# [Numpy-discussion] Tensor contraction

David Goldsmith d.l.goldsmith@gmail....
Sun Jun 13 14:59:51 CDT 2010

Is this not what
core.numeric.tensordot<http://docs.scipy.org/numpy/docs/numpy.core.numeric.tensordot/>does?

DG

On Sun, Jun 13, 2010 at 12:37 PM, Friedrich Romstedt <
friedrichromstedt@gmail.com> wrote:

> 2010/6/13 Alan Bromborsky <abrombo@verizon.net>:
> > I am writing symbolic tensor package for general relativity.  In making
> > symbolic tensors concrete
> > I generate numpy arrays stuffed with sympy functions and symbols.
>
> That sound's interesting.
>
> > The
> > operations are tensor product
> > (numpy.multiply.outer), permutation of indices (swapaxes),  partial and
> > covariant (both vector operators that
> > increase array dimensions by one) differentiation, and contraction.
>
> I would like to know more precisely what this differentiations do, and
> how it comes that they add an index to the tensor.
>
> > I think I need to do the contraction last
> > to make sure everything comes out correctly.  Thus in many cases I would
> > be performing multiple contractions
> > on the tensor resulting from all the other operations.
>
> Hm, ok, so I guess I shall give my 1 cent now.
>
> Ok.
>
>    # First attempt (FYI, failed):
>
>    # The general procedure is, to extract a multi-dimensional diagonal
> array.
>    # The sum \sum_{ij = 0}^{M} \sum_{kl = 0}^{N} is actually the sum over a
>    # 2D array with indices I \equiv i \equiv j and K \equiv k \equiv
> l.  Meaning:
>    # \sum_{(I, K) = (0, 0)}^{(M, N)}.
>    # Thus, if we extract the indices with 2D arrays [[0], [1], ...,
> [N - 1]] for I and
>    # [[0, 1, ..., M - 1]] on the other side for K, then numpy's
>    # mechanism will broadcast them to the same shape, yielding (N, M)
> arrays.
>    # Then finally we sum over this X last dimensions when there were X
>    # contractions, and we're done.
>
>    # Hmmm, when looking closer at the problem, it seems that this isn't
>    # adequate.  Because we would have to insert open slices, but cannot
>    # create them outside of the [] operator ...
>
>    # So now follows second attemt:
>
> def contract(arr, *contractions):
>    """*CONTRACTIONS is e.g.:
>        (0, 1), (2, 3)
>    meaning two contractions, one of 0 & 1, and one of 2 & 2,
>    but also:
>        (0, 1, 2),
>    is allowed, meaning contract 0 & 1 & 2."""
>
>    # First, we check if we can contract using the *contractions* given ...
>
>    for contraction in contractions:
>        # Extract the dimensions used.
>        dimensions = numpy.asarray(arr.shape)[list(contraction)]
>
>        # Check if they are all the same.
>        dimensionsdiff = dimensions - dimensions[0]
>        if numpy.abs(dimensionsdiff).sum() != 0:
>            raise ValueError('Contracted indices must be of same
> dimension.')
>
>    # So now, we can contract.
>    #
>    # First, pull the contracted dimensions all to the front ...
>
>    # The names of the indices.
>    names = range(arr.ndim)
>
>    # Pull all of the contractions.
>    names_pulled = []
>    for contraction in contractions:
>        names_pulled = names_pulled + list(contraction)
>        # Remove the pulled index names from the pool:
>        for used_index in contraction:
>            # Some more sanity check
>            if used_index not in names:
>                raise ValueError('Each index can only be used in one
> contraction.')
>            names.remove(used_index)
>
>    # Concatenate the pulled indices and the left-over indices.
>    names_final = names_pulled + names
>
>    # Perform the swap.
>    arr = arr.transpose(names_final)
>
>    # Perform the contractions ...
>
>    for contraction in contractions:
>        # The respective indices are now, since we pulled them, the
> frontmost indices:
>        ncontraction = len(contraction)
>        # The index array:
>        # shape[0] = shape[1] = ... = shape[ncontraction - 1]
>        I = numpy.arange(0, arr.shape[0])
>        # Perform contraction:
>        index = [I] * ncontraction
>        arr = arr[tuple(index)].sum(axis=0)
>
>    # If I made no mistake, we should be done now.
>    return arr
>
> Ok, it didn't get much shorter than Pauli's solution, so you decide ...
>
> > One question to
> > ask would be considering that I am stuffing
> > the arrays with symbolic objects and all the operations on the objects
> > would be done using the sympy modules,
> > would using numpy operations to perform the contractions really save any
> > time over just doing the contraction in
> > python code with a numpy array.
>
> I don't know anything about sympy.  I think there's some typo around:
> I guess you mean creating some /sympy/ array and doing the operations
> using that instead of using a numpy array having sympy dtype=object
> content?
>
> Friedrich
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