
aeiouaeiou  ->  aeiou

Answered By: itprorh66

Answer 4

I feel this is pretty concise:

def del_rep_vow(s):
    for ch in 'eaiou':
        if ch in s:
            i = s.index(ch) + 1
            s = s[:i] + s[i:].replace(ch, '')
    return s


print(del_rep_vow('adaptation'))
print(del_rep_vow('repetitions'))
print(del_rep_vow('kingkong'))
print(del_rep_vow('aeiouaeiou'))

Result:

adpttion
reptitons
kingkong
aeiou

Answered By: Grismar

Answer 5

The str.partition method closely matches your need. It splits the string at the first occurrence of a separator and returns a 3-tuple containing the part before the separator, the separator itself or "" if not found, and the part after the separator. All that remains is to remove the vowel from the after-part.

def del_rep_vow(s):
    for v in 'aeiou':
        before, vowel, after = s.partition(v)
        s = before + vowel + after.replace(v, '')
    return s

print(del_rep_vow('adaptation'))  # 'adpttion'
print(del_rep_vow('repetitions')) # 'reptitons'
print(del_rep_vow('kingkong'))    # 'kingkong'
print(del_rep_vow('aeiouaeiou'))  # 'aeiou'

Answered By: Olivier Melançon

Answer 6

You could also keep your initial code, adding just a little condition to the test:

def del_rep_vow(s):
    final_list = []
    vowels = 'aeiou'
    for i in s:
        if i not in vowels or i not in final_list: # allowed consonants to be added every time
            final_list.append(i)
    return ''.join(final_list)

Answered By: Swifty

Answer 7

recursive solution:

def del_rep_vow(s):
    return del_rep_vow(s[:-1]) + ('' if s[-1] in 'aeiou' and s[-1] in s[:-1] else s[-1]) if s else ''

del_rep_vow('repetitions')  # reptitons'

Answered By: SergFSM

Delete repeated vowels in string

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