Python 'String Index Out of Range Error'?

Question

I am 15 and currently doing a GCSE in Computing. My knowledge is very basic and I've had a bit of trouble with a piece of code I have to write for a 'Vowel Worth Calculator' which is supposed to check through a word and give it a vowel score depending on how many and which vowels it has. I keep getting an error, and I am completely stumped, any help would be appreciated. Here is my source code :

Vowel Worth Counter

print('Welcome to the Vowel Worth Counter!')

word = input('Please input your word, in lower-case, or type Q to quit.')

if word == 'Q' :
    quit()

def vowelcount(word) :
    lettercount = int(len(word))
    vowelscore = 0
    checkcount = 1
    position = 0
    while lettercount != checkcount :
        if word[position] == str('a') :
            vowelscore = vowelscore + 5
        if word[position] == str('e') :
            vowelscore = vowelscore + 4
        if word[position] == str('i') :
            vowelscore = vowelscore + 5
        if word[position] == str('o') :
            vowelscore = vowelscore + 5
        if word[position] == str('u') :
            vowelscore = vowelscore + 5
        position = position + 1
    if lettercount == checkcount :
        print('I have finished calculatiing your Vowel Score.')
        print('Your Vowel score is ' + str(vowelscore) + '!')
        for x in range (0,1) :
            break
vowelcount(word)

As I said, any help would be appreciated, thank you.

You never change checkcount, so your while loop can never end. — Peter DeGlopper
– Peter DeGlopper, Commented Nov 20, 2013 at 22:01
while lettercount != checkcount. You never alter either in your while loop, while you still increment position. This is why you're code is dying — inspectorG4dget
– inspectorG4dget, Commented Nov 20, 2013 at 22:02
Also, in Python it is far more usual to loop directly over data structures like lists or strings than it is to use an index. for letter in word: would be the Pythonic idiom, at least if I weren't going to use more advanced constructions. — Peter DeGlopper
– Peter DeGlopper, Commented Nov 20, 2013 at 22:02

Óscar López · Accepted Answer · 2013-11-20 22:00:42Z

3

The exit contidion in the loop should be:

while position < lettercount:

Or even simpler, you can iterate over each character in a string like this:

for c in word:
    if c == 'a':
        # and so on

answered Nov 20, 2013 at 22:00

Óscar López

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Comments

C.B. · Accepted Answer · 2013-11-20 22:23:30Z

2

It might be more pythonic to use the dictionary data structure

vowels = { 'a' : 5, 'e' : 4, 'i' : 5, 'o' : 5, 'u' : 5}

vowelscore = 0

for letter in word:
    if letter in vowels:
        vowelscore += vowels[letter]

answered Nov 20, 2013 at 22:23

C.B.

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4 Comments

tmj Over a year ago

Haha! I had written the exact same code, but had resisted from answering the question, because I thought it was a little too cocky. Plus, it did not answer the question the OP asked, "Why his code is not working?". Anyway! :)

C.B. Over a year ago

Yeah. I figured since @ÓscarLópez had already answered the OP, I figured I would suggest looking at a data structure he/shre might not have encountered before :). That's funny though

Peter DeGlopper Over a year ago

OK, since we're getting into alternate constructions - define the dict as above, then vowelscore = sum(vowels.get(letter, 0) for letter in word)

user3015131 Over a year ago

Thank you all so much, was not expecting a response!

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