How to find the sum of number layer in ndarray

Question

I have a 10*10 array. I need to find the sum of all layer. Following figure will clear my question:

How can i do this easily?

Answer 1

Just to add another answer... While the general idea of subtracting inner squares from outer ones is probably the best approach, if you wanted performance none of the implementations presented would do too good, since they keep adding the same numbers over and over. To speed this calculation up, you could use what in image processing is called an integral image:

>>> int_arr = np.cumsum(np.cumsum(arr, axis=0), axis=1)
>>> int_arr
array([[  8,   8,  15,  19,  22,  22,  22,  29,  35,  44],
       [ 16,  17,  24,  34,  40,  44,  46,  54,  64,  79],
       [ 24,  30,  46,  62,  76,  80,  82,  91, 104, 119],
       [ 27,  35,  59,  84, 103, 115, 123, 137, 156, 178],
       [ 33,  48,  77, 103, 123, 135, 148, 163, 191, 213],
       [ 41,  65,  98, 129, 158, 177, 190, 205, 233, 259],
       [ 48,  81, 120, 158, 191, 217, 235, 251, 286, 316],
       [ 54,  95, 136, 180, 214, 249, 275, 296, 333, 364],
       [ 63, 105, 147, 194, 235, 276, 310, 331, 376, 414],
       [ 65, 115, 163, 213, 260, 306, 340, 364, 410, 456]])

From this helper array you can compute the area of any rectangular subarray adding two entries and subtracting two others, e.g.:

>>> np.sum(arr[1:-1, 1:-1])
286
>>> int_arr[-2,-2] + int_arr[0, 0] - int_arr[-2, 0] - int_arr[0, -2]
286

With this, you could compute your sums easily as, e.g.:

sums = [int_arr[-1, -1]]
top = 0
bot = len(arr) - 2
while top < bot:
    new_sum = (int_arr[bot, bot] + int_arr[top, top] -
               int_arr[top, bot] - int_arr[bot, top])
    sums[-1] -= new_sum
    sums.append(new_sum)
    top += 1
    bot -= 1

>>> sums
[170, 122, 85, 62, 17]

Answer 2

Here's a solution without repeated summation of the same elements and without an intermediate array (just brute-force indexing fun) which works for square n by n arrays for odd or even n:

import numpy as np

def sum_shells(a):
    n = len(a)
    no2 = n // 2
    shell_sums = []
    for i in range(no2):
        shell_sums.append(np.sum(a[i,i:n-i]) + np.sum(a[n-i-1,i:n-i]) +
                          np.sum(a[i+1:n-i-1,i]) + np.sum(a[i+1:n-i-1,n-i-1]))
    if n % 2:
        shell_sums.append(a[no2, no2])
    return shell_sums

a = np.array([
  [8, 0, 7, 4, 3, 0, 0, 7, 6, 9],
  [8, 1, 0, 6, 3, 4, 2, 1, 4, 6],
  [8, 5, 9, 6, 8, 0, 0, 1, 3, 0],
  [3, 2, 8, 9, 5, 8, 6, 5, 6, 7],
  [6, 7, 5, 1, 1, 0, 5, 1, 9, 0],
  [8, 9, 4, 5, 9, 7, 0, 0, 0, 4],
  [7, 9, 6, 7, 4, 7, 5, 1, 7, 4],
  [6, 8, 2, 6, 1, 9, 8, 5, 2, 1],
  [9, 1, 1, 3, 7, 6, 8, 0, 8, 7],
  [2, 8, 6, 3, 6, 5, 0, 3, 1, 8] ])

b = np.array([[9, 5, 8, 6, 5],
              [1, 1, 0, 5, 1],
              [5, 9, 7, 0, 0],
              [7, 4, 7, 5, 1],
              [9, 5, 0, 2, 3] ])
print(sum_shells(a))
print(sum_shells(b))

Produces:

[170, 122, 85, 62, 17]
[67, 31, 7]