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
On Wed, 19 Jun 2019 at 17:39, Lucy Jackson <lucy.jackson....@gmail.com> wrote:
>
> 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
>
> On Wed, 19 Jun 2019 at 16:51, Oscar Benjamin <oscar.j.benja...@gmail.com>
> wrote:
>>
>> 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
>>
>> On Wed, 19 Jun 2019 at 15:15, Lucy Jackson <lucy.jackson....@gmail.com>
>> wrote:
>> >
>> > 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
>> >
>> >
>> >
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