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 <[email protected]> wrote: >> >>> Friedrich Romstedt wrote: >>> >>>> 2010/6/13 Pauli Virtanen <[email protected]>: >>>> >>>> >>>>> 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 >>>> [email protected] >>>> 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 >>> [email protected] >>> http://mail.scipy.org/mailman/listinfo/numpy-discussion >>> >>> >> _______________________________________________ >> NumPy-Discussion mailing list >> [email protected] >> 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 > [email protected] > http://mail.scipy.org/mailman/listinfo/numpy-discussion -- Dag Sverre _______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
