code: def nlargest(self, n=None):
  		
  		if n is None:
  			return sorted(self.counts(), key=itemgetter(1), reverse=True)
  		else:
  			return heapq.nlargest(n, self.counts(), key=itemgetter(1))
docs: List the n most common elements and their counts.
  
  		List is from the most
  		common to the least.  If n is None, the list all element counts.
  
  		Run time should be O(m log m) where m is len(self)
  		Args:
  			n (int): The number of elements to return