从两个列表快速创建对角矩阵

Diagonal matrix
一张单子就可以填得满满的 diagonal-constant 另一个将填充矩阵。

例如:

fast_matrix([1,2], [6,7,8])

应输出2个矩阵:

[2] # unused
[1, 6, 7, 8]
[6, 1, 7, 8]
[6, 7, 1, 8]
[6, 7, 8, 1]

和

[1] # unused
[2, 6, 7, 8]
[6, 2, 7, 8]
[6, 7, 2, 8]
[6, 7, 8, 2]

code makes 10000 transformations

from pprint import pprint
import timeit

def not_so_fast_matrix(A, B):
    rt_obj = []
    for i,_ in enumerate(A):
        for z in range(len(B) + 1):
            new_from = A.copy()
            new_from.remove(A[i])
            new_list = B.copy()
            new_list.insert(z, A[i])
            rt_obj.append({'remain': new_from, 'to_list': new_list})
    return rt_obj

# pprint(not_so_fast_matrix([1,2], [6,7,8]))

A = ([1,2,3,4,5,6,7,8,9,10])
B = ([60,70,80,90,100,200,300])
t = timeit.Timer(lambda: not_so_fast_toeplitz(A, B))
print("not_so_fast_matrix took: {:.3f}secs for 10000 iterations".format(t.timeit(number=10000)))

我想知道用另一种方法是否能更快。

from scipy.linalg import circulant

print(circulant([1, 8,7,6])) # <- Should be inverted
outputs:
[[1 6 7 8]
 [8 1 6 7]
 [7 8 1 6]
 [6 7 8 1]]

元素被向右移动(推动)。