How to get a sublist of all other elements while iterating?

Question

Using list comprehension I can easily do the following:

l = [0, 1, 2]
for i in l:
    subl = [j for j in l if j != i]

Is there a non-list comprehension way to do this by manipulating indices with the % operator?

Edit

To clarify: the indices match the list elements

Answer 1

Here's a solution that doesn't use list comprehension:

l = [0, 1, 2]
for i in range(len(l)):
    subl = l[:]
    subl.remove(i)

Output

[1, 2]
[0, 2]
[0, 1]

Answer 2

This is completely artificial and there is really no reason for doing it this way:

>>> l = [0, 1, 2]
>>> n = len(l)
>>> [[l[(j+i)%n] for j in range(n-1)] for i in range(n)]
[[0, 1], [1, 2], [2, 0]]

It rotates around the list which is what % will do, perhaps more obvious with a longer list:

>>> l = [0, 1, 2, 4, 5]
>>> n = len(l)
>>> [[l[(j+i)%n] for j in range(n-1)] for i in range(n)]
[[0, 1, 2, 4], [1, 2, 4, 5], [2, 4, 5, 0], [4, 5, 0, 1], [5, 0, 1, 2]]

If you really do need this rotational output then suggest looking into collections.deque().