Find location of the largest sum in 2d numpy array

Approach #1 : We can use SciPy's 2D convolution to get summations in sliding windows of shape (n,n) and choose the index of the window with the biggest sum with argmax and translate to row, col indices with np.unravel_index, like so -

from scipy.signal import convolve2d as conv2

def largest_sum_pos_app1(a, n):
    idx = conv2(a, np.ones((n,n),dtype=int),'same').argmax()
    return np.unravel_index(idx, a.shape)

Sample run -

In [558]: a
Out[558]: 
array([[0, 0, 0, 0, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 0, 0, 0, 0]])

In [559]: largest_sum_pos_app1(a, n=3)
Out[559]: (2, 2)

Approach #1S (Super charged) : We can boost it further by using uniform filter, like so -

from scipy.ndimage.filters import uniform_filter as unif2D

def largest_sum_pos_app1_mod1(a, n):
    idx = unif2D(a.astype(float),size=n, mode='constant').argmax()
    return np.unravel_index(idx, a.shape)

Approach #2 : Another based on scikit-image's sliding window creating tool view_as_windows, we would create sliding windows of shape (n,n) to give us a 4D array with the last two axes of shape (n,n) corresponding to the search window size. So, we would sum along those two axes and get the argmax index and translate it to the actual row, col positions.

Hence, the implementation would be -

from skimage.util.shape import view_as_windows

def largest_sum_pos_app2(a, n):
    h = (n-1)//2 # half window size
    idx = view_as_windows(a, (n,n)).sum((-2,-1)).argmax()
    return tuple(np.array(np.unravel_index(idx, np.array(a.shape)-n+1))+h)

As also mentioned in the comments, a search square with an even n would be confusing given that it won't have its center at any element coordinate.

Runtime test

In [741]: np.random.seed(0)

In [742]: a = np.random.randint(0,1000,(1000,1000))

In [743]: largest_sum_pos_app1(a, n= 5)
Out[743]: (966, 403)

In [744]: largest_sum_pos_app1_mod1(a, n= 5)
Out[744]: (966, 403)

In [745]: largest_sum_pos_app2(a, n= 5)
Out[745]: (966, 403)

In [746]: %timeit largest_sum_pos_app1(a, n= 5)
     ...: %timeit largest_sum_pos_app1_mod1(a, n= 5)
     ...: %timeit largest_sum_pos_app2(a, n= 5)
     ...: 
10 loops, best of 3: 57.6 ms per loop
100 loops, best of 3: 10.1 ms per loop
10 loops, best of 3: 47.7 ms per loop