How to make index() method return an integer instead of None?

Question

I am trying to return the index value of the first vowel in a string using these functions:

def is_vowel(c):
    v = 'aeiou'
    for l in v:
        if l == c:
            return True
        else:
            return False

def index_of_first_vowel(string):
    for l in string:
        if is_vowel(l):
            index = int(string.index(l))
            return index

So the string 'hello' should return 1.

I have tried many different ways of writing the second function and every time it returns None. Am I using the index() method correctly? Or is my logic off somewhere?

Does this answer your question? [What is a debugger and how can it help me diagnose problems?](https://stackoverflow.com/questions/25385173/what-is-a-debugger-and-how-can-it-help-me-diagnose-problems) — mkrieger1, Nov 23 '20 at 20:27
the is_vowel function returns true if a letter is a vowel and false otherwise. is that where the problem is? — Garrett, Nov 23 '20 at 20:31
The problem is that it *doesn't* do this. What do you get when you print `is_vowel('e')`? — mkrieger1, Nov 23 '20 at 20:31
You should also read this: https://ericlippert.com/2014/03/05/how-to-debug-small-programs/ — mkrieger1, Nov 23 '20 at 20:40

score 1 · Accepted Answer · answered Nov 23 '20 at 20:34

This should solve your problem, you don't need to loop in the list of letters in the function is_vowel, you can just look if c is in it.

Also, your loop would be interrupted at the first pass every time you don't pass an 'a'

>>> def is_vowel(c):
...     if c in "aeiou":
...             return True
...     return False
... 
>>> def index_of_first_vowel(string):
...     for l in string:
...             if is_vowel(l):
...                     index = int(string.index(l))
...                     return index
... 
>>> print(index_of_first_vowel("hello"))
1

norok2 · Answer 2 · 2020-11-23T23:14:48.513

Note that your approach can be improved quite a bit:

The whole is_vowel() can be rewritten in one line (and considering also the case):

def is_vowel(c):
    return c in 'aeiouAEIOU'

and that can be easily inlined.

Instead of using str.index(), you can just keep track of the index at each iteration. The idiomatic way in Python is by using enumerate().
If you do not want your function to return None, you have to explicitly use return after your loop to handle the case of no character actually being a vowel.
The code can be made more general by inputing the "vowel" definition as a paremeter.

Putting all these together, one gets:

def find_any(s, chars='aeiouAEIOU'):
    for i, c in enumerate(s):
        if c in chars:
            return i
    return -1

For larger input strings, and small character sets a much more efficient approach is to use the start and stop parameters of str.index() or, possibly faster, str.find() which does not require a try/catch to handle missing matches:

def find_any_other(s, chars='aeiouAEIOU'):
    result = -1
    for char in chars:
        idx = s.find(char, 0, result)
        if idx > 0:
            result = idx
        if not idx:
            break
    return result if result < len(s) else -1

A simple test to show the timings indicate that the second approach can be much faster:

n = 1
s = 'x' * n + 'hello'
%timeit find_any(s)
# 1.35 µs ± 174 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit find_any_other(s)
# 6.04 µs ± 332 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

n = 1000000
s = 'x' * n + 'hello'
%timeit find_any(s)
# 172 ms ± 6.18 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit find_any_other(s)
# 790 µs ± 36.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

How to make index() method return an integer instead of None?

2 Answers2