1
$\begingroup$

I would like to write a function, that matricizes a higher order tensor according to the following rule:

Let $\mathcal{A} \in \mathbb{C}^{I_{1} \times I_{2} \times \ldots \times I_{N}}$ be a tensor of order $N$. The matrix unfolding $\mathbf{A}_{(n)} \in \mathbb{C}^{I_{n} \times\left(I_{n+1} I_{n+2} \ldots I_{N} I_{1} I_{2} \ldots I_{n-1}\right)}$ contains the element $a_{i_{1} i_{2} \ldots i_{N}}$ at the position with row number $i_n$ and column number equal to $$\begin{array}{l}{\left(i_{n+1}-1\right) I_{n+2} I_{n+3} \ldots I_{N} I_{1} I_{2} \ldots I_{n-1}+\left(i_{n+2}-1\right) I_{n+3} I_{n+4} \ldots I_{N} I_{1} I_{2} \ldots I_{n-1}+\cdots} \\ {\quad+\left(i_{N}-1\right) I_{1} I_{2} \ldots I_{n-1}+\left(i_{1}-1\right) I_{2} I_{3} \ldots I_{n-1}+\left(i_{2}-1\right) I_{3} I_{4} \ldots I_{n-1}+\cdots+i_{n-1}}\end{array}$$

How do a write a function, that will accept a higher order tensor of variable rank $N$ as an input?

Thanks

$\endgroup$
1
$\begingroup$

This should do what you want (you might want to double-check that):

unfold[ten_, n_] := Flatten /@ Transpose[ten, RotateRight[Range@TensorRank@ten, n - 1]]

This works by first rearranging the levels of the tensor from $\{1,\dots,N\}$ to $\{n,n+1,\dots,N,1,\dots,n-1\}$ and the flattening everything but the first level.

An example:

A = Array[Subscript[a, ##] &, {3, 3, 3}];

Column@Table[MatrixForm@unfold[A, n], {n, 3}]

enter image description here

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer! $\endgroup$ – Emil_M Jul 29 '19 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.