Removing elements that have consecutive duplicates

Question

>>> 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]

A "lazy" approach would be to use itertools.groupby.

import itertools

list1 = [1, 2, 3, 3, 4, 3, 5, 5]
list1 = [g for g, _ in itertools.groupby(list1)]
print(list1)

outputs

[1, 2, 3, 4, 3, 5]

Here is a solution without dependence on outside packages:

list = [1,1,1,1,1,1,2,3,4,4,5,1,2] 
L = list + [999]  # append a unique dummy element to properly handle -1 index
[l for i, l in enumerate(L) if l != L[i - 1]][:-1] # drop the dummy element

Then I noted that Ulf Aslak's similar solution is cleaner :)