List comprehension to get a 2D list of the same values

Question

I have the following digits:

arr = [4, 5, 5, 5, 6, 6, 4, 1, 4, 4, 3, 6, 6, 3, 6, 1, 4, 5, 5, 5]

I want to create a list comprehension that will match me all the same values in a 2d array of lists like:

[[1, 1], [3, 3], [4, 4, 4, 4, 4], [5, 5, 5, 5, 5, 5], [6, 6, 6, 6, 6]]

I tried something like:

listArr = sorted(arr)

unfortunately I don't know how to put the sorted numbers into a 2D array of lists.

Answer 1

After sorting, you can use itertools.groupby.

>>> L = [4, 5, 5, 5, 6, 6, 4, 1, 4, 4, 3, 6, 6, 3, 6, 1, 4, 5, 5, 5]
>>> from itertools import groupby
>>> [list(group) for _key, group in groupby(sorted(L))]
[[1, 1], [3, 3], [4, 4, 4, 4, 4], [5, 5, 5, 5, 5, 5], [6, 6, 6, 6, 6]]

However, it may be more efficient to sort after, like in Andrej's answer, since it would mean fewer swaps.