On Tue, Apr 6, 2010 at 9:22 AM, Ken Basye <kbas...@jhu.edu> wrote:
> From: Vincent Schut <sc...@sarvision.nl>
>
> On 04/05/2010 06:06 PM, Keith Goodman wrote:
>
>
> On Mon, Apr 5, 2010 at 8:44 AM, Ken Basye<kbas...@jhu.edu> wrote:
>
>
> Hi Folks,
> I have two arrays, A and B, with the same shape. I want to find the
> highest values in A along some axis, then extract the corresponding
> values from B. I can get the highest values in A with A.max(axis=0) and
> the indices of these highest values with A.argmax(axis=0). I'm trying
> to figure out a loop-free way to extract the corresponding elements from
> B using these indices. Here's code with a loop that will do what I want
> for two-dimensional arrays:
>
> >>> a
> array([[ 100., 0., 0.],
> [ 0., 100., 100.],
> [ 0., 0., 0.]])
>
> >>> a.max(axis=0)
> array([ 100., 100., 100.])
>
> >>> sel = a.argmax(axis=0)
> >>>sel
> array([0, 1, 1])
>
> >>> b = np.arange(9).reshape((3,3))
> >>> b
> array([[0, 1, 2],
> [3, 4, 5],
> [6, 7, 8]])
>
> >>> b_best = np.empty(3)
> >>> for i in xrange(3):
> ... b_best[i] = b[sel[i], i]
> ...
> >>> b_best
> array([ 0., 4., 5.])
>
>
> Here's one way:
>
>
>
> b[a.argmax(axis=0), range(3)]
>
>
> array([0, 4, 5])
>
>
> Which does not work anymore when your arrays become more-dimensional
> (like in my case: 4 or more) and the axis you want to select on is not
> the first/last one. If I recall correctly, I needed to construct the
> full index arrays for the other dimensions too (like with ogrid I
> think). So: create the ogrid, replace the one for the dimensions you
> want the argmax selection to take place on with the argmax parameter,
> and use those index arrays to index your b array.
> I'd need to look up my source code to be more sure/precise. If anyone
> would like me to, please let me know. If anyone knows a less elaborate
> way, also please let us know! :-)
>
>
> Hi Vincent,
> I'd like to see more about your solution. For my present purposes,
> Keith's solution was sufficient, but I'm still very interested in a solution
> that's independent of dimension and axis.
> Thanks (and thanks, Keith),
> Ken
an alternative to Vincent's general solution, if you have unique max
or want all argmax is using a mask
>>> a
array([[ 100., 0., 0.],
[ 0., 100., 100.],
[ 0., 0., 0.]])
>>> b
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
>>> ax=0; b[a==np.expand_dims(a.max(ax),ax)]
array([0, 4, 5])
>>> ax=1; b[a==np.expand_dims(a.max(ax),ax)]
array([0, 4, 5, 6, 7, 8])
>>> aa=np.eye(3)
>>> ax=1; b[aa==np.expand_dims(aa.max(ax),ax)]
array([0, 4, 8])
>>> ax=0; b[aa==np.expand_dims(aa.max(ax),ax)]
array([0, 4, 8])
Josef
>
>
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