How to sort a Python list descending based on item count, but return a list with sorted items only, and not a count also?
Question: I am generating some large lists in python, and I need to figure out the
absolute fastest way to return a sort list from something like this:

    
    
    myList = ['a','b','c','a','c','a']
    

and return a list of it in descending order, based on the count of the items,
so it looks like this.

    
    
    sortedList = ['a','c','b']
    

I have been using Counter().most_common(), but this returns a tuple in
descending order with the item, and the number of times in appears in the
list. I really just need a tuple or list in descending order based off count,
with just the items, and not the count amounts. Any ideas?

Edit: So would doing something like this be faster?

    
    
    myList = ['a','b','c','a','c','a']
    count = Counter(myList).most_common()
    res = [k for k,v in count]
    


Answer: 
    from collections import Counter
    
    myList = ['a','b','c','a','c','a']
    res = [k for k,v in Counter(myList).most_common()]
    # ['a', 'c', 'b']
    

