Count vowels from raw input

Question

I have a homework question which asks to read a string through raw input and count how many vowels are in the string. This is what I have so far but I have encountered a problem: ``` def vowels(): vowels = ["a","e","i","o","u"] count = 0 string = raw_input ("Enter a string: ") for i in range(0, len(string)): if string[i] == vowels[i]: count = count+1 print count vowels() ``` It counts the vowels fine, but due to `if string[i] == vowels[i]:`, it will only count one vowel once as `i` keeps increasing in the range. How can I change this code to check the inputted string for vowels without encountering this problem?

Accepted Answer

`in` operator

You probably want to use the `in` operator instead of the `==` operator - the `in` operator lets you check to see if a particular item is in a sequence/set.

```
1 in [1,2,3] # True
1 in [2,3,4] # False
'a' in ['a','e','i','o','u'] # True
'a' in 'aeiou' # Also True
```

Some other comments:

Sets

The `in` operator is most efficient when used with a `set`, which is a data type specifically designed to be quick for "is item X part of this set of items" kind of operations.*

```
vowels = set(['a','e','i','o','u'])
```

*`dict`s are also efficient with `in`, which checks to see if a key exists in the dict.

Iterating on strings

A string is a sequence type in Python, which means that you don't need to go to all of the effort of getting the length and then using indices - you can just iterate over the string and you'll get each character in turn:

E.g.:

```
for character in my_string:
    if character in vowels:
        # ...
```

Initializing a set with a string

Above, you may have noticed that creating a set with pre-set values (at least in Python 2.x) involves using a list. This is because the `set()` type constructor takes a sequence of items. You may also notice that in the previous section, I mentioned that strings are sequences in Python - sequences of characters.

What this means is that if you want a set of characters, you can actually just pass a string of those characters to the `set()` constructor - you don't need to have a list one single-character strings. In other words, the following two lines are equivalent:

```
set_from_string = set('aeiou')
set_from_list = set(['a','e','i','o','u'])
```

Neat, huh? :) Do note, however, that this can also bite you if you're trying to make a set of strings, rather than a set of characters. For instance, the following two lines are not the same:

```
set_with_one_string = set(['cat'])
set_with_three_characters = set('cat')
```

The former is a set with one element:

```
'cat' in set_with_one_string # True
'c' in set_with_one_string # False
```

Whereas the latter is a set with three elements (each one a character):

```
'c' in set_with_three_characters` # True
'cat' in set_with_three_characters # False
```

Case sensitivity

Comparing characters is case sensitive. `'a' == 'A'` is False, as is `'A' in 'aeiou'`. To get around this, you can transform your input to match the case of what you're comparing against:

```
lowercase_string = input_string.lower()
```

Count vowels from raw input

Solution

Problem