could someone explain in detail on this ((sum(x[i-n:i+n+1]))..where does i-n:i+n+1 start and end?

Question

could someone explain in detail on this ((sum(x[i-n:i+n+1]))..where does i-n:i+n+1 start and end?? i just don't want to copy the code`

The function should create a return a new list r where

def smooth_a(x, n):
    x = [x[0]]*n+x+[x[-1]]*n #creates a copy of x  
    res = [] #empty list     
    for i in range(n, len(x)-n):
        res.append(sum(x[i-n:i+n+1])/(2*n+1)) 
    return res 

x = [1, 2, 6, 4, 5, 0, 1, 2] #def of x
print('smooth_a(x, 1): ', smooth_a(x, 1)) #prints new list with n=1

Does this answer your question? [Understanding slicing](https://stackoverflow.com/questions/509211/understanding-slicing) — Tony Beta Lambda, Sep 20 '22 at 15:20

Bastián Bas · Answer 1 · 2022-09-15T16:52:17.150

0

Let's say x = [1, 2, 6, 4, 5, 0, 1, 2] and n = 1

Then, if you take i = 3 for example:

sum(x[i-n:i+n+1]) = sum(x[2:5])
                  = sum([6, 4, 5]) <- slices x from 2 to 4, both inclusive
                  = 15

For your specific problem, be careful with the lower slice index, if it's negative, the slicing will return an empty list, so I would write it as sum(x[max(i-n, 0):i+n+1]).

edited Sep 15 '22 at 16:52

answered Sep 15 '22 at 16:46

Bastián Bas

36
3

so if n=1 and i=0 it will be sum(x[max(-1,0): 0+1+1])????.......or n=1 and i=8???? – Kwame Holland Sep 15 '22 at 19:52
In my answer, it would be the first option. The minimum Index for an array slicing on python is 0. – Bastián Bas Sep 19 '22 at 19:22

could someone explain in detail on this ((sum(x[i-n:i+n+1]))..where does i-n:i+n+1 start and end?

1 Answers1