[Numpy-discussion] Tensor contraction

Sebastian Walter sebastian.walter@gmail....
Mon Jun 14 07:37:16 CDT 2010


On Sun, Jun 13, 2010 at 8:11 PM, Alan Bromborsky <abrombo@verizon.net> wrote:
> Friedrich Romstedt wrote:
>> 2010/6/13 Pauli Virtanen <pav@iki.fi>:
>>
>>> def tensor_contraction_single(tensor, dimensions):
>>>    """Perform a single tensor contraction over the dimensions given"""
>>>    swap = [x for x in range(tensor.ndim)
>>>            if x not in dimensions] + list(dimensions)
>>>    x = tensor.transpose(swap)
>>>    for k in range(len(dimensions) - 1):
>>>        x = np.diagonal(x, axis1=-2, axis2=-1)
>>>    return x.sum(axis=-1)
>>>
>>> def _preserve_indices(indices, removed):
>>>    """Adjust values of indices after some items are removed"""
>>>    for r in reversed(sorted(removed)):
>>>        indices = [j if j <= r else j-1 for j in indices]
>>>    return indices
>>>
>>> def tensor_contraction(tensor, contractions):
>>>    """Perform several tensor contractions"""
>>>    while contractions:
>>>        dimensions = contractions.pop(0)
>>>        tensor = tensor_contraction_single(tensor, dimensions)
>>>        contractions = [_preserve_indices(c, dimensions)
>>>                        for c in contractions]
>>>    return tensor
>>>
>>
>> Pauli,
>>
>> I choke on your code for 10 min or so.  I believe there could be some
>> more comments.
>>
>> Alan,
>>
>> Do you really need multiple tensor contractions in one step?  If yes,
>> I'd like to put in my 2 cents in coding such one using a different
>> approach, doing all the contractions in one step (via broadcasting).
>> It's challenging.  We can generalise this problem as much as we want,
>> e.g. to contracting three instead of only two dimensions.  But first,
>> in case you have only two dimensions to contract at one single time
>> instance, then Josef's first suggestion would be fine I think.  Simply
>> push out the diagonal dimension to the end via .diagonal() and sum
>> over the last so created dimension.  E.g.:
>>
>> # First we create some bogus array to play with:
>>
>>>>> a = numpy.arange(5 ** 4).reshape(5, 5, 5, 5)
>>>>>
>>
>> # Let's see how .diagonal() acts (just FYI, I haven't verified that it
>> is what we want):
>>
>>>>> a.diagonal(axis1=0, axis2=3)
>>>>>
>> array([[[  0, 126, 252, 378, 504],
>>         [  5, 131, 257, 383, 509],
>>         [ 10, 136, 262, 388, 514],
>>         [ 15, 141, 267, 393, 519],
>>         [ 20, 146, 272, 398, 524]],
>>
>>        [[ 25, 151, 277, 403, 529],
>>         [ 30, 156, 282, 408, 534],
>>         [ 35, 161, 287, 413, 539],
>>         [ 40, 166, 292, 418, 544],
>>         [ 45, 171, 297, 423, 549]],
>>
>>        [[ 50, 176, 302, 428, 554],
>>         [ 55, 181, 307, 433, 559],
>>         [ 60, 186, 312, 438, 564],
>>         [ 65, 191, 317, 443, 569],
>>         [ 70, 196, 322, 448, 574]],
>>
>>        [[ 75, 201, 327, 453, 579],
>>         [ 80, 206, 332, 458, 584],
>>         [ 85, 211, 337, 463, 589],
>>         [ 90, 216, 342, 468, 594],
>>         [ 95, 221, 347, 473, 599]],
>>
>>        [[100, 226, 352, 478, 604],
>>         [105, 231, 357, 483, 609],
>>         [110, 236, 362, 488, 614],
>>         [115, 241, 367, 493, 619],
>>         [120, 246, 372, 498, 624]]])
>> # Here, you can see (obviously :-) that the last dimension is the
>> diagonal ... just believe in the semantics ....
>>
>>>>> a.diagonal(axis1=0, axis2=3).shape
>>>>>
>> (5, 5, 5)
>>
>> # Sum over the diagonal shape parameter:
>> # Again I didn't check this result's numbers.
>>
>>>>> a.diagonal(axis1=0, axis2=3).sum(axis=-1)
>>>>>
>> array([[1260, 1285, 1310, 1335, 1360],
>>        [1385, 1410, 1435, 1460, 1485],
>>        [1510, 1535, 1560, 1585, 1610],
>>        [1635, 1660, 1685, 1710, 1735],
>>        [1760, 1785, 1810, 1835, 1860]])
>>
>> The .diagonal() approach has the benefit that one doesn't have to care
>> about where the diagonal dimension ends up, it's always the last
>> dimension of the resulting array.  With my solution, this was not so
>> fine, because it could also become the first dimension of the
>> resulting array.
>>
>> For the challenging part, I'll await your response first ...
>>
>> Friedrich
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>>
>>
> I am writing symbolic tensor package for general relativity.  In making
> symbolic tensors concrete

Does that mean you are only interested in the numerical values of the tensors?
I mean, is the final goal to obtain a numpy.array(...,dtype=float)
which contains
the wanted coefficients?
Or do you need the symbolic representation?

> I generate numpy arrays stuffed with sympy functions and symbols.  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
> 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.  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.

Not 100% sure. But for/while loops are really slow in Python and the
numpy.ndarray.__getitem__ and ndarray.__setitem__ cause also a lot of
overhead.
I.e., using Python for loops on an element by element basis  is going
to take a long time if you have big tensors.

You could write a small benchmark and post the results here. I'm also
curious what the result is going to be ;).

As to your original question:
I think it may be helpful to look at numpy.lib.stride_tricks

There is a really nice advanced tutoria by Stéfan van der Walt
http://mentat.za.net/numpy/numpy_advanced_slides/

E.g.  to get a view of the diagonal elements of a matrix you can do
something like:


In [44]: from numpy.lib import stride_tricks

In [45]: x = numpy.arange(4*4)

In [46]: x
Out[46]: array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15])

In [47]: y = stride_tricks.as_strided(x, shape=(4,4),strides=(8*4,8))

In [48]: y
Out[48]:
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15]])


In [54]: z = stride_tricks.as_strided(x, shape=(4,),strides=(8*5,))

In [55]: z
Out[55]: array([ 0,  5, 10, 15])

In [56]: sum(z)
Out[56]: 30

As you can see, you get the diagonal elements without having to copy any memory.

Sebastian


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