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: d

Solution 1:

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

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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'])

*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:

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()

Solution 2:

You can simplify this code:

def vowels():
    vowels = 'aeiou'
    count = 0string = raw_input ("Enter a string: ")
    for i instring:
        if i in vowels:
            count += 1
    print count

Strings are iterable in Python.

Solution 3:

for i in range(0, len(string)):
    ifstring[i] == vowels[i]:

This actually has a subtler problem than only counting each vowel once - it actually only tests if the first letter of the string is exactly a, if the second is exactly e and so on.. until you get past the fifth. It will try to test string[5] == vowels[5] - which gives an error.

You don't want to use i to look into vowels, you want a nested loop with a second index that will make sense for vowels - eg,

for i in range(len(string)):
   for j in range(len(vowels)):
       ifstring[i] == vowels[j]:
          count += 1

This can be simplified further by realising that, in Python, you very rarely want to iterate over the indexes into a sequence - the for loop knows how to iterate over everything that you can do string[0], string[1] and so on, giving:

for s in string:for v in vowels:ifs==v:count+=1

The inner loop can be simplified using the in operation on lists - it does exactly the same thing as this code, but it keeps your code's logic at a higher level (what you want to do vs. how to do it):

for s in string:if s in vowels:count+=1

Now, it turns out that Python lets do math with booleans (which is what s in vowels gives you) and ints - True behaves as 1, False as 0, so True + True + False is 2. This leads to a one liner using a generator expression and sum:

sum(s in vowels for s in string)

Which reads as 'for every character in string, count how many are in vowels'.

Solution 4:

you can use filter for a one liner

printlen(filter(lambda ch:ch.lower() in"aeiou","This is a String"))

Solution 5:

Here's a more condensed version using sum with a generator:

def vowels():
    string = raw_input("Enter a string: ")
    print sum(1for x instringif x.lower() in 'aeiou')

vowels()