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

[Sebastian Berg]

On Mi, 2015-04-08 at 19:34 -0400, Neil Girdhar wrote:
> Numpy's outer product works fine with vectors.  However, I seem to
> always want len(outer(a, b).shape) to be equal to len(a.shape) +
> len(b.shape).  Wolfram-alpha seems to agree
> https://reference.wolfram.com/language/ref/Outer.html with respect to
> matrix outer products.  My suggestion is to define outer as defined
> below.  I've contrasted it with numpy's current outer product.
> 

Actually this behaviour already exists (I was not aware it was
different) for all ufuncs, though it is not that well known, and I tend
to broadcast manually which may be a pity ;):

np.multiply.outer(a, b)

does exactly what you want, though I am not sure if np.outer might
leverage blas or not.

- Sebastian

> 
> In [36]: def a(n): return np.ones(n)
> 
> 
> In [37]: b = a(())
> 
> 
> In [38]: c = a(4)
> 
> 
> In [39]: d = a(5)
> 
> 
> In [40]: np.outer(b, d).shape
> Out[40]: (1, 5)
> 
> 
> In [41]: np.outer(c, d).shape
> Out[41]: (4, 5)
> 
> 
> In [42]: np.outer(c, b).shape
> Out[42]: (4, 1)
> 
> 
> In [43]: def outer(a, b):
>     return a[(...,) + len(b.shape) * (np.newaxis,)] * b
>    ....:
> 
> 
> In [44]: outer(b, d).shape
> Out[44]: (5,)
> 
> 
> In [45]: outer(c, d).shape
> Out[45]: (4, 5)
> 
> 
> In [46]: outer(c, b).shape
> Out[46]: (4,)
> 
> 
> Best,
> 
> 
> Neil
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