[Numpy-discussion] array of matrices

Anne Archibald peridot.faceted@gmail....
Sat Mar 28 23:15:57 CDT 2009

2009/3/28 Geoffrey Irving <irving@naml.us>:
> On Sat, Mar 28, 2009 at 12:47 AM, Robert Kern <robert.kern@gmail.com> wrote:
>> 2009/3/27 Charles R Harris <charlesr.harris@gmail.com>:
>>> On Fri, Mar 27, 2009 at 4:43 PM, Robert Kern <robert.kern@gmail.com> wrote:
>>>> On Fri, Mar 27, 2009 at 17:38, Bryan Cole <bryan@cole.uklinux.net> wrote:
>>>> > I have a number of arrays of shape (N,4,4). I need to perform a
>>>> > vectorised matrix-multiplication between pairs of them I.e.
>>>> > matrix-multiplication rules for the last two dimensions, usual
>>>> > element-wise rule for the 1st dimension (of length N).
>>>> >
>>>> > (How) is this possible with numpy?
>>>> dot(a,b) was specifically designed for this use case.
>>> I think maybe he wants to treat them as stacked matrices.
>> Oh, right. Sorry. dot(a, b) works when a is (N, 4, 4) and b is just
>> (4, 4). Never mind.
> It'd be great if this operation existed as a primitive.  What do you
> think would be the best way in which to add it?  One option would be
> to add a keyword argument to "dot" giving a set of axes to map over.
> E.g.,
>    dot(a, b, map=0) = array([dot(u,v) for u,v in zip(a,b)]) # but in C
> "map" isn't a very good name for the argument, though.

I think the right long-term solution is to make dot (and some other
linear algebra functions) into "generalized ufuncs", so that when you
dot two multidimensional objects together, they are treated as arrays
of two-dimensional arrays, broadcasting is done on all but the last
two dimensions, and then the linear algebra is applied "elementwise".
This covers basically all "stacked matrices" uses in a very general
way, but would require some redesigning of the linear algebra system -
for example, dot() currently works on both two- and one-dimensional
arrays, which can't work in such a setting.

The infrastructure to support such generalized ufuncs has been added
to numpy, but as far as I know no functions yet make use of it.


> Geoffrey
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