[Numpy-discussion] Tensor contraction
Dag Sverre Seljebotn
dagss at student.matnat.uio.no
Tue Jun 15 06:54:56 EDT 2010
Alan Bromborsky wrote:
> Dag Sverre Seljebotn wrote:
>
>> 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 at verizon.net> wrote:
>>>>
>>>>
>>>>
>>>>> Friedrich Romstedt wrote:
>>>>>
>>>>>
>>>>>
>>>>>> 2010/6/13 Pauli Virtanen <pav at 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
>>>>>> _______________________________________________
>>>>>> NumPy-Discussion mailing list
>>>>>> NumPy-Discussion at scipy.org
>>>>>> 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
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>> _______________________________________________
>>>>> NumPy-Discussion mailing list
>>>>> NumPy-Discussion at scipy.org
>>>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>>>>>
>>>>>
>>>>>
>>>>>
>>>> _______________________________________________
>>>> NumPy-Discussion mailing list
>>>> NumPy-Discussion at scipy.org
>>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>>>>
>>>>
>>>>
>>>>
>>> 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.
>>> _______________________________________________
>>> NumPy-Discussion mailing list
>>> NumPy-Discussion at scipy.org
>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>>>
>>>
>>
>>
> Can dtype in Theano tensor be a sympy expression
>
I don't use Theano myself so I don't know. I know it is based on
building expressions, couples to SymPy to do derivatives and so on,
although how that works for tensors I don't know, you'll have to read
the docs.
Dag Sverre
More information about the NumPy-Discussion
mailing list