Htykuut

Given 2 lists I would like to create a Diagonal matrix.

One list will fill the diagonal-constant and the other will fill the matrix.

For example:

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

Should output 2 matrices:

[2] # unused

[1, 6, 7, 8]

[6, 1, 7, 8]

[6, 7, 1, 8]

[6, 7, 8, 1]

and

[1] # unused

[2, 6, 7, 8]

[6, 2, 7, 8]

[6, 7, 2, 8]

[6, 7, 8, 2]

My code makes 10000 transformations in 2.5secs in my pc.

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)))

I was wondering if it can go faster using another approach.

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]]

Elements were shifted (pushed) right.

Transposed and flatten matrix has a structure where elements from B are repeated. This approach uses that property to create blueprint of matrix with wrong values on diagonal, and than fill diagonal with correct value.

import numpy



def create_matrices(A, B):

    # Create blueprint of result matrix

    n = len(B) + 1

    b = numpy.empty((n * n,), dtype=numpy.int32)

    b[:-1] = numpy.repeat(B, n + 1)

    b = b.reshape((n, n)).T  # <- added transposition

    # Change diagonal elements

    for a in A:

        m = b.copy()

        numpy.fill_diagonal(m, a)

        print(m)



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