[Numpy-discussion] Consider improving numpy.outer's behavior with zero-dimensional vectors

[Neil Girdhar]

On Fri, Apr 17, 2015 at 10:47 AM, <josef.pktd at gmail.com> wrote:

> On Fri, Apr 17, 2015 at 10:07 AM, Sebastian Berg
> <sebastian at sipsolutions.net> wrote:
> > On Do, 2015-04-16 at 15:28 -0700, Matthew Brett wrote:
> >> Hi,
> >>
> > <snip>
> >>
> >> So, how about a slight modification of your proposal?
> >>
> >> 1) Raise deprecation warning for np.outer for non 1D arrays for a few
> >> versions, with depraction in favor of np.multiply.outer, then
> >> 2) Raise error for np.outer on non 1D arrays
> >>
> >
> > I think that was Neil's proposal a bit earlier, too. +1 for it in any
> > case, since at least for the moment I doubt outer is used a lot for non
> > 1-d arrays. Possible step 3) make it work on higher dims after a long
> > period.
>
> sounds ok to me
>
> Some random comments of what I remember or guess in terms of usage
>
> I think there are at most very few np.outer usages with 2d or higher
> dimension.
> (statsmodels has two models that switch between 2d and 1d
> parameterization where we don't use outer but it has similar
> characteristics. However, we need to control the ravel order, which
> IIRC is Fortran)
>
> The current behavior of 0-D scalars in the initial post might be
> useful if a numpy function returns a scalar instead of a 1-D array in
> size=1. np.diag which is a common case, doesn't return a scalar (in my
> version of numpy).
>
> I don't know any use case where I would ever want to have the 2d
> behavior of np.multiply.outer.
>

My use case is pretty simple.  Given an input vector x, and a weight matrix
W, and a model y=Wx, I calculate the gradient of the loss L with respect
W.  It is the outer product of x with the vector of gradients dL/dy.  So
the code is simply:

W -= outer(x, dL_by_dy)

Sometimes, I have some x_indices and y_indices.  Now I want to do:

W[x_indices, y_indices] -= outer(x[x_indices], dL_by_dy[y_indices])

Unfortunately, if x_indices or y_indices are "int" or slice in some way
that removes a dimension, the left side will have fewer dimensions than the
right.  np.multipy.outer does the right thing without the ugly cases:

if isinstance(x_indices, int): … # ugly hacks follow.

I guess we will or would have applications for outer along an axis,
> for example if x.shape = (100, 10), then we have
> x[:,None, :] * x[:, :, None]     (I guess)
> Something like this shows up reasonably often in econometrics as
> "Outer Product". However in most cases we can avoid constructing this
> matrix and get the final results in a more memory efficient or faster
> way.
> (example an array of covariance matrices)
>

Not sure I see this.  outer(a, b) should return something that has shape:
(a.shape + b.shape).  If you're doing it "along an axis", you mean you're
reshuffling the resulting shape vector?

>
> Josef
>
>
>
>
> >
> > - Sebastian
> >
> >
> >> Best,
> >>
> >> Matthew
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> >>
> >
> >
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