Python: Pack consecutive duplicates of a given list elements into sublists
Python List: Exercise - 74 with Solution
Write a Python program to pack consecutive duplicates of a given list of elements into sublists.
from itertools import groupby def pack_consecutive_duplicates(l_nums): return [list(group) for key, group in groupby(l_nums)] n_list = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4 ] print("Original list:") print(n_list) print("\nAfter packing consecutive duplicates of the said list elements into sublists:") print(pack_consecutive_duplicates(n_list))
Original list: [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4] After packing consecutive duplicates of the said list elements into sublists: [[0, 0], , , , [4, 4], , [6, 6, 6], , , , [4, 4]]
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Python: Tips of the Day
Creates a list of elements, grouped based on the position in the original lists:
def tips_zip(*args, fill_value=None): max_length = max([len(lst) for lst in args]) result =  for i in range(max_length): result.append([ args[k][i] if i < len(args[k]) else fillvalue for k in range(len(args)) ]) return result print(tips_zip(['a', 'b'], [1, 2], [True, False])) print(tips_zip(['a'], [1, 2], [True, False])) print(tips_zip(['a'], [1, 2], [True, False], fill_value = '1'))
[['a', 1, True], ['b', 2, False]] [['a', 1, True], [None, 2, False]] [['a', 1, True], ['1', 2, False]]
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