Removing elements that have consecutive duplicates
>>> L = [1,1,1,1,1,1,2,3,4,4,5,1,2]>>> from itertools import groupby>>> [x[0] for x in groupby(L)][1, 2, 3, 4, 5, 1, 2]If you wish, you can use map instead of the list comprehension
>>> from operator import itemgetter>>> map(itemgetter(0), groupby(L))[1, 2, 3, 4, 5, 1, 2]For the second part
>>> [x for x, y in groupby(L) if len(list(y)) < 2][2, 3, 5, 1, 2]If you don't want to create the temporary list just to take the length, you can use sum over a generator expression
>>> [x for x, y in groupby(L) if sum(1 for i in y) < 2][2, 3, 5, 1, 2]
Oneliner in pure Python
[v for i, v in enumerate(your_list) if i == 0 or v != your_list[i-1]]
If you use Python 3.8+, you can use assignment expression :=:
list1 = [1, 2, 3, 3, 4, 3, 5, 5]prev = object()list1 = [prev:=v for v in list1 if prev!=v]print(list1)Prints:
[1, 2, 3, 4, 3, 5]