how to get a function to read input and define it

Question

I'm trying to make code that allows me to take an input string and return how many vowels are in that input string. I had it working with built in functions, but my teacher asked me to then convert it to a user defined function and I've been having a ton of problems. I managed to fix most of them, and have spent several hours trying to search for the answer but I don't know enough terminology yet to search well.

I've tried(and have been fiddling with it for a couple of hours now but no luck.):

def count_vowels(string):
    string = input('Enter string: ')

def count_vowels(i):
    return i in string
    if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u' or i=='A' or i=='E' or i=='I' or i=='O' or i=='U' ):
        vowels = vowels + 1

print(count_vowels(string))

Current error is 'i' is undefined.

can you post the full error as the code you hav given when run prints true dosnt give error i is undefined — Chris Doyle, Oct 04 '19 at 10:26
`return sum([item.lower() in 'aeiou' for item in i])`. You've already got answers to your problem, that gives you the next step in your learning: _list comprehensions_ and how booleans are a subclass of integers, so can be summed — roganjosh, Oct 04 '19 at 10:40

score 0 · Answer 1 · edited Jun 20 '20 at 09:12

0

UPDATE

First of all you use the input function inside your count_vowels so you do not pass string as parameter. Try the following:

def count_vowels():
    string = input('Enter string: ')
    vowels = 0
    for i in string:
        if i.lower() in 'aieou':
            vowels += 1
    return vowels

print(count_vowels())

could also be done in one line:

print(len([x for x in input('Enter string: ') if x.lower() in 'aieou']))

edited Jun 20 '20 at 09:12

Community

1
1

answered Oct 04 '19 at 10:27

Kostas Charitidis

2,991
1
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`for i in string: if i.lower() in ('a', 'e', 'i', 'o', 'u'):`? – roganjosh Oct 04 '19 at 10:31
Thank you so much, now I know what works I'll spend some time working out why it works like that. Thanks dude. – Alden Tank Oct 04 '19 at 10:32
@roganjosh yeap of course. I was a bit lazy to change it entirely so copied his part of `if` statement. :) – Kostas Charitidis Oct 04 '19 at 10:32
@AldenTank Updated with roganjosh 's observation – Kostas Charitidis Oct 04 '19 at 10:33
Wait, I missed the main problem here. There is no need to have the `input()` inside the function _at all_. There was nothing wrong with how the OP was doing it; it's in fact more correct than your approach. The only side issue is that `string` is a module too, so there could be a name collision, but it would be rare – roganjosh Oct 04 '19 at 10:47
@roganjosh of course there is no need but it's not a problem too. I said about the `input` because he used a `string` parameter in his function and then inside he instantiated the `string` with `input()` which is redundant and wrong. – Kostas Charitidis Oct 04 '19 at 10:49
No they didn't. There is no call to `input()` in the body of that function. There is a broken `return`, but no `input` – roganjosh Oct 04 '19 at 10:51
Ok. We're going to disagree about having `input` inside the function -> I don't think it belongs there and it makes the function less extensible. For 1-liners, I already posted one as a comment under the question – roganjosh Oct 04 '19 at 10:56
@roganjosh i didn't see your comment. Also the extensibility is very objective issue while we're talking about a very specific example with very specific needs but ok! :) – Kostas Charitidis Oct 04 '19 at 10:59

score 0 · Answer 2 · answered Oct 04 '19 at 10:28

You should read an String parameter as input, i named it input_string, then character by character i've checked whether it's a vowel or not, i'm counting it using variable vowels, finally i return it as an answer

def count_vowels(input_string):
    vowels = 0
    for i in input_string:
    if(i=='a' or i=='e' or i=='i' or i=='o' or i=='u' or i=='A' or i=='E' or i=='I' or i=='O' or i=='U' ):
        vowels = vowels + 1
    return vowels

string = input('Enter string: ')
print(count_vowels(string))

score 0 · Answer 3 · answered Oct 04 '19 at 10:33

0

A quick and clear solution is the following:

string = input('Enter string: ')

def count_vowels(i):
  counter = 0
  for letter in i:
    if letter.lower() in ['a', 'e', 'i', 'o', 'u']:
      counter += 1
  return counter

print(count_vowels(string))

The lower() function transforms each letter to undercase, so it can be compared more easily.

Please don't hesitate to ask if you have any question about it.

Regards

answered Oct 04 '19 at 10:33

carlesgg97

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a string is in its self a list so instead of `['a', 'e', 'i', 'o', 'u']` you can just have `'aeiou'` – Chris Doyle Oct 04 '19 at 10:34
1

@ChrisDoyle it's not a list, it's a sequence. And for longer sequences, it might make sense to make it a `set()` – roganjosh Oct 04 '19 at 10:35
1

A string is not a list - but rather an iterable. So yes, you are in fact right. ['a', 'e', 'i', 'o', 'u'] could easily be substituted by 'aeiou'. – carlesgg97 Oct 04 '19 at 10:35
yes sorry you are right list was a poor choice of word as you said its an iterable – Chris Doyle Oct 04 '19 at 10:37

score 0 · Answer 4 · edited Oct 04 '19 at 10:46

You can use the fact that a string is actually an iterable of chars and get rid of all your explicit i == 'a' or i == 'A' stuff and instead just write a string of the letters you want to match and check if i is in it.

def count_vowels(string):
    vowels = 0
    for i in string:
        if i in 'AEIOUaeiou':
            vowels += 1
    return vowels


string = input('Enter string: ')
print(count_vowels(string))

INPUT

The quick brown fox jumps over the lazy dog

OUTPUT

how to get a function to read input and define it

4 Answers4

UPDATE