Python empty string index error

You can use the str.startswith() method to test if a string starts with a specific character; the method takes either a single string, or a tuple of strings:

if s.lower().startswith(tuple('aeiou')):

The str.startswith() method doesn't care if s is empty:

>>> s = ''
>>> s.startswith('a')
False

By using str.lower() you can save yourself from having to type out all vowels in both lower and upper case; you can just store vowels into a separate variable to reuse the same tuple everywhere you need it:

vowels = tuple('aeiou')
if s.lower().startswith(vowels):

In that case I'd just include the uppercase characters; you only need to type it out once, after all:

vowels = tuple('aeiouAEIOU')
if s.startswith(vowels):

This will check the boolean value of s first, and only if it's True it will try to get the first character. Since a empty string is boolean False it will never go there unless you have at least a one-character string.

if s and s[0] in ["a","e","i","o","u","A","E","I","O","U"]:
    print("an", s)

General solution using try/except:

s = input("Enter a phrase: ")
try:
    if s[0].lower() in "aeiou":
        print("an", s)
    else:
        print("a", s)
except IndexError:
    # stuff you want to do if string is empty

Another approach:

s = ""
while len(s) == 0:
    s = input("Enter a phrase: ")

if s[0].lower() in "aeiou":
    print("an", s)
else:
    print("a", s)

Even shorter:if s[0:][0] in vowels: but of course this not pass as 'pythonic' I guess. What it does: a slice (variable[from:to]) may be empty without causing an error. We just have to make sure that only the first element is returned in case the input is longer than 1 character.

Edit: no, sorry, this will not work if s=''. You have to use 'if s[0:][0:] in vowels:' but this clearly crosses a line now. Ugly.

Just use

if s: if s[0] in vowels:

as suggested before.