Removing elements that have consecutive duplicates

Question

I was curios about the question: Eliminate consecutive duplicates of list elements, and how it should be implemented in Python.

What I came up with is this:

list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
i = 0

while i < len(list)-1:
    if list[i] == list[i+1]:
        del list[i]
    else:
        i = i+1

Output:

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

Which I guess is ok.

So I got curious, and wanted to see if I could delete the elements that had consecutive duplicates and get this output:

[2, 3, 5, 1, 2]

For that I did this:

list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
i = 0
dupe = False

while i < len(list)-1:
    if list[i] == list[i+1]:
        del list[i]
        dupe = True
    elif dupe:
        del list[i]
        dupe = False
    else:
        i += 1

But it seems sort of clumsy and not pythonic, do you have any smarter / more elegant / more efficient way to implement this?

Answer 1

Oneliner in pure Python

[v for i, v in enumerate(your_list) if i == 0 or v != your_list[i-1]]

Answer 2

To Eliminate consecutive duplicates of list elements; as an alternative, you may use itertools.izip_longest() with list comprehension as:

>>> from itertools import izip_longest

>>> my_list = [1,1,1,1,1,1,2,3,4,4,5,1,2]
>>> [i for i, j in izip_longest(my_list, my_list[1:]) if i!=j]
[1, 2, 3, 4, 5, 1, 2]

Answer 3

>>> 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]