查找字符串中的最后一个元音

Question

I cant seem to find the proper way to search a string for the last vowel, and store any unique consonants after that last vowel. I have it set up like this so far.

word = input('Input a word: ')
wordlow = word.lower()
VOWELS = 'aeiou'
last_vowel_index = 0

for i, ch in enumerate(wordlow):
    if ch == VOWELS:
        last_vowel_index += i

print(wordlow[last_vowel_index + 1:])

Answer 1

I like COLDSPEED's approach, but for completeness, I will suggest a regex based solution:

import re
s = 'sjdhgdfgukgdk'
re.search(r'([^AEIOUaeiou]*)$', s).group(1)
# 'kgdk'

# '[^AEIOUaeiou]'  matches a non-vowel (^ being the negation)
# 'X*'  matches 0 or more X
# '$' matches the end of the string
# () marks a group, group(1) returns the first such group

See the docs on python regular expression syntax . Further processing is also needed for the uniqueness part ;)

Answer 2

You can reverse your string, and use itertools.takewhile to take everything until the "last" (now the first after reversal) vowel:

from itertools import takewhile

out = ''.join(takewhile(lambda x: x not in set('aeiou'), string[::-1]))[::-1]
print(out)
'ng'

If there are no vowels, the entire string is returned. Another thing to note is that, you should convert your input string to lower case using a str.lower call, otherwise you risk not counting uppercase vowels.

---

If you want unique consonants only (without any repetition), a further step is needed:

from collections import OrderedDict
out = ''.join(OrderedDict.fromkeys(out).keys())

Here, the OrderedDict lets us keep order while eliminating duplicates, since, the keys must be unique in any dictionary.

Alternatively, if you want consonants that only appear once, use:

from collections import Counter

c = Counter(out)
out = ''.join(x for x in out if c[x] == 1)

Answer 3

You can simply write a function for that:

def func(astr):
    vowels = set('aeiouAEIOU')

    # Container for all unique not-vowels after the last vowel
    unique_notvowels = set()

    # iterate over reversed string that way you don't need to reset the index
    # every time a vowel is encountered.
    for idx, item in enumerate(astr[::-1], 1):  
        if item in vowels:
            # return the vowel, the index of the vowel and the container
            return astr[-idx], len(astr)-idx, unique_notvowels
        unique_notvowels.add(item)

    # In case no vowel is found this will raise an Exception. You might want/need
    # a different behavior...
    raise ValueError('no vowels found')

For example:

>>> func('asjhdskfdsbfkdes')
('e', 14, {'s'})

>>> func('asjhdskfdsbfkds')
('a', 0, {'b', 'd', 'f', 'h', 'j', 'k', 's'})

It returns the last vowel, the index of the vowel and all unique not-vowels after the last vowel.

In case the vowels should be ordered you need to use an ordered container instead of the set, for example a list (could be much slower) or collections.OrderedDict (more memory expensive but faster than the list).

Answer 4

You can just reverse your string and loop over each letter until you encounter the first vowel:

for i, letter in enumerate(reversed(word)):
    if letter in VOWELS:
        break
print(word[-i:])

Answer 5

last_vowel will return the last vowel in the word

last_index will give you the last index of this vowel in the input

Python 2.7

input = raw_input('Input a word: ').lower()
last_vowel = [a for a in input if a in "aeiou"][-1]
last_index = input.rfind(last_vowel)
print(last_vowel)
print(last_index)

Python 3.x

input = input('Input a word: ').lower()
last_vowel = [a for a in input if a in "aeiou"][-1]
last_index = input.rfind(last_vowel)
print(last_vowel)
print(last_index)