Python numpy 编制矩阵

Question

Answer 1

a = [[1, 2, 5, 7],
     [4, 1, 8, 0],
     [2, 0, 5, 1],
     [0, 2, 1, 1]]

n = int(len(a) / 2)

s = [[0, 0], [0, 0]]

for i in range(0, 2 * n):
    r = int(i/n)

    s[r][0] += sum(a[i][0:n])
    s[r][1] += sum(a[i][n:2*n])
    
for i in range(0, 2):
    for j in range(0, 2):
        s[i][j] /= n * n
        
print(s)


# Output
[[2.0, 5.0], [1.0, 2.0]]