Like Pascal Jasmin pointed out, fold is overkill for this example. For example:
(1 vec} A)-:vec fold A
1
But this does suggest a way of illustrating the use of a fold
conjunction, if that's what you want to focus on:
(A ]F..{{1 (<x)} y}} vec) -: vec fold A
1
(Folds are "popular" right now, which is kind of interesting. But
using them isn't really a substitute for understanding math nor for
understanding the operations we use with folds.)
Anyways... I hope this helps,
--
Raul
On Sat, Apr 23, 2022 at 2:02 PM Pawel Jakubas <jakubas.pa...@gmail.com> wrote:
>
> Dear J enthusiast,
>
> I try to play with fold functionality F...
> and here solve more general cases, so when the accumulator structure is
> different from the list structure that is iterated.
>
> To be concrete let's say we have the array, A, and a list of 2-elem
> vectors, vec,
> which institute coordinates (x,y) which I want to update in A.
> ```
> A=: 10 10 $ 0
> ]vec=: 3 2 $ 1 2 3 4 5 6
> 1 2
> 3 4
> 5 6
> ```
> Now, I can of course solve this recursively, eg:
> ```
> fold=: 4 : 0
> 'r c' =. ,"0 $ x
> if. (r=1) do.
> 'x0 y0'=. {. x
> 1 (<(x0),(y0)) } y
> else.
> 'x0 y0'=. {. x
> y1=.1 (<(x0),(y0)) } y
> (}. x) fold y1
> end.
> )
> vec fold A
> 0 0 0 0 0 0 0 0 0 0
> 0 0 1 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 1 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 1 0 0 0
> 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0
> ```
>
> How to start with A, use vec as iterator list in one of folds F.. and end
> up with updated A as above?
> Thanks in advance!
> Cheers,
> Pawel Jakubas
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