Putting together a larger matrix from smaller matrices

[Matjaz Bezovnik]

Dear all,

I'm but a layman so do not take offence at this maybe over simple
question.

This is something which is done often in FEM methods, and the alike.

I have matrix A of 3x3 elements, and B, of the same number of
elements, 3x3.

What would be the most obvious way to assemble a matrix which:
a11 a12 a13
a21 a22 a23
a31 a32 a33+b11 b12 b13
b21 b22 b23
b31 b32 b33

(the missing elements = zero)

(you see the pattern - if there was a third matrix C, it would
"connect" to b33 on the main diagonal)


Matjaz

 

[Peter Otten]

This is something which is done often in FEM methods, and the alike.

I have matrix A of 3x3 elements, and B, of the same number of
elements, 3x3.

What would be the most obvious way to assemble a matrix which:
a11 a12 a13
a21 a22 a23
a31 a32 a33+b11 b12 b13
b21 b22 b23
b31 b32 b33

(the missing elements = zero)

(you see the pattern - if there was a third matrix C, it would
"connect" to b33 on the main diagonal)

Unless there is a dedicated function:

partial = [...] # list of matrices of shape NxN

N = partial[0].shape[0]
width = N*len(partial)
b = numpy.zeros((width, width))

off = 0
for m in partial:
b[offff+N, offff+N] = m
off += N
print b

Peter

 

[Matjaz Bezovnik]

Matjaz Bezovnik wrote:

If you are using numpy (which it sounds like you are):

IDLE 2.6.2

import numpy as np
v = np.array([[0,1,2],[3,4,5],[6,7,8]], dtype=float)
v

array([[ 0., 1., 2.],
[ 3., 4., 5.],
[ 6., 7., 8.]])

w = np.array([[10,11,12],[13,14,15],[16,17,18]], dtype=float)
w

array([[ 10., 11., 12.],
[ 13., 14., 15.],
[ 16., 17., 18.]])array([[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.]])

r[:3,:3] = v
r

array([[ 0., 1., 2., 0., 0., 0.],
[ 3., 4., 5., 0., 0., 0.],
[ 6., 7., 8., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0., 0.]])

r[3:,3:] = w
r

array([[ 0., 1., 2., 0., 0., 0.],
[ 3., 4., 5., 0., 0., 0.],
[ 6., 7., 8., 0., 0., 0.],
[ 0., 0., 0., 10., 11., 12.],
[ 0., 0., 0., 13., 14., 15.],
[ 0., 0., 0., 16., 17., 18.]])
In general, make the right-sized array of zeros, and at various points:
and you can ssign to subranges of the result array:

N = 3
result = np.zeros((len(parts) * N, len(parts) * N), dtype=float)
for n, chunk in enumerate(parts):
base = n * 3
result[base : base + 3, base : base + 3] = chunk

--Scott David Daniels
(e-mail address removed)

Scott, thank you very much for the snippet.

It is exactly what I looked for; simple to read and obvious as to what
it does even a month later to a non-pythonist!

Matjaz

 

[sturlamolden]

Scott, thank you very much for the snippet.

It is exactly what I looked for; simple to read and obvious as to what
it does even a month later to a non-pythonist!

Since you were talking about matrices, observe that numpy has a matrix
subclass of ndarray, which e.g. changes the meaning of the * operator
mean indicate matrix multiplication. Thus,
array([[ 1., 0., 0., 0.],
[ 0., 1., 0., 0.],
[ 0., 0., 1., 0.],
[ 0., 0., 0., 1.]])matrix([[ 1., 0., 0., 0.],
[ 0., 1., 0., 0.],
[ 0., 0., 1., 0.],
[ 0., 0., 0., 1.]])
array([[ 1., 1., 1., 1.],
[ 1., 1., 1., 1.],
[ 1., 1., 1., 1.],
[ 1., 1., 1., 1.]])matrix([[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.]])matrix([[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.]])matrix([[ 64., 64., 64., 64.],
[ 64., 64., 64., 64.],
[ 64., 64., 64., 64.],
[ 64., 64., 64., 64.]])

In Matlab, you use .* vs. * and .^ vs. ^ to obtain the same effect. In
NumPy we use different classes for arrays and matrices.



Sturla Molden

 

[Robert Kern]

Dear all,

I'm but a layman so do not take offence at this maybe over simple
question.

This is something which is done often in FEM methods, and the alike.

I have matrix A of 3x3 elements, and B, of the same number of
elements, 3x3.

What would be the most obvious way to assemble a matrix which:
a11 a12 a13
a21 a22 a23
a31 a32 a33+b11 b12 b13
b21 b22 b23
b31 b32 b33

(the missing elements = zero)

(you see the pattern - if there was a third matrix C, it would
"connect" to b33 on the main diagonal)

You will certainly want to use numpy for this and ask questions on the numpy
mailing list.

http://www.scipy.org/Mailing_Lists

I believe that someone recently posted a recipe for constructing such "block
diagonal" arrays. I think it will be included in a future release of numpy.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco