Henry,

I can't quite understand your comment.
I think you are saying that you transpose the RGB with 0 1&|: but I don't
know where to go with u;._3 from there because the RGB data arrives as
shape R C 3, I believe.

In light of Robs trimalt (see my comment about using the right tine as a
pre-process constant), I'm not even sure whether it is worth pursuing u;._3
any further for my case. But I am still interested in your observations.

Thanks,

On Thu, Jun 1, 2017 at 2:44 PM, Henry Rich <[email protected]> wrote:

> When I do image work I use RGB planes (shape 3 512 512 for example).
>  u;.n is much faster on rank 2 than on higher ranks.
>
> Henry Rich
>
>
> On 6/1/2017 2:30 PM, robert therriault wrote:
>
>> Brian,
>>
>> I thought that as well, but I just finished a slightly different approach
>> where I went directly to the selection of the triples from the matrix. It
>> looks more complicated but is at least twice as fast and a third of the
>> space.
>>
>>      trim1 =: >@(tess &(cp;._3))@:(<"1)
>>     100000 timespacex 'trim1 i. 7 10 3'
>> 5.78836e_6 21632
>>
>>      trimalt=: {~ <@:(>:@:(3 * i.)@:<.@:(%&3)&.>)@:}:@:$    NB. Hook with
>> {~ the left tine
>>      100000 timespacex 'trimalt i. 7 10 3'
>> 2.41935e_6 7040
>>
>> 6.17228e_6 21632
>>
>>     (trim1-:trimh) i."1 _3[\ , >,&3 each {3&+@:i. each 15 ;15 NB.
>> combinations from 3 3 3 to 15 15 3
>> 1
>>
>> For indices of less than 3 there is a difference in the shape produced,
>> so watch out for that boundary.
>>
>>      $ trimalt i. 2 2 3
>> 0 0 3
>>     $ trim1 i. 2 2 3
>> 0 0
>>
>> Cheers, bob
>>
>> On Jun 1, 2017, at 11:18 AM, Brian Schott <[email protected]> wrote:
>>>
>>> That's very thorough and appears to suggest that no dramatic improvements
>>> can be made
>>>
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>
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