在Python中找到元音的第一个出现

Question

I have a function isvowel that returns either True or False , depending on whether a character ch is a vowel.

    def isvowel(ch):
          if "aeiou".count(ch) >= 1:
              return True
          else:
              return False

I want to know how to use that to get the index of the first occurrence of any vowel in a string. I want to be able to take the characters before the first vowel and add them end of the string. Of course, I can't do s.find(isvowel) because isvowel gives a boolean response. I need a way to look at each character, find the first vowel, and give the index of that vowel.

How should I go about doing this?

Answer 1

You can always try something like this:

import re

def first_vowel(s):
    i = re.search("[aeiou]", s, re.IGNORECASE)
    return -1 if i == None else i.start()

s = "hello world"
print first_vowel(s)

Or, if you don't want to use regular expressions:

def first_vowel(s):
    for i in range(len(s)):
        if isvowel(s[i].lower()):
            return i
    return -1

s = "hello world"
print first_vowel(s)

Answer 2

[isvowel(ch) for ch in string].index(True)

Answer 3

(ch for ch in string if isvowel(ch)).next()

or for just the index (as asked):

(index for ch, index in itertools.izip(string, itertools.count()) if isvowel(ch)).next()

This will create an iterator and only return the first vowel element. Warning: a string with no vowels will throw StopIteration , recommend handling that.

Answer 4

Here is my take:

>>> vowel_str = "aeiou"

>>> def isVowel(ch,string):
...     if ch in vowel_str and ch in string:
...             print string.index(ch)
...     else:
...             print "notfound"
... 
>>> isVowel("a","hello")
not found

>>> isVowel("e","hello")
1
>>> isVowel("l","hello")
not found

>>> isVowel("o","hello")
4

Answer 5

Using next for a generator is quite efficient, it means you don't iterate though the entire string (once you have found a string).

first_vowel(word):
    "index of first vowel in word, if no vowels in word return None"
    return next( (i for i, ch in enumerate(word) if is_vowel(ch), None)
is_vowel(ch):
    return ch in 'aeiou'

Answer 6

my_string = 'Bla bla'
vowels = 'aeyuioa'

def find(my_string):
    for i in range(len(my_string)):
        if my_string[i].lower() in vowels:
            return i
            break 

print(find(my_string))