On my system plain fancy indexing is fastest

Hardly surprising, since take_along_axis is doing that under the hood,
after constructing the index for you :)
https://github.com/numpy/numpy/blob/v1.17.0/numpy/lib/shape_base.py#L58-L172
I deliberately didn't expose the internal function that constructs the
slice, since leaving it private frees us to move those functions to c or in
the distant future gufuncs.
On Fri, Nov 1, 2019, 15:54 Allan Haldane

my thought was to try `take` or `take_along_axis`:

ind = np.argmin(a, axis=1) np.take_along_axis(a, ind[:,None], axis=1)

But those functions tend to simply fall back to fancy indexing, and are pretty slow. On my system plain fancy indexing is fastest:

%timeit a[np.arange(N),ind] 1.58 µs ± 18.1 ns per loop %timeit np.take_along_axis(a, ind[:,None], axis=1) 6.49 µs ± 57.3 ns per loop %timeit np.min(a, axis=1) 9.51 µs ± 64.1 ns per loop

Probably `take_along_axis` was designed with uses like yours in mind, but it is not very optimized.

(I think numpy is lacking a category of efficient indexing/search/reduction functions, like 'findfirst', 'groupby', short-circuiting any_*/all_*/nonzero, the proposed oindex/vindex, better gufunc broadcasting. There is slow but gradual infrastructure work towards these, potentially).

Cheers, Allan

On 10/30/19 11:31 PM, Daniele Nicolodi wrote:

On 30/10/2019 19:10, Neal Becker wrote:

max(axis=1)?

Hi Neal,

I should have been more precise in stating the problem. Getting the values in the array for which I'm looking at the maxima is only one step in a more complex piece of code for which I need the indexes along the second axis of the array. I would like to avoid to have to iterate the array more than once.

Thank you!

Cheers, Dan

On Wed, Oct 30, 2019, 7:33 PM Daniele Nicolodi

mailto:daniele@grinta.net> wrote: Hello,

this is a very basic question, but I cannot find a satisfying answer. Assume a is a 2D array and that I get the index of the maximum value along the second dimension:

i = a.argmax(axis=1)

Is there a better way to get the value of the maximum array entries along the second axis other than:

v = a[np.arange(len(a)), i]

??

Thank you.

Cheers, Daniele _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@python.org mailto:NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion

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