2008/4/7, Pearu Peterson <pearu@xxxxxxxxx>:
Martin Sandve Alnæs wrote:
> I don't agree with the x.T stuff. There is no differentiation between
> columns and rows in tensor algebra, so the transpose has no meaning
> for a rank 1 tensor. This has to generalize to where x is a rank 1
> expression and not only an explicitly constructed vector. If you have
> a rank 4 tensor and pick a sub-tensor of rank 1 from it, is it a
> column or row? Columns and rows doesn't generalize, and special cases
> are evil.
Transpose is well-defined also for arbitrary rank tensors. It swaps
the first two indices:
transpose(t_ijkl..) -> t_jikl..
One can also defined generalised transpose:
transpose(t_ijkl.., [p_i, p_j, p_k, p_l, ..]) -> t_(p_i)(p_j)(p_k)(p_l)..
We _could_ support something like that, but I don't think it's worth
the trouble since the more general index notation covers this without
making detailed definitions:
# reference components
Aijkl = A[i,j,k,l]
# make tensor with different axis ordering
B = Tensor(Aijkl, (j,i,k,l))
