Python Exercise 19:advanced list sort

advanced_sort([1,2,1,2]) -> [[1,1],[2,2]]
advanced_sort([2,1,2,1]) ->  [[2,2],[1,1]]
advanced_sort([3,2,1,3,2,1]) -> [[3,3],[2,2],[1,1]]
advanced_sort([5,5,4,3,4,4]) ->  [[5,5],[4,4,4],[3]]
advanced_sort([80,80,4,60,60,3])-> [[80,80],[4],[60,60],[3]]
advanced_sort(['c','c','b','c','b',1,1])-> [['c','c','c'],['b','b'],[1,1]]
advanced_sort([1234, 1235, 1234, 1235, 1236, 1235])-> [[1234, 1234],[1235, 1235, 1235],[1236]]
advanced_sort(['1234', '1235', '1234', '1235', '1236', '1235'])-> [['1234', '1234'],['1235', '1235', '1235'],['1236']]

>>separate the original list to different sublist
  initialist a empty list: new_list 
  add the first number of orginal list into new_list
  for each number in the original list:
    if number equals to the first element of any sublist:
        add number to that sublist
    else:
        add number to an new empty sublist

def advanced_sort(original_list: list) -> list:  
    # create a list with the first element of the original_list  
    new_list = [[original_list[0]]]  
    for index, item in enumerate(original_list):  
        if index == 0:  
            continue  
        not_added = True
        # To iterate over every sublist in the new_list
        for sublist_index, sublist in enumerate(new_list):  
            # try:  
            # if item is the same with first element of the sublist of new list and not already added            
            if item == sublist[0] and not_added:  
                # add item to that sublist  
                sublist.append(item)  
                not_added = False  
            # if no same item appear after checking the whole new list and not already added  
            if item != sublist[0] and sublist_index == len(new_list) - 1 and not_added:  
                # add the item the end of the new_list  
                new_list.append([item])  
                not_added = False  
    return new_list

def advanced_sort(lst):
    return [[i] * lst.count(i) for i in sorted(set(lst), key=lst.index)]

>>transform the original list into a set, thus only unique item will appear
>>rearrange the transformed set by the index order
>>count how many time each unique appear in the original list
>>form a new list by putting in each item times their occurance in a sublist