[Numpy-discussion] Tensor contraction
Dag Sverre Seljebotn
dagss@student.matnat.uio...
Mon Jun 14 16:23:02 CDT 2010
Did you have a look at the tensors in Theano? They seem to merge tensor
algebra, SymPy, NumPy and (optional) GPU computing etc. Even if it
doesn't fill your needs it could perhaps be a better starting point?
http://deeplearning.net/software/theano/library/tensor/basic.html
Dag Sverre
Alan Bromborsky wrote:
> Sebastian Walter wrote:
>> 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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>>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>>>>
>>>>
>>>>
>>> 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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>>>
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>>
>>
> Thank you for your reply. All my array entries are symbolic. The use
> of the abstract tensor module will be to generate the equations (finite
> difference for finite element) required for the solution of General
> Relativistic systems. The resulting equations would then be translated
> into C++, C, or Fortran (which ever is most appropriate). The array
> would initially be stuffed with appropriate linear combinations of basis
> functions with symbolic coefficients. I will save the contraction
> problem for last and try to implement both solutions, compare them, and
> let you know the results. Note that the dimension of any axis in a real
> problem is four (4-dimensional space-time) and the highest tensor rank
> is four (256 components), although the rank of the products before
> contraction could be significantly higher.
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--
Dag Sverre
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