[Numpy-discussion] Tuple outer product?

[Keith Goodman]

On Fri, Sep 25, 2009 at 11:53 AM, Keith Goodman <kwgoodman at gmail.com> wrote:
> On Fri, Sep 25, 2009 at 11:05 AM, Mads Ipsen <mpi at comxnet.dk> wrote:
>> Sure, and you could do the same thing using two nested for loops. But
>> its bound to be slow when the arrays become large (and they will). The
>> question is whether it can be done in pure numpy. An output like
>>
>> [[[1,4],[1,5],[1,6]],[[2,4],[2,5],[2,6]],[[3,4],[3,5],[3,6]]]
>>
>> is also OK.
>
> This doesn't work. But maybe someone can make it work:
>
>>> x = np.array([1,2,3])
>>> y = np.array([4,5,6])
>>>
>>> a = np.ones((3,3,2), dtype=np.int)
>>> a[:,:,0] = x
>>> a[:,:,1] = y
>>> a.tolist()
>   [[[1, 4], [2, 5], [3, 6]], [[1, 4], [2, 5], [3, 6]], [[1, 4], [2,
> 5], [3, 6]]]

How about this?

>> x = np.array([1,2,3])
>> y = np.array([4,5,6])
>>
>> a = np.ones((3,3,2), dtype=np.int)
>> a[:,:,0] = x
>> a.T[1] = y
>> a.tolist()
   [[[1, 4], [2, 4], [3, 4]], [[1, 5], [2, 5], [3, 5]], [[1, 6], [2,
6], [3, 6]]]