user1679073 user1679073 - 1 year ago 71
Python Question

Count vowels from raw input

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


It counts the vowels fine, but due to
if string[i] == vowels[i]:
, it will only count one vowel once as
keeps increasing in the range. How can I change this code to check the inputted string for vowels without encountering this problem?

Answer Source

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:


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'])

*dicts 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:


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()
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