Comparing vowel index to word

[winnar]

Hey guys! I wrote a small program, which works. Unfortunately, when things get too complicated, it breaks down.

The objective is to list the occurrences of vowels in a given word, such that for (word = yellow) and index spits out (e = 1, o = 1)

The code KIND OF does it, but seems to reiterate its objective. Can someone take a look and tell me what I did wrong?

def duplicount():
    import string

    word = raw_input('Please enter your word:')
    vowels = ['a','e','i','o','u']
    count = {}

    for index,letters in enumerate(word):
        if letters in vowels:
            count[letters] = index
            print count


duplicount()

Yes, I searched google. Yes, I searched the forums.

:wub:

[TrustyTony]

You are printing inside loop instead of out of it, also you only keep the position of last occurance of vowel, use the setdefault method or defaultdict from collections module:

def duplicount():
    import string

    word = raw_input('Please enter your word:')
    vowels = ['a','e','i','o','u']
    count = {}

    for index,letter in enumerate(word):
        if letter in vowels:
            # D.setdefault(k[,d]) -> D.get(k,d), also set D[k]=d if k not in D
            # http://docs.python.org/release/2.5.2/lib/typesmapping.html
            count.setdefault(letter, []).append(index)

    print count
    # one liner for the fancy solution
    print dict((v, [ind for ind, c in enumerate(word) if c==v]) for v in set(word) if v in vowels)


duplicount()
"""Example:
Please enter your word:average
{'a': [0, 4], 'e': [2, 6]}
{'a': [0, 4], 'e': [2, 6]}
"""

[woooee]

The objective is to list the occurrences of vowels in a given word, such that for (word = yellow) and index spits out (e = 1, o = 1)

This (e = 1, o = 1) is not the position/occurrences so I am assuming you want to count the number of times each vowel occurs.

if letters in vowels:
            if letters not in count:
                count[letters]=0     ## add to dictionary, initialize counter to zero

            count[letters] += 1  ## add one to a counter