Placing Which Lyrics More Is Repeated And In Order

The class Counter basically iterates over the string and for each character it increments the value in the dictionary:

texto = 'anderson carlos woss'
quantidade = {}

for caractere in texto:
  quantidade[caractere] = quantidade.get(caractere, 0) + 1

print(quantidade)
# {'a': 2, 'n': 2, 'd': 1, 'e': 1, 'r': 2, 's': 4, 'o': 3, ' ': 2, 'c': 1, 'l': 1, 'w': 1}

In fact, the Counter class is a specialization of dict that uses the update method:

class Counter(dict):

    def __init__(self, iterable=None, **kwds):
        '''Create a new, empty Counter object.  And if given, count elements
        from an input iterable.  Or, initialize the count from another mapping
        of elements to their counts.

        >>> c = Counter()                           # a new, empty counter
        >>> c = Counter('gallahad')                 # a new counter from an iterable
        >>> c = Counter({'a': 4, 'b': 2})           # a new counter from a mapping
        >>> c = Counter(a=4, b=2)                   # a new counter from keyword args

        '''
        self.update(iterable, **kwds)

The update method is overridden by the class as:

def update(self, iterable=None, **kwds):
    if iterable is not None:
        if isinstance(iterable, Mapping):
            if self:
                self_get = self.get
                for elem, count in iterable.iteritems():
                    self[elem] = self_get(elem, 0) + count
            else:
                dict.update(self, iterable) # fast path when counter is empty
        else:
            self_get = self.get
            for elem in iterable:
                self[elem] = self_get(elem, 0) + 1
    if kwds:
        self.update(kwds)

That, in a very simplified way, is what I presented initially.

To return the items in order of frequency, simply check the implementation of the most_common method:

def most_common(self, n=None):
    if n is None:
        return sorted(self.iteritems(), key=_itemgetter(1), reverse=True)
    return _heapq.nlargest(n, self.iteritems(), key=_itemgetter(1))

That is, just use the sorted function.

All source code for the class can be viewed at link