[ll[i-4:i+4] for i in range(4, len(ll)-4)]

does the trick, I should think.

Here is the quick code I wrote

lst = [1, 2, 3, 4, 5, 6, 7, 8]
sliding_window_size = 3

def get_sliding_list(l, index):
    l_list = []
    r_list = []

    min_range = 0
    if index > sliding_window_size:
        min_range = index - sliding_window_size

    max_range = len(l)
    if index + sliding_window_size < len(l):
        max_range = index + sliding_window_size + 1

    return (l[min_range:index], l[index + 1:max_range])

print(get_sliding_list(lst, 0))
print(get_sliding_list(lst, 1))
print(get_sliding_list(lst, 2))
print(get_sliding_list(lst, 3))
print(get_sliding_list(lst, 4))
print(get_sliding_list(lst, 5))
print(get_sliding_list(lst, 6))
print(get_sliding_list(lst, 7))

Output

([], [2, 3, 4])
([1], [3, 4, 5])
([1, 2], [4, 5, 6])
([1, 2, 3], [5, 6, 7])
([2, 3, 4], [6, 7, 8])
([3, 4, 5], [7, 8])
([4, 5, 6], [8])
([5, 6, 7], [])

Pass index of the element for which you want to retrieve sliding window

Code:

arr = [1, 2, 3, 4, 5, 6, 7, 8]
window = 3
for backward, current in enumerate(range(len(arr)), start = 0-window):
    if backward < 0:
        backward = 0
    print(arr[current+1:current+1+window], arr[backward:current])

Output:

[2, 3, 4], []
[3, 4, 5], [1]
[4, 5, 6], [1, 2]
[5, 6, 7], [1, 2, 3]
[6, 7, 8], [2, 3, 4]
[7, 8], [3, 4, 5]
[8], [4, 5, 6]
[], [5, 6, 7]

One Liner:

print(dict([(e, (lst[i+1:i+4], lst[max(i-3,0):i])) for i,e in enumerate(last)]))

Output:

{1: ([2, 3, 4], []),
 2: ([3, 4, 5], [1]),
 3: ([4, 5, 6], [1, 2]),
 4: ([5, 6, 7], [1, 2, 3]),
 5: ([6, 7, 8], [2, 3, 4]),
 6: ([7, 8], [3, 4, 5]),
 7: ([8], [4, 5, 6]),
 8: ([], [5, 6, 7])}

Credit: thanks to suggestions from @FeRD and @Androbin, the solution now looks better

Your sliding window reminds me of another data structure: fixed-size stacks. If you think about it, what you want is actually a fixed-size stack of 7 elements where the right three are the forward window elements and the back three are the back window elements. The 4th element is the current element. Here's how I would do it:

import collections

my_list = [1, 2, 3, 4, 5, 6, 7, 8]

window = collections.deque([], 7)

for i in my_list:
    window.append(i)

    # Get the back three elements
    forward_window = list(window)[-3:]
    # Get the front three elements
    backward_window = list(window)[:len(window)-4]

    print()
    print(list(forward_window))
    print(list(backward_window))

Of course, the code is not exactly what you want as the stack needs to be primed with some starting elements but that can be done with a bit more work:

import collections

my_list = [1, 2, 3, 4, 5, 6, 7, 8]

# Start with the first three elements
window = collections.deque(my_list[:3], 7)

# Iterate from the fourth element
for i in my_list[3:]:
    window.append(i)
    forward_window = list(window)[-3:]
    backward_window = list(window)[:len(window)-4]
    print()
    print(list(forward_window))
    print(list(backward_window))

After that you just need to clear the stack by adding some empty elements:

while len(window) != 4:
    window.popleft()
    forward_window = list(window)[4:]
    backward_window = list(window)[:3]
    print()
    print(list(forward_window))
    print(list(backward_window))

This should get you started:

from dataclasses import dataclass
from typing import List

@dataclass
class Window:
  index: int
  backward: List[int]
  forward: List[int]

def window(iterable, window_size, index):
  backward = iterable[max(0, index - window_size):index]
  forward = iterable[index + 1:index + 1 + window_size]
  return Window(index, backward, forward)

>>> window([1,2,3,4,5,6], 3, 0)
Window(index=0, backward=[], forward=[2, 3, 4])
>>> window([1,2,3,4,5,6], 3, 5)
Window(index=5, backward=[3, 4, 5], forward=[])

I would also suggest adding some checks whether the index and window size make sense.

If you are stuck with an older Python version that doesn't have dataclasses yet, you can use Named Tuples instead.

Here's a short and neat code based on list comprehension.

forward = [lst[i+1:i+1+window] for i in range(len(lst)]
backward = [lst[::-1][i+1:i+1+window] for i in range(len(lst)] # first reverse the input list and do same as did in forward
out = zip(forward,backward[::-1]) # first reverse the backward list and zip two list into one

Output

>>> forward
[[2, 3, 4], [3, 4, 5], [4, 5, 6], [5, 6, 7], [6, 7, 8], [7, 8], [8], []]
>>> backward
[[7, 6, 5], [6, 5, 4], [5, 4, 3], [4, 3, 2], [3, 2, 1], [2, 1], [1], []]
>>> out
[([2, 3, 4], []), ([3, 4, 5], [1]), ([4, 5, 6], [2, 1]), ([5, 6, 7], [3, 2, 1]), ([6, 7, 8], [4, 3, 2]), ([7, 8], [5, 4, 3]), ([8], [6, 5, 4]), ([], [7, 6, 5])]

This will work with more_itertools.windowed if you adjust the window size. Since you want 7 items (3 backward, 1 current, 3 forward), set the window size to 7.

from itertools import chain
from more_itertools import windowed
n = 3
iterable = [1, 2, 3, 4, 5, 6, 7, 8]
# pad the iterable so you start with an empty backward window
it = chain([None] * n, iterable, [None] * n)

for window in windowed(it, n * 2 + 1):
    print(window[n])
    print('forward =', [x for x in window[n + 1:] if x is not None])
    print('backward =', [x for x in window[:n] if x is not None])

The output is:

1
forward = [2, 3, 4]
backward = []

2
forward = [3, 4, 5]
backward = [1]

3
forward = [4, 5, 6]
backward = [1, 2]

4
forward = [5, 6, 7]
backward = [1, 2, 3]

5
forward = [6, 7, 8]
backward = [2, 3, 4]

6
forward = [7, 8]
backward = [3, 4, 5]

7
forward = [8]
backward = [4, 5, 6]

8
forward = []
backward = [5, 6, 7]

You can just use min and max to make sure you stay within the list (no loops needed) .

lst = [1, 2, 3, 4, 5, 6, 7, 8]
ws = 3 # window

st = 3 # starting point

mn = max(st-ws-1, 0)
mx = min(st+ws, len(lst))
print('Forward = ',lst[st:mx])
print('Backward = ', lst[mn:st-1])

Output:

Forward =  [4, 5, 6]
Backward =  [1, 2]