Best to file an issue. -viral
On Sunday, February 22, 2015 at 12:52:17 PM UTC+5:30, Spencer Lyon wrote: > > Bump on this thread. I need this functionality again... > > On Friday, December 6, 2013 at 11:04:44 AM UTC-5, Spencer Lyon wrote: >> >> I am working with tensors with more than 2 dimensions. I would like to >> find a Julia equivalent to the numpy function tensordot. The docstring >> for the function explains its use well, so I will just copy/paste it here: >> >> def tensordot(a, b, axes=2): >> """ >> Compute tensor dot product along specified axes for arrays >= 1-D. >> >> Given two tensors (arrays of dimension greater than or equal to one), >> ``a`` and ``b``, and an array_like object containing two array_like >> objects, ``(a_axes, b_axes)``, sum the products of ``a``'s and ``b``'s >> elements (components) over the axes specified by ``a_axes`` and >> ``b_axes``. The third argument can be a single non-negative >> integer_like scalar, ``N``; if it is such, then the last ``N`` >> dimensions of ``a`` and the first ``N`` dimensions of ``b`` are summed >> over. >> >> Parameters >> ---------- >> a, b : array_like, len(shape) >= 1 >> Tensors to "dot". >> >> axes : variable type >> >> * integer_like scalar >> Number of axes to sum over (applies to both arrays); or >> >> * array_like, shape = (2,), both elements array_like >> Axes to be summed over, first sequence applying to ``a``, second >> to ``b``. >> >> See Also >> -------- >> dot, einsum >> >> Notes >> ----- >> When there is more than one axis to sum over - and they are not the last >> (first) axes of ``a`` (``b``) - the argument ``axes`` should consist of >> two sequences of the same length, with the first axis to sum over given >> first in both sequences, the second axis second, and so forth. >> >> Examples >> -------- >> A "traditional" example: >> >> >>> a = np.arange(60.).reshape(3,4,5) >> >>> b = np.arange(24.).reshape(4,3,2) >> >>> c = np.tensordot(a,b, axes=([1,0],[0,1])) >> >>> c.shape >> (5, 2) >> >>> c >> array([[ 4400., 4730.], >> [ 4532., 4874.], >> [ 4664., 5018.], >> [ 4796., 5162.], >> [ 4928., 5306.]]) >> >>> # A slower but equivalent way of computing the same... >> >>> d = np.zeros((5,2)) >> >>> for i in range(5): >> ... for j in range(2): >> ... for k in range(3): >> ... for n in range(4): >> ... d[i,j] += a[k,n,i] * b[n,k,j] >> >>> c == d >> array([[ True, True], >> [ True, True], >> [ True, True], >> [ True, True], >> [ True, True]], dtype=bool) >> """ >> >> I think the functionality laid out in the explicit for loops would be >> quite easy to replicate using the @nloops and @nref macros in Cartesian, >> but I am not sure if that will be most performant way to implement such a >> function. Also, I wanted to check here to see if this has already been done >> before I spend some time working on this. >> >> Thanks, >> >
