Is there a better way to check for vowels in the first position of a word?

Question

I'm trying to check for a vowel as the first character of a word. For my code I currently have this:

if first == 'a' or first == 'e' or first == 'i' or first == 'o' or first == 'u':

I was wondering is there a much better way to do this check or is this the best and most efficient way?

Answer 1

You can try like this using the in:

if first.lower() in 'aeiou':

or better like

if first.lower() in ('a', 'e', 'i', 'o', 'u'):

Answer 2

Better create a set of vowels, like this

>>> vowels = set('aeiouAEIOU')
>>> vowels
set(['a', 'A', 'e', 'i', 'o', 'I', 'u', 'O', 'E', 'U'])

and then check if first is one of them like this

>>> if first in vowels:
...

---

Note: The problem with

if first in 'aeiouAEIOU':

approach is, if your input is wrong, for example, if first is 'ae', then the test will fail.

>>> first = 'ae'
>>> first in 'aeiouAEIOU'
True

But ae is clearly not a vowel.

---

Improvement:

If it is just a one-time job, where you don't care to create a set beforehand, then you can use if first in 'aeiouAEIOU': itself, but check the length of first first, like this

>>> first = 'ae'
>>> len(first) == 1 and first in 'aeiouAEIOU'
False

Answer 3

Here is the regex approach:

from re import match

if match(r'^[aieou]', first):
    ...

This regular expression will match if the first character of "first" is a vowel.

Answer 4

If your function is returning boolean value then easiest and simplest way will be

`bool(first.lower() in 'aeiou')`

Or

return first.lower() in 'aeiou'