How do I build this block matrix in python?

Question

I want to build a block matrix for a system of equation with n points. The result is an (2n+2)x(2n+2) matrix. In example, for 2 points the matrix is:

1 0 0 0 0 0
a b c d 0 0
e f g h 0 0
0 0 a b c d
0 0 e f g h
0 0 0 0 0 1

For 3 points the matrix is

1 0 0 0 0 0 0 0
a b c d 0 0 0 0
e f g h 0 0 0 0
0 0 a b c d 0 0
0 0 e f g h 0 0
0 0 0 0 a b c d
0 0 0 0 e f g h
0 0 0 0 0 0 0 1

Here, a, b, c, d, e, f, g, h are float values known ahead of time. But I don't know value of n ahead of time. Is there a library in python to do this? I've looked at scipy.sparse.diag, scipy.linalg.block_diag and numpy.bat, but these do not achieve what I want.

This link might be helpful: stackoverflow.com/questions/26506204/… — Ohad Rubin
– Ohad Rubin, Commented Jul 6, 2017 at 19:43

Ofer Sadan · Accepted Answer · 2017-07-06 20:15:24Z

We can use np.identity to give us a "square" array (same dimensions in both axis) with the ones and zeros as you specified:

myarr = np.identity(2*n+2)

Then, we define our little subset values for a-h:

subset = np.array([[a,b,c,d],[e,f,g,h]])

Now to replace the corresponding values in our larger array:

for i in range(1, 2*n+2-1, 2):
    myarr[i:i+2, i-1:i+3] = subset

EDIT: this is the output for some random values i chose for a-h:

>>> myarr = np.identity(2*n+2)
>>> subset = np.array([[a,b,c,d],[e,f,g,h]])
>>> for i in range(1, 2*n+2-1, 2):
...     myarr[i:i+2, i-1:i+3] = subset
... 
>>> myarr
array([[  1.,   0.,   0.,   0.,   0.,   0.,   0.,   0.],
       [ 11.,   2.,   3.,   4.,   0.,   0.,   0.,   0.],
       [  5.,   6.,   7.,   9.,   0.,   0.,   0.,   0.],
       [  0.,   0.,  11.,   2.,   3.,   4.,   0.,   0.],
       [  0.,   0.,   5.,   6.,   7.,   9.,   0.,   0.],
       [  0.,   0.,   0.,   0.,  11.,   2.,   3.,   4.],
       [  0.,   0.,   0.,   0.,   5.,   6.,   7.,   9.],
       [  0.,   0.,   0.,   0.,   0.,   0.,   0.,   1.]])

Divakar · Accepted Answer · 2017-07-06 20:07:10Z

Here's an approach upon computing those linear indices in a broadcasted manner and then assigning into an zeros-initialized array -

def block_mat(list_, n = 2):
    input_arr = np.array(list_).reshape(-1,4)
    N = 2*n + 2
    M = 2*N + 2

    p,q = input_arr.shape
    I,J,K = np.ix_(M*np.arange(n), N*np.arange(p), np.arange(q))
    idx = I + J + K + N

    out = np.zeros((N,N),dtype=input_arr.dtype)
    out.flat[idx] = input_arr
    out.flat[[0,-1]] = 1
    return out

Sample runs -

1) Input elements :

In [497]: a,b,c,d,e,f,g,h = range(3,11)

In [498]: a,b,c,d,e,f,g,h
Out[498]: (3, 4, 5, 6, 7, 8, 9, 10)

2) Various n cases :

In [499]: block_mat([a,b,c,d,e,f,g,h], n = 2)
Out[499]: 
array([[ 1,  0,  0,  0,  0,  0],
       [ 3,  4,  5,  6,  0,  0],
       [ 7,  8,  9, 10,  0,  0],
       [ 0,  0,  3,  4,  5,  6],
       [ 0,  0,  7,  8,  9, 10],
       [ 0,  0,  0,  0,  0,  1]])

In [500]: block_mat([a,b,c,d,e,f,g,h], n = 3)
Out[500]: 
array([[ 1,  0,  0,  0,  0,  0,  0,  0],
       [ 3,  4,  5,  6,  0,  0,  0,  0],
       [ 7,  8,  9, 10,  0,  0,  0,  0],
       [ 0,  0,  3,  4,  5,  6,  0,  0],
       [ 0,  0,  7,  8,  9, 10,  0,  0],
       [ 0,  0,  0,  0,  3,  4,  5,  6],
       [ 0,  0,  0,  0,  7,  8,  9, 10],
       [ 0,  0,  0,  0,  0,  0,  0,  1]])

In [501]: block_mat([a,b,c,d,e,f,g,h], n = 4)
Out[501]: 
array([[ 1,  0,  0,  0,  0,  0,  0,  0,  0,  0],
       [ 3,  4,  5,  6,  0,  0,  0,  0,  0,  0],
       [ 7,  8,  9, 10,  0,  0,  0,  0,  0,  0],
       [ 0,  0,  3,  4,  5,  6,  0,  0,  0,  0],
       [ 0,  0,  7,  8,  9, 10,  0,  0,  0,  0],
       [ 0,  0,  0,  0,  3,  4,  5,  6,  0,  0],
       [ 0,  0,  0,  0,  7,  8,  9, 10,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  3,  4,  5,  6],
       [ 0,  0,  0,  0,  0,  0,  7,  8,  9, 10],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  1]])

Ajax1234 · Accepted Answer · 2017-07-07 13:39:45Z

Here is code that will work for you:

import string
n = 3
letters = string.ascii_lowercase
block1 = letters[:4]
block2 = letters[4:8]
matrix = [[1 if b == 0 and i == 0 or b == 2*n+1 and i == 2*n + 1 else 0 for b in range(2*n+2)] for i in range(2*n+2)]
count = 0

for i, a in enumerate(matrix[1:-1]): #fix to account for the fact that we are starting at 2, not 0

    if i%2 == 0:
        matrix[i+1][count:count+4] = list(block1)


    else:
        matrix[i+1][count:count+4] = list(block2)


        count += 2

for i in matrix:
    print i

Output for when n == 3:

[1, 0, 0, 0, 0, 0, 0, 0]
['a', 'b', 'c', 'd', 0, 0, 0, 0]
['e', 'f', 'g', 'h', 0, 0, 0, 0]
[0, 0, 'a', 'b', 'c', 'd', 0, 0]
[0, 0, 'e', 'f', 'g', 'h', 0, 0]
[0, 0, 0, 0, 'a', 'b', 'c', 'd']
[0, 0, 0, 0, 'e', 'f', 'g', 'h']
[0, 0, 0, 0, 0, 0, 0, 1]

Output for when n == 2:

[1, 0, 0, 0, 0, 0]
['a', 'b', 'c', 'd', 0, 0]
['e', 'f', 'g', 'h', 0, 0]
[0, 0, 'a', 'b', 'c', 'd']
[0, 0, 'e', 'f', 'g', 'h']
[0, 0, 0, 0, 0, 1]

I realize that you are not actually using the strings "a", "b", "c", etc., but when you want to use variables, replace block1 with a list of your first four variables and block2 with a list of your last four variables.

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