Codewars 4 Kyu kata: "Most frequently used words in a text"

Question

This was the first kata I found hard so I wanted to ask about it. Any ways I could I have done it better?

here's the question -> Most frequently used words in a text

def check_alphabet(string): #checks if the given string contains at least one alphabet
    for i in string:
        if i.isalpha():
            return True
    return False

def top_3_words(text):    
    text += " " # the for loop ends before it can check the final word
    text = text.lower() # given in the problem
    
    current_word = ""
    
    words = {
    }
    
    letter_index = 0
    
    for letter in text:
        if letter.isalpha() or letter == "'": # make the word
            current_word += letter
            letter_index += 1
            
        elif letter_index == 0: # this happens if there is more than one whitespace
            pass
        
        else: # add the word into the dictionary
            
            if not check_alphabet(current_word): # filter out words only made out of '
                current_word = ""
                continue
            
            if current_word in words.keys(): # if the word has already come out before
                words[current_word] += 1
            else: #if the word is new
                words.update({current_word : 1})
                
            current_word = "" #reset
            letter_index = 0
            
    words = sorted(words.items(), key = lambda x: x[1], reverse= True) # sort the dictionary
    
    if len(words) == 0:
        return []
    elif len(words) == 1:
        return [words[0][0]]
    elif len(words) == 2:
        return [words[0][0], words[1][0]]
    elif len(words) >= 3:
        return [words[0][0], words[1][0], words[2][0]]

Answer 1

The good: No globals, use of functions, reasonable variable names.

You can make better use of the standard library - regexes and Counter will make this easier.

+= on what will already be a long string may be expensive, because Python needs to recreate the entire string as a new object (strings are immutable).

If you wanted to keep the dictionary approach, defaultdict(int) would make this easier; but again Counter does all of this for you.

The question also asks

Bonus points (not really, but just for fun): [...] Avoid sorting the entire array of unique words.

You're doing this in your words =, but this can be avoided (again by the use of Counter).

words.update({current_word : 1}) doesn't need to call update, nor does it need to form a literal dictionary; just words[current_word] = 1.

Add the test cases from the problem statement to your question.

Suggested

import re
from collections import Counter

r'''
A word is a string of letters (A to Z) optionally containing one or more apostrophes (') in ASCII
Apostrophes can appear at the start, middle or end of a word ('abc, abc', 'abc', ab'c are all valid)
Any other characters (e.g. #, \, / , . ...) are not part of a word and should be treated as whitespace.
Matches should be case-insensitive, and 
'''
WORD_PAT = re.compile(
    r"(?ai)"     # ASCII, case-insensitive
    r"[a-z']+"   # a-z or ', one or more
)


def top_n_words(text: str, n: int = 3) -> list[str]:
    # the words in the result should be lowercased
    words = WORD_PAT.findall(text.lower())
    counter = Counter(words)
    return [
        word
        for word, freq in counter.most_common(n)
    ]


def test() -> None:
    assert top_n_words(
        '''
        In a village of La Mancha, the name of which I have no desire to call to
        mind, there lived not long since one of those gentlemen that keep a lance
        in the lance-rack, an old buckler, a lean hack, and a greyhound for
        coursing. An olla of rather more beef than mutton, a salad on most
        nights, scraps on Saturdays, lentils on Fridays, and a pigeon or so extra
        on Sundays, made away with three-quarters of his income.
        '''
    ) == ['a', 'of', 'on']

    assert top_n_words(
        'e e e e DDD ddd DdD: ddd ddd aa aA Aa, bb cc cC e e e'
    ) == ['e', 'ddd', 'aa']

    assert top_n_words(
        "  //wont won't won't"
    ) == ["won't", "wont"]


if __name__ == '__main__':
    test()