Concatenate Matrix

[Lucy Jackson]

Hi,

I am looking to concatenate three matrices (shown below) using BlockMatrix, however, the output has the wrong dimensions.

theta = sym.MatrixSymbol('theta', 5, 1)

phi = sym.MatrixSymbol('phi', 3, 1)

a = sym.Matrix([[0],[0],[0]])

X = sym.BlockMatrix([a, phi, theta])

With this code I get the following output:

 X = Matrix([

[          0],
[          0],
[          0],
[  phi[0, 0]],
[theta[0, 0]]])

However, I am expecting an output thats 11 x 1. I am also unable to check the shape of X as X.shape yields the following error:

AttributeError: 'Zero' object has no attribute 'shape'

Any help would be greatly appreciated!

Many Thanks,

Lucy 

[Oscar Benjamin]

Hi Lucy,

Running on master I get this:

In [13]: sym.BlockMatrix([[a], [phi], [theta]])
Out[13]:
⎡⎡0⎤⎤
⎢⎢ ⎥⎥
⎢⎢0⎥⎥
⎢⎢ ⎥⎥
⎢⎣0⎦⎥
⎢ ⎥
⎢ φ ⎥
⎢ ⎥
⎣ θ ⎦

In [14]: sym.BlockMatrix([[a], [phi], [theta]]).shape
Out[14]: (11, 1)

Note that I've used extra square brackets to indicate that I want
these combined column-wise. Without those I get:

In [11]: sym.BlockMatrix([a, phi, theta])
Out[11]:
⎡⎡0⎤ ⎤
⎢⎢ ⎥ ⎥
⎢⎢0⎥ φ θ⎥
⎢⎢ ⎥ ⎥
⎣⎣0⎦ ⎦

In [12]: sym.BlockMatrix([a, phi, theta]).shape
Out[12]: (3, 3)

It's unfortunate that the sizes aren't checked there since those are
inconsistent.

It looks like the main problem you're having is fixed in master. What
version of SymPy are you using?


--
Oscar

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[Lucy Jackson]

Hi Oscar,

I am using version 1.3.

The introduction of the extra [] brackets now makes the shape 11,1. However, when the variable is viewed it is as follows:

Matrix([
[Matrix([
[0],
[0],
[0]])],
[                     phi],
[                   theta]])

What I really want is a single matrix with:

Matrix([[0]

[0]  

[0] 

phi[0,0]

phi[1,0]

phi[2,0]

theta[0.0]

theta[1,0]

theta[2,0]

theta[3,0]

theta[4,0] ])

Is this how sympy has stored it even though it does not display it like this?

Thanks,

Lucy

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[Oscar Benjamin]

Hi Lucy,

The point of a block-matrix is that it is composed of smaller matrices
that we keep as whole objects. If you want to flatten this to an
explicit matrix you can use as_explicit:

In [6]: sym.BlockMatrix([[a], [phi], [theta]]).as_explicit()
Out[6]:
⎡ 0 ⎤
⎢ ⎥
⎢ 0 ⎥
⎢ ⎥
⎢ 0 ⎥
⎢ ⎥
⎢φ₀₀⎥
⎢ ⎥
⎢φ₁₀⎥
⎢ ⎥
⎢φ₂₀⎥
⎢ ⎥
⎢θ₀₀⎥
⎢ ⎥
⎢θ₁₀⎥
⎢ ⎥
⎢θ₂₀⎥
⎢ ⎥
⎢θ₃₀⎥
⎢ ⎥
⎣θ₄₀⎦

Your calculations should come out correctly with or without flattening
though e.g.:

In [7]: A = MatrixSymbol('A', 2, 2)

In [8]: B = MatrixSymbol('B', 2, 2)

In [9]: A
Out[9]: A

In [10]: A.as_explicit()
Out[10]:
⎡A₀₀ A₀₁⎤
⎢ ⎥
⎣A₁₀ A₁₁⎦

In [11]: A*B
Out[11]: A⋅B

In [12]: (A*B).as_explicit()
Out[12]:
⎡A₀₀⋅B₀₀ + A₀₁⋅B₁₀ A₀₀⋅B₀₁ + A₀₁⋅B₁₁⎤
⎢ ⎥
⎣A₁₀⋅B₀₀ + A₁₁⋅B₁₀ A₁₀⋅B₀₁ + A₁₁⋅B₁₁⎦

In [13]: A.as_explicit() * B.as_explicit()
Out[13]:
⎡A₀₀⋅B₀₀ + A₀₁⋅B₁₀ A₀₀⋅B₀₁ + A₀₁⋅B₁₁⎤
⎢ ⎥
⎣A₁₀⋅B₀₀ + A₁₁⋅B₁₀ A₁₀⋅B₀₁ + A₁₁⋅B₁₁⎦

--
Oscar

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