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

>>> L = [1,1,1,1,1,1,2,3,4,4,5,1,2]
>>> from itertools import groupby
>>> [key for key, _group in groupby(L)]
[1, 2, 3, 4, 5, 1, 2]

---

For the second part

>>> [k for k, g in groupby(L) if len(list(g)) < 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

>>> [k for k, g in groupby(L) if sum(1 for i in g) < 2]
[2, 3, 5, 1, 2]