Re: [Numpy-discussion] Indexing arrays with arrays

[Pau G. <pau...@gm...>]

On 5/18/06, Bill Baxter <wb...@gm...> wrote:
> One thing I haven't quite managed to figure out yet, is what the heck
> indexing an array with an array is supposed to give you.
> This is sort of an offshoot of the "nonzero()" discussion.
> I was wondering why nonzero() and where() return tuples at all instead of
> just an array, till I tried it.
>
> >>> b
> array([[0, 1, 2],
>        [3, 4, 5],
>        [6, 7, 8]])
> >>> b[ num.asarray(num.where(b>4)) ]
> array([[[3, 4, 5],
>         [6, 7, 8],
>         [6, 7, 8],
>         [6, 7, 8]],
>
>        [[6, 7, 8],
>         [0, 1, 2],
>         [3, 4, 5],
>         [6, 7, 8]]])
>
> Whoa, not sure what that is.  Can someone explain the current rule and in
> what situations is it useful?
>
> Thanks
>  --bb


i think that is b and x are arrays then,
   b[x]_ijk =3D b[ x_ijk ]
where ijk is whatever x needs to be indexed with.
Note that the indexing on b is _only_ on its first dimension.

>>> from numpy import *
>>> b =3D arange(9).reshape(3,3)
>>> b
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])
>>> x,y =3D where(b>4)
>>> xy =3D asarray( (x,y) )
>>> xy
array([[1, 2, 2, 2],
       [2, 0, 1, 2]])
>>> b[x,y]
array([5, 6, 7, 8])
>>> b[xy]
array([[[3, 4, 5],
        [6, 7, 8],
        [6, 7, 8],
        [6, 7, 8]],

       [[6, 7, 8],
        [0, 1, 2],
        [3, 4, 5],
        [6, 7, 8]]])
>>> asarray( (b[x],b[y]) )
array([[[3, 4, 5],
        [6, 7, 8],
        [6, 7, 8],
        [6, 7, 8]],

       [[6, 7, 8],
        [0, 1, 2],
        [3, 4, 5],
        [6, 7, 8]]])