Taking middle X+3 and X-3 elements from array

Question

I am trying to find the middle element of a list and then take the next 3 items forwards and backwards and make a new list in python.

For example:

lst = [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q]
x = my_func(lst)
print(x)
[f, g, h, i, j, k, l]

How would one accomplish this?

I know I can find the middle by doing this: middle_index = (len(lst) - 1)//2 but I am a bit stuck on how to slice it afterwards.

I know i can find the middle index with `middle_index = (len(lst) - 1)/2` but I am kinda stuck there on how to slice it afterwards — Mojo713, Jan 14 '21 at 15:32
[how to slice an array](https://stackoverflow.com/questions/509211/understanding-slice-notation) ? google search ? — sahasrara62, Jan 14 '21 at 15:36
[sliciing array in python](https://stackoverflow.com/questions/509211/understanding-slice-notation) i think you should go through this and it will help you — sahasrara62, Jan 14 '21 at 15:48

Timur Shtatland · Accepted Answer · 2021-01-14T15:53:03.757

1

Use list slices and avoid index out of range errors like so:

def middle_slice(lst):
    middle = len(lst) // 2
    return lst[max(0, (middle - 3)) : min(len(lst), (middle + 4))]

for length in [9, 8, 2, 1, 0]:
    print(length, middle_slice(list(range(length))))
# 9 [1, 2, 3, 4, 5, 6, 7]
# 8 [1, 2, 3, 4, 5, 6, 7]
# 2 [0, 1]
# 1 [0]
# 0 []

edited Jan 14 '21 at 15:53

answered Jan 14 '21 at 15:47

Timur Shtatland

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score 0 · Answer 2 · answered Jan 14 '21 at 15:39

Something like this ?

def my_func(lst):
    middle = len(lst)//2
    nlst = lst[middle-3:middle]
    nlst += lst[middle:middle+3]
    return nlst

lst = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q']
x = my_func(lst)
print(x) #==> [f, g, h, i, j, k, l]

Taking middle X+3 and X-3 elements from array

2 Answers2