How do I return the earliest letter in the alphabet from a choice of letters?

Question

I am doing an exercise the requires me to return the most frequent letter in a string of letters. In the event that there are two letters that appear with the same frequency, the 'tiebreaker' is whichever appears first in the alphabet. For example:

'aaaabbbb' #should return a.

My code to return the letter is below. I recognize that it may not be the most efficient. I'll worry about that later.

def mostWantedLetter(text):
    text = text.lower()
    mwl = {}
    for letter in text:
        if letter not in mwl:
            mwl[letter] = 1
        else:
            mwl[letter] += 1

    for letter in sorted(mwl, key = mwl.get, reverse = True):
        if letter.isalpha():
            #letter = sorted(list(letter))
            return letter      #[0]

Thank you to the Stack Overflow community for your help in getting me this far!

Answer 1

Congratulations on making it thus far. I'll offer two hints:

Hint 1: modify the key argument to sorted() to take into account both the letter's count and its position in the alphabet.

Hint 2: look into how Python compares tuples.

On a separate note, collections.Counter and collections.defaultdict are worth knowing about.

Answer 2

Right, I found this and it was pretty damn cool:

>>> import collections
>>>
>>> ordered1 = collections.Counter('aaaabbbb').most_common(1))
>>> ordered1[0][0]
'a'
>>>
>>> ordered2 = collections.Counter('abb').most_common(1))
>>> ordered2[0][0]
'b'

This puts letters in order primarily of occurrence, and then if two letters occur equal times then they are sorted alphabetically.

Answer 3

def mostWantedLetter(text):
    # count how many times each letter occurs
    mwl = {}
    for ch in text.lower():
        if ch.isalpha():
            mwl[ch] = mwl.get(ch, 0) + 1

    # ordered descending by count (highest first) then ascending by letter
    least = min(mwl.items(), key=lambda x:(-x[1], x[0]))

    # return the letter from the least item
    return least[0]

Answer 4

def most_wanted(s):
    return chr(min(set([ord(x) for x in list(s)])))

It breaks the string into a list and then converts them all to their ascii representation. It makes this a set so that duplicates don't need to be compared. It takes the minimum value from these and then converts it back to a letter.