[Numpy-discussion] return index of maximum value in an array easily?
Chao YUE
chaoyuejoy@gmail....
Fri Jan 11 17:26:13 CST 2013
Hi,
I don't know how others think about this. Like you point out, one can use
np.nonzero(a==np.max(a)) as a workaround.
For the second point, in case I have an array:
a = np.arange(24.).reshape(2,3,4)
suppose I want to find the index for maximum value of each 2X3 array along
the 3rd dimension, what I can think of will be:
index_list = []
for i in range(a.shape[-1]):
data = a[...,i]
index_list.append(np.nonzero(data==np.max(data)))
In [87]:
index_list
Out[87]:
[(array([1]), array([2])),
(array([1]), array([2])),
(array([1]), array([2])),
(array([1]), array([2]))]
If we want to make the np.argmax function doing the job of this part of
code,
could we add another some kind of boolean keyword argument, for example,
"exclude" to the function?
[this is only my thinking, and I am only a beginner, maybe it's stupid!!!]
np.argmax(a,axis=2,exclude=True) (default value for exclude is False)
it will give the index of maximum value along all other axis except the
axis=2
(which is acutally the 3rd axis)
The output will be:
np.array(index_list).squeeze()
array([[1, 2],
[1, 2],
[1, 2],
[1, 2]])
and one can use a[1,2,i] (i=1,2,3,4) to extract the maximum value.
I doubt this is really useful...... too complicated......
Chao
On Fri, Jan 11, 2013 at 11:00 PM, Nathaniel Smith <njs@pobox.com> wrote:
> On Thu, Jan 10, 2013 at 9:40 AM, Chao YUE <chaoyuejoy@gmail.com> wrote:
> > Dear all,
> >
> > Are we going to consider returning the index of maximum value in an array
> > easily
> > without calling np.argmax and np.unravel_index consecutively?
>
> This does seem like a good thing to support somehow. What would a good
> interface look like? Something like np.nonzero(a == np.max(a))? Should
> we support vectorized operation (e.g. argmax of each 2-d subarray of a
> 3-d array along some axis)?
>
> -n
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--
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Chao YUE
Laboratoire des Sciences du Climat et de l'Environnement (LSCE-IPSL)
UMR 1572 CEA-CNRS-UVSQ
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Tel: (33) 01 69 08 29 02; Fax:01.69.08.77.16
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