Incorrect operation of a character counter within strings in Python

The error is in the level of indentation of the else. Antonio's response is corrected but the cause is not explained. It must always be remembered that the blocks of code in Python are defined by the levels of indentation. As the code is, the logic is as follows:

An indentation level must be added to else so that it is at the same level of if letter in count.keys(): , so that if this is not fulfilled, its content is executed and the key is added.

import string
alphabet = string.ascii_letters

def contador(str_):
    count= {}
    for letter in str_:
        if letter in alphabet:
            if letter in count.keys():
               count[letter] += 1

       #else  <<< A este nivel está en tu código, y debe ser:
            else:
                count[letter] = 1

    return count

Keep in mind that does not include capital letters, "ñ", accented vowels or "ü" .

On the other hand, to count appearances in an iterable collections.Counter offers a simple and much more efficient solution. This returns us a dictionary counting all the characters (not just letters), we can filter it if we want it without problems:

from collections import  Counter
import string

alphabet = set(string.ascii_letters + ("áéíóúÁÉÍÓÚñÑÜü"))
def contador(str_):
    count = Counter(oracion)
    return {key: count[key] for key in alphabet & count.keys()}

oracion = 'Cigüeña, camión, 123'
print(contador(oracion))

Exit:

{'n': 1, 'i': 2, 'C': 1, 'a': 2, 'c': 1, 'e': 1, 'm': 1, 'ó': 1, 'ü': 1, 'y': 1, 'g': 1, 'ñ': 1}

It differs between uppercase and lowercase. If you want to count both cases in one, it is not possible either.

Good morning: Your script is almost correct because it worked for me. I write you here:     import string     alphabet = string.ascii_letters

oracion = 'Mi perro bonito come todas las noches'

def contador(str):
    count= {}
    for letter in str:
        if letter in alphabet:
            if letter in count.keys():
                count[letter]+=1
            else:
                count[letter]=1
    return count

print(contador(oracion))

That's how it worked for me. He has shown me on the screen an associative array of letters and their repetitions.

You needed to use the print () function . I tested it in an editor that I have on my mobile and it worked correctly