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HHMathewChan
HHMathewChan

Posted on • Originally published at rebirthwithcode.tech

Python Exercise 19:advanced list sort

#beginners #python #challenge

Question

  • Create a function that
    • takes a list of numbers or strings 
    • and returns a list with the items from the original list stored into sublists.
    • Items of the same value should be in the same sublist.
  • The sublists should be returned in the order of each element's first appearance in the given list.

Example 

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']]
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My solution

  • algorithm 
>>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
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  • code 
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
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Other solution 

def advanced_sort(lst): return [[i] * lst.count(i) for i in sorted(set(lst), key=lst.index)]
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  • the algorithm 
>>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
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My reflection

  • about while loop
    • the loop will not instantly end when the boolean condition is changed to false
    • need to use if condition to control the execution of code
  • Just admire people can write short code in elegant way

Credit

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