Question

I’m looking for a clean, Pythonic, way to eliminate from the following list:

li = [0, 1, 2, 3, 3, 4, 3, 2, 2, 2, 1, 0, 0]

all contiguous repeated elements (runs longer than one number) so as to obtain:

re = [0, 1, 2, 4, 3, 1]

but although I have working code, it feels un-Pythonic and I am quite sure there must be a way out there (maybe some lesser known itertools functions?) to achieve what I want in a far more concise and elegant way.

Answer 1

Here is a version based on Karl’s answer which doesn’t requires copies of the list (tmp, the slices, and the zipped list). izip is significantly faster than (Python 2) zip for large lists. chain is slightly slower than slicing but doesn’t require a tmp object or copies of the list. islice plus making a tmp is a bit faster, but requires more memory and is less elegant.

from itertools import izip, chain
[y for x, y, z in izip(chain((None, None), li),
                       chain((None,), li),
                       li) if x != y != z]

A timeit test shows it to be approximately twice as fast as Karl’s answer or my fastest groupby version for short groups.

Make sure to use a value other than None (like object()) if your list can contain Nones.

Use this version if you need it to work on an iterator / iterable that isn’t a sequence, or your groups are long:

[key for key, group in groupby(li)
        if (next(group) or True) and next(group, None) is None]

timeit shows it’s about ten times faster than the other version for 1,000 item groups.

Earlier, slow versions:

[key for key, group in groupby(li) if sum(1 for i in group) == 1]
[key for key, group in groupby(li) if len(tuple(group)) == 1]