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What's wrong with this code?
I tried to create a program to count vowels on Python. My Idea was to create a list of vowels, then I'd convert the user input into a list and use a function to count the vowels like this: vowels = ["a", "e", "i", "o", "u"] det count_vowels(x): i = 0 for letter in x and vowels: i += 1 print(i) while True: word = input("Enter a word: ") letters = list(word) count_vowels(letters) The program just outputs 1 2 3 4 5 for every word I input. What's wrong?
6 Antworten
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vowels = ["a","e","i","o","u"]
def count_vowels(x)
i = 0
for l in x:
if l in vowels:
i += 1
elif x[-1] == l
return i
while True:
word = input ("Enter a word: ")
letters = list(word)
count_vowels(letters)
print(count_vowels(letters))
That elif statement should help. It’s checking if x[-1] which equals the last letter of the string and seeing if it is equal to whatever is in l. Only on the last iteration will the statement execute, returning the value of i which is the number of vowels in the string
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I think the problem is that the function is only counting the elements of the vowels list, so there's something to do with the use of the boolean operator 'and' but Idk how to solve it.
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Try doing:
for letter in x:
if letter in vowels:
i += 1
Also, it’s printing that because you’re printing i after every iteration in the for loop
You’ll want to do return i after iterating all the way through x
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Thank you very much, Travis. Coding is addictive even to people who can't do much more than couting vowels.
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Well, the output now is only 1 :/
Doing it as:
vowels = ["a","e","i","o","u"]
def count_vowels(x)
i = 0
for l in x:
if l in vowels:
i += 1
return i
while True:
word = input ("Enter a word: ")
letters = list(word)
count_vowels(letters)
print(count_vowels(letters))
0
No problem man, everyone starts somewhere and I’m pretty new myself to python.