Найти произведение ненулевых диагональных элементов матрицы (Паскаль -> Питон)

import numpy as np
# get matrix, named A
diag = A.diagonal()
diag_no_0 = diag[diag != 0]
mult = np.multiply.reduce(diag_no_0)

matr = [[1,2,3,4,5,6,7,8,9,0],
        [1,0,3,4,5,6,7,8,9,1],
        [1,2,3,4,5,6,7,8,9,2],
        [1,2,3,0,5,6,0,8,9,3],
        [1,2,3,4,5,0,7,8,9,4],
        [1,2,3,4,0,6,7,8,9,5],
        [1,2,3,4,5,6,7,8,9,6],
        [1,2,3,4,5,6,7,0,9,7],
        [1,0,3,4,5,6,7,8,9,8],
        [1,2,3,4,5,6,7,8,9,9]]
 
print reduce(lambda res, x: res*x, [el[i] for i, el in enumerate(matr) if el[i] > 0], 1)
print reduce(lambda res, x: res*x, [el[i] for i, el in enumerate(matr[::-1])  if el[i] > 0], 1)
 
>>51030
864

А.flat()[n-1:n*n-1:n-1]

from numpy.lib.stride_tricks import as_strided
buf = np.frombuffer(arr, dtype=np.int32, offset=8)
as_strided(buf, shape=(n,), strides=(8, ))

def three_ones(a):
    d1 = np.diff(a)
    d2 = np.diff(d1)
    n_d2 = np.arange(d2.size)
    zero_d2 = d2 == 0
    idx_d2 = n_d2[zero_d2]
    result = []
    for idx in idx_d2:
        if d1[idx] == d1[idx+1] == 0 and a[idx] == a[idx+1] == a[idx+2]==1:
            result.append(idx)
            
    return result

def three_ones(a):
    d1 = np.diff(a)
    d2 = np.diff(d1)
    n_d2 = np.arange(d2.size)
    zero_d2 = d2 == 0
    idx_d2 = n_d2[zero_d2]
    result = []
    idx_d1 = np.logical_and(d1[idx_d2]==0, d1[idx_d2+1]==0)
    idx_a = np.logical_and(a[idx_d2]==1, np.logical_and(a[idx_d2+1]==1, a[idx_d2+2]==1))
    result = n_d2[idx_d2[np.logical_and(idx_d1, idx_a)]]
            
    return result

np.diagonal(np.fliplr(field))