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Background Information:

I recently started using the Einstein summation notation to express certain operations over an "image" $\mathbf{A}$ where to each pixel a square matrix is attached. That is, an array of the shape $\left[n_{row}\times n_{col}\times n \times n\right]$. Now, if I want to represent the multiplication of each "submatrix" $A_{pq}$ of size $n \times n$ by a matrix $\mathbf{B}$ of compatible size, I can express it as:

$A^{'}_{pqij} = B_{ik}A_{pqkj}$.

The problem: I have an array $\mathbf{B}$ of the same size as $\mathbf{A}$ and I want to compute a third array $\mathbf{A}^{'}$ where each pixel $\mathbf{A}_{pix}^{'}$ at the first two indices $p$ and $q$ corresponds to the matrix multiplication of matrix $\mathbf{A}_{pix}$ with $\mathbf{B}_{pix}$ in array $\mathbf{B}$ at the same indices $p$ and $q$.

In the notation of linear algebra, for each $n \times n$ matrix indicized by $p$ and $q$ I want:

$A_{pix}^{'} = B_{pix}A_{pix}$

Attempt:

I initially tried something of the form: $A_{mnloij}^{'} = B_{mnik}A_{lokj}$ but if I think this will compute the submatrix multiplication at all combinations of indices $m,l,n,o$.

The Question in Brief: Is there a way, using the Einstein summation notation, to compute the sum:

$A_{pqij}^{'} = B_{mnik}A_{lokj}$

for the combinations of indices where $(m,n)=(l,o)=\left(p,q\right)$ only?

Additional Attempt: In the worst case, I suppose that I have to first compute the expression above and then only slice the array in order to select the "diagonals" where $(m,n)=(l,o)$.

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  • $\begingroup$ Correct: "Background". $\endgroup$
    – user93957
    Dec 13, 2013 at 16:01
  • $\begingroup$ You're using the Einstein notation incorrectly: your tensors need both upper and lower indices and your implied sums should have one of each. $\endgroup$
    – Ryan Reich
    Dec 13, 2013 at 16:18
  • $\begingroup$ You are right, but for my purposes, it does not matter. I'm using this "pseudo-notation" to express some multiplication that I will later perform using Pythons "einsum" notation ,which does not distinguish between lower and upper indices. In general, I'm not interested in the difference between covariance and contravariance and I'm using this notation as a proxy to think about some transformations I am performing on data. $\endgroup$
    – Bafe
    Dec 13, 2013 at 16:26
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    $\begingroup$ Mathematically, if $P$ is your set of pixels, indexed however you like, then the objects you're considering are most naturally viewed as honest functions $P \to M_N(\mathbb{R})$. In particular, from this perspection, $A^\prime = BA$ is the pointwise product of the "constant" $B$ with $A$, i.e., $A^\prime(p) = BA(p)$ for each pixel (label) $p \in P$. At the risk of asking a stupid question, is there any way of encoding things this way instead? Because if so, this should obviate all your problems, since you're just doing honest matrix multiplication "pointwise". $\endgroup$ Dec 14, 2013 at 1:11
  • $\begingroup$ No, $\mathbf{B}\left(p\right)$ is indexed by $p$ too. I'd like to use the implicit sum notation because it allows me to express the problem in a coincise way and to implement it later on. $\endgroup$
    – Bafe
    Dec 14, 2013 at 16:17

2 Answers 2

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Numpy.einsum has a shortcut for this. You can specify the axes you want out.

Normally you line up, and sum repeated indices.

If you specify the output indices, it alines things the same way, but summs the indices that aren't in the specified output.

#Hadamard product
c = np.einsum('ij,ij->ij',a,b)

#select the diagonal of a matrix
c = np.einsum('ii->i',a)

#sum
c = np.einsum('ij->',a)

#squared length of each vector in a 2d array
c = np.einsum('ijk,ijk->ij',a,a)

I think your application can be said as:

#matrix product of each matrix in a 2d array
c = np.einsum('ijkl,ijlm->ijkm',a,b)

I find this is much simpler and more direct. I don't know if it matters, but I also think that this will be much more efficient than a direct implementation of the accepted answer.

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Define the tensor

$${E_{ijk}} = \mathrm{1\ \ if\ i=j=k,\ 0\ otherwise}$$

$\mathbf{E}$ has some interesting properties: $$ \mathbf{E.a = Diag(a)} = \mathrm{matrix\ (\mathbf{a}\ along\ the\ diagonal)} $$ $$ \mathbf{E:A = diag(A)} = \mathrm{vector\ (diagonal\ of\ \mathbf{A})}$$ $$ \mathbf{a.E.b = E:ab} = \mathrm{Schur\ product\ (\mathbf{a \odot b})}$$

Your question is similar to the Schur product case

$${A'_{pqij} = E_{pml}\ E_{qno}\ B_{mnik}\ A_{lokj}}$$

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  • $\begingroup$ Thanks, that's exactly what I needed! $\endgroup$
    – Bafe
    Dec 14, 2013 at 16:14

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