CODEWITHSUNDEEP
julia
Sakurai.JJ
Sakurai.JJ

Reputation: 684

Tensor mode n product in Julia

I need tensor mode n product. The defination of tenosr mode n product can be seen here. https://www.alexejgossmann.com/tensor_decomposition_tucker/

I found python code. I would like to convert this code into julia.

def mode_n_product(x, m, mode):
   x = np.asarray(x)
   m = np.asarray(m)
   if mode <= 0 or mode % 1 != 0:
       raise ValueError('`mode` must be a positive interger')
   if x.ndim < mode:
       raise ValueError('Invalid shape of X for mode = {}: {}'.format(mode, x.shape))
   if m.ndim != 2:
       raise ValueError('Invalid shape of M: {}'.format(m.shape))
   return np.swapaxes(np.swapaxes(x, mode - 1, -1).dot(m.T), mode - 1, -1)

I have found another answer using Tensortoolbox.jl

using TensorToolbox
X=rand(5,4,3);
A=rand(2,5);
ttm(X,A,n)  #X times A[1] by mode n

Upvotes: 1

Views: 634

Answers (1)

mcabbott
mcabbott

Reputation: 2580

One way is:

using TensorOperations
@tensor y[i1, i2, i3, out, i5] := x[i1, i2, i3, s, i5] * a[out, s]

This is literally the formula given at your link to define this, except that I changed the name of the summed index to s; you can you any index names you like, they are just markers. The sum is implicit, because s does not appear on the left.

There is nothing very special about putting the index out back in the same place. Like your python code, @tensor permutes the dimensions of x in order to use ordinary matrix multiplication, and then permutes again to give y the requested order. The fewer permutations needed, the faster this will be.

Alternatively, you can try using LoopVectorization, Tullio; @tullio y[i1, i2, ... with the same notation. Instead of permuting in order to call a library matrix multiplication function, this writes a pure-Julia version which works with the array as it arrives.

Upvotes: 2

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