[Numpy-discussion] argmax() indexes to value

[Daniele Nicolodi]

On 30/10/2019 22:42, Elliot Hallmark wrote:
> I wouldn't be surprised at all if calling max in addition to argmax
> wasn't as fast or faster than indexing the array using argmax.
> Regardless, just use that then profile when you're done with the
> whole thing and see if there's any gains to be made. Very likely not here.

Hi Elliot,

how do you arrive at this conclusion? np.argmax() and np.max() are O(N)
while indexing is O(1) thus I don't see how you can conclude that
running both np.argmax() and np.max() on the input array is going to
incur in a small penalty compared to running np.argmax() and then indexing.

Cheers,
Dan


> 
> -elliot
> 
> On Wed, Oct 30, 2019, 10:32 PM Daniele Nicolodi <daniele at grinta.net
> <mailto:daniele at grinta.net>> 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 <daniele at grinta.net
>     <mailto:daniele at grinta.net>
>     > <mailto:daniele at grinta.net <mailto:daniele at 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
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