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For the most part your code is okay, however, reading what it is that you want it to do here's what I came up with...

#!/usr/bin/env python3 import re import typing # def most_frequent_letter(text: str) -> typing.Tuple[int, list]: ## Use above if returning an int and list instead of int and string def most_frequent_letter(text: str) -> typing.Tuple[int, str]: text = re.sub('[^a-z]', '', text.replace(' ', '').lower()) letter_counts = {} for letter in text: letter_counts[letter] = letter_counts.get(letter, 0) + 1 counted_letters = {} for letter, counter in letter_counts.items(): counted_letters[counter] = counted_letters.get(counter, []) + [letter] max_key = max(counted_letters.keys()) return max_key, sorted(counted_letters[max_key]) if __name__ == '__main__': counted, letters = most_frequent_letter('spam ham jello spam') print("{counted} -> {letter}".format( counted = counted, letter = letters[0])) 

... while I don't know that it's an improvement overall it does return results that may be of use...

3 -> a 

Note if ya want the whole list use the following instead under that last if statement...

# ... print("{counted} -> {letters}".format( counted = counted, letters = letters)) # ... Should output... # 3 -> ['a', 'm'] 

... while also not counting spaces as letters; which is the only thing that seemed missing from your implementation.

I didn't see any need to use if statements (within the function) so didn't include'em, because dictionaries have a handy get method that allows for defaulting nonexistent key value pares. And I didn't see any use in using collections though it was tempting to use @200_success's suggestion.

One thing about your code that I do still question though is why are ya wrapping the returned value within parentheses?

For the most part your code is okay, however, reading what it is that you want it to do here's what I came up with...

#!/usr/bin/env python3 import re import typing def most_frequent_letter(text: str) -> typing.Tuple[int, list]: text = re.sub('[^a-z]', '', text.replace(' ', '').lower()) letter_counts = {} for letter in text: letter_counts[letter] = letter_counts.get(letter, 0) + 1 counted_letters = {} for letter, counter in letter_counts.items(): counted_letters[counter] = counted_letters.get(counter, []) + [letter] max_key = max(counted_letters.keys()) return max_key, sorted(counted_letters[max_key]) if __name__ == '__main__': counted, letters = most_frequent_letter('spam ham jello spam') print("{counted} -> {letter}".format( counted = counted, letter = letters[0])) 

... while I don't know that it's an improvement overall it does return results that may be of use...

3 -> a 

Note if ya want the whole list use the following instead under that last if statement...

# ... print("{counted} -> {letters}".format( counted = counted, letters = letters)) # ... Should output... # 3 -> ['a', 'm'] 

... while also not counting spaces as letters; which is the only thing that seemed missing from your implementation.

I didn't see any need to use if statements (within the function) so didn't include'em, because dictionaries have a handy get method that allows for defaulting nonexistent key value pares. And I didn't see any use in using collections though it was tempting to use @200_success's suggestion.

One thing about your code that I do still question though is why are ya wrapping the returned value within parentheses?

For the most part your code is okay, however, reading what it is that you want it to do here's what I came up with...

#!/usr/bin/env python3 import re def most_frequent_letter(text: str) -> str: text = re.sub('[^a-z]', '', text.replace(' ', '').lower()) letter_counts = {} for letter in text: letter_counts[letter] = letter_counts.get(letter, 0) + 1 counted_letters = {} for letter, counter in letter_counts.items(): counted_letters[counter] = counted_letters.get(counter, []) + [letter] max_key = max(counted_letters.keys()) return max_key, sorted(counted_letters[max_key]) if __name__ == '__main__': counted, letters = most_frequent_letter('spam ham jello spam') print("{counted} -> {letter}".format( counted = counted, letter = letters[0])) 

... while I don't know that it's an improvement overall it does return results that may be of use...

3 -> a 

Note if ya want the whole list use the following instead under that last if statement...

# ... print("{counted} -> {letters}".format( counted = counted, letters = letters)) # ... Should output... # 3 -> ['a', 'm'] 

... while also not counting spaces as letters; which is the only thing that seemed missing from your implementation.

I didn't see any need to use if statements (within the function) so didn't include'em, because dictionaries have a handy get method that allows for defaulting nonexistent key value pares. And I didn't see any use in using collections though it was tempting to use @200_success's suggestion.

One thing about your code that I do still question though is why are ya wrapping the returned value within parentheses?

For the most part your code is okay, however, reading what it is that you want it to do here's what I came up with...

#!/usr/bin/env python3 import re import typing # def most_frequent_letter(text: str) -> typing.Tuple[int, list]: ## Use above if returning an int and list instead of int and string def most_frequent_letter(text: str) -> typing.Tuple[int, str]: text = re.sub('[^a-z]', '', text.replace(' ', '').lower()) letter_counts = {} for letter in text: letter_counts[letter] = letter_counts.get(letter, 0) + 1 counted_letters = {} for letter, counter in letter_counts.items(): counted_letters[counter] = counted_letters.get(counter, []) + [letter] max_key = max(counted_letters.keys()) return max_key, sorted(counted_letters[max_key]) if __name__ == '__main__': counted, letters = most_frequent_letter('spam ham jello spam') print("{counted} -> {letter}".format( counted = counted, letter = letters[0])) 

... while I don't know that it's an improvement overall it does return results that may be of use...

3 -> a 

Note if ya want the whole list use the following instead under that last if statement...

# ... print("{counted} -> {letters}".format( counted = counted, letters = letters)) # ... Should output... # 3 -> ['a', 'm'] 

... while also not counting spaces as letters; which is the only thing that seemed missing from your implementation.

I didn't see any need to use if statements (within the function) so didn't include'em, because dictionaries have a handy get method that allows for defaulting nonexistent key value pares. And I didn't see any use in using collections though it was tempting to use @200_success's suggestion.

One thing about your code that I do still question though is why are ya wrapping the returned value within parentheses?

For the most part your code is okay, however, reading what it is that you want it to do here's what I came up with...

#!/usr/bin/env python3 import re def most_frequent_letter(text: str) -> str: text = re.sub('[^a-z]', '', text.replace(' ', '').lower()) letter_counts = {} for letter in text: letter_counts[letter] = letter_counts.get(letter, 0) + 1 counted_letters = {} for letter, counter in letter_counts.items(): counted_letters[counter] = counted_letters.get(counter, []) + [letter] max_key = max(counted_letters.keys()) return max_key, sorted(counted_letters[max_key]) if __name__ == '__main__': counted, letters = most_frequent_letter('spam ham jello spam') print("{counted} -> {letter}".format( counted = counted, letter = letters[0])) 

... while I don't know that it's an improvement overall it does return results that may be of use...

3 -> a 

Note if ya want the whole list use the following instead under that last if statement...

# ... print("{counted} -> {letters}".format( counted = counted, letters = letters)) # ... Should output... # 3 -> ['a', 'm'] 

... while also not counting spaces as letters; which is the only thing that seemed missing from your implementation.

I didn't see any need to use if statements so didn't include'em, because dictionaries have a handy get method that allows for defaulting nonexistent key value pares. And I didn't see any use in using collections though it was tempting to use @200_success's suggestion.

For the most part your code is okay, however, reading what it is that you want it to do here's what I came up with...

#!/usr/bin/env python3 import re def most_frequent_letter(text: str) -> str: text = re.sub('[^a-z]', '', text.replace(' ', '').lower()) letter_counts = {} for letter in text: letter_counts[letter] = letter_counts.get(letter, 0) + 1 counted_letters = {} for letter, counter in letter_counts.items(): counted_letters[counter] = counted_letters.get(counter, []) + [letter] max_key = max(counted_letters.keys()) return max_key, sorted(counted_letters[max_key]) if __name__ == '__main__': counted, letters = most_frequent_letter('spam ham jello spam') print("{counted} -> {letter}".format( counted = counted, letter = letters[0])) 

... while I don't know that it's an improvement overall it does return results that may be of use...

3 -> a 

Note if ya want the whole list use the following instead under that last if statement...

# ... print("{counted} -> {letters}".format( counted = counted, letters = letters)) # ... Should output... # 3 -> ['a', 'm'] 

... while also not counting spaces as letters; which is the only thing that seemed missing from your implementation.

I didn't see any need to use if statements (within the function) so didn't include'em, because dictionaries have a handy get method that allows for defaulting nonexistent key value pares. And I didn't see any use in using collections though it was tempting to use @200_success's suggestion.

One thing about your code that I do still question though is why are ya wrapping the returned value within parentheses?