Number of vowels in the Python function

Question

I want to do a function in Python that makes a count of how many vowels there are in a string in which the user type.

My code:

def cacavogais():
    i = 0
    j = 0
    string = str (input("Digite alguma coisa: "))
    for i in string:
        if i.lower() in string:
             j+=1
    print ("A string digitada foi: {}".format(string))
    print ("A quantidade de vogais que possui é {}".format(j))

cacavogais()

Answer 1

One way to solve it is to use a comprehensilist on (much more succinct and pythonic):

vogais = 'aeiou'
s = 'abcdefghijABCEx'
qtd_vogais = len([c for c in s.lower() if c in vogais])
print(qtd_vogais) # 5

The line qtd_vogais = len([c for c in s.lower() if c in vogais]) is equivalent to loop suggested by the other answers:

vogais = 'aeiou'
s = 'abcdefghijABCEx'
qtd_vogais = 0
for c in s.lower():
    if c in vogais:
        qtd_vogais += 1

print(qtd_vogais) # 5

Basically, s.lower() turns the string into lower-case letters, and for c in s.lower() makes a loop for all characters in the string. At each iteration, c will be a character. Then just test if c is a vowel and increment the counter.

In the case of comprehensilist on, a list with all vowels (because the expression is inside brackets), and the function len returns the size of this list.

---

If you also want to consider accented characters, you can use the module unicodedata:

from unicodedata import normalize

vogais = 'aeiou'
s = 'ábcdefghijABCEx'
qtd_vogais = len([c for c in normalize('NFD', s.lower()) if c in vogais])
print(qtd_vogais) # 5

In a nutshell, the normalization in NFD "breaks" the "á" in two characters: the a (without accent) and the accent itself. Thus, it is possible to check the vowels in the same way as before. (has a more detailed explanation of the standardisation here, here and here - although not in Python, the idea is the same)

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Another option is to use a Counter:

from collections import Counter
from unicodedata import normalize

vogais = 'aeiou'
s = 'ábcdefghijABCEx'
c = Counter(normalize('NFD', s.lower()))
qtd_vogais = sum(qtd for letra, qtd in c.items() if letra in vogais)
print(qtd_vogais) # 5

The Counter resulting is a dictionary, whose keys are string characters and values are the amount of times each character occurs in the string.

Then just go through it, see which keys are vowels and add up the amounts.

Answer 2

You can do it like this:

def cacavogais():
    i = 0
    j = 0
    string = str (input("Digite alguma coisa: "))
    for i in string:
        if (i == 'A' or i == 'a'
        or i == 'E' or i == 'e'
        or i == 'I' or i == 'i'
        or i == 'O' or i == 'o'
        or i == 'U' or i == 'u'):
             j+=1
    print("vogais: ", j)

cacavogais()

But he’s got better ways of doing it here. Another way to do this would be like this, for example:

def cacavogais():
    i = 0
    j = 0
    string = str (input("Digite alguma coisa: "))
    string = string.lower()
    for i in string:
        if (i == 'a'
        or i == 'e'
        or i == 'i'
        or i == 'o'
        or i == 'u'):
             j+=1
    print("vogais: ", j)

cacavogais()

Answer 3

To count vowels we will follow the following steps:

1. Read a user-typed string
2. Read letter by string typed by user
3. Check if this letter is a vowel (for educational purposes we will ignore accents)
4. If it’s a vowel, we increment a counter
5. After reading the whole word, we will show the contents of the counter

---

Step by step

1. Read a user-typed string

   We will use the method str.input() to receive the user string

   palavra = input("Digite uma palavra: ")
   

2. Read letter by string typed by user

   We will use the for to iterate over each letter of the word typed.

   for letra in palavra:
       # ...
   

3. Check if this letter is a vowel (for educational purposes we will ignore accents)

   We’ll use a if to test whether the letter is a vowel or not.

   For this we will test if the letter exists inside the string "aeiouAEIOU", if it exists is a vowel (uppercase or lowercase).

   for letra in palavra:
       if letra in "aeiouAEIOU":
           # ...
   

4. If it’s a vowel, we increment a counter

   contador = 0
   for letra in palavra:
       if letra in "aeiouAEIOU":
           contador += 1
   

5. After reading the whole word, we will show the contents of the counter

   print("A palavra '{}' tem {} vogais.".format(palavra, contador))
   

---

The final code will remain:

palavra = input("Digite uma palavra: ")
contador = 0

for letra in palavra:
    if letra in "aeiouAEIOU":
        contador += 1

print("A palavra '{}' tem {} vogais.".format(palavra, contador))

^{Repl.it with the code running}

---

Extra

You can use the function len() and list-comprehensions to do the same thing, but with the leanest code:

palavra = input("Digite uma palavra: ")
contador = len([letra for letra in palavra if letra in "aeiouAEIOU"])
print("A palavra '{}' tem {} vogais.".format(palavra, contador))

Answer 4

Alex, follow the example below that solves your problem.

def contarVogais(a):
    vogais = "aeuioAEUIO"
    result = 0
    for char in a:
        if char in vogais:
            result = result + 1
    return result

contarVogais("ajdgejfifhou")

I hope I’ve helped.