Unfortunately, Pawel wants 2 f d to be 3 11. However, I find that 3 f d IS 3
11.
Other results are a bit strange, too:
4 f d
2 12
8 f d
|index error: f
| y ({~,])x i.~+/\(i.@#~:i.~)y
I wasn’t going to post my effort, but it might interest Pawel. This version
works on the slightly more intuitive (for me at least, here) origin 1 value of
“x”:
g =: ] ({~ , ]) (i.~ >./@:(+/\"1)@:(~. =/ ]))
2 g d
2 5
3 g d
3 11
4 g d. NB. also not error-checked, though!
|index error: g
| 4 g d
This is a quick get-around to act as Pawel asks, and to give an empty result
rather than an error if nothing satisfies the left argument:
h =: (g~ >:)~ :: ‘’
2 h d
3 11
5 h d
#5 h d
0
FWIW,
Mike
Sent from my iPad
> On 26 Jan 2022, at 21:00, Raul Miller <[email protected]> wrote:
>
> Here's a variation that works:
>
> d=: 1 _1 2 3 4 2 5 6 3 8 10 3 2
> f=: {{y ({~,]) x i.~ +/\(i.@#~:i.~)y }}
> 1 f d
> 2 5
> 2 f d
> 3 8
>
> The phrase (i.@# ~: i.~) finds the locations of duplicates
>
> (i.@#~:i.~) 1 _1 2 3 4 2 5 6 3 8 10 3 2
> 0 0 0 0 0 1 0 0 1 0 0 1 1
>
> And, +/\ computes a running sum
> +/\(i.@#~:i.~) d
> 0 0 0 0 0 1 1 1 2 2 2 3 4
>
> With this, we can find the index of the first occurrence of a
> duplicate count number using x i.~ ...
>
> Once we have the index of a duplicate, we can return that index and
> the corresponding value from the list.
>
> --
> Raul
>
>> On Wed, Jan 26, 2022 at 3:49 PM Pawel Jakubas <[email protected]>
>> wrote:
>>
>> It should be of course
>> 1 f d
>> 2 5
>>
>> Would be great if you could decompose your solution and the idea behind the
>> solution. Many thanks.
>>
>> Cheers,
>> Pawel
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